1000 Solved Problems In Modern Physics

1000 Solved Problems in Modern Physics

Ahmad A. Kamal
1000 Solved Problems
in Modern Physics
123

Dr. Ahmad A. Kamal
425 Silversprings Lane
Murphy, TX 75094, USA
anwarakamal@yahoo.com
ISBN 978-3-642-04332-1 e-ISBN 978-3-642-04333-8
DOI 10.1007/978-3-642-04333-8
Springer Heidelberg Dordrecht London New York
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Preface
This book is targeted mainly to the undergraduate students of USA, UK and other
European countries, and the M.Sc of Asian countries, but will be found useful for the
graduate students, Graduate Record Examination (GRE), Teachers and Tutors. This
is a by-product of lectures given at the Osmania University, University of Ottawa
and University of Tebrez over several years, and is intended to assist the students in
their assignments and examinations. The book covers a wide spectrum of disciplines
in Modern Physics, and is mainly based on the actual examination papers of UK and
the Indian Universities. The selected problems display a large variety and conform to
syllabi which are currently being used in various countries. The book is divided into
ten chapters. Each chapter begins with basic concepts containing a set of formulae
and explanatory notes for quick reference, followed by a number of problems and
their detailed solutions.
The problems are judiciously selected and are arranged section-wise. The solu-
tions are neither pedantic nor terse. The approach is straight forward and step-by-
step solutions are elaborately provided. More importantly the relevant formulas used
for solving the problems can be located in the beginning of each chapter. There are
approximately 150 line diagrams for illustration.
Basic quantum mechanics, elementary calculus, vector calculus and Algebra are
the pre-requisites. The areas of Nuclear and Particle physics are emphasized as rev-
olutionary developments have taken place both on the experimental and theoretical
fronts in recent years. No book on problems can claim to exhaust the variety in the
limited space. An attempt is made to include the important types of problems at the
undergraduate level.
Chapter 1 is devoted to the methods of Mathematical physics and covers such
topics which are relevant to subsequent chapters. Detailed solutions are given to
problems under Vector Calculus, Fourier series and Fourier transforms, Gamma and
Beta functions, Matrix Algebra, Taylor and Maclaurean series, Integration, Ordinary
differential equations, Calculus of variation Laplace transforms, Special functions
such as Hermite, Legendre, Bessel and Laguerre functions, complex variables, sta-
tistical distributions such as Binomial, Poisson, Normal and interval distributions
and numerical integration.
Chapters 2 and 3 focus on quantum physics. Chapter 2 is basically concerned
with the old quantum theory. Problems are solved under the topics of deBroglie
vii

viii Preface
waves, Bohr’s theory of hydrogen atom and hydrogen-like atoms, positronium and
mesic atoms, X-rays production and spectra, Moseley’s law and Duan–Hunt law,
spectroscopy of atoms and molecules, which include various quantum numbers and
selection rules, and optical Doppler effect.
Chapter 3 is concerned with the quantum mechanics of Schrodinger and
Hesenberg. Problems are solved on the topics of normalization and orthogonality
of wave functions, the separation of Schrodinger’s equation into radial and angu-
lar parts, 1-D potential wells and barriers, 3-D potential wells, Simple harmonic
oscillator, Hydrogen-atom, spatial and momentum distribution of electron, Angular
momentum, Clebsch–Gordon coefficients ladder operators, approximate methods,
scattering theory-phase-shift analysis and Ramsuer effect, the Born approximation.
Chapter 4 deals with problems on Thermo–dynamic relations and their applica-
tions such a specific heats of gases, Joule–Thompson effect, Clausius–Clapeyron
equation and Vander waal’s equation, the statistical distributions of Boltzmann
and Fermi distributions, the distribution of rotational and vibrational states of gas
molecules, the Black body radiation, the solar constant, the Planck’s law and Wein’s
law.
Chapter 5 is basically related to Solid State physics and material science. Prob-
lems are covered under the headings, crystal structure, Lattice constant, Electrical
properties of crystals, Madelung constant, Fermi energy in metals, drift velocity, the
Hall effect, the Debye temperature, the intrinsic and extrinsic semiconductors, the
junction diode, the superconductor and the BCS theory, and the Josephson effect.
Chapter 6 deals with the special theory of Relativity. Problems are solved under
Lorentz transformations of length, time, velocity, momentum and energy, the invari-
ance of four-momentum vector, transformation of angles and Doppler effect and
threshold of particle production.
Chapters 7 and 8 are concerned with problems in low energy Nuclear physics.
Chapter 7 covers the interactions of charged particles with matter which include
kinematics of collisions, Rutherford Scattering, Ionization, Range and Straggiling,
Interactions of radiation with matter which include Compton scattering, photoelec-
tric effect, pair production and nuclear resonance fluorescence, general radioactivity
which includes problems on chain decays, age of earth, Carbon dating, alpha decay,
Beta decay and gamma decay.
Chapter 8 is devoted to the static properties of nuclei such as nuclear masses,
nuclear spin and parity, magnetic moments and quadrupole moments, the Nuclear
models, the Fermi gas model, the shell model, the liquid drop model and the optical
model, problems on fission and fusion and Nuclear Reactors.
Chapters 9 and 10 are concerned with high energy physics. Chapter 9 covers
the problems on natural units, production, interactions and decays of high energy
unstable particles, various types of detectors such as ionization chambers, propror-
tional and G.M. counters, Accelerators which include Betatron, Cyclotron, Synchro-
Cyclotron, proton and electron Synchrotron, Linear accelerator and Colliders.
Chapter 10 deals with the static and dynamic properties of elementary particles
and resonances, their classification from the point of view of the Fermi–Dirac and
Bose–Einstein statistics as well as the three types of interactions, strong, Electro-

Preface ix
magnetic and weak, the conservation laws applicable to the three types of interac-
tions, Gell-mann’s formula, the properties of quarks and classification into super-
multiplets, the types of weak decays and Cabibbo’s theory, the neutrino oscillations,
Electro–Weak interaction, the heavy bosons and the Standard model.
Acknowledgements
It is a pleasure to thank Javid for the bulk of typing and suggestions and Maryam
for proof reading. I am indebted to Muniba for the line drawings, to Suraiya, Maq-
sood and Zehra for typing and editing. I am grateful to the Universities of UK and
India for permitting me to use their question papers cited in the text to CERN photo
service for the cover page to McGraw-Hill and Co: for a couple of diagrams from
Quantum Mechanics, L.I. Schiff, 1955, to Cambridge University Press for using
some valuable information from Introduction to High Energy Physics, D.H. Perkins
and to Ginn and Co: and Pearson and Co: for access to Differential and Integral
Calculus, William A. Granville, 1911. My thanks are due to Springer-Verlag, in
particular Claus Ascheron, Adelheid Duhm and Elke Sauer for constant encourage-
ment.
Murphy, Texas Ahmad A. Kamal
February 2010

Contents
1 Mathematical Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Basic Concepts and Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.2.1 Vector Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.2.2 Fourier Series and Fourier Transforms . . . . . . . . . . . . . . . . . 22
1.2.3 Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.2.4 Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.2.5 Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.2.6 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.2.7 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.2.8 Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . 26
1.2.9 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.2.10 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.2.11 Complex Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1.2.12 Calculus of Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.2.13 Statistical Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.2.14 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.3.1 Vector Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.3.2 Fourier Series and Fourier Transforms . . . . . . . . . . . . . . . . . 39
1.3.3 Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . . . . . . . 42
1.3.4 Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
1.3.5 Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
1.3.6 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
1.3.7 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
1.3.8 Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . 57
1.3.9 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
1.3.10 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
1.3.11 Complex Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
1.3.12 Calculus of Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
1.3.13 Statistical Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
1.3.14 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
xi

xii Contents
2 Quantum Mechanics – I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
2.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
2.2.1 de Broglie Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
2.2.2 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
2.2.3 X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
2.2.4 Spin and μ and Quantum Numbers – Stern–Gerlah’s
Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
2.2.5 Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.2.6 Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
2.2.7 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
2.2.8 Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
2.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
2.3.1 de Broglie Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
2.3.2 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
2.3.3 X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
2.3.4 Spin and μ and Quantum Numbers – Stern–Gerlah’s
Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
2.3.5 Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
2.3.6 Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
2.3.7 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
2.3.8 Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
3 Quantum Mechanics – II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
3.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
3.2.1 Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
3.2.2 Schrodinger Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
3.2.3 Potential Wells and Barriers . . . . . . . . . . . . . . . . . . . . . . . . . 140
3.2.4 Simple Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 146
3.2.5 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
3.2.6 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
3.2.7 Approximate Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
3.2.8 Scattering (Phase-Shift Analysis) . . . . . . . . . . . . . . . . . . . . . 153
3.2.9 Scattering (Born Approximation) . . . . . . . . . . . . . . . . . . . . . 154
3.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
3.3.1 Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
3.3.2 Schrodinger Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
3.3.3 Potential Wells and Barriers . . . . . . . . . . . . . . . . . . . . . . . . . 168
3.3.4 Simple Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 199
3.3.5 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
3.3.6 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
3.3.7 Approximate Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
3.3.8 Scattering (Phase Shift Analysis) . . . . . . . . . . . . . . . . . . . . . 233
3.3.9 Scattering (Born Approximation) . . . . . . . . . . . . . . . . . . . . . 240

Contents xiii
4 Thermodynamics and Statistical Physics . . . . . . . . . . . . . . . . . . . . . . . . . . 247
4.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
4.2.1 Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
4.2.2 Maxwell’s Thermodynamic Relations . . . . . . . . . . . . . . . . . 253
4.2.4 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
4.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
4.3.1 Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
4.3.2 Maxwell’s Thermodynamic Relations . . . . . . . . . . . . . . . . . 266
4.3.4 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
5 Solid State Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
5.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
5.2.1 Crystal Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
5.2.2 Crystal Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
5.2.3 Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
5.2.4 Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
5.2.5 Superconductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
5.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
5.3.1 Crystal Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
5.3.2 Crystal Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
5.3.3 Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
5.3.4 Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
5.3.5 Superconductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
6 Special Theory of Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
6.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
6.2.1 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
6.2.2 Length, Time, Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
6.2.3 Mass, Momentum, Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 323
6.2.4 Invariance Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326
6.2.5 Transformation of Angles and Doppler Effect . . . . . . . . . . 328
6.2.6 Threshold of Particle Production . . . . . . . . . . . . . . . . . . . . . 330
6.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
6.3.1 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
6.3.2 Length, Time, Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
6.3.3 Mass, Momentum, Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 342
6.3.4 Invariance Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
6.3.5 Transformation of Angles and Doppler Effect . . . . . . . . . . 355
6.3.6 Threshold of Particle Production . . . . . . . . . . . . . . . . . . . . . 365

xiv Contents
7 Nuclear Physics – I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
7.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
7.2.1 Kinematics of Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
7.2.2 Rutherford Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
7.2.3 Ionization, Range and Straggling . . . . . . . . . . . . . . . . . . . . . 385
7.2.4 Compton Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
7.2.5 Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
7.2.6 Pair Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390
7.2.7 Cerenkov Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390
7.2.8 Nuclear Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390
7.2.9 Radioactivity (General) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
7.2.10 Alpha-Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
7.2.11 Beta-Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
7.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
7.3.1 Kinematics of Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
7.3.2 Rutherford Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399
7.3.3 Ionization, Range and Straggling . . . . . . . . . . . . . . . . . . . . . 404
7.3.4 Compton Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
7.3.5 Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411
7.3.6 Pair Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
7.3.7 Cerenkov Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
7.3.8 Nuclear Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416
7.3.9 Radioactivity (General) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
7.3.10 Alpha-Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422
7.3.11 Beta-Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423
8 Nuclear Physics – II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427
8.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434
8.2.1 Atomic Masses and Radii . . . . . . . . . . . . . . . . . . . . . . . . . . . 434
8.2.2 Electric Potential and Energy . . . . . . . . . . . . . . . . . . . . . . . . 435
8.2.3 Nuclear Spin and Magnetic Moment . . . . . . . . . . . . . . . . . . 435
8.2.4 Electric Quadrupole Moment . . . . . . . . . . . . . . . . . . . . . . . . 435
8.2.5 Nuclear Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
8.2.6 Fermi Gas Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
8.2.7 Shell Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
8.2.8 Liquid Drop Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438
8.2.9 Optical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439
8.2.10 Nuclear Reactions (General) . . . . . . . . . . . . . . . . . . . . . . . . . 440
8.2.11 Cross-sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
8.2.12 Nuclear Reactions via Compound Nucleus . . . . . . . . . . . . . 443
8.2.13 Direct Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443
8.2.14 Fission and Nuclear Reactors . . . . . . . . . . . . . . . . . . . . . . . . 444
8.2.15 Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

Contents xv
8.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447
8.3.1 Atomic Masses and Radii . . . . . . . . . . . . . . . . . . . . . . . . . . . 447
8.3.2 Electric Potential and Energy . . . . . . . . . . . . . . . . . . . . . . . . 449
8.3.3 Nuclear Spin and Magnetic Moment . . . . . . . . . . . . . . . . . . 450
8.3.4 Electric Quadrupole Moment . . . . . . . . . . . . . . . . . . . . . . . . 451
8.3.5 Nuclear Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454
8.3.6 Fermi Gas Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
8.3.7 Shell Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
8.3.8 Liquid Drop Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458
8.3.9 Optical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460
8.3.10 Nuclear Reactions (General) . . . . . . . . . . . . . . . . . . . . . . . . . 462
8.3.11 Cross-sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468
8.3.12 Nuclear Reactions via Compound Nucleus . . . . . . . . . . . . . 469
8.3.13 Direct Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
8.3.14 Fission and Nuclear Reactors . . . . . . . . . . . . . . . . . . . . . . . . 471
8.3.15 Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
9 Particle Physics – I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
9.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
9.2.1 System of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
9.2.2 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
9.2.3 Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
9.2.4 Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491
9.2.5 Ionization Chamber, GM Counter and Proportional
Counters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
9.2.6 Scintillation Counter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495
9.2.7 Cerenkov Counter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
9.2.8 Solid State Detector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
9.2.9 Emulsions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
9.2.10 Motion of Charged Particles in Magnetic Field . . . . . . . . . 497
9.2.11 Betatron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498
9.2.12 Cyclotron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
9.2.13 Synchrotron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
9.2.14 Linear Accelerator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
9.2.15 Colliders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
9.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
9.3.1 System of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
9.3.2 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504
9.3.3 Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505
9.3.4 Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508
9.3.5 Ionization Chamber, GM Counter and Proportional
Counters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512
9.3.6 Scintillation Counter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515

xvi Contents
9.3.7 Cerenkov Counter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518
9.3.8 Solid State Detector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520
9.3.9 Emulsions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520
9.3.10 Motion of Charged Particles in Magnetic Field . . . . . . . . . 521
9.3.11 Betatron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
9.3.12 Cyclotron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
9.3.13 Synchrotron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
9.3.14 Linear Accelerator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530
9.3.15 Colliders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531
10 Particle Physics – II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
10.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544
10.2.1 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544
10.2.2 Strong Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
10.2.3 Quarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550
10.2.4 Electromagnetic Interactions . . . . . . . . . . . . . . . . . . . . . . . . . 551
10.2.5 Weak Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552
10.2.6 Electro-Weak Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 556
10.2.7 Feynman Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 556
10.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
10.3.1 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
10.3.2 Strong Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563
10.3.3 Quarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572
10.3.4 Electromagnetic Interactions . . . . . . . . . . . . . . . . . . . . . . . . . 577
10.3.5 Weak Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578
10.3.6 Electro-weak Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 590
10.3.7 Feynman Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592
Appendix: Problem Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633

Chapter 1
Mathematical Physics
1.1 Basic Concepts and Formulae
Vector calculus
Angle between two vectors, cos θ = A.B
|A||B|
Condition for coplanarity of vectors, A.B × C = 0
Del
∇ =
∂
∂x
î +
∂
∂y
ĵ +
∂
∂z
k̂
Gradient
∇φ =

∂φ
∂x
î +
∂φ
∂y
ĵ +
∂φ
∂z
k̂

Divergence
If V (x, y, z) = V1î + V2 ĵ + V3k̂, be a differentiable vector field, then
∇.V = ∂
∂x
V1 + ∂
∂y
V2 + ∂
∂z
V3
Laplacian
∇2
=
∂2
∂x2
+
∂2
∂y2
+
∂2
∂z2
(Cartesian coordinates x, y, z)
∇2
=
1
r2
∂
∂r

r2 ∂
∂r

+
1
r2 sin θ
∂
∂θ

sin θ
∂
∂θ

+
1
r2 sin2
θ
∂2
∂Φ2
(Spherical coordinates r, θ, Φ)
∇2
=
∂2
∂r2
+
1
r
∂
∂r
+
1
r2
∂2
∂θ2
+
∂2
∂z2
(Cylindrical coordinates r, θ, z)
Line integrals
(a)

C φ dr
1

2 1 Mathematical Physics
(b)

C A . dr
(c)

C A × dr
where φ is a scalar, A is a vector and r = xî + y ĵ + zk̂, is the positive vector.
Stoke’s theorem

C
A . dr =

S
(∇ × A) . n ds =

S
(∇ × A) . ds
The line integral of the tangential component of a vector A taken around a simple
closed curve C is equal to the surface integral of the normal component of the curl
of A taken over any surface S having C as its boundary.
Divergence theorem (Gauss theorem)

V
∇ . A dv =

S
A.n̂ ds
The volume integral is reduced to the surface integral.
Fourier series
Any single-valued periodic function whatever can be expressed as a summation of
simple harmonic terms having frequencies which are multiples of that of the given
function. Let f (x) be defined in the interval (−π, π) and assume that f (x) has
the period 2π, i.e. f (x + 2π) = f (x). The Fourier series or Fourier expansion
corresponding to f (x) is defined as
f (x) =
1
2
a0 +
∞
n=1
(a0 cos nx + bn sin nx) (1.1)
where the Fourier coefficient an and bn are
an =
1
π
π
−π
f (x) cos nx dx (1.2)
bn =
1
π
π
−π
f (x) sin nx dx (1.3)
where n = 0, 1, 2, . . .
If f (x) is defined in the interval (−L, L), with the period 2L, the Fourier series
is defined as
f (x) =
1
2
a0 +
∞
n=1
(an cos(nπx/L) + bn sin(nπx/L)) (1.4)
where the Fourier coefficients an and bn are

1.1 Basic Concepts and Formulae 3
an =
1
L
L
−L
f (x) cos(nπx/L) dx (1.5)
bn =
1
L
L
−L
f (x) sin(nπx/L) dx (1.6)
Complex form of Fourier series
Assuming that the Series (1.1) converges at f (x),
f (x) =
∞
n=−∞
Cneinπx/L
(1.7)
with
Cn =
1
L
C+2L
C
f (x)e−iπnx/L
dx =
⎧
⎪
⎨
⎪
⎩
1
2
(an − ibn) n 0
1
2
(a−n + ib−n) n 0
1
2
ao n = 0
(1.8)
Fourier transforms
The Fourier transform of f (x) is defined as
ℑ( f (x)) = F(α) =
∞
−∞
f (x)eiαx
dx (1.9)
and the inverse Fourier transform of F(α) is
ℑ−1
( f (α)) = F(x) =
1
2π
∞
−∞
F(α)ei∝x
dα (1.10)
f (x) and F(α) are known as Fourier Transform pairs. Some selected pairs are given
in Table 1.1.
Table 1.1
f (x) F(α) f (x) F(α)
1
x2 + a2
πe−aα
a
e−ax a
α2 + a2
x
x2 + a2
−
πiα
a
e−aα
e−ax2 1
2
π
a
e−α2/4a
1
x
π
2
xe−ax2
√
π
4a3/2
αe−α2/4a
Gamma and beta functions
The gamma function Γ(n) is defined by

Γ(n) =
∞
0
e−x
xn−1
dx (Re n 0) (1.11)
Γ(n + 1) = nΓ(n) (1.12)
If n is a positive integer
Γ(n + 1) = n! (1.13)
Γ

1
2

=
√
π; Γ

3
2

=
√
π
2
; Γ

5
2

=
3
4
√
π (1.14)
Γ

n +
1
2

=
1.3.5 . . . (2n − 1)
√
π
2n
(n = 1, 2, 3, . . .) (1.15)
Γ

−n +
1
2

=
(−1)n
2n
√
π
1.3.5 . . . (2n − 1)
(n = 1, 2, 3, . . .) (1.16)
Γ(n + 1) = n! ∼
=
√
2πn nn
e−n
(Stirling’s formula) (1.17)
n → ∞
Beta function B(m, n) is defined as
B(m, n) =
Γ(m)Γ(n)
Γ(m + n)
(1.18)
B(m, n) = B(n, m) (1.19)
B(m, n) = 2
π/2
0
sin2m−1
θ cos2n−1
θ dθ (1.20)
B(m, n) =
∞
0
tm−1
(1 + t)m+n
dt (1.21)
Special funtions, properties and differential equations
Hermite functions:
Differential equation:
y′′
− 2xy′
+ 2ny = 0 (1.22)
when n = 0, 1, 2, . . . then we get Hermite’s polynomials Hn(x) of degree n, given
by
Hn(x) = (−1)n
ex2 dn
dxn
e−x2

(Rodrigue’s formula)

First few Hermite’s polynomials are:
Ho(x) = 1, H1(x) = 2x, H2(x) = 4x2
− 2
H3(x) = 8x3
− 12x, H4(x) = 16x4
− 48x2
+ 12 (1.23)
Generating function:
e2tx−t2
=
∞
n=0
Hn(x)tn
n!
(1.24)
Recurrence formulas:
H′
n(x) = 2nHn−1(x)
Hn+1(x) = 2x Hn(x) − 2nHn−1(x) (1.25)
Orthonormal properties:
∞
−∞
e−x2
Hm(x)Hn(x) dx = 0 m = n (1.26)
∞
−∞
e−x2
{Hn(x)}2
dx = 2n
n!
√
π (1.27)
Legendre functions:
Differential equation of order n:
(1 − x2
)y′′
− 2xy′
+ n(n + 1)y = 0 (1.28)
when n = 0, 1, 2, . . . we get Legendre polynomials Pn(x).
Pn(x) =
1
2nn!
dn
dxn
(x2
− 1)n
(1.29)
First few polynomials are:
Po(x) = 1, P1(x) = x, P2(x) =
1
2
(3x2
− 1)
P3(x) =
1
2
(5x3
− 3x), P4(x) =
1
8
(35x4
− 30x2
+ 3) (1.30)
1
√
1 − 2tx + t2
=
∞
n=0
Pn(x)tn
(1.31)

x P′
n(x) − P′
n−1(x) = nPn(x)
P′
n+1(x) − P′
n−1(x) = (2n + 1)Pn(x) (1.32)
1
−1
Pm(x)Pn(x) dx = 0 m = n (1.33)
1
−1
{Pn(x)}2
dx =
2
2n + 1
(1.34)
Other properties:
Pn(1) = 1, Pn(−1) = (−1)n
, Pn(−x) = (−1)n
Pn(x) (1.35)
Associated Legendre functions:
(1 − x2
)y′′
− 2xy′
+

l(l + 1) −
m2
1 − x2

y = 0 (1.36)
Pm
l (x) = (1 − x2
)m/2 dm
dxm
Pl (x) (1.37)
where Pl(x) are the Legendre polynomials stated previously, l being the positive
integer.
Po
l (x) = Pl(x) (1.38)
and Pm
l (x) = 0 if m n (1.39)
1
−1
Pm
n (x)Pm
l (x) dx = 0 n = l (1.40)
1
−1
{Pm
l (x)}2
dx =
2
2l + 1
(l + m)!
(l − m)!
(1.41)
Laguerre polynomials:
xy′′
+ (1 − x)y′
+ ny = 0 (1.42)
if n = 0, 1, 2, . . . we get Laguerre polynomials given by

Ln(x) = ex dn
dxn
(xn
e−x
) (Rodrigue’s formula) (1.43)
The first few polynomials are:
Lo(x) = 1, L1(x) = −x + 1, L2(x) = x2
− 4x + 2
L3(x) = −x3
+ 9x2
− 18x + 6, L4(x) = x4
− 16x3
+ 72x2
− 96x + 24 (1.44)
e−xs/(1−s)
1 − s
=
∞
n=0
Ln(x)sn
n!
(1.45)
Ln+1(x) − (2n + 1 − x)Ln(x) + n2
Ln−1(x) = 0
xL′
n(x) = nLn(x) − n2
Ln−1(x) (1.46)
∞
0
e−x
Lm(x)Ln(x) dx = 0 m = n (1.47)
∞
0
e−x
{Ln(x)}2
dx = (n!)2
(1.48)
Bessel functions: (Jn(x))
Differential equation of order n
x2
y′′
+ xy′
+ (x2
− n2
)y = 0 n ≥ 0 (1.49)
Expansion formula:
Jn(x) =
∞
k=0
(−1)k
(x/2)2k−n
k!Γ(k + 1 − n)
(1.50)
Properties:
J−n(x) = (−1)n
Jn(x) n = 0, 1, 2, . . . (1.51)
J′
o(x) = −J1(x) (1.52)
Jn+1(x) =
2n
x
Jn(x) − Jn−1(x) (1.53)
ex(s−1/s)/2
=
∞
n=−∞
Jn(x)tn
(1.54)

Laplace transforms:
Definition:
A Laplace transform of the function F(t) is
∞
0
F(t)e−st
dt = f (s) (1.55)
The function f (s) is the Laplace transform of F(t). Symbolically, L{F(t)} =
f (s) and F(t) = L−1
{ f (s)} is the inverse Laplace transform of f (s). L−1
is called
the inverse Laplace operator.
Table of Laplace transforms:
F(t) f (s)
aF1(t) + bF2(t) af1(s) + bf2(s)
aF(at) f (s/a)
eat
F(t) f (s − a)
F(t − a) t a
0 t a
e−as
f (s)
1
1
s
t
1
s2
tn−1
(n − 1)!
1
sn
n = 1, 2, 3, . . .
eat 1
s − a
sin at
a
1
s2 + a2
cos at
s
s2 + a2
sinh at
a
1
s2 − a2
cosh at
s
s2 − a2
Calculus of variation
The calculus of variation is concerned with the problem of finding a function y(x)
such that a definite integral, taken over a function shall be a maximum or minimum.
Let it be desired to find that function y(x) which will cause the integral
I =
x2
x1
F(x, y, y′
) dx (1.56)

to have a stationary value (maximum or minimum). The integrand is taken to be
a function of the dependent variable y as well as the independent variable x and
y′
= dy/dx. The limits x1 and x2 are fixed and at each of the limits y has definite
value. The condition that I shall be stationary is given by Euler’s equation
∂F
∂y
−
d
dx
∂F
∂y′
= 0 (1.57)
When F does not depend explicitly on x, then a different form of the above
equation is more useful
∂F
∂x
−
d
dx

F − y′ ∂F
∂y′

= 0 (1.58)
which gives the result
F − y′ ∂ F
∂y′
= Constant (1.59)
Statistical distribution
Binomial distribution
The probability of obtaining x successes in N-independent trials of an event for
which p is the probability of success and q the probability of failure in a single trial
is given by the binomial distribution B(x).
B(x) =
N!
x!(N − x)!
px
qN−x
= CN
x px
qN−x
(1.60)
B(x) is normalized, i.e.
N
x=0
B(x) = 1 (1.61)
It is a discrete distribution.
The mean value,
x = Np (1.62)
The S.D.,
σ =

Npq (1.63)

Poisson distribution
The probability that x events occur in unit time when the mean rate of occurrence
is m, is given by the Poisson distribution P(x).
P(x) =
e−m
mx
x!
(x = 0, 1, 2, . . .) (1.64)
The distribution P(x) is normalized, that is
∞
x=0
p(x) = 1 (1.65)
This is also a discrete distribution.
When N P is held fixed, the binomial distribution tends to Poisson distribution
as N is increased to infinity.
The expectation value, i.e.
x = m (1.66)
The S.D.,
σ =
√
m (1.67)
Properties:
pm−1 = pm (1.68)
px−1 =
x
m
pm and px+1 =
m
m + 1
px (1.69)
Normal (Gaussian distribution)
When p is held fixed, the binomial distribution tends to a Normal distribution as N
is increased to infinity. It is a continuous distribution and has the form
f (x) dx =
1
√
2πσ
e−(x−m)2
/2σ2
dx (1.70)
where m is the mean and σ is the S.D.
The probability of the occurrence of a single random event in the interval m − σ
and m + σ is 0.6826 and that between m − 2σ and m + 2σ is 0.973.
Interval distribution
If the data contains N time intervals then the number of time intervals n between t1
and t2 is
n = N(e−at1
− e−at2
) (1.71)
where a is the average number of intervals per unit time. Short intervals are more
favored than long intervals.

Two limiting cases:
(a) t2 = ∞; N = Noe−λt
(Law of radioactivity) (1.72)
This gives the number of surviving atoms at time t.
(b) t1 = 0; N = No(1 − e−λt
) (1.73)
For radioactive decays this gives the number of decays in time interval 0 and t.
Above formulas are equally valid for length intervals such as interaction lengths.
Moment generating function (MGF)
MGF = Ee(x−μ)t
= E

1 + (x − μ)t + (x − μ)2 t2
2!
+ . . .

= 1 + 0 + μ2
t2
2!
+ μ3
t3
3!
+ . . . (1.74)
so that μn, the nth moment about the mean is the coefficient of tn
/n!.
Propagation of errors
If the error on the measurement of f (x, y, . . .) is σ f and that on x and y, σx and σy,
respectively, and σx and σy are uncorrelated then
σ2
f =

∂ f
∂x
2
σ2
x +

∂ f
∂y
2
σ2
y + · · · (1.75)
Thus, if f = x ± y, then σf =

σ2
x + σ2
y
1/2
And if f = x
y
then σ f
f
=
σ2
x
x2 +
σ2
y
y2
1/2
Least square fit
(a) Straight line: y = mx + c
It is desired to fit pairs of points (x1, y1), (x2, y2), . . . , (xn, yn) by a straight line
Residue: S =
n
i=1(yi − mxi − C)2
Minimize the residue: ∂s
∂m
= 0; ∂s
∂c
= 0
The normal equations are:
m
n
i=1
x2
i + C
n
i=1
xi −
n
i=1
xi yi = 0
m
n
i=1
xi + nC −
n
i=1
yi = 0

which are to be solved as ordinary algebraic equations to determine the best
values of m and C.
(b) Parabola: y = a + bx + cx2
Residue: S =
n
i=1(yi − a − bxi − cx2
i )2
Minimize the residue: ∂s
∂a
= 0; ∂s
∂b
= 0; ∂s
∂c
= 0
The normal equations are:

yi − na − b

xi − c

x2
i = 0

xi yi − a

xi − b

x2
i − c

x3
i = 0

x2
i yi − a

x2
i − b

x3
i − c

x4
i = 0
which are to be solved as ordinary algebraic equations to determine the best
value of a, b and c.
Numerical integration
Since the value of a definite integral is a measure of the area under a curve, it follows
that the accurate measurement of such an area will give the exact value of a definite
integral; I =
x2
x1
y(x)dx. The greater the number of intervals (i.e. the smaller Δx is),
the closer will be the sum of the areas under consideration.
Trapezoidal rule
area =

1
2
y0 + y1 + y2 + · · · yn−1 +
1
2
yn

Δx (1.76)
Simpson’s rule
area =
Δx
3
(y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · yn), n being even. (1.77)
Fig. 1.1 Integration by
Simpson’s rule and
Trapezoidal rule

Matrices
Types of matrices and definitions
Identity matrix:
I2 =

1 0
0 1

; I3 =
⎛
⎝
1 0 0
0 1 0
0 0 1
⎞
⎠ (1.78)
Scalar matrix:

a11 0
0 a22

;
⎛
⎝
a11 0 0
0 a22 0
0 0 a33
⎞
⎠ (1.79)
Symmetric matrix:

aji = ai j

;
⎛
⎝
a11 a12 a13
a12 a22 a23
a13 a23 a33
⎞
⎠ (1.80)
Skew symmetric:

aji = −ai j

;
⎛
⎝
a11 a12 a13
−a12 a22 a23
−a13 −a23 a33
⎞
⎠ (1.81)
The Inverse of a matrix B = A−1
(B equals A inverse):
if AB = B A = I and further, (AB)−1
= B−1
A−1
A commutes with B if AB = B A
A anti-commutes with B if AB = −B A
The Transpose (A′
) of a matrix A means interchanging rows and columns.
Further, (A + B)′
= A′
+ B′
(A′
)′
= A, (k A)′
= k A′
(1.82)
The Conjugate of a matrix. If a matrix has complex numbers as elements, and if
each number is replaced by its conjugate, then the new matrix is called the conjugate
and denoted by A∗
or A (A conjugate)
The Trace (Tr) or Spur of a matix is the num of the diagonal elements.
Tr =

aii (1.83)

Hermetian matrix
If A′ = A, so that ai j = a ji for all values of i and j. Diagonal elements of an
Hermitian matrix are real numbers.
Orthogonal matrix
A square matrix is said to be orthogonal if AA′
= A′
A = I, i.e. A′
= A−1
The column vector (row vectors) of an orthogonal matrix A are mutually orthog-
onal unit vectors.
The inverse and the transpose of an orthogonal matrix are orthogonal.
The product of any two or more orthogonal matrices is orthogonal.
The determinant of an Orthogonal matrix is ±1.
Unitary matrix
A square matrix is called a unitary matrix if (A)′
A = A(A)′
= I, i.e. if (A)′
= A−1
.
The column vectors (row vectors) of an n-square unitary matrix are an orthonor-
mal set.
The inverse and the transpose of a unitary matrix are unitary.
The product of two or more unitary matrices is unitary.
The determinant of a unitary matrix has absolute value 1.
Unitary transformations
The linear transformation Y = AX (where A is unitary and X is a vector), is called
a unitary transformation.
If the matrix is unitary, the linear transformation preserves length.
Rank of a matrix
If |A| = 0, it is called non-singular; if |A| = 0, it is called singular.
A non-singular matrix is said to have rank r if at least one of its r-square minors
is non-zero while if every (r + 1) minor, if it exists, is zero.
Elementary transformations
(i) The interchange of the ith rows and jth rows or ith column or jth column.
(ii) The multiplication of every element of the ith row or ith column by a non-zero
scalar.
(iii) The addition to the elements of the ith row (column) by k (a scalar) times the
corresponding elements of the jth row (column). These elementary transfor-
mations known as row elementary or row transformations do not change the
order of the matrix.

Equivalence
A and B are said to be equivalent (A ∼ B) if one can be obtained from the other by
a sequence of elementary transformations.
The adjoint of a square matrix
If A = [ai j ] is a square matrix and αi j the cofactor of ai j then
adj A =
⎡
⎢
⎢
⎣
α11 α21 · · · αn1
α12 α22 · · · · · ·
· · · · · · · · · · · ·
· · · · · · · · · αnn
⎤
⎥
⎥
⎦
The cofactor αi j = (−1)i+ j
Mi j
where Mi j is the minor obtained by striking off the ith row and jth column and
computing the determinant from the remaining elements.
Inverse from the adjoint
A−1
=
adj A
|A|
Inverse for orthogonal matrices
A−1
= A′
Inverse of unitary matrices
A−1
= (A)′
Characteristic equation
Let AX = λX (1.84)
be the transformation of the vector X into λX, where λ is a number, then λ is called
the eigen or characteristic value.
From (1.84):
(A − λI)X =
⎡
⎢
⎢
⎢
⎣
a11 − λ a12 · · · a1n
a21 a22 − λ · · · a2n
.
.
. · · · · · · · · ·
an1 · · · · · · ann − λ
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
x1
x2
.
.
.
xn
⎤
⎥
⎥
⎥
⎦
= 0 (1.85)

The system of homogenous equations has non-trivial solutions if
|A − λI| =
⎡
⎢
⎢
⎢
⎣
a11 − λ a12 · · · a1n
a21 a22 − λ · · · a2n
.
.
. · · · · · · · · ·
an1 · · · · · · ann − λ
⎤
⎥
⎥
⎥
⎦
= 0 (1.86)
The expansion of this determinant yields a polynomial φ(λ) = 0 is called the
characteristic equation of A and its roots λ1, λ1, . . . , λn are known as the charac-
teristic roots of A. The vectors associated with the characteristic roots are called
invariant or characteristic vectors.
Diagonalization of a square matrix
If a matrix C is found such that the matrix A is diagonalized to S by the
transformation
S = C−1
AC (1.87)
then S will have the characteristic roots as the diagonal elements.
Ordinary differential equations
The methods of solving typical ordinary differential equations are from the book
“Differential and Integral Calculus” by William A. Granville published by Ginn
Co., 1911.
An ordinary differential equation involves only one independent variable, while
a partial differential equation involves more than one independent variable.
The order of a differential equation is that of the highest derivative in it.
The degree of a differential equation which is algebraic in the derivatives is the
power of the highest derivative in it when the equation is free from radicals and
fractions.
Differential equations of the first order and of the first degree
Such an equation must be brought into the form Mdx+Ndy = 0, in which M and N
are functions of x and y.
Type I variables separable
When the terms of a differential equation can be so arranged that it takes on the
form
(A) f (x) dx + F(y) dy = 0
where f (x) is a function of x alone and F(y) is a function of y alone, the process is
called separation of variables and the solution is obtained by direct integration.
(B)

f (x) dx +

F(y) dy = C
where C is an arbitrary constant.

Equations which are not in the simple form (A) can be brought into that form by
the following rule for separating the variables.
First step: Clear off fractions, and if the equation involves derivatives, multiply
through by the differential of the independent variable.
Second step: Collect all the terms containing the same differential into a single
term. If then the equation takes on the form
XY dx + X′
Y′
dy = 0
where X, X′
are functions of x alone, and Y, Y′
are functions of y alone, it may be
brought to the form (A) by dividing through by X′
Y′
.
Third step: Integrate each part separately as in (B).
Type II homogeneous equations
The differential equation
Mdx + Ndy = 0
is said to be homogeneous when M and N are homogeneous function of x and y
of the same degree. In effect a function of x and y is said to be homogenous in the
variable if the result of replacing x and y by λx and λy (λ being arbitrary) reduces
to the original function multiplied by some power of λ. This power of λ is called
the degree of the original function. Such differential equations may be solved by
making the substitution
y = vx
This will give a differential equation in v and x in which the variables are sepa-
rable, and hence we may follow the rule (A) of type I.
Type III linear equations
A differential equation is said to be linear if the equation is of the first degree in the
dependent variables (usually y) and its derivatives. The linear differential equation
of the first order is of the form
dy
dx
+ Py = Q
where P, Q are functions of x alone, or constants, the solution is given by
ye

Pdx
=

Qe

Pdx
dx + C

Type IV equations reducible to linear form
Some equations that are not linear can be reduced to the linear form by a suitable
substitution, for example
(A)
dy
dx
+ Py = Qyn
where P, Q are functions of x alone, or constants. Equation (A) may be reduced to
the linear form (A), Type III by means of the substitution x = y−n+1
.
Differential equations of the nth Order and of the nth degree
Consider special cases of linear differential equations.
Type I – The linear differential equation
(A)
dn
y
dxn
+ P1
dn−1
y
dxn−1
+ P2
dn−2
y
dxn−2
+ · · · + Pn y = 0
in which coefficients P1, P2, . . . Pn are constants.
Consider the differential equation of third order
(B)
d3
y
dx3
+ P1
d2
y
dx2
+ P2
dy
dx
+ P3 y = 0
where P1, P2 and P3 are constants. The corresponding auxiliary equation is
r3
+ P1r2
+ P2r + P3 = 0
Let the roots be r1,r2,r3.
If r1,r2,r3 are real and distinct,
y = C1er1x
+ C2er2x
+ C3er3x
If r1,r2,r3 are real and equal
y = C1e−r1x
+ C2xe−r2x
+ C3x2
e−r3x
In case a + bi and a − bi are each multiple roots of the auxiliary equation occur-
ring s times, the solutions would be
C1eax
cos bx, C2xeax
cos bx, C3x2
eax
cos bx, . . . Cs xs−1
eax
cos bx
C′
1eax
sin bx, C′
2xeax
sin bx, C′
3x2
eax
sin bx, . . . C′
s xs−1
eax
sin bx
Summary for the rule for solving differential equations of the type
dn
y
dxn
+ P1
dn−1
y
dxn−1
+ P2
dn−2
y
dxn−2
+ · · · + Pn y = 0
where P1, P2, . . . Pn are constants.

First step: Write down the corresponding auxiliary equation
Dn
+ p1 Dn−1
+ p2 Dn−2
+ · · · + pn = 0
Second step: Solve completely the auxiliary equation.
Third step: From the roots of the auxiliary equation, write down the correspond-
ing particular solutions of the differential equation as follows
Auxiliary equation Differential equation
(a) Each distinct real root r1 Gives a particular solution er1x
(b) Each distinct pair of imaginary
roots a ± bi
Gives two particular solutions
eax
cos bx, eax
sin bx
(c) A multiple root occurring s times Gives s particular solutions obtained by
multiplying the particular solutions
(a) or (b) by 1, x, x2
, . . . , xn−1
Fourth step: Multiple each of the n independent solutions by an arbitrary constant
and add the results. This gives the complete solution.
Type II
(I)
dn
y
dxn
+ P1
dn−1
y
dxn−1
+ P2
dn−2
y
dxn−2
+ · · · + Pn y = X
where X is a function of x alone, or constant, and P1, P2, . . . Pn are constants.
When X = 0, (I) reduces to (A) Type I.
(J)
dn
y
dxn
+ P1
dn−1
y
dxn−1
+ P2
dn−2
y
dxn−2
+ · · · + Pn y = 0
The complete solution of (J) is called the complementary function of (I).
Let u be the complete solution of (J), i.e. the complementary function of (I), and
v any particular solution of (I). Then
dn
v
dxn
+ P1
dn−1
v
dxn−1
+ P2
dn−2
v
dxn−2
+ · · · + Pnv = X
and
dn
u
dxn
+ P1
dn−1
u
dxn−1
+ P2
dn−2
u
dxn−2
+ · · · + Pnu = 0
Adding we get
dn
(u + v)
dxn
+ P1
dn−1
(u + v)
dxn−1
+ P2
dn−2
(u + v)
dxn−2
+ · · · + Pn(u + v) = X
showing that u + v is a solution of I.

Procedure:
First step: Replace the RHS member of the given equation (I) by zero and solve
the complimentary function of I to get y = u.
Second step: Differentiate successively the given equation (I) and obtain, either
directly or by elimination, a differential equation of a higher order of type I.
Third step: Solving this new equation by the previous rule we get its complete
solution
y = u + v
where the part u is the complimentary function of (I) already found in the first step,
and v is the sum of additional terms found
Fourth step: To find the values of the constants of integration in the particular
solution v, substitute
y = u + v
and its derivatives in the equation (I). In the resulting identity equation equate the
coefficients of like terms, solve for constants of integration, substitute their values
back in
y = u + v
giving the complete solution of (I).
Type III
dn
y
dxn
= X
where X is a function of x alone, or constant
Integrate n times successively. Each integration will introduce one arbitrary
constant.
Type IV
d2
y
dx2
= Y
where Y is a function of y alone
Multiply the LHS member by the factor 2dy
dx
dx and the RHS member by k equiv-
alent factor 2dy
2
dy
dx
d2
y
dx2
dx = d

dy
dx
2
= 2Y dy

d

dy
dx
2
=

dy
dx
2
=

2Y dy + C1

1.2 Problems 21
Extract the square root of both members, separate the variables, and integrate
again, introducing the second arbitrary constant C2.
Complex variables
Complex number z = r(cos θ + i sin θ), where i =
√
−1
zn
= cos nθ + i sin nθ
Analytic functions
A function f of the complex variable z is analytic at a point zo if its derivative f ′
(z)
exists not only at zo but at every point z in some neighborhood of zo. As an example
if f (z) = 1
z
then f ′
(z) = − 1
z2 (z = 0). Thus f is analytic at every point except the
point z = 0, where it is not continuous, so that f ′
(0) cannot exist. The point z = 0
is called a singular point.
Contour
A contour is a continuous chain of finite number of smooth arcs. If the contour
is closed and does not intersect itself, it is called a closed contour. Boundaries of
triangles and rectangles are examples. Any closed contour separates the plane into
two domains each of which have the points of C as their only boundary points. One
of these domains is called the interior of C, is bounded; the other, the exterior of C,
is unbounded.
Contour integral is similar to the line integral except that here one deals with the
complex plane.
The Cauchy integral formula
Let f be analytic everywhere within and on a closed contour C. If zo is any point
interior to C, then
f (zo) =
1
2πi

C
f (z) dz
z − zo
where the integral is taken in the positive sense around C.
1.2 Problems
1.2.1 Vector Calculus
1.1 If φ = 1
r
, where r = (x2
+ y2
+ z2
)1/2
, show that ∇φ = r
r3 .
1.2 Find a unit vector normal to the surface xy2
+ xz = 1 at the point (−1, 1, 1).
1.3 Show that the divergence of the Coulomb or gravitational force is zero.
1.4 If A and B are irrotational, prove that A×B is Solenoidal that is div (A×B) =
0

1.5 (a) If the field is centrally represented by F = f (x, y, z), r = f (r)r, then it
is conservative conditioned by curl F = 0, that is the field is irrotational.
(b) What should be the function F(r) so that the field is solenoidal?
1.6 Evaluate

c A . dr from the point P(0, 0, 0) to Q(1, 1, 1) along the curve
r = ît + ĵt2
+ k̂r3
with x = t, y = t2
, z = t3
, where A = yî + xz ĵ + xyzk̂
1.7 Evaluate
!
c A . dr around the closed curve C defined by y = x2
and y2
= 8x,
with A = (x + y)î + (x − y) ĵ
1.8 (a) Show that F = (2xy + z2
)î + x2
ĵ + xyzk̂, is a conservative force field.
(b) Find the scalar potential.
(c) Find the work done in moving a unit mass in this field from the point
(1, 0, 1) to (2, 1, −1).
1.9 Verify Green’s theorem in the plane for

c(x + y)dx + (x − y) dy, where C is
the closed curve of the region bonded by y = x2
and y2
= 8x.
1.10 Show that

s A . ds = 12
5
π R2
, where S is the sphere of radius R and
A = îx3
+ ĵ y3
+ k̂z3
1.11 Evaluate

r A . dr around the circle x2
+ y2
= R2
in the xy-plane, where
A = 2yî − 3x ĵ + zk̂
1.12 (a) Prove that the curl of gradient is zero.
(b) Prove that the divergence of a curl is zero.
1.13 If φ = x2
y − 2xz3
, then:
(a) Find the Gradient.
(b) Find the Laplacian.
1.14 (a) Find a unit vector normal to the surface x2
y + xz = 3 at the point
(1, −1, 1).
(b) Find the directional derivative of φ = x2
yz + 2xz3
at (1, 1, −1) in the
direction 2î − 2 ĵ + k̂.
1.15 Show that the divergence of an inverse square force is zero.
1.16 Find the angle between the surfaces x2
+ y2
+ z2
= 1 and z = x2
+ y2
− 1 at
the point (1, +1, −1).
1.2.2 Fourier Series and Fourier Transforms
1.17 Develop the Fourier series expansion for the saw-tooth (Ramp) wave f (x) =
x/L, −L x L, as in Fig. 1.2.
1.18 Find the Fourier series of the periodic function defined by:
f (x) = 0, if − π ≤ x ≤ 0
f (x) = π, if 0 ≤ x ≤ π

1.2 Problems 23
Fig. 1.2 Saw-tooth wave
1.19 Use the result of Problem 1.18 for the Fourier series for the square wave to
prove that:
1 −
1
3
+
1
5
−
1
7
+ · · · =
π
4
1.20 Find the Fourier transform of f (x) =

1, |x| a
0, |x| a
1.21 Use the Fourier integral to prove that:
∞
0
cos axdx
1 + a2
=
π
2
e−x
1.22 Show that the Fourier transform of the normalized Gaussian distribution
f (t) =
1
τ
√
2π
e
−t2
2τ2
, −∞ t ∞
is another Gaussian distribution.
1.2.3 Gamma and Beta Functions
1.23 The gamma function is defined by:
Γ(z) =
∞
0
e−x
xz−1
dx, (Re z 0)
(a) Show that Γ(z + 1) = zΓ(z)
(b) And if z is a positive integer n, then Γ(n + 1) = n!
1.24 The Beta function B(m, n) is defined by the definite integral:
B(m, n) =
1
0
xm−1
(1 − x)n−1
dx
and this defines a function of m and n provided m and n are positive. Show
that:
B(m, n) =
T (m)T (n)
T (m + n)

1.25 Use the Beta functions to evaluate the definite integral
π/2
0 (cos θ)r
dθ
1.26 Show that:
(a) Γ(n)Γ(1 − n) = π
sin(nπ)
; 0 n 1
(b) |Γ(in)|2
= π
n sin h(nπ)
1.2.4 Matrix Algebra
1.27 Prove that the characteristic roots of a Hermitian matrix are real.
1.28 Find the characteristic equation and the Eigen values of the matrix:
⎛
⎝
1 −1 1
0 3 −1
0 0 2
⎞
⎠
1.29 Given below the set of matrices:
A =

−1 0
0 −1

, B =

0 1
1 0

, C =

2 0
0 2

, D =
√
3
2
1
2
−1
2
√
3
2
#
what is the effect when A, B, C and D act separately on the position vector

x
y

?
1.30 Find the eigen values of the matrix:
⎛
⎝
6 −2 2
−2 3 −1
2 −1 3
⎞
⎠
1.31 Diagonalize the matrix given in Problem 1.30 and find the trace (Tr = λ1 +
λ2 + λ3)
1.32 In the Eigen vector equation AX = λX, the operator A is given by
A =

3 2
4 1

.
Find:
(a) The Eigen values λ
(b) The Eigen vector X
(c) The modal matrix C and it’s inverse C−1
(d) The product C−1
AC
1.2.5 Maxima and Minima
1.33 Solve the equation x3
− 3x + 3 = 0, by Newton’s method.
1.34 (a) Find the turning points of the function f (x) = x2
e−x2
.
(b) Is the above function odd or even or neither?

1.2 Problems 25
1.2.6 Series
1.35 Find the interval of convergence for the series:
x −
x2
22
+
x3
32
−
x4
42
+ · · ·
1.36 Expand log x in powers of (x − 1) by Taylor’s series.
1.37 Expand cos x into an infinite power series and determine for what values of x
it converges.
1.38 Expand sin(a + x) in powers of x by Taylor’s series.
1.39 Sum the series s = 1 + 2x + 3x2
+ 4x3
+ · · · , |x| 1
1.2.7 Integration
1.40 (a) Evaluate the integral:

sin3
x cos6
x dx
(b) Evaluate the integral:

sin4
x cos2
x dx
1.41 Evaluate the integral:

1
2x2 − 3x − 2
dx
1.42 (a) Sketch the curve in polar coordinates r2
= a2
sin 2θ
(b) Find the area within the curve between θ = 0 and θ = π/2.
1.43 Evaluate:

(x3
+ x2
+ 2)
(x2 + 2)2
dx
1.44 Evaluate the definite integral:
+∞
0
4a3
x2 + 4a2
dx
1.45 (a) Evaluate:

tan6
x sec4
x dx
(b) Evaluate:

tan5
x sec3
x dx

1.46 Show that:
4
2
2x + 4
x2 − 4x + 8
dx = ln 2 + π
1.47 Find the area included between the semi-cubical parabola y2
= x3
and the
line x = 4
1.48 Find the area of the surface of revolution generated by revolving the hypocy-
cloid x2/3
+ y2/3
= a2/3
about the x-axis.
1.49 Find the value of the definite double integral:
a
0
√
a2−x2
0
(x + y) dy dx
1.50 Calculate the area of the region enclosed between the curve y = 1/x, the
curve y = −1/x, and the lines x = 1 and x = 2.
1.51 Evaluate the integral:

dx
x2 − 18x + 34
1.52 Use integration by parts to evaluate:
1
0
x2
tan−1
x dx
[University of Wales, Aberystwyth 2006]
1.53 (a) Calculate the area bounded by the curves y = x2
+ 2 and y = x − 1 and
the lines x = −1 to the left and x = 2 to the right.
(b) Find the volume of the solid of revolution obtained by rotating the area
enclosed by the lines x = 0, y = 0, x = 2 and 2x + y = 5 through 2π
radians about the y-axis.
1.54 Consider the curve y = x sin x on the interval 0 ≤ x ≤ 2π.
(a) Find the area enclosed by the curve and the x-axis.
(b) Find the volume generated when the curve rotates completely about the
x-axis.
1.2.8 Ordinary Differential Equations
1.55 Solve the differential equation:
dy
dx
=
x3
+ y3
3xy2

1.2 Problems 27
1.56 Solve:
d3
y
dx3
− 3
d2
y
dx2
+ 4y = 0 (Osmania University)
1.57 Solve:
d4
y
dx4
− 4
d3
y
dx3
+ 10
d2
y
dx2
− 12
dy
dx
+ 5y = 0 (Osmania University)
1.58 Solve:
d2
y
dx2
+ m2
y = cos ax
1.59 Solve:
d2
y
dx2
− 5
dy
dx
+ 6y = x
1.60 Solve the equation of motion for the damped oscillator:
d2
x
dt2
+ 2
dx
dt
+ 5x = 0
subject to the condition, x = 5, dx/dt = −3 at t = 0.
1.61 Two equal masses are connected by means of a spring and two other identical
springs fixed to rigid supports on either side (Fig. 1.3), permit the masses
to jointly undergo simple harmonic motion along a straight line, so that the
system corresponds to two coupled oscillations.
Assume that m1 = m2 = m and the stiffness constant is k for both the
oscillators.
(a) Form the differential equations for both the oscillators and solve the
coupled equations and obtain the frequencies of oscillations.
(b) Discuss the modes of oscillation and sketch the modes.
Fig. 1.3 Coupled oscillator
1.62 A cylinder of mass m is allowed to roll on a spring attached to it so that it
encounters simple harmonic motion about the equilibrium position. Use the
energy conservation to form the differential equation. Solve the equation and
find the time period of oscillation. Assume k to be the spring constant.

1.63 Solve:
d2
y
dx2
− 8
dy
dx
= −16y
1.64 Solve:
x2 dy
dx
+ y(x + 1)x = 9x2
1.65 Find the general solution of the differential equation:
d2
y
dx2
+
dy
dx
− 2y = 2cosh(2x)
1.66 Solve:
x
dy
dx
− y = x2
1.67 Find the general solution of the following differential equations and write
down the degree and order of the equation and whether it is homogenous or
in-homogenous.
(a) y′
− 2
x
y = 1
x3
(b) y′′
+ 5y′
+ 4y = 0
1.68 Find the general solution of the following differential equations:
(a) dy
dx
+ y = e−x
(b) d2
y
dx2 + 4y = 2 cos(2x)
1.69 Find the solution to the differential equation:
dy
dx
+
3
x + 2
y = x + 2
which satisfies y = 2 when x = −1, express your answer in the form y =
f (x).
1.70 (a) Find the solution to the differential equation:
d2
y
dx2
− 4
dy
dx
+ 4y = 8x2
− 4x − 4
which satisfies the conditions y = −2 and dy
dx
= 0 when x = 0.
(b) Find the general solution to the differential equation:
d2
y
dx2
+ 4y = sin x

1.2 Problems 29
1.71 Find a fundamental set of solutions to the third-order equation:
d3
y
dx3
−
d2
y
dx2
+
dy
dx
− y = 0
1.2.9 Laplace Transforms
1.72 Consider the chain decay in radioactivity A
λA
→ B
λB
→ C, where λA and λB are
the disintegration constants. The equations for the radioactive decays are:
dNA(t)
dt
= −λA NA(t), and
dNB(t)
dt
= −λ2 NB(t) + λA NA(t)
where NA(t) and NB(t) are the number of atoms of A and B at time t, with
the initial conditions NA(0) = N0
A; NB(0) = 0. Apply Laplace transform to
obtain NA(t) and NB(t), the number of atoms of A and B as a function of time
t, in terms of N0
A, λA and λB.
1.73 Consider the radioactive decay:
A
λA
→ B
λB
→ C (Stable)
The equations for the chain decay are:
dNA
dt
= −λA NA (1)
dNB
dt
= −λB NB + λA NA (2)
dNC
dt
= +λB NB (3)
with the initial conditions NA(0) = N0
A; NB(0) = 0; NC (0) = 0, where various
symbols have the usual meaning. Apply Laplace transforms to find the growth
of C.
1.74 Show that:
(a) £(eax
) = 1
s−a
, i f s a
(b) £(cos ax) = s
s2+a2 , s 0
(c) £(sin ax) = a
s2+a2
where £ means Laplacian transform.
1.2.10 Special Functions
1.75 The following polynomial of order n is called Hermite polynomial:
H′′
n − 2ξ H′
n + 2nHn = 0
Show that:
(a) H′
n = 2nHn−1
(b) Hn+1 = 2ξ Hn − 2nHn−1

1.76 The Bessel function Jn (x) is given by the series expansion
Jn (x) =

(−1)k
(x/2)n+2k
k!Γ(n + k + 1)
Show that:
(a) d
dx
[xn
Jn(x) ] = xn
Jn−1(x)
(b) d
dx
[x−n
Jn(x)] = −x−n
Jn+1(x)
1.77 Prove the following relations for the Bessel functions:
(a) Jn−1(x) − Jn+1(x) = 2 d
dx
Jn(x)
(b) Jn−1(x) + Jn+1(x) = 2n
x
Jn(x)
1.78 Given that Γ
1
2

=
√
π, obtain the formulae:
(a) J1/2(x) =
$
2
πx
sin x
(b) J−1/2(x) =
$
2
πx
cos x
1.79 Show that the Legendre polynomials have the property:
l
−l
Pn(x)Pm(x) dx =
2
2n + 1
, if m = n
= 0, if m = n
1.80 Show that for large n and small θ, Pn(cos θ) ≈ J0(nθ)
1.81 For Legendre polynomials Pl (x) the generating function is given by:
T (x, s) = (1 − 2sx + s2
)−1/2
=
∞
l=0
Pl(x)sl
, s 1
Use the generating function to show:
(a) (l + 1)Pl+1 = (2l + 1)x Pl − l Pl−1
(b) Pl(x) + 2x P′
l (x) = P′
l+1(x) + P′
l−1(x), Where prime means differentiation
with respect to x.
1.82 For Laguerre’s polynomials, show that Ln(0) = n!. Assume the generating
function:
e−xs/(1−s)
1 − s
=
∞
n=0
Ln(x)sn
n!
1.2.11 Complex Variables
1.83 Evaluate
!
c
dz
z−2
where C is:
(a) The circle |z| = 1
(b) The circle |z + i| = 3

1.2 Problems 31
1.84 Evaluate
!
c
4z2
−3z+1
(z−1)3 dz when C is any simple closed curve enclosing z = 1.
1.85 Locate in the finite z-plane all the singularities of the following function and
name them:
4z3
− 2z + 1
(z − 3)2(z − i)(z + 1 − 2i)
1.86 Determine the residues of the following function at the poles z = 1 and
z = −2:
1
(z − 1)(z + 2)2
1.87 Find the Laurent series about the singularity for the function:
ex
(z − 2)2
1.88 Evaluate I =
∞
0
dx
x4+1
1.2.12 Calculus of Variation
1.89 What is the curve which has shortest length between two points?
1.90 A bead slides down a frictionless wire connecting two points A and B as in
the Fig. 1.4. Find the curve of quickest descent. This is known as the Brachis-
tochrome, discovered by John Bernoulli (1696).
Fig. 1.4 Brachistochrome
1.91 If a soap film is stretched between two circular wires, both having their planes
perpendicular to the line joining their centers, it will form a figure of revolution
about that line. At every point such as P (Fig. 1.5), the horizontal component
of the surface of revolution acting around a vertical section of the film will be
constant. Find the equation to the figure of revolution.
1.92 Prove that the sphere is the solid figure of revolution which for a given surface
area has maximum volume.

Fig. 1.5 Soap film stretched
between two parallel circular
wires
1.2.13 Statistical Distributions
1.93 Poisson distribution gives the probability that x events occur in unit time
when the mean rate of occurrence is m.
Px =
e−m
mx
x!
(a) Show that Px is normalized.
(b) Show that the mean rate of occurrence or the expectation value x , is
equal to m.
(c) Show that the S.D., σ =
√
m
(d) Show that Pm−1 = Pm
(e) Show that Px−1 = x
m
Pm and Px+1 = m
x+1
Px
1.94 The probability of obtaining x successes in N-independent trials of an event
for which p is the probability of success and q the probability of failure in a
single trial is given by the Binomial distribution:
B(x) =
N!
x!(N − x)!
px
qN−x
= CN
x px
qN−x
(a) Show that B(x) is normalized.
(b) Show that the mean value is Np
(c) Show that the S.D. is
√
Npq
1.95 A G.M. counter records 4,900 background counts in 100 min. With a radioac-
tive source in position, the same total number of counts are recorded in
20 min. Calculate the percentage of S.D. with net counts due to the source.
[Osmania University 1964]
1.96 (a) Show that when p is held fixed, the Binomial distribution tends to a nor-
mal distribution as N is increased to infinity.
(b) If Np is held fixed, then binomial distribution tends to Poisson distribu-
tion as N is increased to infinity.
1.97 The background counting rate is b and background plus source is g. If the
background is counted for the time tb and the background plus source for a
time tg, show that if the total counting time is fixed, then for minimum sta-
tistical error in the calculated counting rate of the source(s), tb and tg should
be chosen so that tb/tg =
√
b/g

1.3 Solutions 33
1.98 The alpha ray activity of a material is measured after equal successive inter-
vals (hours), in terms of its initial activity as unity to the 0.835; 0.695; 0.580;
0.485; 0.405 and 0.335. Assuming that the activity obeys an exponential
decay law, find the equation that best represents the activity and calculate
the decay constant and the half-life.
1.99 Obtain the interval distribution and hence deduce the exponential law of
radioactivity.
1.100 In a Carbon-dating experiment background counting rate = 10 C/M. How
long should the observations be made in order to have an accuracy of 5%?
Assume that both the counting rates are measured for the same time.
14
C + background rate = 14.5 C/M
1.101 Make the best fit for the parabola y = a0 + a1x + a2x2
, with the given pairs
of values for x and y.
x −2 −1 0 +1 +2 +3
y 9.1 3.5 0.5 0.8 4.6 11.0
1.102 The capacitance per unit length of a capacitor consisting of two long concen-
tric cylinder with radii a and b, (b a) is C = 2πε0
ln(b/a)
. If a = 10 ± 1 mm and
b = 20 ± 1 mm, with what relative precision is C measured?
[University of Cambrdige Tripos 2004]
1.103 If f (x) is the probability density of x given by f (x) = xe−x/λ
over the
interval 0 x ∞, find the mean and the most probable values of x.
1.2.14 Numerical Integration
1.104 Calculate
10
1 x2
dx, by the trapezoidal rule.
1.105 Calculate
10
1 x2
dx, by the Simpson’s rule.
1.3 Solutions
1.1 ∇Φ =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

(x2
+ y2
+ z2
)−1/2
=

−
1
2
.2xî −
1
2
.2y ĵ −
1
2
.2zk̂

(x2
+ y2
+ z2
)−3/2
= −(xî + y ĵ + zk̂)(x2
+ y2
+ z2
)− 3
2 = −
r
r3

1.2 ∇(xy2
+ xz) =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

(xy2
+ xz)
= (y2
+ z)î + (2xy) ĵ + xk̂
= 2î − 2 ĵ − k̂, at(−1, 1, 1)
A unit vector normal to the surface is obtained by dividing the above vector
by its magnitude. Hence the unit vector is
(2î − 2 ĵ − k̂)[(2)2
+ (−2)2
+ (−1)2
]−1/2
=
2
3
î −
2
3
ĵ −
1
3
k̂
1.3 F ∝ 1/r2
∇ . (r−3
r) = r−3
∇ . r + r . ∇r−3
But ∇ . r =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

.

îx + ĵ y + k̂z

=
∂x
∂x
+
∂y
∂y
+
∂z
∂z
= 3
r . ∇r−3
= (xî + y ĵ + zk̂) .

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

(x2
+ y2
+ z2
)−3/2
= (xî + y ĵ + zk̂) .

−
3
2

. (2xî + 2y ĵ + 2zk̂)(x2
+ y2
+ z2
)−5/2
= −3(x2
+ y2
+ z2
)(x2
+ y2
+ z2
)− 5
2 = −3r−3
Thus ∇ . (r−3
r) = 3r−3
− 3r−3
= 0
1.4 By problem ∇ × A = 0 and ∇ × B = 0, it follows that
B . (∇ × A) = 0
A. (∇ × B) = 0
Subtracting, B . (∇ × A) − A . (∇ × B) = 0
Now ∇ . (A × B) = B . (∇ × A) − A . (∇ × B)
Therefore ∇ . (A × B) = 0, so that (A × B) is solenoidal.
1.5 (a) Curl {r f (r)} = ∇ × {r f (r)} = ∇ × {x f (r)î + y f (r) ĵ + zf (r)k̂}
=
%
%
%
%
%
%
%
%
î ĵ k̂
∂
∂x
∂
∂y
∂
∂z
x f (r) y f (r) zf (r)
%
%
%
%
%
%
%
%
=

z
∂ f
∂y
− y
∂ f
∂z

î +

x
∂ f
∂z
− z
∂ f
∂x

ĵ +

y
∂ f
∂x
− x
∂ f
∂y

k̂
But ∂ f
∂x
= ∂ f
∂r
∂r
∂x

= ∂ f
∂r
∂(x2
+y2
+z2
)1/2
∂x
= x f ′
r
Similarly ∂ f
∂y
= yf ′
r
and ∂ f
∂z
= zf ′
r
, where prime means differentiation with
respect to r.
Thus,
curl{r f (r)} =

zy f ′
r
−
yzf ′
r

î +

xzf ′
r
−
zx f ′
r

ĵ +

yx f ′
r
−
xy f ′
r

k̂ = 0

1.3 Solutions 35
(b) If the field is solenoidal, then, ∇.rF(r) = 0
∂(x F(r))
∂x
+
∂(yF(r))
∂y
+
∂(zF(r))
∂z
= 0
F + x
∂ F
∂x
+ F + y
∂F
∂y
+ F + z
∂ F
∂z
= 0
3F(r) + x
∂F
∂r
x
r
+ y
∂ F
∂r
y
r
+ z
∂ F
∂r
z
r
= 0
3F(r) +

∂ F
∂r

x2
+ y2
+ z2
r

= 0
But (x2
+ y2
+ z2
) = r2
, therefore, ∂ F
∂r
= −3F(r)
r
Integrating, ln F = −3 lnr + ln C where C = constant
ln F = − lnr3
+ ln C = ln
C
r3
Therefore F = C/r3
. Thus, the field is A =
r
r3
(inverse square law)
1.6 x = t, y = t2
, z = t3
Therefore, y = x2
, z = x3
, dy = 2xdx, dz = 3x2
dx

c
A.dr =

(yî + xz ĵ + xyzk̂).(îdx + ĵdy + k̂dz)
=
1
0
x2
dx + 2
1
0
x5
dx + 3
1
0
x8
dx
=
1
3
+
1
3
+
1
3
= 1
1.7 The two curves y = x2
and y2
= 8x intersect at (0, 0) and (2, 4). Let us
traverse the closed curve in the clockwise direction, Fig. 1.6.

c
A.dr =

c
[(x + y)î + (x − y) ĵ].(î dx + ĵ dy)
=

c
[(x + y)dx + (x − y)dy]
=
0
2
[(x + x2
)dx + (x − x2
)2xdx] (along y = x2
)
Fig. 1.6 Line integral for a
closed curve

+
4
0

y2
8
+ y

ydy
4
+

y2
8
− y

dy (along y2
= 8x)
= +
16
3
1.8 (a) It is sufficient to show that Curl F = 0
∇ × F =
%
%
%
%
%
%
%
%
i j k
∂
∂x
∂
∂y
∂
∂z
2xy + z2
x2
2xz
%
%
%
%
%
%
%
%
= î.0 − ĵ(2z − 2z) + k̂(2x − 2x) = 0
(b) dΦ = F. dr = ((2xy + z2
)î + x2
ĵ + 2xzk̂)).(îdx + ĵdy + k̂dz)
= (2xy + z2
) dx + x2
dy + 2xzdz
= (2xydx + x2
dy) + (z2
dx + 2xzdz)
= d(x2
y) + d(z2
x) = d(x2
y + xz2
)
Therefore Φ = x2
y + xz2
+ constant
(c) Work done = Φ2 − Φ1 = 5.0
1.9 Let U = x + y; V = x − y
∂U
∂x
= 1;
∂V
∂y
= −1
The curves y = x2
and y2
= 8x intersect at (0, 0) and (2, 4).

∂U
∂x
−
∂V
∂x

dx dy =

S
(1 − (−1))dx dy = 2
2
x=0
2
√
2x
y=x2
dxdy
= 2
2
0
2
√
2x
x2
dy
'
dx = 2
2
0
(2
√
2
√
x − x2
) dx = 2

4
√
2
3
x3/2
−
x3
3
'2
0
=
16
3
This is in agreement with the value obtained in Problem 1.7 for the line inte-
gral.
1.10 Use the divergence theorem

A . ds =

∇. A dν
But ∇. A =
∂
∂x
x3
+
∂
∂y
y3
+
∂
∂z
z3
= 3x2
+ 3y2
+ 3z2
= 3(x2
+ y2
+ z2
) = 3R2

A . ds =

3R2
dν =

(3R2
)(4π R2
dR)
= 12π

R4
dR =
12
5
π R5
1.11

c
A . dr =

(2yî − 3x ĵ + zk̂).(dxî + dy ĵ + dzk̂)
=

(2ydx − 3xdy + zdz)

1.3 Solutions 37
Put x = R cos θ, dx = −R sin θ dθ, y = R sin θ, dy = R cos θ, z = 0, 0
θ 2π

A . dr = −2R2

sin2
θ dθ − R2

cos2
θ dθ
= −2π R2
− π R2
= −3π R2
1.12 (a) ∇ × (∇Φ) =
%
%
%
%
%
%
%
i j k
∂
∂x
∂
∂y
∂
∂z
∂Φ
∂x
∂Φ
∂y
∂Φ
∂z
%
%
%
%
%
%
%
= i

∂2
Φ
∂y∂z
−
∂2
Φ
∂z∂y

− j

∂2
Φ
∂x∂z
−
∂2
Φ
∂z∂x

+ k

∂2
Φ
∂x∂y
−
∂2
Φ
∂y∂x

= 0
because the order of differentiation is immaterial and terms in brackets
cancel in pairs.
(b) To show ∇.(∇ × V) = 0

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

.
%
%
%
%
%
%
i j k
∂
∂x
∂
∂y
∂
∂z
Vx Vy Vz
%
%
%
%
%
%
=

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

·

î
%
%
%
%
∂
∂y
∂
∂z
Vy Vz
%
%
%
% − ĵ
%
%
%
%
∂
∂x
∂
∂z
Vx Vz
%
%
%
% + k̂
%
%
%
%
∂
∂x
∂
∂y
Vx Vy
%
%
%
%

=
∂
∂x
%
%
%
%
∂
∂y
∂
∂z
Vy Vz
%
%
%
% −
∂
∂y
%
%
%
%
∂
∂x
∂
∂z
Vx Vz
%
%
%
% +
∂
∂z
%
%
%
%
∂
∂x
∂
∂y
Vx Vy
%
%
%
%
=
%
%
%
%
%
%
∂
∂x
∂
∂y
∂
∂z
∂
∂x
∂
∂y
∂
∂z
Vx Vy Vz
%
%
%
%
%
%
= 0
The value of the determinant is zero because two rows are identical.
1.13 Φ = x2
y − 2xz3
(a) ∇Φ =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

(x2
y − 2xz3
)
= 2(xy − z3
)î + x2
ĵ + 6xz2
k̂
(b) ∇2
Φ =

∂2
∂x2
+
∂2
∂y2
+
∂2
∂z2

(x2
y − 2xz3
)
= 2y − 12xz
1.14 (a) ∇(x2
y + xz) =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

(x2
y + xz)
= (2xy + z)î + x2
ĵ + xk̂
= −î + ĵ + k̂

A unit vector normal to the surface is obtained by dividing the above
vector by its magnitude. Hence the unit vector is
−î + ĵ + k̂
[(−1)2 + 12 + 12]1/2
= −
î
√
3
+
ĵ
√
3
+
k̂
√
3
(b) ∇Φ =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

(x2
yz + 2xz2
)
= (2xyz + 2z2
)î + x2
z ĵ + 4xzk̂
= − ĵ − 4k̂ at (1, 1, −1)
The unit vector in the direction of 2î − 2 ĵ + k̂, is
n̂ =
2î − 2 ĵ + k̂
[22 + (−2)2 + 12]1/2
= 2î/3 − 2 ĵ/3 + k̂/3
The required directional derivative is
∇Φ.n = (− ĵ − 4k̂).

2î
3
−
2 ĵ
3
+
2k̂
3
#
=
2
3
−
4
3
= −
2
3
Since this is negative, it decreases in this direction.
1.15 The inverse square force can be written as
f α
r
r3
∇. f = ∇.r−3
r = r−3
∇. r + r .∇r−3
But ∇. r =

î
∂
∂x
+ ĵ
∂
∂y
+ k̂
∂
∂z

· (îx + ĵ y + k̂z)
=
∂x
∂x
+
∂y
∂y
+
∂z
∂z
= 3
Now ∇rn
= nrn−2
r
so that ∇r−3
= −3r−5
r
∴ ∇. (r−3
r) = 3r−3
− 3r−5
r . r = 3r−3
− r−3
= 0
Thus, the divergence of an inverse square force is zero.
1.16 The angle between the surfaces at the point is the angle between the normal to
the surfaces at the point.
The normal to x2
+ y2
+ z2
= 1 at (1, +1, −1) is
∇Φ1 = ∇(x2
+ y2
+ z2
) = 2xî + 2y ĵ + 2zk̂ = 2î + 2 ĵ − 2k̂
The normal to z = x2
+ y2
− 1 or x2
+ y2
− z = 1 at (1, 1, −1) is
∇Φ2 = ∇(x2
+ y2
− z) = 2xî + 2y ĵ − k̂ = 2î + 2 ĵ − k̂
(∇Φ1).(∇Φ2) = |∇Φ1||∇Φ2| cos θ
where θ is the required angle.
(2î + 2 ĵ − 2k̂).(2î + 2 ĵ − k̂) = (12)1/2
(9)1/2
cos θ
∴ cos θ =
10
6
√
3
= 0.9623
θ = 15.780

1.3 Solutions 39
1.17 f (x) =
1
2
a0 +
∞
n=1
an cos
nπx
L

+ bn sin
nπx
L

(1)
an = (1/L)
L
−L
f (x) cos
nπx
L

dx (2)
bn = (1/L)
L
−L
f (x) sin
nπx
L

dx (3)
As f (x) is an odd function, an = 0 for all n.
bn = (1/L)
L
−L
f (x) sin
nπx
L

dx
= (2/L)
L
0
x sin
nπx
L

dx
= −

2
nπ

cos nπ = −

2
nπ

(−1)n
=

2
nπ

(−1)n+1
Therefore,
f (x) = (2/π)
∞
1
(−1)n+1
n
sin
nπx
L

= (2/π)[sin
πx
L

−
1
2
sin

2πx
L

+

1
3

sin

3πx
L

− · · · ]
Figure 1.7 shows the result for first 3 terms, 6 terms and 9 terms of the
Fourier expansion. As the number of terms increases, a better agreement with
the function is reached. As a general rule if the original function is smoother
compared to, say the saw-tooth function the convergence of the Fourier series
is much rapid and only a few terms are required. On the other hand, a highly
discontinuous function can be approximated with reasonable accuracy only
with large number of terms.
Fig. 1.7 Fourier expansion of the saw-tooth wave

1.18 The given function is of the square form. As f (x) is defined in the interval
(−π, π), the Fourier expansion is given by
f (x) =
1
2
a0 +
∞
n=1
(an cos nx + bn sin nx) (1)
where an = (1/π)
π
−π
f (x) cos nx dx (2)
a0 = (1/π)
π
−π
f (x) dx (3)
bn =

1
π
π
−π
f (x) sin nx dx (4)
By (3)
a0 = (1/π)
0
−π
0dx +
π
0
πdx

= π (5)
By (2)
an = (1/π)
π
0
cos nx dx = 0, n ≥ 1 (6)
By (4)
bn = (1/π)
π
0
π sin nx dx =

1
n

(1 − cos nπ) (7)
Using (5), (6) and (7) in (1)
f (x) =
π
2
+ 2

sin(x) +

1
3

sin 3x +

1
5

sin 5x + · · ·

The graph of f (x) is shown in Fig. 1.8. It consists of the x-axis from −π to 0
and of the line AB from 0 to π. A simple discontinuity occurs at x = 0 at
which point the series reduces to π/2.
Now,
π/2 = 1/2[ f (0−) + f (0+)]
Fig. 1.8 Fourier expansion of
a square wave

1.3 Solutions 41
which is consistent with Dirichlet’s theorem. Similar behavior is exhibited at
x = π, ±2π . . . Figure 1.8 shows first four partial sums with equations
y = π/2
y = π/2 + 2 sin x
y = π/2 + 2(sin x + (1/3) sin 3x)
y = π/2 + 2(sin x + (1/3) sin 3x + (1/5) sin 5x)
1.19 By Problem 1.18,
y =
π
2
+ 2

sin x +

1
3

sin 3x +

1
5

sin 5x +

1
7

sin 7x + · · ·

Put x = π/2 in the above series
y = π =
π
2
+ 2

1 −
1
3
+
1
5
−
1
7
+ · · ·

Hence π
4
= 1 − 1
3
+ 1
5
− 1
7
+ · · ·
1.20 The Fourier transform of f (x) is
T (u) =
1
√
2π
a
−a
eiux
f (x)dx
=
1
√
2π
a
−a
1.eiux
dx =
1
√
2π
eiux
iu
%
%
%
%
a
−a
=
1
√
2π

eiua
− e−iua
iu

=
2
π
sin ua
u
, u = 0
For u = 0, T (u) =
$
2
π
u.
The graphs of f (x) and T (u) for u = 3 are shown in Fig. 1.9a, b, respec-
tively
Note that the above transform finds an application in the FraunHofer
diffraction.
f̃ (ω) = A sin α/α
This is the basic equation which describes the Fraunhofer’s diffraction pat-
tern due to a single slit.
Fig. 1.9 Slit function and its Fourier transform

1.21 Consider the Fourier integral theorem
f (x) =
2
π
∞
0
cos ax da
∞
0
e−u
cos au du
Put f (x) = e−x
. Now the definite integral
∞
0
e−bu
cos(au) du =
b
b2 + a2
Here
∞
0
e−u
cos au du =
1
1 + a2
∴
2
π
∞
0
cos ax
1 + a2
dx = f (x) or
∞
0
cos ax
1 + a2
=
π
2
e−x
1.22 The Gaussian distribution is centered on t = 0 and has root mean square
deviation τ.
f̃ (ω) =
1
√
2π
∞
−∞
f (t)e−iωt
dt
=
1
√
2π
∞
−∞
1
τ
√
2π
e−t2
/2τ2
e−iωt
dt
=
1
√
2π
∞
−∞
1
τ
√
2π
e−[t2
+2τ2
iωt+(τ2
iω)2
−(τ2
iω)2
]/2τ2
dt
=
1
√
2π
e− τ2ω2
2

1
τ
√
2π
∞
−∞
e
−(t+iτ2ω2)2
2τ2
dt

The expression in the Curl bracket is equal to 1 as it is the integral for a
normalized Gaussian distribution.
∴ f̃ (ω) =
1
√
2π
e− τ2ω2
2
which is another Gaussian distribution centered on zero and with a root mean
square deviation 1/τ.
1.3.3 Gamma and Beta Functions
1.23 Γ(z + 1) = limT →∞
T
0 e−x
xz
dx
Integrating by parts
Γ(z + 1) = lim
T →∞
[−xz
e−x
|T
0 + z
T
0
e−x
xz−1
dx]
= z limT →∞
T
0
e−x
xz−1
dx = zΓ (z)
because T z
e−T
→ 0 as T → ∞
Also, since Γ(1) =
∞
0 e−x
dx = 1
If z is a positive integer n,
Γ(n + 1) = n!

1.3 Solutions 43
Thus the gamma function is an extension of the factorial function to numbers
which are not integers.
1.24 B(m, n) =
1
0
xm−1
(1 − x)n−1
dx (1)
With the substitution x = sin 2Φ (1) becomes
B(m, n) = 2
π/2
0
(sin Φ)2m−1
(cos Φ)2n−1
dΦ (2)
Now Γ(n) = 2
∞
0 y2n−1
e−y2
dy
Γ (m) = 2
∞
0
y2m−1
e−x2
dx
∴ Γ (m)Γ (n) = 4
∞
0
∞
0
x2m−1
y2n−1
exp −(x2
+ y2
)dxdy (3)
The double integral may be evaluated as a surface integral in the first
quadrant of the xy-plane. Introducing the polar coordinates x = r cos θ and
y = r sin θ, the surface element ds = rdrdθ, (3) becomes
Γ (m)Γ (n) = 4
π/2
0
∞
0
r2m−1
(cos θ)2m−1
(sin θ)2n−1
e−r2
rdrdθ
Γ (m)Γ (n) = 2
π/2
0
(cos θ)2m−1
(sin θ)2n−1
dθ.2
∞
0
r2(m+n)−1
e−r2
dr (4)
In (4), the first integral is identified as B(m, n) and the second one as
Γ (m + n). It follows that
B(m, n) =
Γ (m)Γ (n)
Γ (m + n)
1.25 One form of Beta function is
2
π/2
0
(cos θ)2m−1
(sin θ)2n−1
dθ = B(m, n) =
Γ (m)Γ (n)
Γ (m + n)
(m 0, n 0)
(1)
Letting 2m − 1 = r, that is m = r+1
2
and 2n − 1 = 0, that is n = 1/2, (1)
becomes
π/2
0
(cos θ)r
dθ =
1
2
Γ
r+1
2

Γ
1
2

Γ
r
2
+ 1
(2)

Now Γ(n) =
∞
0 xn−1
e−x
dx, put x = y2
, dx = 2ydy, so that
Γ(n) = 2
∞
0
y2n−1
e−y2
dy
Γ(1/2) = 2
∞
0
e−y2
dy =
2
√
π
2
=
√
π
So that
π
2
0
(cos θ)r
dθ =
√
π
2
Γ
r+1
2

Γ
r
2
+ 1

1.26 (a) B(m, n) =
1
0
xm−1
(1 − x)n−1
dx (1)
Put x =
y
1 + y
(2)
B(m, n) =
∞
0
yn−1
dy
(1 + y)m+n
=
Γ (m)Γ (n)
Γ (m + n)
Letting m = 1 − n; 0 n 1
∞
0
yn−1
(1 + y)
dy =
Γ (1 − n)Γ (n)
Γ (1)
But Γ (1) = 1 and
∞
0
yn−1
(1 + y)
dy =
π
sin(nπ)
; 0 n 1
Γ (n)Γ (1 − n) =
π
sin(nπ)
(3)
(b) |Γ (in)|2
= Γ (in)Γ (−in)
Now Γ (n) = Γ (n+1)
n
Γ (−in) =
Γ (1 − in)
−in
∴ |Γ (in)|2
=
Γ (in)Γ (1 − in)
−in(sin iπn)
by (3)
Further sinh(πn) = i sin iπn
∴ |Γ (in)|2
=
π
n sinh(πn)
1.27 Let H be the hermitian matrix with characteristic roots λi . Then there exists a
non-zero vector Xi such that

1.3 Solutions 45
H Xi = λi Xi (1)
Now X̄′
ι H Xi = X̄′
ιλi Xi = λi X̄′
ι Xi (2)
is real and non-zero. Similarly the conjugate transpose
X̄′
ι H Xi = λ̄ι X̄′
ι Xi (3)
Comparing (2) and (3),
λ̄i = λi
Thus λi is real
1.28 The characteristic equation is given by
%
%
%
%
%
%
1 − λ −1 1
0 3 − λ −1
0 0 2 − λ
%
%
%
%
%
%
= 0
(1 − λ)(3 − λ)(2 − λ) + 0 + 0 = 0 (1)
or λ3
− 6λ2
+ 11λ − 6 = 0 (characteristic equation) (2)
The eigen values are λ1 = 1, λ2 = 3, and λ3 = 2.
1.29 Let X =

x
y

AX =

−1 0
0 1

x
y

=

−x
−y

It produces reflection through the origin, that is inversion. A performs the
parity operation, Fig. 1.10a.
Fig. 1.10a Parity operation
(inversion through origin)
BX =

0 1
1 0

x
y

=

y
x

Here the x and y coordinates are interchanged. This is equivalent to a reflec-
tion about a line passing through origin at θ = 45◦
, Fig. 1.10b
C X =

2 0
0 2

x
y

=

2x
2y

Fig. 1.10b Reflection about a
line passing through origin at
45◦
Fig. 1.10c Elongating a
vector in the same direction
Here the magnitude becomes double without changing its orientation.
DX =
⎛
⎜
⎜
⎝
√
3
2
1
2
−
1
2
√
3
2
⎞
⎟
⎟
⎠

x
y

=

cos 30◦
sin 30◦
− sin 30◦
cos 30◦

x
y

=
⎛
⎜
⎜
⎝
√
3
2
x +
y
2
−
x
2
+
√
3
2
y
⎞
⎟
⎟
⎠
The matrix D is a rotation matrix which rotates the vector through 30◦
about
the z-axis,. Fig.1.10d.
Fig. 1.10d Rotation of a
vector through 30◦
1.30 The matrix A =
⎛
⎝
6 −2 2
−2 3 −1
2 −1 3
⎞
⎠
The characteristic equation is
|A − λI| =
%
%
%
%
%
%
6 − λ −2 2
−2 3 − λ −1
2 −1 3 − λ
%
%
%
%
%
%
= 0
This gives −λ3
+ 12λ2
− 36λ + 32 = 0
or (λ − 2)(λ − 2)(λ − 8) = 0
The characteristic roots (eigen values) are
λ1 = 2, λ2 = 2 and λ3 = 8

1.3 Solutions 47
1.31 A =
⎛
⎝
6 −2 2
−2 3 −1
2 −1 3
⎞
⎠
In Problem 1.30, the characteristic roots are found to be λ = 2, 2, 8. With
λ = 2, we find the invariant vectors.
⎛
⎝
6 − 2 −2 2
−2 3 − 2 −1
2 −1 3 − 2
⎞
⎠
⎛
⎝
x1
x2
x3
⎞
⎠ = 0
The two vectors are X1 = (1, 1, −1)′
and X2 = (0, 1, 1)′
. The third vector
can be obtained in a similar fashion. It can be chosen as X3 = (2, −1, 1)′
. The
three column vectors can be normalized and arranged in the form of a matrix.
The matrix A is diagnalized by the similarity transformation.
S−1
AS = diag A
S =
⎛
⎜
⎝
1
√
3
0 2
√
6
1
√
3
1
√
2
− 1
√
6
− 1
√
3
1
√
2
1
√
6
⎞
⎟
⎠
As the matrix S is orthogonal, S−1
= S′
. Thus
⎛
⎜
⎝
1
√
3
1
√
3
− 1
√
3
0 1
√
2
1
√
2
2
√
6
− 1
√
6
1
√
6
⎞
⎟
⎠
⎛
⎝
6 −2 2
−2 3 −1
2 −1 3
⎞
⎠
⎛
⎜
⎝
1
√
3
0 2
√
6
1
√
3
1
√
2
− 1
√
6
− 1
√
3
1
√
2
1
√
6
⎞
⎟
⎠ =
⎛
⎝
2 0 0
0 2 0
0 0 8
⎞
⎠
1.32 H =

a11 a12
a21 a22

A =

3 2
4 1

(a)
%
%
%
%
3 − λ 2
4 1 − λ
%
%
%
% = 0, characteristic equation is
(3 − λ)(1 − λ) − 8 = 0
λ2
− 4λ − 5 = 0, (λ − 5)(λ + 1) = 0
The eigen values are λ1 = 5 and λ2 = −1
(b) and (c) The desired matrix has the form
C =

C11 C12
C21 C22

The columns which satisfy the system of equations
(ai j − δi j λk)Cjk = 0, no sum on k (1)
yielding
(a11 − λk)C1k + a12C2k = 0, no sum on k
a21C1k + (a22 − λk)C2k = 0, k = 1, 2
Since a11 = 3, a21 = 4, a12 = 2, a22 = 1, we get

on setting k = 1 and λ1 = 5
−2C11 + 2C21 = 0 (2)
4C11 − 4C21 = 0 (3)
Thus C21 = C11 = a = 1
The substitution of k = 2 and λ2 = −1 yields
(3 + 1)C12 + 2C22 = 0
4C12 + 2C22 = 0
or C22 = −2C12
We may set C12 = 1 so that C22 = −2
Thus C =

1 1
1 −2

(modal matrix)
The inverse of C is easily found to be
C−1
=

2
3
1
3
1
3
−1
3
#
Eigen vectors:

a11 − λ a12
a21 a22 − λ

x1
x2

= 0
Put λ = λ1 = 5;

3 − 5 2
4 1 − 5

x1
x2

= 0→−2x1 + 2x2 = 0→x1 =x2
The normalized invariant vector is 1
√
2

1
1

Put λ = λ2 = −1;

3 − (−1) 2
4 1 − (−1)

x1
x2

=0 → 4x1 +2x2 =0 →
x2 = −2x1
The second invariant eigen normalized eigen vector is 1
√
5

1
−2

(d) C−1
AC =

2/3 1/3
1/3 −1/3

3 2
4 1

1 1
1 −2

=

5 0
0 −1

1.33 Let f = y = x3
− 3x + 3 = 0
Let the root be a
If x = a = −2, y = +1
If x = a = −3, y = −15
Thus x = a lies somewhere between −2 and −3.
For x = a − 2.1, y = 0.039, which is close to zero.
Assume as a first approximation, the root to be a = v + h
Put v = −2.1

1.3 Solutions 49
h = −
f (v)
f ′(v)
; f (v) = f (−2.1)
f ′
(v) =
dy
dx
|v; f ′
(v) = 10.23
h = −
0.039
10.23
= −0.0038
To a first approximation the root is −2.1 − 0.0038123 or −2.1038123. As a
second approximation, assume the root to be
a = −2.1038123 + h,
Put v1 = −2.1038123
h1 = − f (v1)/f ′
(v1) = −0.000814/6.967 = −0.0001168
The second approximation, therefore, gives a = −2.1039291.
The third and higher approximations can be made in this fashion. The first
approximation will be usually good enough in practice.
1.34 y(x) = x2
exp(−x2
) (1)
Turning points are determined from the location of maxima and minima. Dif-
ferentiating (1) and setting dy/dx = 0
dy/dx = 2x(1 − x2
) exp(−x2
) = 0
x = 0, +1, −1. These are the turning points.
We can now find whether the turning points are maxima or minima.
dy
dx
= 2(x − x3
)e−x2
y
′′
= 2(2x4
− 5x2
+ 1)e−x2
For x = 0, d2
y
dx2 = +2 → minimum
For x = +1, , d2
y
dx2 = −4e−1
→ maximum
For x = −1, , d2
y
dx2 = −4e−1
→ maximum
y(x) = x2
e−x2
is an even function because y(−x) = +y(x)
1.3.6 Series
1.35 The given series is x −
x2
22
+
x3
32
−
x4
42
+ · · · (A)
The series formed by the coefficients is
1 −
1
22
+
1
32
−
1
42
+ · · · (B)
lim
n=∞

an+1
an

= lim
n=∞

−
n2
(n + 1)2

=
∞
∞
Apply L’Hospital rule.

Differentiating, limn=∞ − 2n
2(n+1)

= ∞
∞
Differentiating again, limn=∞

−2
2

= −1(= L)
%
%
%
%
1
L
%
%
%
% =
%
%
%
%
1
−1
%
%
%
% = 1
The series (A) is
I. Absolutely convergent when |Lx| 1 or |x|
%
% 1
L
%
% i.e. −
%
% 1
L
%
% x
+
%
% 1
L
%
%
II. Divergent when |Lx| 1, or |x|
%
% 1
L
%
%
III. No test when |Lx| = 1, or |x| =
%
% 1
L
%
%.
By I the series is absolutely convergent when x lies between −1 and +1
By II the series is divergent when x is less than −1 or greater than +1
By III there is no test when x = ±1.
Thus the given series is said to have [−1, 1] as the interval of convergence.
1.36 f (x) = log x; f (1) = 0
f ′
(x) =
1
x
; f ′
(1) = 1
f ′′
(x) = −
1
x2
; f ′′
(1) = −1
f ′′′
(x) =
2
x3
; f ′′′
(1) = 2
Substitute in the Taylor series
f (x) = f (a) +
(x − a)
1!
f ′
(a) +
(x − a)2
2!
f ′′
(a) +
(x − a)3
3!
f ′′′
(a) + · · ·
log x = 0 + (x − 1) −
1
2
(x − 1)2
+
1
3
(x − 1)2
− · · ·
1.37 Use the Maclaurin’s series
f (x) = f (0) +
x
1!
f ′
(0) +
x2
2!
f ′′
(0) +
x3
3!
f ′′′
(0) + · · · (1)
Differentiating first and then placing x = 0, we get
f (x) = cos x, f (0) = 1
f ′
(x) = − sin x f ′
(0) = 0
f ′′
(x) = − cos x, f ′′
(0) = −1
f ′′′
(x) = sin x, f ′′′
(0) = 0
f iv
(x) = cos x, f iv
(0) = 1
etc.
Substituting in (1)
cos x = 1 −
x2
2!
+
x4
4!
−
x6
6!
+ · · ·
The series is convergent with all the values of x.

1.3 Solutions 51
1.38 f (a + x) = sin(a + x)
Put x = 0
f (a) = sin a
f ′
(a) = cos a
f ′′
(a) = − sin a
f ′′′
(a) = − cos a
Substitute in
f (x) = f (a) +
(x − a)
1!
f ′
(a) +
(x − a)2
2!
f ′′
(a) +
(x − a)3
3!
f ′′′
(a) + · · ·
sin(a + x) = sin a +
x
1
cos a −
x2
2!
sin a −
x3
3!
cos a + · · ·
1.39 We know that
y = 1 + x + x2
+ x3
+ x4
+ · · · = 1/(1 − x)
Differentiating with respect to x,
dy/dx = 1 + 2x + 3x2
+ 4x3
+ · · · = 1/(1 − x)2
= S
1.3.7 Integration
1.40 (a)

sin3
x cos6
xdx =

sin2
x cos6
x sin xdx
= −

(1 − cos2
x) cos6
x d(cos x) =

cos8
x d(cos x) −

cos6
x d(cos x)
=
cos9
x
9
−
cos7
x
7
+ C
(b)

sin4
x cos2
xdx =

(sin x cos x)2
sin2
xdx
=

1
4
sin2
2x(
1
2
−
1
2
cos 2x)dx
=
1
8

sin2
2xdx −
1
8

sin2
2x cos 2x dx
=
1
8

(
1
2
−
1
2
cos 4x)dx −
1
8

sin2
2x cos 2xdx
=
x
16
−
sin 4x
64
−
sin3
2x
48
+ C
1.41 Express the integrand as sum of functions.
Let
1
2x2 − 3x − 2
=
1
(2x + 1)(x − 2)
=
A
2x + 1
+
B
x − 2
=
A(x − 2) + B(2x + 1)
(2x + 1)(x − 2)
B − 2A = 1
A + 2B = 0
Solving, A = −2
5
and B = 1
5
I = −
2
5

dx
2x + 1
+
1
5

dx
x − 2
+ C

= −
1
5
ln(2x + 1) +
1
5
ln(x − 2) + C
=
1
5
ln

x − 2
2x + 1

+ C
1.42 r2
= a2
sin 2θ
Elementary area
dA =
1
2
r2
dθ
A =
1
2
π/2
0
r2
dθ =
a2
2
π/2
0
sin 2θdθ =
a2
π/2
0
sin θd(sin θ) =
a2
2
sin2
θ|1
0 =
a2
2
Fig. 1.11 Polar diagram of
the curve r2
= a2
sin 2θ
1.43 Since x2
+ 2 occurs twice as a factor, assume
x3
+ x2
+ 2
(x2 + 2)2
=
Ax + B
(x2 + 2)2
+
Cx + D
x2 + 2
On clearing off the fractions, we get
x3
+ x2
+ 2 = Ax + B + (Cx + D)(x2
+ 2)
or x3
+ x2
+ 2 = Cx3
+ Dx2
+ (A + 2C)x + B + 2D
Equating the coefficients of like powers of x
C = 1, D = 1, A + 2C = 0, B + 2D = 2
This gives A = −2, B = 0, C = 1, D = 1
Hence,
x3
+ x2
+ 2
(x2 + 2)2
= −
2x
(x2 + 2)2
+
x
x2 + 2
+
1
x2 + 2

(x3
+ x2
+ 2)dx
(x2 + 2)2
= −

2xdx
(x2 + 2)2
+

xdx
x2 + 2
+

dx
x2 + 2
=
1
x2 + 2
+
1
2
ln(x2
+ 2) +
1
√
2
tan−1

x
√
2

+ C
1.44
∞
0
4a3
dx
x2 + 4a2
= lim
b=∞
b
0
4a3
dx
x2 + 4a2
= lim
b=∞
*
2a2
tan−1 x
2a
+ b
0
= limb=∞

2a2
tan−1

b
2a

= 2a2
.
π
2
= πa2

1.3 Solutions 53
1.45 (a)

tan6
x sec4
xdx =

tan6
x(tan2
x + 1) sec2
xdx
=

(tan x)8
sec2
xdx +

tan6
x sec2
xdx
=

(tan x)8
d(tan x) +

(tan x)6
d(tan x)
=
tan9
x
9
+
tan7
x
7
+ C
(b)

tan5
x sec3
xdx =

tan4
x sec2
x sec x tan x dx
=

(sec2
x − 1)2
sec2
x sec x tan x dx
=

(sec6
x − 2 sec4
x + sec2
x)d(sec x)
=
sec7
x
7
− 2
sec5
x
5
+
sec3
x
3
+ C
1.46
4
2
2x + 4
x2 − 4x + 8
dx =
4
2
2x − 4 + 8
(x − 2)2 + 4
dx
=
4
2
2x − 4
(x − 2)2 + 4
dx + 8
4
2
dx
(x − 2)2 + 4
= ln [(x − 2)2
+ 4]4
2 + (8/2) tan−1
1
= ln 2 + π
1.47 Let us first find the area OMP which is half of the required area OPP′
. For the
upper branch of the curve, y = x3/2
, and summing up all the strips between
the limits x = 0 and x = 4, we get
Area OMP =
4
0 ydx =
4
0 x3/2
dx = 64
5
.
Hence area OPP′
= 2x 64
5
= 25.6 units.
Note: for the lower branch y = x3/2
and the area will be −64/5. The area
will be negative simply because for the lower branch the y-coordinates are
negative. The result for the area OPP′
pertains to total area regardless of sign.
Fig. 1.12 Semi-cubical
parabola

1.48 x2/3
+ y2/3
= a2/3
(1)
The arc AB generates only one half of the surface.
Sx
2
= 2π
b
a
y

1 +

dy
dx
2
'1/2
dx (2)
From (1) we find
dy
dx
= −
y1/3
x1/3
; y = a
2
3 − x
2
3
3/2
(3)
Substituting (3) in (2)
Sx
2
= 2π
a
0
(a2/3
− x2/3
)

1 +
y2/3
x2/3
1/2
dx
= 2π
a
0
(a2/3
− x2/3
)3/2

a2/3
x2/3
1/2
dx
= 2πa1/3
a
0
(a2/3
− x2/3
)3/2
x−1/3
dx
=
6πa2
5
∴ Sx =
12πa2
5
Fig. 1.13 Curve of
hypocycloid
x2/3
+ y2/3
= a2/3
1.49
a
0
√
a2−x2
0
(x + y)dy dx =
a
0
√
a2−x2
0
(x + y)dy
'
dx
=
a
0

xy +
y2
2

dx
√
a2−x2
0
=
a
0

x

a2 − x2 +
a2
− x2
2

dx
=
2a3
3
1.50 Area to be calculated is
A = ACFD = 2 × ABED

1.3 Solutions 55
= 2

ydx
= 2
2
1
1
x
dx = 2 ln 2
= 1.386 units
Fig. 1.14 Area enclosed
between the curves y = 1/x
and y = −1/x and the lines
x = 1 and x = 2
1.51

1
x2 − 18x + 34
dx =

1
(x − 3)2 + 25
dx
= (1/5) tan−1

x − 3
5

1.52
1
0
x2
tan−1
x dx = (x3
/3) tan−1
x|1
0 − 1/3
1
0
x3
(x2 + 12)
dx
=
π
12
−
1
3
1
0

x −
x
(x2 + 1)

dx
=
π
12
−
x2
6
%
%
%
%
1
0
+

1
6

ln(x2
+ 1)
%
%
%
%
1
0
=
π
12
−
1
6
+
1
6
ln 2
1.53 (a) The required area is for the figure formed by ABDGEFA. This area is equal
to the area under the curve y = x2
+2, that is ACEFA, minus ΔBCD, plus
ΔDGE (Fig 1.15a)
=
2
−1
ydx −
1
2
BC . CD +
1
2
DE . EG
=
2
−1
(x2
+ 2)dx −
1
2
. 2.2 +
1
2
. 1.1
= 7.5 units
(b) The required volume V = Volume of cylinder BDEC of height H and
radius r and the cone ABC. (Fig 1.15b)
V = πr2
H +
1
3
πr2
h = πr2

H +
h
3

Fig. 1.15a Area bounded by ABGFA (see the text, Prob 1.53a )
Fig. 1.15b Volume of the
cylinder plus the cone
(See Prob 1.15b)
= π. 22

5 +
4
3

= 25
1
3
π units
1.54 (a) Area =
2π
0
ydx =
π
0
x sin x dx +
2π
π
x sin x dx
= −x cos x + sin x|π
0 − x cos x + sin x|2π
π = π + 3π = 4π
The area refers to the magnitude
(b) Volume, V = 2π

y2
dx = 2π

x2
sin2
x dx
=
4π4
3
−
π2
2

1.3 Solutions 57
1.55
dy
dx
=
x3
+ y3
3xy2
The equation is homogenous because f (λx, λy) = f (x, y). Use the trans-
formation
y = Ux, dy = U dx + x dU
Udx + xdu
dx
=
x3
+ U3
x3
3x.U2x2
=
1 + U3
3U2
3U3
dx + 3xU2
du = (1 + U3
)dx
or (2U3
− 1)dx + 3xU2
du = 0
Dividing by x(2U2
− 1),
dx
x
+
3U2
du
2U3 − 1
= 0
Integrating, ln x + 1
2
ln(2U3
− 1) = C
2 ln x + ln

2y3
x3
− 1

= C
or 2y3
− x3
= Cx
1.56
d3
y
dx3
− 3
d2
y
dx2
+ 4y = 0
The auxiliary equation is
D3
− 3D2
+ 4 = 0
Solving, the roots are −1, 2, 2.
The root −1 gives the solution e−x
.
The double root 2 gives two solutions e2x
, x e2x
.
The general solution is
y = C1e−x
+ C2e2x
+ C3xe2x
1.57
d4
y
dx4
− 4
d3
y
dx3
+ 10
d2
y
dx2
− 12
dy
dx
+ 5y = 0
The auxiliary equation is
D4
− 4D3
+ 10D2
− 12D + 5 = 0
Solving, the roots are 1, 1, 1 ± 2i
The pair of imaginary roots 1 ± 2i gives the two solutions ex
cos 2x and
ex
sin 2x.
The double root gives the two solutions ex
, xex
.
The general solution is
Y = C1ex
+ C2xex
+ C3ex
cos 2x + C4ex
sin 2x
or, y = (C1 + C2x + C3 cos 2x + C4 sin 2x)ex
.

1.58
d2
y
dx2
+ m2
y = cos bx (1)
Replacing the right-hand member by zero,
d2
y
dx2
+ m2
y = 0. (2)
Solving, we get the complimentary function
y = C1 sin mx + C2 cos mx = U. (3)
Differentiating (1) twice, we get
d4
y
dx4
+ m2 d2
y
dx2
= −b2
cos bx. (4)
Multiply (1) by b2
and adding the result to (4) gives
d4
y
dx4
+ (m2
+ b2
)
d2
y
dx2
+ b2
m2
y = 0. (5)
The complete solution of (5) is
y = C1 sin mx + C2 cos mx + C3 sin bx + C4 cos bx
or y = U + C3 sin bx + C4 cos bx = U + V
We shall now determine C3 and C4 so that C3 sin bx + C4 cos bx shall be a
particular solution V of (1)
Substituting
y = C3 sin bx + C4 cos bx,
dy
dx
= C3 b cos bx − C4b sin bx,
d2
y
dx2 = −C3b2
sin bx − C4b2
cos bx in (1), we get
C4(m2
− b2
) cos bx + C3(m2
− b2
) sin bx = cos bx
Equating the coefficients of like terms in this identity we get
C4(m2
− b2
) = 1 → C4 =
1
m2 − b2
C3(m2
− b2
) = 0 → C3 = 0
Hence a particular solution of (1) is
V =
cos bx
m2 − b2
and the complete solution is
y = 0 + V = C1 sin mx + C2 cos mx +
cos bx
m2 − b2
1.59
d2
y
dx2
− 5
dy
dx
+ 6y = x (1)
Replace the right-hand member by zero to form the auxiliary equation
D2
− 5D + 6 = 0 (2)
The roots are D = 2 and 3. The solution is
y = C1e2x
+ C2e3x
= 0 (3)

1.3 Solutions 59
The complete solution is
y = U + V (4)
where V = C3x + C4 (5)
In order that V be a particular solution of (1), substitute y = C3x + C4, in
(1) in order to determine C3 and C4.
−5C3 + 6(C3x + C4) = x
Equating the like coefficients
6C3 = 1 → C3 = 1/6
6C4 − 5C3 = 0 → C4 = 5/36
Hence the complete solution is
y = C1e2x
+ C2e3x
+
x
6
+
5
36
1.60
d2
x
dt2
+ 2
dx
dt
+ 5x = 0 (1)
Put x = eλt
, dx
dt
= λeλt
, d2
x
dt2 = λ2
eλt
in (1)
λ2
+ 2λ + 5 = 0 (2)
its roots being, λ = −1 ± 2i
x = Ae−t(1−2i)
+ Be−t(1+2i)
x = e−t
[C cos 2t + D sin 2t]
where C and D are constants to be determined from the initial conditions.
At t = 0, x = 5. Hence C = 5. Further
dx
dt
= −e−t
[C cos 2t + D sin 2t + 2C sin 2t − 2D cos 2t]
At t = 0, dx/dt = −3
−3 = −C + 2D = −5 + 2D
whence D = 1. Therefore the complete solution is
x = e−t
(5 cos 2t + sin 2t)
1.61 (a) Let the mass 1 be displaced by x1 and mass m2 by x2. The force due to the
spring on the left acting on mass 1 is −kx1 and that due to the coupling is
−k(x1 − x2).
The net force
F1 = −kx1 − k(x1 − x2) = −k(2x1 − x2)
The equation of motion for mass 1 is
mẍ1 + k(2x1 − x2) = 0 (1)
Similarly, for mass 2, the spring on the right exerts a force −kx2, and the
coupling spring exerts a force −k(x2 − x1). The net force
F2 = −kx2 − k(x2 − x1) = −k(2x2 − x1)

The equation of motion for mass 2 is
mẍ2 + k(2x2 − x1) = 0 (2)
The two Eqs. (1) and (2) are coupled equations.
Let x1 = A1 sin ωt (3)
x2 = A2 sin ωt (4)
ẍ1 = −ω2
A1 sin ωt = −ω2
x1 (5)
ẍ2 = −ω2
A2 sin ωt = −ω2
x2 (6)
Inserting (5) and (6) in (1) and (2)
−mω2
x1 + k(2x1 − x2) = 0
− mω2
x2 + k(2x2 − x1) = 0
Rearranging
(2k − mω2
)x1 − kx2 = 0 (7)
−kx1 + (2k − mω2
)x2 = 0 (8)
In order that the above equations may have a non-trivial solution, the
determinant formed from the coefficients of x1 and x2 must vanish.
%
%
%
%
%
2k − mω2
−k
−k 2k − mω2
%
%
%
%
%
= 0 (9)
(2k − mω2
)2
− k2
= 0
or (mω2
− k)(mω2
− 3k) = 0
The solutions are
ω1 =
k
m
(10)
ω2 =
3k
m
(11)
(b) The two solutions to the problem are
x1 = A1 sin ω1t; x2 = A2 sin ω1t (12)
x1 = B1 sin ω2t; x2 = B2 sin ω2t (13)
In (12) and (13) the amplitudes are not all independent as we can verify
with the use of (7) and (8). Substituting (10) and (12) in (7), yields A2 =
A1. Substitution of (11) and (13) in (7), gives B2 = −B1.
Dropping off the subscripts on A′
s and B′
s the solutions can be
written as
x1 = A sin ω1t = x2 (14)
x1 = B sin ω2t = −x2 (15)

1.3 Solutions 61
Fig. 1.16 Two modes of Oscillation
If initially x1 = x2, the masses oscillate in phase with frequency ω1 (sym-
metrical mode) as in Fig. 1.16(a). If initially x2 = −x2 then the masses
oscillate out of phase (asymmetrical) as in Fig. 1.16(b)
1.62 Sum of translational + rotational + potential energy = constant
1
2
mv2
+
1
2
Iω2
+
1
2
kx2
= const.
But I =
1
2
mR2
and ω = v/R
Therefore
3
4
mv2
+
1
2
kx2
= const.
3
4
m(dx/dt)2
+
1
2
kx2
= const.
Differentiating with respect to time,

3
2

md2
x
dt2

.
dx
dt
+ kx.
dx
dt
= 0
Cancelling dx/dt through and simplifying d2
x/dt2
+ (2k/3m)x = 0. This
is an equation to SHM.
Writing ω2
= 2k
3m
, time period T = 2π
ω
= 2π
$
3m
2k
1.63
d2
y
dx2
− 8
dy
dx
= −16y
d2
y
dx2
− 8
dy
dx
+ 16y = 0
Auxiliary equation:
D2
− 8D + 16 = 0
(D − 4)(D − 4) = 0
The roots are 4 and 4.
Therefore y = C1e4x
+ C2xe4x
1.64 x2 dy
dx
+ y(x + 1)x = 9x2
(1)
Put the above equation in the form

dy
dx
+ Py = Q (2)
dy
dx
+
y(x + 1)
x
= 9 (3)
Let y = Uz (4)
dy
dx
=
Udz
dx
+
zdU
dx
(5)
Substituting (4) and (5) in (3)
Udz
dx
+

dU
dx
+
U(x + 1)
x

z = 9 (6)
Now to determine U, we place the coefficients of z equal to zero. This gives
dU
dx
+
U(x + 1)
x
= 0
dU
U
= −

1 +
1
x

dx
Integrating, ln U = −x − ln x or
U = e−x
/x (7)
As the term in z drops off, Eq. (6) becomes
U
dz
dx
= 9 (8)
Eliminating U between (7) and (8)
dz = 9x ex
dx
Integrating z = 9

xex
dx = 9ex
(x − 1) (9)
Substituting U and z in y = Uz,
y =
9(x − 1)
x
1.65 d2
y
dx2
+
dy
dx
− 2y = 2 cosh 2x (1)
The complimentary solution is found from
d2
y
dx2
+
dy
dx
− 2y = 0
D2
+ D − 2 = 0
(D − 1)(D + 2) = 0
D = 1, −2
Y = U = C1ex
+ C2e−2x
(2)

1.3 Solutions 63
Differentiating (1) twice
d4
y
dx4
+
d3
y
dx3
− 2
d2
y
dx2
= 8 cosh 2x (3)
Multiply (1) by (4) and subtract the resulting equation from (3)
d4
y
dx4
+
d3
y
dx3
−
6d2
y
dx2
−
4dy
dx
+ 8y = 0 (4)
D4
+ D3
− 6D2
− 4D + 8 = 0
(D − 1)(D − 2)(D + 2)2
= 0
D = 1, 2, −2, −2
The complete solution of (4) is
y = C1ex
+ C3e2x
+ C2e−2x
+ C4xe−2x
= U + C3e2x
+ C4xe−2x
= U + V
V = C3e2x
+ C4xe−2x
(5)
Inserting (5) in (1), writing 2 cosh 2x = e2x
+ e−2x
and comparing the
coefficients of e2x
and e−2x
, we find C3 = 1
4
and C4 = −1
3
. Thus the complete
solution of (1) is
y = C1ex
+ C2e−2x
+
1
4
e2x
−
1
3
x e−2x
1.66
xdy
dx
− y = x2
dy
dx
−
y
x
= x
The standard equation is
dy
dx
+ Py = Q
∴ P = −
1
x
; Q = x
y exp

pdx

=

Q exp

pdx

dx + C
y exp

−

1
x
dx

=

x exp

−

1
x
dx

+ C
y exp (− ln x) =

x exp (− ln x)

+ C
yx−1
=

x x−1
dx + C
y = x2
+ Cx

1.67 (a)y′
−
2y
x
=
1
x3
(1)
Let y = px, y′
= p + xp′
Then (1) becomes
xp′
− p = 1/x3
Now d
dx
p
x

= xp−p
x2
∴ xp′
− p = x2 d
dx
p
x

=
1
x3
d
dx
p
x

=
1
x5
or d
p
x

=
dx
x5
Integrating
p
x
= −
1
4x4
+ C
or
y
x2
= −
1
4x4
+ C
y = −
1
4x2
+ Cx2
It is inhomogeneous, first order.
(b) y′′
+ 5y′
+ 4y = 0
D2
+ 5D + 4 = 0
(D + 4)(D + 1) = 0
D = −4, −1
y = A e−4x
+ B e−x
It is inhomogeneous, second order.
1.68 (a)
dy
dx
+ y = e−x
Compare with the standard equation
dy
dx
+ py = Q
P = 1; Q = e−x
y exp

pdx

=

Q exp

p dx

dx + C
y exp

1 dx

=

e−x
exp

1 dx

dx + C
yex
= x + C
y = xe−x
+ Ce−x
(b) d2
y
dx2
+ 4y = 2 cos(2x) (1)
The complimentary function is obtained from y′′
+ 4y = 0
y = U = C1 sin 2x + C2 cos 2x
Differentiate (1) twice

1.3 Solutions 65
d4
y
dx4
+
4d2
y
dx2
= −8 cos(2x) (2)
Multiply (1) by 4 and add to (2),
d4
y
dx4
+
8d2
y
dx2
+ 16y = 0
D4
+ 8D2
+ 16 = 0
(D2
+ 4)2
= 0
D = ±2i
y = C1 sin 2x + C3 x sin 2x + C2 cos 2x + C4x cos 2x
= U + C3x sin 2x + C4x cos 2x
= U + V
Y = V = C3x sin 2x + C4x cos 2x (3)
Use (3) in (1) and compare the coefficients of sin 2x and cos 2x to find
C3 = 1/2 and C4 = 0. Thus the complete solution is
y = C1 sin 2x + C2 cos 2x +
1
2
x sin 2x
1.69
dy
dx
+
3y
x + 2
= x + 2
This equation is of the form
dy
dx
+ yp(x) = Q(x)
with P = 3
x+2
and Q = x + 2
The solution is obtained from
y exp

p(x)dx =

Q(x) exp

p(x)dx

dx + C
Now

p(x)dx = 3

dx
x + 2
= 3 ln(x + 2) = ln(x + 2)3
∴ y exp (ln(x + 2)3
) =

(x + 2) exp (ln (x + 2)3
)

dx + C
y(x + 2)3
=

(x + 2)4
dx + C
=
(x + 2)5
5
+ C
or y =
(x + 2)2
5
+ C
y = 2 when x = −1
Therefore C = 9
5
y =
(x + 2)2
5
+
9
5
=
x2
+ 4x + 13
5

1.70 (i) d2
y
dx2
−
4dy
dx
+ 4y = 8x2
− 4x − 4 (1)
Replace the RHS member by zero to get the auxiliary solution.
D2
− 4D + 4 = 0
The roots are D = 2 and 2. Therefore the auxiliary solution is
y = Ae2x
+ Bxe2x
(2)
Complete solution is
y = (A + Bx)e2x
+ Cx2
+ Dx + E (3)
The derivatives are
dy
dx
= (2A + 2Bx + B)e2x
+ 2Cx + D (4)
d2
y/dx2
= 4(A + B + Bx)e2x
+ 2C (5)
Use (3), (4) and (5) in (1) and compare the coefficients of like terms. We
get three equations. Two more equations are obtained from the conditions
y = −2 and dy
dx
= 0 when x = 0.
Solving the five equations we get, A = −3, B = 3, C = 2, D = 3 and
E = 1. Hence the complete solution is y = 3(x − 1)e2x
+ 2x2
+ 3x + 1
(ii) d2
y
dx2
+ 4y = sin x (1)
Replace the RHS member by zero and write down the auxiliary equation
D2
+ 4 = 0
The roots are ±2i. The auxiliary solution is
Y = A cos 2x + B sin 2x
Y = A cos 2x + B sin 2x + C sin x (2)
d2
y
dx2
= −4(A cos 2x + B sin 2x) − C sin x (3)
Substitute (2) and (3) in (1) to find C = 1/3. Thus
y = A cos 2x + b sin 2x +
1
3
sin x
1.71 y′′′
− y′′
+ y′
− y = 0
Auxiliary equation is
D3
− D2
+ D − 1 = 0
(D − 1)(D2
+ 1) = 0 The roots are D = 1, ±i
The solution is
y = A sin x + B cos x + cex

1.3 Solutions 67
1.72
dNA(t)
dt
= −λA NA(t) (1)
dNB(t)
dt
= −λB NB(t) + λA NA(t) (2)
Applying Laplace transform to (1)
sL(NA) − NA(0) = −λAL(NA)
or L(NA) =
N0
A
s + λA
=
N0
A
s − (−λA)
(3)
∴ NA = N0
A exp(–λAt) (4)
Applying the Laplace transform to (2)
sL(NB) − NB(0) = −λBL(NB) + λAL(NA) (5)
Using (3) in (4) and putting N2(0) = 0
L(NB)(s + λB) =
λA N0
A
s + λA
or L(NB) =
λA N0
A
(s + λA)(s + λB)
=
λA N0
A
λB − λA

1
s + λA
−
1
s + λB

=
λA N0
A
λB − λA

1
s − (−λA)
−
1
s − (−λB)

∴ NB =
λA N0
A
λB − λA
,
e−λAt
− e−λB t
-
1.73
dNA
dt
= −λA NA (1)
dNB
dt
= −λB NB + λA NA (2)
dNC
dt
= +λB NB (3)
Applying the Laplace transform to (3)
sL{NC } − NC (0) = λBL{NB} =
λBλA N0
A
(s + λA)(s + λB)
Given Nc(0) = 0
L{Nc} =
λAλB N0
A
s(s + λA)(s + λB)
=
λAλB N0
A
(λB − λA)s

1
s + λA
−
1
s + λB

=
λAλB N0
A
(λB − λA)

1
λA

1
s
−
1
s + λA

−
1
λB

1
s
−
1
s + λB

= N0
1

1
s
−
λB
(λB − λA)
1
(s + λA)
+
λA
(λB − λA)
1
(s + λB)

∴ Nc = N0
A

1 +
1
λB − λA
(λA exp (−λBt) − λB exp (−λAt))

1.74 (a) L{eax
} =
∞
0
e−sx
eax
dx =
∞
0
e−(s−a)x
dx
=
1
s − a
, if s a
(b) and (c). From part (a), L(eax
) = 1
s−a
Replace a by ai
L(eiax
) = L{cos ax + i sin ax}
= L{cos ax} + iL{sin ax}
=
1
s − ai
=
s + ai
s2 + a2
=
s
s2 + a2
+
ia
s2 + a2
Equating real and imaginary parts:
L{cos ax} =
s
s2 + a2
; L{sin ax} =
a
x2 + a2
1.75 Express Hn in terms of a generating function T (ξ, s).
T (ξ, s) = exp[ξ2
− (s − ξ)2
] = exp[−s2
+ 2sξ]
=
∞

n=0
Hn(ξ)sn
n!
(1)
Differentiate (1) first with respect to ξ and then with respect to s.
∂T
∂ξ
= 2s exp(−s2
+ 2sξ) =

n
2sn+1
Hn(ξ)
n!
=

n
sn
H′
n(ξ)
n!
(2)
Equating equal powers of s
H′
n = 2nHn−1 (3)
∂T
∂s
=ξ(−2s + 2ξ) exp(−s2
+2sξ)=

n
(−2s+2ξ)sn
Hn(ξ)=

n
sn−1
Hn(ξ)
(n − 1)!
(4)
Equating equal powers of s in the sums of equations
Hn+1 = 2ξ Hn − 2nHn−1 (5)
It is seen that (5) satisfies the Hermite’s equation

1.3 Solutions 69
H
′′
n − 2ξ H′
n + 2nHn = 0
1.76 Jn(x) =

k
(−1)k
x
2
n+2k
k!Γ (n + k + 1)
(a) Differentiate
d
dx
[xn
Jn(x)] = Jn(x)nxn−1
+ xn dJn(x)
dx
=

k
(−1)k
x
2
n+2k
nxn−1
k!Γ (n + k + 1)
+
xn
(n + 2k)(−1)k
xn+2k−1
k!Γ (n + k + 1).2n+2k
=

k
(−1)k
x
2
n+2k−1
(n + k)xn
k!Γ (n + k + 1)
=
(−1)k
x
2
n+2k−1
xn
Γ (n + k)
= xn
Jn−1(x)
(b) A similar procedure yields
d
dx
[x−n
Jn(x)] = −x−n
Jn+1(x)
1.77 From the result of 1.76(a)
d
dx
[xn
Jn(x)] = Jn(x)nxn−1
+ xn d Jn(x)
dx
= xn
Jn−1(x)
Divide through out by xn
n
x
Jn(x) +
dJn(x)
dx
= Jn−1(x)
Similarly from (b)
−
n
x
Jn(x) +
dJn(x)
dx
= −Jn+1(x)
Add and subtract to get the desired result.
1.78 Jn(x) =
∞

k=0
(−1)k
x
2
n+2k
k!Γ (n + k + 1)
(a) Therefore, with n = 1/2
J1
2
(x) =
x
2
1/2
Γ
3
2
−
x
2
5/2
1.Γ
5
2
+
x
2
9/2
2!Γ
7
2
− · · ·
Writing Γ
3
2

=
√
π
2
, Γ
5
2

= 3
√
π
4
, Γ
7
2

= 15
8
√
π
J1
2
(x) =
2
πx

x −
x3
3!
+
x5
5!
+ · · ·

=
2
πx
sin x
(b) With n = −1/2
J− 1
2
(x) =
x
2
−1/2
Γ
1
2
−
x
2
3/2
1.Γ
3
2
+
x
2
7/2
2!Γ
5
2
− · · ·

=
2
πx

1 −
x2
2!
+
x4
4!
− · · ·

=
2
πx
cos x
1.79 The normalization of Legendre polynomials can be obtained by l – fold inte-
gration by parts for the conventional form
Pl(x) =
1
2ll!
dl
dxl
(x2
− 1)l
(Rodrigues’s formula)
+1
−1
[Pl(x)]2
dx =

1
2ll!
2 +1
−1

dl
(x2
− 1)l
dxl

dl
(x2
− 1)l
dxl

dx
= (−1)l
(
1
2ll!
)2
+1
−1

d2l
(x2
− 1)
dx2l

(x2
− 1)l
dx
= (−1)l

(2l)!
2ll!
2 +1
−1
(x2
− 1)l
dx =
2
2l + 1
Put l = n to get the desired result.
The orthogonality can be proved as follows. Legendre’s differential equation
d
dx

(1 − x2
)
dPn(x)
dx

+ n(n + 1)Pn(x) = 0 (1)
can be recast as
[(1 − x2
)P′
n]′
= −n(n + 1)Pn(x) (2)
[(1 − x2
)P′
m]′
= −m(m + 1)Pm(x) (3)
Multiply (2) by Pm and (3) by Pn and subtract the resulting expressions.
Pm[(1 − x2
)P′
n]′
− Pn[(1 − x2
)P′
m]′
= [m(m + 1) − n(n + 1)]Pm Pn (4)
Now, LHS of (4) can be written as
Pm[(1 − x2
)P′
n]′
− Pn[(1 − x2
)P′
m]′
= Pm[(1 − x2
)P′
n]′
+ P′
m[(1 − x2
)P′
n] − Pn[(1 − x2
)P′
m] − Pn[(1 − x2
)P′
m]′
(4) can be integrated
d
dx
[(1 − x2
)(Pm P′
n − Pn P′
m) = [m(m + 1) − n(n + 1)]Pm Pn
(1 − x2
)

Pm P′
n − Pn P′
m

|1
−1 = [m(m + 1) − n(n + 1)]
1
−1
Pm Pndx
Since (1 − x2
) vanishes at x = ±1, the LHS is zero and the orthogonality
follows.
1
−1
Pm(x)Pn(x)dx = 0; m = n

1.3 Solutions 71
1.80 Legendre’s equation is
(1 − x2
)
∂2
Pn(x)
∂x2
− 2x
∂ Pn(x)
∂x
+ n(n + 1)Pn(x) = 0 (1)
Put x = cos θ, Eq. (1) then becomes
sin2
θ
∂2
Pn
∂ cos2 θ
− 2 cos θ
∂ Pn
∂ cos θ
+ n(n + 1)Pn = 0 (2)
For large n, n(n + 1) → n2
, and cos θ → 1 for small θ,
sin2
θ
∂2
Pn
∂ cos2 θ
− 2
∂ Pn
∂ cos θ
+ n2
Pn = 0 (3)
Now, Bessel’s equation of zero order is
x2 d2
J0(x)
dx2
+ x
d
dx
J0(x) + x2
J0(x) = 0 (4)
Letting x = 2n sin θ/2 = n sin θ, in (4) for small θ, and noting that cos θ →
1, after simple manipulation we get
sin2
θ
d2
J0(n sin θ)
d cos2 θ
− 2d
d J0(n sin θ)
d cos θ
+ n2
J0(n sin θ) = 0 (5)
Comparing (5) with (3), we conclude that
Pn(cos θ) → J0(n sin θ)
1.81 T (x, s) = (1 − 2sx + s2
)−1/2
=

pl(x)sl
(1)
(a) Differentiate (1) with respect to s.
∂T
∂s
= (x − s)(1 − 2sx + s2
)− 3
2
=

(x − s)(1 − 2sx + s2
)−1
pl(x)sl
=

lpl(x)sl−1
Multiply by (1 − 2sx + s2
)

(x − s)pl sl
=

lpl sl−1
(1 − 2sx + s2
)
Equate the coefficients of sl
xpl − pl−1 = (l + 1)pl+1 − 2xlpl + (l − 1)pl−1
or (l + 1)pl+1 = (2l + 1)xpl − lpl−1

(b) Differentiate with respect to x
∂T
∂x
= s(1 − 2sx + s2
)− 3
2
=

(1 − 2sx + s2
)−1
plsl+1
=

p′
lsl
Multiply by (1 − 2sx + s2
)

sl+1
pl =

(sl
− 2xsl+1
+ sl+2
)p′
l
Equate coefficients of sl+1
pl = p′
l+1 − 2xp′
l + p′
l−1
or pl (x) + 2xp′
l(x) = p′
l+1 + p′
l−1
1.82
e− xs
1−s
1 − s
=
∞

n=0
Ln(x)sn
n!
Put x = 0
∞

n=0
Ln(0)
sn
n!
=
1
1 − s
= 1 + s + s2
+ · · · sn
+ · · ·
=
∞

n=0
sn
Therefore Ln(0) = n!
1.83 (a) Since the pole at z = 2 is not interior to |z| = 1, the integral equals zero
(b) Since the pole at z = 2 is interior to |z + i| = 3, the integral equals 2πi.
1.84 Method 1

c
4z2
− 3z + 1
(z − 1)3
dz =

c
4(z − 1)2
+ 5(z − 1) + 2
(z − 1)3
dz
= 4

c
dz
z − 1
+ 5

c
dz
(z − 1)2
+ 2

c
dz
(z − 1)3
= 4(2πi) + 5(0) + 6(0) = 8πi
where we have used the result

1.3 Solutions 73

c
dz
(z − a)n
= 2πi if n = 1
= 0 if n 1
Method 2
By Cauchy’s integral formula
f (n)(a) =
n!
2πi

c
f (z)
(z − a)n+1
dz
If n = 2 and f (z) = 4z2
− 3z + 1, then f ′′
(1) = 8. Hence
8 =
2!
2πi

c
4z2
− 3z + 1
(z − 1)3
or

4z2
− 3z + 1
(z − 1)3
= 8πi
1.85 z = 3 is a pole of order 2 (double pole);
z = i and z = −1 + 2i are poles of order 1 (simple poles).
1.86 z = 1 is a simple pole, z = −2 is a pole of order 2 or double pole.
Residue at z = 1 is limz→1(z − 1)
.
1
(z−1)(z+2)2
/
= 1
9
Residue at z = −2 is limz→−2
d
dz
.
(z+2)2
(z−1)(z+2)2
/
=
d
dz2

1
z − 1

=
2
(z − 1)2
=
2
9
1.87 The singularity is at z = 2
Let z − 2 = U. Then z = 2 + U.
ez
(z − 1)2
=
e2+U
U2
= e2
.
eU
U2
=
e2
U2

1 + U +
U2
2!
+
U3
3!
+ · · ·

=
e2
(z − 2)2
+
e2
z − 2
+
e2
2!
+
e2
(z − 2)
3!
+
e2
(z − 2)2
4!
+ · · ·
The series converges for all values of z = 2
1.88 Consider
!
c
dz
z4+1
, where C is the closed contour consisting of line from −R
to R and the semi-circle Γ, traversed in the counter clock-wise direction.
The poles for Z4
+ 1 = 0, are z = exp(πi/4), exp(3πi/4), exp(5πi/4),
exp(7πi/4). Only the poles exp(πi/4) and exp(3πi/4) lie within C. Using
L’Hospital’s rule

Residue at exp(πi/4) = limz→exp(πi
4 )

z − exp

πi
4

1
z4 + 1

=
1
4z3
=
1
4
exp

−
3πi
4

Residue at exp(3πi/4) = limz→exp(3πi
4 )

z − exp

3πi
4

1
z4 + 1

=
1
4z3
=
1
4
exp

−
3πi
4

Thus

c
dz
z4 + 1
= 2πi

1
4
exp

−
3πi
4

+
1
4
exp

−
3πi
4

=
π
√
2
Thus
R
−R
dx
x4 + 1
+

dz
z4 + 1
=
π
√
2
Taking the limit of both sides as R → ∞
limR→∞
+R
−R
dx
x4 + 1
=
∞
−∞
dx
x4 + 1
=
π
√
2
It follows that
∞
0
dx
x4 + 1
=
π
2
√
2
Fig. 1.17 Closed contour
consisting of line from −R
to R and the semi-circle Γ
1.89 Let I =
x1
x0
F(x, y, y′
)dx (1)
Here I =
x1
x0
0
1 +

dy
dx
2
dx (2)
Now the Euler equation is
∂ F
∂y
−
d
dx

∂F
∂y′

= 0 (3)

1.3 Solutions 75
But in (2), F = F(y′
).
Hence
∂ F
∂y
= 0
∂F
∂y′
=
∂
∂y′
(1 + y′2
)
1
2 = y′
(1 + y′2
)−1/2
d
dx
,
y′
(1 + y′2
)−1/2
-
= 0
or y′
(1 + y′2
)−1/2
= C = constant
or y′2
(1 − C2
) = C2
or y′
=
dy
dx
= a = constant
Integrating y = ax + b which is the equation to a straight line. The constants
a and b can be found from the coordinates P0(x0, y0) and P1(x1, y1)
1.90 The velocity of the bead which starts from rest is
ds
dt
=

2gy (1)
The time of descent is therefore
I = t =

ds
√
2gy
=
1
√
2g
0
dx2 + dy2
y
=
1
√
2g

0
1 + y′2
y
dx (2)
F =
0
(1 + y′2)
y
(3)
Here F involves only y and y′
. The Euler equation is
dF
dx
−
d
dx
(
∂F
∂y′
) = 0 (4)
which does not contain x explicitly. In that case F(y, y′
) is given by
dF
dx
=
∂ F
∂y
dy
dx
+
∂ F
∂y′
dy′
dx
(5)
Multiply (4) by dy
dx
dy
dx
.
dF
dy
−
dy
dx
d
dx

dF
dy′

= 0 (6)
Combining (5) and (6)
dF
dx
=
d
dx

dF
dy′
dy
dx

(7)
Integrating F = dF
dy′
dy
dx
+ C

F =
0
1 + y′2
y
=
y′2

1 + y′2
+ C
Simplifying 1
√
y(1+y′2)
= C where we have used (3)
Or y(1 + y′2
) = constant, say 2a
∴

dy
dx
2
=
2a − y
y
∴
dx
dy
=

y
2a − y
1/2
=
y
(2ay − y2)1/2
This equation can be easily solved by a change of variable y = 2a sin2
θ and
direct integration. The result is
x = 2a sin−1 y
2a

−

2ay − y2 + b
which is the equation to a cycloid.
1.91 Irrespective of the function y, the surface generated by revolving y about the
x-axis has an area
2π
x2
x1
yds = 2π

y(1 + y′2
)1/2
dx (1)
If this is to be minimum then Euler’s equation must be satisfied.
∂F
∂x
−
d
dx
(F − y′ ∂F
∂y′
) = 0 (2)
Here
F = y(1 + y′2
)1/2
(3)
Since F does not contain x explicitly, ∂F
∂x
= 0. So
F − y′ ∂ F
∂y′
= a = constant (4)
Use (3) in (4)
y(1 + y′2
)
1
2 − yy′2
(1 + y′2
)− 1
2 = a
Simplifying
y
(1 + y′2)1/2
= a
or
dy
dx
=
y2
a2
− 1, y = a cosh
x
a
+ b

This is an equation to a Catenary.
1.92 The area is
A = 2π

yds = 2π
a
0
y(1 + y′2
)1/2
dx

1.3 Solutions 77
The volume:
V = π
a
0
y2
dx
Therefore, dropping off the constant factors
K = y2
+ λy(1 + y′2
)1/2
which must satisfy the Euler’s equation
∂K
∂x
−
d
dx
(K − y′ ∂K
∂y′
) = 0
It is convenient to use the above form as K does not explicitly contain x, and
∂K
∂x
= 0. Therefore,
K − y′ ∂K
∂y′
= y2
+ λy(1 + λy′2
)
1
2 − λyy′2
(1 + y′2
)− 1
2 = 0
Now y = 0 at x = 0 and at x = a which can be true if C = 0. Hence
y2
+ λy(1 + y′2
)−1/2
= 0
Or y = −λ(1 + y′2
)−1/2
Solving for y′
,
dy
dx
=
1
y
(λ2
− y2
)1/2
Integrating,
− (λ2
− y2
)
1
2 = x − x0
Or (x − x0)2
+ y2
= λ2
This is the equation to a sphere with the centre on the x-axis at x0, and of
radius λ.
1.3.13 Statistical Distribution
1.93 (a)
∞
x=0
Px =
∞
x=0
e−m
mx
x!
= e−m

1 +
m
1!
+
m2
2!
+ · · ·

= e−m
× e+m
= 1
Thus the distribution is normalized.
(b) x =
∞
x=0
x Px =
∞
x=0
xe−m
mx
x!
=
∞
x=0
e−m
mx
(x − 1)!
= e−m

m +
m2
1!
+
m3
2!
+ · · ·

(∵ (−1)! = ∞)
= m e−m

1 +
m
1!
+
m2
2!
+ · · ·

= m e−m
× em
= m

(c) x2
=

x2 e−m
mx
x!
=

[x(x − 1) + x]
e−m
mx
x!
=
∞
x=0
e−m
mx
(x − 2)!
+
∞
x=0
x e−m mx
x!
= e−m

m2
+
m3
1!
+
m4
2!
+ · · ·

+ m
= m2
e−m
em
+ m = m2
+ m
σ2
= (x − x̄)2
= x2
−2 x x̄+ x̄ 2
= x2
−m2
σ2
= m or σ =
√
m
(d) Pm−1 =
e−m
mm−1
(m − 1)!
=
e−m
mm
(m − 1)!m
=
e−m
mm
m!
= Pm
That is the probability for the occurrence of the event at x = m − 1 is
equal to that at x = m
(e) Px−1 =
e−m
mx−1
(x − 1)!
=
e−m
mx
x!
x
m
=
x
m
Px
Px+1 =
e−m
mx+1
(x + 1)!
=
m e−m
mx
x!(x + 1)
=
m
x + 1
Px
1.94 (a) (q + p)N
= qN
+ NqN−1
P +
N(N − 1)qN−2
2!
P2
+ · · ·
N!
x!(N − x)!
Px
qN−x
+ · · · PN
=
N
x=0
N!
x!(N − x)!
Px
qN−x
= 1(∵ q + p = 1)
(b) We can use the moment generating function Mx (t) about the mean μ
which is given as
Mx (t) = Ee(x−μ)t
= E

1 + (x − μ)t + (x − μ)2 t2
2!
+ · · ·

= 1 + 0 + μ2
t2
2!
+ μ3
t3
3!
+ · · ·
So that μn is the coefficient of tn
n!

1.3 Solutions 79
Mx (t) = Eext
=
∞
x=0
ext
B(x)
=
∞
x=0

N
r

(pet
)r
(1 − p)N−r
= (pet
+ 1 − p)N

N
r

pr
qN−r
= (pet
+ 1 − p)N
μ0
n =
∂n
Mx (t)
∂tn
|t=0
Therefore μ0
1 = ∂M
∂t
|t=0 = Npet
(q + pet
)N−1
|t=0 = Np
Thus the mean = Np
(c) μ0
2 =
∂2
M
∂t2
= [N(N − 1)p2
et
(q + pet
)N−2
+ Npet
(q + pet
)N−1
]t=0
= N(N − 1)p2
+ Np
But μ2 = μ0
2 − (μ0
1)2
= N(N − 1)p2
+ Np − N2
p2
= Np − Np2
= Np(1 − p) = Npq
or σ =

Npq
1.95 Total counting rate/minute, m1 = 245
Background rate/minute, m2 = 49
Counting rate of source, m = m1 − m2 = 196
m1 =
n1
t1
±
√
n1
t1
; m2 =
n2
t2
±
√
n2
t2
; Net count m = m1 − m2
σ = (σ2
1 + σ2
2 )1/2
=

n1
t2
1
+
n2
t2
2
1/2
=

m1
t1
+
m2
t2
1/2
σ =

m1
t1
+
m2
t2
1/2
=

49
100
+
245
20
1/2
= 3.57
Percentage S.D. = σ
m
× 100 = 3.57
196
× 100 = 1.8%
1.96 (a) B(x) =
N!
x!(N − x)!
px
qN−x
Using Sterling’s theorem

N! →
√
2π N N N
e−N
x! →
√
2πxxx
e−x
(N − x)! →

2π(N − x)(N − x)N−x
e−N+x
B(x) → f (x) =
0
N
2πx(N − x)
px
qN−x
N N
xx (N − x)N−x
=
0
N
2πx(N − x)

Np
x
x
Nq
(N − x)
N−x
Let δ denote the deviation of x from the expected value Np, that is
δ = x − Np
then N − x = N − Np − δ = Nq − δ
f (x) =

2π Npq

1 +
δ
Np

1 −
δ
Nq
−1/2

1 +
δ
Np
−(Np+δ)
·

1 −
δ
Nq
−(Nq−δ)
Assume that |δ| ≪ Npq so that
%
%
%
%
δ
Np
%
%
%
% ≪ 1 and
%
%
%
%
δ
Nq
%
%
%
% ≪ 1
The first bracket reduces to (2π Npq)− 1
2 . Take loge on both sides.
ln
*
f (x)(2π Npq)
1
2
+
= −(Np + δ) ln

1+
δ
Np

−(Nq − δ) ln

1−
δ
Nq

= − (Np + δ)

δ
Np
−
δ2
2N2 p2
+
δ3
3N3 p3
− · · ·

− (Nq − δ)

−
δ
Nq
−
δ2
2N2q2
−
δ3
3N3q3
− · · ·

= −
δ2
2Npq
−
δ3
(p2
− q2
)
6N2 p2q2
Neglect the δ3
term and putting σ =
√
Npq and σ = x − Np = x̄.
f (x) =
1
σ
√
2π
exp

−(x − x̄)2
2σ2

(Normal or Gaussian distribution)

1.3 Solutions 81
(b) B(x) =
N!px
qN−x
x!(N − x)!
=
N(N − 1) . . . (N − x − 1)px
(1 − p)N−x
x!
=
N(N − 1) . . . (N − x + 1)(Np)x
(1 − p)N−x
Nx x!
=
mx
x!

1 −
1
N

1 −
2
N

· · ·

1 −
x − 1
N

(1 − p)N−x
=
mx
x!

1 − 1
N

1 − 2
N

· · ·

1 − x−1
N

(1 − p)N
(1 − p)x
The poisson distribution can be deduced as a limiting case of the bino-
mial distribution, for those random processes in which the probability of
occurrence is very small, p ≪ 1, while the number of trials N becomes
very large and the mean value m = pn remains fixed. Then m ≪ N and
x ≪ N, so that approximately
(1 − p)N−x
≈ e−p(N−x)
≈ e−pN
= e−m
Thus B(x) → P(x) = e−m
mx
x!
1.97 S = (g − b) ±
0
g
tg
+
b
tb
t = tb + tg = constant
tg = t − tb
σ2
= σ2
g + σ2
b =
g
t − tb
+
b
tb
must be minimum. Therefore, ∂(σ2
)
∂tb
= 0
g
(t − tb)2
−
b
t2
b
= 0
t2
b
(t − tb)2
=
b
g
→
tb
tg
=
0
b
g
1.98 A = A0e−λt
ln

A0
A

= λt
y = λt
S = y − λt

Normal equation gives
λ =

tn yn

t2
n
6

n=1
tn yn = 1 × ln(1/0.835) + 2 × ln(1/0.695)
+ 3 × ln(1/0.58) + 4 × ln(1/0.485)
+ 5 × ln(1/0.405) + 6 × ln(1/0.335)
= 16.5175
6

n=1
t2
n = 12
+ 22
+ 32
+ 42
+ 52
+ 62
= 91
∴ λ =
16.5175
91
= 0.1815 h−1
T1/2 =
0.693
λ
=
0.693
0.1815
= 3.82 h
1.99 We determine the probability P(t) that a given counter records no pulse
during a period t. We divide the interval t into two parts t = t1 + t2. The
probabilities that no pulses are recorded in either the first or the second period
are given by P(t1) and P(t2), respectively, while the probability that no pulse
is recorded in the whole interval is P(t) = P1(t1 + t2).
Since the two events are independent,
P(t1 + t2) = P(t1)p(t2)
The above equation has the solution
P(t) = e−at
where a is a positive constant. The reason for using the minus sign for a is
that P(t) is expected to decrease with increasing t.
The probability that there will be an event in the time interval dt is c dt. The
combined probability that there will be no events during time interval t, but
one event between time t and t + dt is c e−at
dt where c = constant. It is
readily shown that c = a. This follows from the normalization condition
∞
0
P(t)dt = c
∞
0
e−at
dt = 1
Thus dp(t) = ae−at
dt
Clearly, small time intervals are more favoured than large time intervals
amongst randomly distributed events. If the data have large number N of
intervals then the number of intervals greater than t1 but less than t2 is
n = N
t2
t1
ae−at
dt = N(e−at1
− e−at2
) (Interval distribution)
a represents the average number of events per unit time.

1.3 Solutions 83
Two limiting cases
(a) t2 = ∞. We find that the number of intervals greater than any duration is
Ne−at
in which at = average number of events in time t. In the case
of radioactivity at time t = 0, let N = N0.
Then the radioactive decay law becomes
N = N0e−λt
where N is the number of surviving atoms at time t, and a = λ is the
decay constant, that is the number of decays per unit time.
(b) t1 = 0, implies that the number of events shorter than any duration t is
N0(1 − e−at
)
For radioactive decay the above equation would read for the number of
decays in time interval 0 to t.
N = N0(1 − e−λt
)
1.100 Ns = N0 − NB = 14.5 − 10 = 4.5
σs =
10
t
+
14.5
t
=
24.5
t
σs
Ns
=
5
100
=
1
4.5
24.5
t
t = 484 min
1.101 The best values of a0, a1 and a2 are found by the Least square fit. The residue
S is given by
S =
6
n=1
(yn − a0 − a1xn − a2x2
n )2
Minimize the residue.
∂S
∂a0
= 0, gives
6
n=1
yn = na0 + a1

xn + a2

x2
n (1)
∂S
∂a1
= 0 gives

xn yn = a0

xn + a1

x2
n + a2

x3
n (2)
∂S
∂a2
= 0 gives

x2
n yn = a0

x2
n + a1

x3
n + a2

x4
n (3)
Equations (1), (2) and (3) are the so-called normal equations which are to be
solved as ordinary simultaneous equations.

N = 6;

yn = 29.5;

xn = 3;

x2
n = 19

x3
n = 27;

x4
n = 115;

xn yn = 21.3;

x2
n yn = 158.1
Solving (1), (2) and (3) we find a0 = 0.582; a1 = −1.182; a2 = 1.556
Fig. 1.18 Least square fit of the parabola
1.102 C =
2πε0
ln
b
a
(1)
From propagation of errors
σc =

∂c
∂b
2
σ2
b +

∂c
∂a
2
σ2
a
'1/2
(2)
∂c
∂b
= −
c
b ln b
a
;
∂c
∂a
=
c
a ln b
a
(3)
Using (1), (2) and (3) and simplifying
σc
c
=

ln
b
a
−1/2
σ2
a
a2
+
σ2
a
b2
1/2
Substituting a = 10 mm, b = 20 mm, σa = 1 mm and σb = 1 mm, σc/c =
0.16

1.3 Solutions 85
1.103
x =
∞
0
x f (x)dx/
∞
0
f (x)dx
=
∞
0
x2
e− x
λ dx/
∞
0
xe− x
λ dx
=
2λ3
λ2
= 2λ
Most probable value of x is obtained by maximizing the function xe−x/λ
d
dx
(xe−x/λ
) = 0
e− x
λ 1 −
x
λ

= 0
∴ x = λ
x(most probable) = λ
1.104 The trapezoidal rule is
Area =

1
2
y0 + y1 + y2 + · · · + yn−1 +
1
2
yn

Δx
Given integral is
10
1 x2
dx. Divide x = 1 to x = 10 into 9 intervals.
Thus b−a
n
= 10−1
9
= 1 = Δx
Substituting the abscissas in the equation y = x2
, we get the ordinates y =
1, 4, 9, 16, · · · 100.
area =

1
2
+ 4 + 9 + 25 + 36 + 49 + 64 + 81 +
1
2
× 100

= 334.5
This may be compared with the value obtained from direct integration,
*
x3
3
+10
1
= 333.
The error is 0.45%.
1.105 For Simpson’s rule
take 10 intervals
Here b−a
n
= 10−0
10
= 1 = Δx
The area under the curve y = x2
is given by
Δx
3
(y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 4yn−1 + yn)

The ordinates are found by substituting x = 0, 1, 2 · · · 10 in the equation
y = x2
. Thus
area =
1
3
(0+4+8+36+32+100+72+196+128+324+100) = 333.3
In this case Simpson’s rule happens to give exactly the same result as that
from direct integration.

Chapter 2
Quantum Mechanics – I
Wave number
ν̃ =
1
λ
(2.1)
1 fm = 10−15
m; 1 Å = 10−10
m; 1 nm = 10−9
m; 1 µm = 10−6
m; c =
197.3 Mev − fm
Photon energy
E = hv (2.2)
Photon momentum
p = hν/c
Photon energy − wavelength conversion
λ(nm) =
1241
E(eν)
(2.3)
de Broglie wavelength
λ = h/p (2.4)
λ(electron) : λ(Å) = (150/V )1/2
(2.5)
λ(neutron) : λ(Å) = 0.286 E−1/2
(E in ev) (2.6)
Atomic units
The Bohr radius 2
/mee2
is frequently used as the unit of length in atomic physics.
In atomic units the energy is measured in multiples of the ionization energy of
hydrogen atom that is mee4
/22
. In these units 2
= 1, e2
= 2 and me = 1
2
in
all equations.
87

88 2 Quantum Mechanics – I
Natural units
= c = 1 (2.7)
Mosley’s law (for characteristic X-rays)
√
ν = A(Z − b) (2.8)
where Z is the atomic number, A and b are constants.
For Kα line
λ =
1,200
(Z − 1)2
Å (2.9)
X-rays absorption
I = I0e−μx
(2.10)
Duane–Hunt law (for continuous X-rays)
λc =
c
νmax
=
hc
eV
=
1,240
V
pm (2.11)
where e is the electron charge and V is the P.D through which the electrons have
been accelerated in the X-ray tube.
Doppler effect (Non-relativistic)
ν = ν0(1 + βc cos θ∗
) (2.12)
where ν is the observed frequency, ν0 the frequency of light in the rest frame of
source emitted at angle θ∗
, ν = βc is the source velocity. The inverse transforma-
tion is
ν0 = ν(1 − βc cos θ) (2.13)
Hydrogen atom (Bohr’s model)
Angular momentum
L = n, n = 1, 2, 3 . . . .. (2.14)
Energy of photon emitted from energy level Ei to final level Ef.
hv = Ei − Ef (2.15)
Radius of the nth orbit
rn =
ε0h2
n2
πµe2z
(2.16)

where μ is the reduced mass given by
μ =
memp
me + mp
(2.17)
and Z = 1 for H-atom.
The radius of the smallest orbit, called the Bohr radius,
a0 = r1 =
ε0h2
πme2
= 0.529 Å (2.18)
Orbital velocity in the nth orbit
vn =
ze2
2εonh
(2.19)
Fine structure constant
α =
v
c
=
e2
2εohc
=
1
137
(2.20)
α is a measure of the electromagnetic interaction
Kinetic Energy of electron
Kn =
1
2
m v2
n =
ze2
8πεorn
(2.21)
Potential energy of electron
Un = −
ze2
4πεorn
(2.22)
Total Energy of electron
En = Kn + Un = −
ze2
8πεorn
= −
mz2
e4
8ε2
oh2n2
=
−13.6
n2
(ev) (2.23)
ν̃if =
1
λif
= R

1
n2
f
−
1
n2
i

(2.24)
Mesic atom
A negatively charged muon or pion when captured by the nucleus forms a bound
system called mesic atom before absorption or decay system.
i. rn ∝ 1/μ; therefore radii are shrunk by a factor of ∼200 for muonic atom
compared to H-atom.
ii. En ∝ μ; therefore energy levels are spaced 200 times greater than those of
H-atom. X-rays are emitted instead of visible light, when the muon cascades
down.

Table 2.1 Hydrogen spectrum
Series Region First line (Å) Series limit (Å)
1. Lyman Ultraviolet 1,215 911
ν̃ = 1
λ
= R 1
12 − 1
n2
i

; ni=2,3,4...
2. Balmer Visible 6,561 3,645
ν̃ = R 1
22 − 1
n2
i

; ni = 3, 4, 5 . . .
3. Pashen Infrared 18,746 8,201
ν̃ = R 1
32 − 1
n2
i

; ni = 4, 5, 6 . . .
4. Brackett Infrared 40,501 14,580
ν̃ = R 1
42 − 1
n2
i

; ni = 5, 6, 7 . . .
5. Pfund Far infra-red 74,558 22,782
ν̃ = R 1
52 − 1
n2
i

; ni = 6, 7, 8 . . .
Positronium
A system of e+
−e−
is called positronium, the reduced mass, μ = 0.5me. Therefore,
the radii are expanded but energy levels are reduced by a factor of 2, compared to
the H-atom, and the entire spectrum is shifted toward the longer wavelength.
Uncertainty principle (Heisenberg)
ΔxΔpx (2.25a)
ΔEΔt (2.25b)
LzΔΦ (2.25c)
Restricted uncertainty principle
ΔxΔpx /2 (2.26)
Bohr magneton
μB =
e
2m
(2.27)
Zeeman effect
The splitting of spectrum lines in a magnetic field is known as Zeeman effect.
Normal Zeeman effect (Strong magnetic field)
Each term is split up into 2J + 1 terms by the magnetic field. When observed trans-
versely (magnetic field at right angle to the light path), the lines are observed to
be split up into three, the middle line linearly polarized parallel to the field, and
the outside lines at right angles to the field; but when observed longitudinally (field

parallel to the light path), they are split into two, which are circularly polarized in
the opposite directions.
The selection rule is Δm = 0, ±1, where m is the magnetic quantum number.
We thus get a simple triplet or doublet. In the former they are equally spaced.
Anamolous Zeeman effect
For not too strong field, one observes splitting into more than three components,
unequally spaced. The additional magnetic energy is given by
Emag = −
eBmg
2μc
(2.28)
where g is Lande’s g-factor. The undisturbed term again splits up into 2J+1 equidis-
tant terms but the lines will not be equidistant because the g-factor for the upper and
lower levels would be different.
g = 1 +
j( j + 1) + s(s + 1) − l(l + 1)
2 j( j + 1)
(2.29)
Degeneracy of H-atom energy levels
The degeneracy = 2
n−1
l=0
(2l + 1) = 2n2
(2.30)
where n is the principal quantum number.
Broadening of spectral lines
The observed spectral lines are not perfectly sharp. The broadening is due to
(i) Natural width explained by the uncertainty principle for time and energy.
(ii) Thermal motion of atoms.
(iii) Molecular collisions.
Spectroscopic notation
2S+1
LJ ,where S is the total electron spin, L is the orbital angular momentum, and
J the total angular momentum.
Stern–Gerlah experiment
In this experiment a collimated beam of neutral atoms emerging from a hot oven is
sent through an inhomogeneous magnetic field. The beam is split up into 2J + 1
components. The experiment affords the determination of the spin of the atoms.

2.2 Problems
2.2.1 de Broglie Waves
2.1 (a) Write down the equation relating the energy E of a photon to its frequency
f . Hence determine the equation relating the energy E of a photon to its
wavelength.
(b) A π0
meson at rest decays into two photons of equal energy. What is the
wavelength (in m) of the photons? (The mass of the π0
is 135 MeV/c)
[University of London 2006]
2.2 Calculate the wavelength in nm of electrons which have been accelerated from
rest through a potential difference of 54 V.
2.3 Show that the deBroglie wavelength for neutrons is given by λ = 0.286 Å/
√
E,
where E is in electron-volts.
[Adapted from the University of New Castle upon Tyne 1966]
2.4 Show that if an electron is accelerated through V volts then the deBroglie wave-
length in angstroms is given by λ =
150
V
1/2
2.5 A thermal neutron has a speed v at temperature T = 300 K and kinetic energy
mnv2
2
= 3kT
2
. Calculate its deBroglie wavelength. State whether a beam of these
neutrons could be diffracted by a crystal, and why?
(b) Use Heisenberg’s Uncertainty principle to estimate the kinetic energy (in
MeV) of a nucleon bound within a nucleus of radius 10−15
m.
2.6 The relation for total energy (E) and momentum (p) for a relativistic particle
is E2
= c2
p2
+ m2
c4
, where m is the rest mass and c is the velocity of light.
Using the relativistic relations E = ω and p = k, where ω is the angular
frequency and k is the wave number, show that the product of group velocity
(vg) and the phase velocity (vp) is equal to c2
, that is vpvg = c2
2.2.2 Hydrogen Atom
2.7 In the Bohr model of the hydrogen-like atom of atomic number Z the atomic
energy levels of a single-electron are quantized with values given by
En =
Z2
mee4
8ε2
0h2n2
where m is the mass of the electron, e is the electronic charge and n is an
integer greater than zero (principal quantum number)

2.2 Problems 93
What additional quantum numbers are needed to specify fully an atomic quan-
tum state and what physical quantities do they quantify? List the allowed quan-
tum numbers for n = 1 and n = 2 and specify fully the electronic quantum
numbers for the ground state of the Carbon atom (atomic number Z = 6)
[Adapted from University of London 2002]
2.8 Estimate the total ground state energy in eV of the system obtained if all the
electrons in the Carbon atom were replaced by π−
particles. (You are given
that the ground state energy of the hydrogen atom is −13.6 eV and that the π−
is a particle with charge −1, spin 0 and mass 270 me
[University of London]
2.9 What are atomic units? In this system what are the units of (a) length (b) energy
(c) 2
(d) e2
(e) me ? (f) Write down Schrodinger’s equation for H-atom in
atomic units
2.10 (a) Two positive nuclei each having a charge q approach each other and elec-
trons concentrate between the nuclei to create a bond. Assume that the
electrons can be represented by a single point charge at the mid-point
between the nuclei. Calculate the magnitude this charge must have to
ensure that the potential energy is negative.
(b) A positive ion of kinetic energy 1 × 10−19
J collides with a stationary
molecule of the same mass and forms a single excited composite molecule.
Assuming the initial internal energies of the ion and neutral molecule were
zero, calculate the internal energy of the molecule.
[Adapted from University of Wales, Aberystwyth 2008]
2.11 (a) By using the deBroglie relation, derive the Bohr condition mvr = n for
the angular momentum of an electron in a hydrogen atom.
(b) Use this expression to show that the allowed electron energy states in
hydrogen atom can be written
En = −
me4
8ε2
0h2n2
(c) How would this expression be modified for the case of a triply ionized
beryllium atom Be(Z = 4)?
(d) Calculate the ionization energy in eV of Be+3
(ionization energy of hydro-
gen = 13.6 eV)
[Adapted from the University of Wales, Aberystwyth 2007]
2.12 When a negatively charged muon (mass 207 me is captured in a Bohr’s orbit
of high principal quantum number (n) to form a mesic atom, it cascades
down to lower orbits emitting X-rays and the radii of the mesic atom are
shrunk by a factor of about 200 compared with the corresponding Bohr’s atom.
Explain.
2.13 In which mu-mesic atom would the orbit with n = 1 just touch the nuclear
surface. Take Z = A/2 and R = 1.3 A1/3
fm.

2.14 Calculate the wavelengths of the first four lines of the Lyman series of the
positronium on the basis of the simple Bohr’s theory
[Saha Institute of Nuclear physics 1964]
2.15 (a) Show that the energy En of positronium is given by En = −α2
mec2
/4n2
where me is the electron mass, n the principal quantum number and α the
fine structure constant
(b) the radii are expanded to double the corresponding radii of hydrogen atom
(c) the transition energies are halved compared to that of hydrogen atom.
2.16 A non-relativistic particle of mass m is held in a circular orbit around the
origin by an attractive force f (r) = −kr where k is a positive constant
(a) Show that the potential energy can be written
U(r) = kr2
/2
Assuming U(r) = 0 when r = 0
(b) Assuming the Bohr quantization of the angular momentum of the particle,
show that the radius r of the orbit of the particle and speed v of the particle
can be written
v2
=

n
m

k
m
1/2
r2
=

n
k

k
m
1/2
where n is an integer
(c) Hence, show that the total energy of the particle is
En = n

k
m
1/2
(d) If m = 3 × 10−26
kg and k = 1180 N m−1
, determine the wavelength of
the photon in nm which will cause a transition between successive energy
levels.
2.17 For high principle quantum number (n) for hydrogen atom show that the spac-
ing between the neighboring energy levels is proportional to 1/n3
.
2.18 In which transition of hydrogen atom is the wavelength of 486.1 nm produced?
To which series does it belong?
2.19 Show that for large quantum number n, the mechanical orbital frequency
is equal to the frequency of the photon which is emitted between adjacent
levels.
2.20 A hydrogen-like ion has the wavelength difference between the first lines of
the Balmer and lyman series equal to 16.58 nm. What ion is it?

2.2 Problems 95
2.21 A spectral line of atomic hydrogen has its wave number equal to the difference
between the two lines of Balmer series, 486.1 nm and 410.2 nm. To which
series does the spectral line belong?
2.2.3 X-rays
2.22 (a) The L → K transition of an X-ray tube containing a molybdenum (Z = 42)
target occurs at a wavelength of 0.0724 nm. Use this information to estimate
the screening parameter of the K-shell electrons in molybdenum.
[Osmania University]
2.23 Calculate the wavelength of the Mo(Z = 42)Kα X-ray line given that the
ionization energy of hydrogen is 13.6 eV
[Adapted from the University of London, Royal Holloway 2002]
2.24 In a block of Cobalt/iron alloy, it is suspected that the Cobalt (Z = 27) is
very poorly mixed with the iron (Z = 26). Given that the ionization energy
of hydrogen is 13.6 eV predict the energies of the K absorption edges of the
constituents of the alloy.
[University of London, Royal Holloway 2002]
2.25 Calculate the minimum wavelength of the radiation emitted by an X-ray tube
operated at 30 kV.
2.26 If the minimum wavelength from an 80 kV X-ray tube is 0.15 × 10−10
m,
deduce a value for Planck’s constant.
[Adapted from the University of New Castle upon Tyne 1964]
2.27 If the minimum wavelength recorded in the continuous X-ray spectrum from
a 50 kV tube is 0.247 Å, calculate the value of Plank’s constant.
[Adapted from the University of Durham 1963]
2.28 The wavelength of the Kα line in iron (Z = 26) is known to be 193 pm. Then
what would be the wavelength of the Kα line in copper (Z = 29)?
2.29 An X-ray tube has nickel as target. If the wavelength difference between the
Kα line and the short wave cut-off wavelength of the continuous X-ray spec-
trum is equal to 84 pm, what is the voltage applied to the tube?
2.30 Consider the transitions in heavy atoms which give rise to Lα line in X-ray
spectra. How many allowed transitions are possible under the selection rule
Δl = ±1, Δ j = 0, ±1.
2.31 When the voltage applied to an X-ray tube increases from 10 to 20 kV the
wavelength difference between the Kα line and the short wave cut-off of the
continous X-ray spectrum increases by a factor of 3.0. Identify the target mate-
rial.
2.32 How many elements have the Kα lines between 241 and 180 pm?

2.33 Moseley’s law for characteristic x-rays is of the form
√
v = a(z−b). Calculate
the value of a for Kα
2.34 The Kα line has a wavelength λ for an element with atomic number Z = 19.
What is the atomic number of an element which has a wavelength λ/4 for the
Kα line?
2.2.4 Spin and µ and Quantum Numbers – Stern–Gerlah’s
Experiment
2.35 Evidence for the electron spin was provided by the Sterrn–Gerlah experiment.
Sketch and briefly describe the key features of the experiment. Explain what
was observed and how this observation may be interpreted in terms of electron
spin.
[Adapted from University of London 2006]
2.36 (i) Write down the allowed values of the total angular momentum quantum
number j, for an atom with spin s and l, respectively (ii) Write down the
quantum numbers for the states described as 2
S1/2, 3
D2 and 5
P3 (iii) Determine
if any of these states are impossible, and if so explain why.
2.37 (a) show that an electron in a classical circular orbit of angular momentum
L around a nucleus has magnetic dipole moment given by µ = −e L/2me
(b) State the quantum mechanical values for the magnitude and the z-compo-
nent of the magnetic moment of the hydrogen atom associated with (i) electron
orbital angular momentum (ii) electron spin
2.38 In a Stern-Gerlach experiment a collimated beam of hydrogen atoms emitted
from an oven at a temperature of 600 K, passes between the poles of a magnet
for a distance of 0.6 m before being detected at a photographic plate a further
1.0 m away. Derive the expression for the observed mean beam separation, and
determine its value given that the magnetic field gradient is 20 Tm−1
(Assume
the atoms to be in the ground state and their mean kinetic energy to be 2 kT;
Bohr magneton μB = 9.27 × 10−24
J T−1
2.39 State the ground state electron configuration and magnetic dipole moment of
hydrogen (Z = 1) and sodium (Z = 11)
2.40 In a Stern–Gerlah experiment a collimated beam of sodium atoms, emit-
ted from an oven at a temperature of 400 K, passes between the poles of
a magnet for a distance of 1.00 m before being detected on a screen a fur-
ther 0.5 m away. The mean deflection detected was 0.14◦
. Assuming that
the magnetic field gradient was 6.0 T m−1
and that the atoms were in the

2.2 Problems 97
ground state and their mean kinetic energy was 2kT, estimate the magnetic
moment.
[Adapted from the University of London, Royal Holloway and Bedford New
College, 2005]
2.41 If the electronic structure of an element is 1s2
2s2
2p6
3s2
3p6
3d10
4s2
4p5
,
why can it not be (a) a transition element (b) a rare-earth element?
[Adopted from the University of Manchester 1958]
2.42 In a Stern–Gerlach experiment the magnetic field gradient is 5.0 V s m−2
mm−1
,
with pole pieces 7 cm long. A narrow beam of silver atoms from an oven at
1, 250 K passes through the magnetic field. Calculate the separation of the
beams as they emerge from the magnetic field, pointing out the assumptions
you have made (Take μ = 9.27 × 10−24
JT−1
]
2.43 (a) The magnetic moment of silver atom is only 1 Bohr magneton although it
has 47 electrons? Explain.
(b) Ignoring the nuclear effects, what is the magnetic moment of an atom in
the 3p0 state?
(c) In a Stern–Gerlah experiment, a collimated beam of neutral atoms is split
up into 7 equally spaced lines. What is the total angular momentum of the
atom?
(d) what is the ratio of intensities of spectral lines in hydrogen spectrum for
the transitions
22
p1/2 → 12
s1/2 and 22
p3/2 → 12
s1/2?
2.44 Obtain an expression for the Bohr magneton.
2.2.5 Spectroscopy
2.45 (a) Given the allowed values of the quantum numbers n,l, m and ms of an
electron in a hydrogen atom (b) What are the allowed numerical values of
l and m for the n = 3 level? (c) Hence show that this level can accept 18
electrons.
[Adopted from University of London 2006]
2.46 State, with reasons, which of the following transitions are forbidden for elec-
tric dipole transitions
3
D1 → 2
F3
2
P3/2 → 2
S1/2
2
P1/2 → 2
S1/2
3
D2 → 3
S1

2.47 The 9
Be+
ion has a nucleus with spin I = 3/2. What values are possible for
the hyperfine quantum number F for the 2
S1/2 electronic level?
[Aligarh University]
2.48 Obtain an expression for the Doppler linewidth for a spectral line of wave-
length λ emitted by an atom of mass m at a temperature T
2.49 For the 2P3/2 → 2S1/2 transition of an alkali atom, sketch the splitting of the
energy levels and the resulting Zeeman spectrum for atoms in a weak exter-
nal magnetic field (Express your results in terms of the frequency v0 of the
transition, in the absence of an applied magnetic field)

The Lande g-factor is given by g = 1 +
j( j + 1) + s(s + 1) − l(l + 1)
2 j( j + 1)

[Adapted from the University of London Holloway 2002]
2.50 The spacings of adjacent energy levels of increasing energy in a calcium triplet
are 30 × 10−4
and 60 × 10−4
eV. What are the quantum numbers of the three
levels? Write down the levels using the appropriate spectroscopic notation.
2.51 An atomic transition line with wavelength 350 nm is observed to be split into
three components, in a spectrum of light from a sun spot. Adjacent compo-
nents are separated by 1.7 pm. Determine the strength of the magnetic field in
the sun spot. μB = 9.17 × 10−24
J T−1
2.52 Calculate the energy spacing between the components of the ground state
energy level of hydrogen when split by a magnetic field of 1.0 T. What fre-
quency of electromagnetic radiation could cause a transition between these
levels? What is the specific name given to this effect.
2.53 Consider the transition 2P1/2 → 2S1/2, for sodium in the magnetic field of
1.0 T, given that the energy splitting ΔE = gμB Bm j , where μB is the Bohr
magneton. Draw the sketch.
2.54 To excite the mercury line 5,461 Å an excitation potential of 7.69 V is required.
If the deepest term in the mercury spectrum lies at 84,181 cm−1
, calculate the
numerical values of the two energy levels involved in the emission of 5,461 Å.
[The University of Durham 1963]
2.55 The mean time for a spontaneous 2p → 1s transition is 1.6 × 10−9
s while
the mean time for a spontaneous 2s → 1s transition is as long as 0.14 s.
Explain.

2.2 Problems 99
2.56 In the Helium-Neon laser (three-level laser), the energy spacing between the
upper and lower levels E2 − E1 = 2.26 in the neon atom. If the optical pump-
ing operation stops, at what temperature would the ratio of the population of
upper level E2 and the lower level E1, be 1/10?
2.2.6 Molecules
2.57 What are the two modes of motion of a diatomic molecule about its centre of
mass? Explain briefly the origin of the discrete energy level spectrum associ-
ated with one of these modes.
2.58 Rotational spectral lines are examined in the HD (hydrogen–deuterium)
molecule. If the internuclear distance is 0.075 nm, estimate the wavelength
of radiation arising from the lowest levels.
2.59 Historically, the study of alternate intensities of spectral lines in the rotational
spectra of homonuclear molecules such as N2 was crucial in deciding the
correct model for the atom (neutrons and protons constituting the nucleus
surrounded by electrons outside, rather than the proton–electron hypothesis
for the Thomas model). Explain.
2.60 The force constant for the carbon monoxide molecule is 1,908 N m−1
. At
1,000 K what is the probability that the molecule will be found in the lowest
excited state?
2.61 At a given temperature the rotational states of molecules are distributed
according to the Boltzmann distribution. Of the hydrogen molecules in the
ground state estimate the ratio of the number in the ground rotational state to
the number in the first excited rotational state at 300 K. Take the interatomic
distance as 1.06 Å.
2.62 Estimate the wavelength of radiation emitted from adjacent vibration energy
levels of NO molecule. Assume the force constant k = 1,550 N m−1
. In which
region of electromagnetic spectrum does the radiation fall?
2.63 Carbon monoxide (CO) absorbs energy at 1.153×1,011 Hz, due to a transition
between the l = 0 and l = 1 rotational states.
(i) What is the corresponding wavelength? In which part of the electro-
magnetic spectrum does this lie?
(ii) What is the energy (in eV)?
(iii) Calculate the reduced mass μ. (C = 12 times, and O = 16 times the
unified atomic mass constant.)
(iv) Given that the rotational energy E = l(l+1)2
2μr2 , find the interatomic distance
r for this molecule.
2.64 Consider the hydrogen molecule H2 as a rigid rotor with distance of separation
of H-atoms r = 1.0 Å. Compute the energy of J = 2 rotational level.

2.65 The J = 0 → J = 1 rotational absorption line occurs at wavelength 0.0026
in C12
O16
and at 0.00272 m in Cx
O16
. Find the mass number of the unknown
Carbon isotope.
2.66 Assuming that the H2
molecule behaves like a harmonic oscillator with force
constant of 573 N/m. Calculate the vibrational quantum number for which the
molecule would dissociate at 4.5 eV.
2.2.7 Commutators
2.67 (a) Show that eipα/
x e−ipα/
= x + α
(b) If A and B are Hermitian, find the condition that the product AB will be
Hermitian
2.68 (a) If A is Hermitian, show that ei A
is unitary
(b) What operator may be used to distinguish between
(a) eikx
and e−ikx
(b) sin ax and cos ax?
2.69 (a) Show that exp (iσ xθ) = cos θ + iσ x sin θ
(b) Show that
d
dx
†
= − d
dx
2.70 Show that
(a) [x, px ] = [y, py] = [z, pz] = i
(b) [x2
, px ] = 2ix
2.71 Show that a hermitian operator is always linear.
2.72 Show that the momentum operator is hermitian
2.73 The operators P and Q commute and they are represented by the matrices

1 2
2 1

and

3 2
2 3

. Find the eigen vectors of P and Q. What do you notice
about these eigen vectors, which verify a necessary condition for commuting
operators?
2.74 An operator Â is defined as Â = αx̂ + iβ p̂, where α , β are real numbers
(a) Find the Hermitian adjoint operator Â†
(b) Calculate the commutators [Â, x̂], [Â, Â] and [Â, P̂]
2.75 A real operator A satisfies the lowest order equation.
A2
− 4A + 3 = 0
(a) Find the eigen values of A (b) Find the eigen states of A (c) Show that A
is an observable.
2.76 Show that (a) [x, H] = ip
μ
(b) [[x, H], x] = 2
μ
where H is the Hamiltonian.
2.77 Show that for any two operators A and B,
[A2
, B] = A[A, B] + [A, B]A

2.3 Solutions 101
2.78 Show that (σ.A)(σ.B) = A.B+iσ.(A×B) where A and B are vectors and σ’s
are Pauli matrices.
2.79 The Pauli matrix σy =

0 −i
i 0

(a) Show that the matrix is real whose eigen values are real.
(b) Find the eigen values of σy and construct the eigen vectors.
(c) Form the projector operators P1 and P2 and show that
P
†
1 P2 =

0 0
0 0

, P1 P
†
1 + P2 P
†
2 = I
2.80 The Pauli spin matrices are σx =

0 1
1 0

, σy =

0 −i
i 0

, σz =

1 0
0 −1

Show that (i) σ2
x = 1 and (ii) the commutator [σx , σy] = 2iσz.
2.81 The condition that must be satisfied by two operators Â and B̂ if they are to
share the same eigen states is that they should commute. Prove the statement.
2.2.8 Uncertainty Principle
2.82 Use the uncertainty principle to obtain the ground state energy of a linear
oscillator
2.83 Is it possible to measure the energy and the momentum of a particle simulta-
neously with arbitrary precision?
2.84 Obtain Heisenberg’s restricted uncertainty relation for the position and momen-
tum.
2.85 Use the uncertainty principle to make an order of magnitude estimate for the
kinetic energy (in eV) of an electron in a hydrogen atom.
2.86 Write down the two Heisenberg uncertainty relations, one involving energy
and one involving momentum. Explain the meaning of each term. Estimate
the kinetic energy (in MeV) of a neutron confined to a nucleus of diameter
10 fm.
2.3 Solutions
2.1 (a) E = h f = hλ/c
(b) Each photon carries an energy
Eγ =
mπ c2
2
=
135
2
= 67.5 MeV

cPγ = Eγ = 67.5 MeV
λ =
h
p
= 2π
c
cp
=
(2π)(197.3 MeV.fm)
67.5 MeV
= 18.36 fm = 1.836 × 10−14
m.
2.2 λ =

150
V
1/2
=

150
54
1/2
= 1.667 Å
2.3 λ =
h
p
=
2πc
(2mc2T )1/2
= (2π) ×
197.3 MeV − fm
(2 × 939 T − MeV)1/2
= 28.6 × 10−5
Å/T 1/2
where T is in MeV. If T is in eV, λ = 0.286 Å/T1/2
2.4 λ =
h
p
=
6.63 × 10−34
J−s
(2 × 9.1 × 10−31 × 1.6 × 10−19)1/2
= 12.286 × 10−10
m/V1/2
=

151
V
1/2
.
2.5 (a)
mnv2
2
=
p2
2mn
=
3kT
2
=

3
2

× 1.38 × 10−23
×
300
1.6 × 10−19
= 0.0388 eV
cCp =
√
2mnc2
· En
1/2
= (2 × 940 × 106 × 0.0388)1/2
= 8,541 eV
λ =
h
p
=
hc
cp
= 3.9 × 10−15
(eV − s) × 3 × 108
m − s−1
/8,541 eV
= 1.37 × 10−10
m = 1.37 Å
Such neutrons can be diffracted by crystals as their deBroglie wavelength
is comparable with the interatomic distance in the crystal.
(b) Δpx .Δx =
Put the uncertainty in momentum equal to the momentum itself, Δpx = p
cp =
c
Δx
=
197.3 MeV − fm
1.0 fm
= 197.3 MeV
E = (c2
p2
+ m2
c4
)1/2
= [(197.3)2
+ (940)2
]1/2
= 960.48 MeV
Kinetic energy T = E − mc2
= 960.5 − 940 = 20.5 MeV
2.6 E2
= c2
p2
+ m2
c4
2
ω2
= c2
2
k2
+ m2
c4
ω =

c2
k2
+
m2
c4
2
1/2

2.3 Solutions 103
νp =
ω
k
=

c2
k2
+
m2
c4
2
1/2
/k
νg =
dω
dk
= kc2

c2
k2
+
m2
c4
2
−1/2
∴ νpνg = c2
2.3.2 Hydrogen Atom
2.7 Apart from the principle quantum number n, three other quantum numbers
are required to specify fully an atomic quantum state viz l, the orbital angular
quantum number, ml the magnetic orbital angular quantum number, and ms
the magnetic spin quantum number.
For n = 1,l = 0, if there is only one electron as in H-atom, then it will be
in 1s orbit. The total angular momentum J = l ± 1/2, so that J = 1/2. In
the spectroscopic notation, 2s+1
LJ , the ground state is therefore a 2
S1/2 state.
For n = 2, the possible states are 2
S and 2
P. if there are two electrons as in
helium atom, both the electrons can go into the K-shell (n = 1) only when
they have antiparallel spin direction (↑↓ ) on account of Pauli’s principle.
This is because if the spins were parallel, all the four quantum numbers would
be the same for both the electrons (n = 1,l = 0, ml = 0, ms = +1/2).
Therefore in the ground state S = 0, and since both electrons are 1s electrons,
L = 0. Thus the ground state is a S state (closed shell). A triplet state is not
given by this electron configuration. An excited state results when an electron
goes to a higher orbit. Then both electrons can have, in addition, the same
spin direction, that is we can have S = 1 as well as S = 0 Excited triplet
and singlet spin states are possible (orthohelium and parahelium). The lowest
triplet has the electron configuration 1s 2s, it is a 3
s1, state. It is a metastable
state. The corresponding singlet state is 21
S0, and lies somewhat higher.
Carbon has six electrons. The Pauli principle requires the ground state con-
figuration 1S2
2S2
2P2
. The superscripts indicate the number of electrons in a
given state.
2.8 A carbon atom has 6 electrons. If all these electrons are replaced by π−
mesons then two differences would arise (i) As π−
mesons are bosons (spin
0) Pauli’s principle does not operate so that all of them can be in the K-shell
(n = 1) (ii) The total energy is enhanced because of the reduced mass μ.
μ =
mcmπ
mc + mπ
=
(12 × 1,840)(270)
[(12 × 1,840) + (270)]
= 266.7 me
For each π−
, E = −13.6 × 266.7 = 3,628 eV
For the 6 pions, E = 3,628 × 6 = 21,766 eV

2.9 In atomic physics the atomic units are as follows:
(i) (a) The Bohr radius 2
/me e2
is used as the unit of length. (b) The energy
is measured in multiples of the ionization energy of hydrogen mee2
/22
(c) 2
= 1 (d) e2
= 2
(ii) In atomic units the Schrodinger equation
−
2
2me
∇2
u −
e2
u
r
= Eu
would read as
−∇2
u −
2u
r
= Eu
2.10 (a) Let the separation between the two nuclei each of charge q be 2d, then the
negative charge Q on the electrons is at a distance d from either nuclei
Total potential energy due to electrostatic interaction between three objects
is
qQ
d
+
qQ
d
+
q2
2d
≤ 0
Taking the equality sign and cancelling q
Q = −
q
4
(b) Let T0 be the initial kinetic energy and p0 the momentum of the ion and T
the kinetic energy and p the momentum of the composite molecule and Q
the excitation energy.
T0 = T + Q Energy conservation (1)
p0 = p Momentum conservation (2)
∴ (2mT0)1/2
= (2.2mT )1/2
(3)
The mass of the composite being 2m as the excitation energy is expected
to be negligible in comparison with the mass of the molecule. From (3) we
get
T =
T0
2
(4)
Using (4) in (1), we find Q = T0
2
= 10−19
2
= 5 × 10−20
J
2.11(a) Stationary orbits will be such that the circumference of a circular orbit is
equal to an integral number of deBroglie wavelength so that constructive
interference may take place i.e. 2πr = nλ
But λ = h/p
∴ L = rp = nh/2π (Bohr’s quantization condition) (1)

2.3 Solutions 105
(b) Equating the coulomb force to the centripetal force
Ze2
/4πεor2
= mv2
/r (2)
Solving (1) and (2)
v = Ze2
/2εonh (3)
r = εon2
h2
/πmZe2
(4)
Total energyE = K + U =
1
2
mv2
− Ze2
/4πεor (5)
Substituting (3) and (4) in (5)
E = −me4
Z2
/8ε2
on2
h2
(6)
(c) E = −mZ2
e4
/8ε2
on2
h2
= −9me4
/8ε2
on2
h2
(for Z = 3)
(d) Ionization energy for Be+3
= 13.6 × 32
= 122.4 eV
2.12 For a hydrogen-like atom the energy in the nth orbit is En = 13.6 μZ2
/n2
For hydrogen atom the reduced mass μ ≈ me, while for muon mesic atom
it is of the order of 200 me. Consequently, the transition energies are enhanced
by a factor of about 200, so that the emitted radiation falls in the x-ray region
instead of U.V., I.R. or visible part of electromagnetic spectrum. The radius is
given by
rn = ε0n2
h2
/πμe2
Here, because of inverse dependence on μ, the corresponding radii are reduced
by a factor of about 200.
2.13 r1 =
a0
μZ
= R = r0 A1/3
= r0(2Z)1/3
Z4/3
≈
0.529 × 10−10
207 × 1.3 × 21/3 × 10−15
= 156
Therefore Z = 44.14 or 44
The first orbit of mu mesic atom will be just grazing the nuclear surface in
the atom of Ruthenium. Actually, in this region A ≈ 2.2 Z so that the answer
would be Z ≈ 43
2.14 The first four lines of the Lyman series are obtained from the transition ener-
gies between n = 2 → 1, 3 → 1, 4 → 1, 5 → 1
Now En = −
α2
mec2
4n2

ΔEn,1 =
α2
mec2
4

1
12
−
1
n2

, n = 2, 3, 4, 5
Thus,
ΔE21 =

1
137
2
0.511 × 106
4

1 −
1
4

= 5.1 eV
λ21 =
1,241
5.1
= 243.3 nm = 2,433 Å
The wavelengths of the other three lines can be similarly computed. They are
2,053, 1,946 and 1,901 Å.
2.15 (a) Using Bohr’s theory of hydrogen atom
En = −
μe4
8ε2
0h2n2
(1)
where μ is the reduced mass. But the fine structure constant
α =
e2
4πε0c
(2)
En = −
α2
μc2
2n2
(3)
For positronium, = me/2. Therefore for positron
En = −
α2
mec2
4n2
(4)
(b) rn = ε0
n2
h2
πμc2
(5)
rn ∝
1
μ
=
2
me
Therefore the radii are doubled.
(c) En ∝ μ =
me
2
Therefore the transition energies are halved.
2.16 (a) U(r) = −

f (r)dr =

kr dr + C
= 1
2
kr2
+ C
U(0) = 0 → C = 0
U(r) = 1
2
kr2
(b) Bohr’s assumption of quantization of angular momentum gives
mvr = n (1)
Equating the attracting force to the centripetal force.

2.3 Solutions 107
mv2
/r = kr (2)
solving (1) and (2)
v2
=

n
m

k
m
1
2
(3)
r2
=
n
√
km
(4)
(c) E = U + T = 1
2
kr2
+ 1
2
mv2
(5)
Substituting (3) and (4) on (5) and simplifying
E = n(k/m)1/2
(d) ΔE = En − En−1 =

k
m
1
2
= 1.05 × 10−34

1,180
3 × 10−26
1
2
J
= 0.13 eV
λ =
1,241
0.13
= 9,546 nm
2.17 En = −
13.6
n2
ΔE = En+1 − En = 13.6

1
n2
−
1
(n + 1)2

=
13.6(2n + 1)
n2(n + 1)2
In the limit n → ∞, ΔE ∝ n
n4 = 1
n3
2.18 The wavelength λ = 486.1 nm corresponds to the transition energy of E =
1241/486.1 = 2.55 eV Looking up Fig. 2.1, for the energy level diagram
for hydrogen atom, the transition n = 4 → 2 gives the energy difference
−0.85 − (−3.4) = 2.55 eV
The line belongs to the Balmer series.
Fig. 2.1 Energy level
diagram for hydrogen atom
2.19 Orbital velocity, v = e2
2nhε0
, a0 = n2
h2
ε0/πe2
m
Orbital frequency f = v/2πa0

f = me4
/4n3
h3
ε2
0
ν =
me4
8ε2
0h3

1
n2
f
−
1
n2
i

=
me4
8ε2
0h3

(ni − nf)(ni + nf)
n2
i n2
f

If both ni and nf are large, and if we let
ni = nf + 1, ν ≈
me4
8ε2
0h3

2
n3
i

= f
2.20 Energy difference for the transitions in the two series ΔE11 − ΔE32 =
1,241/16.58 = 74.85 eV
13.6Z2

1
12
−
1
22

−

1
22
−
1
32

= 74.85
Solving for Z, we get Z = 3.
The ion is Li++
2.21 Note that wave number is proportional to energy. The wavelength 486.1 nm
in the Balmer series to the energy difference of 2.55 eV, and is due to the
transition between n = 4(E4 = −0.85 eV) and n = 2(E2 = −3.4 eV).
ΔE42 = −0.85 − (−3.4) = 2.55 eV. The wavelength 410.2 nm in the
Balmer series corresponds to the energy difference of 3.0 eV and is due to
the transition between n = 6(E6 = −0.38 eV) and n = 2(E2 = −3.4 eV).
ΔE62 = −0.38 − (−3.4) = 3.02 eV
Thus ΔE62 − ΔE42 = 3.02 − 2.55 = 0.47 eV
The difference of 0.47 eV is also equal to difference in E6 = −0.38 eV
(n = 6) and E4 = −0.85 eV (n = 4). Thus the line arising from the transition
n = 6 → n = 4, must belong to Bracket series.
Note that in the above analysis we have used the well known law of spec-
troscopy, ṽmn − ṽkn = ṽmk
2.3.3 X-rays
2.22 The wavelength λLK = 0.0724 nm corresponds to the energy
Eγ = 1,241/λLK = 1,241/0.0724 = 17,141 eV
Now 17,141 = 13.6 × 3
4
(Z − σ)2
The factor 3/4 is due to the L → K transition. Substituting Z = 42, and
solving for σ we obtain σ = 1.0
2.23 Eγ = 13.6 × 3(Z − σ)2
/4 = 13.6 × 3(42 − 1)2
/4 = 17,146.2 eV.
λLK = 1,241/17,146.2 = 0.07238 nm
= 0.7238 Å
2.24 Cobalt: EK = 13.6(Z − σ)2
= 13.6(27 − 1)2
= 9,193.6 eV
λK =
1,241
9,193.6
nm = 0.135 nm = 1.35 Å

2.3 Solutions 109
Iron: EK = 13.6(26 − 1)2
= 8,500 eV
λK =
1,241
8,500
= 0.146 nm = 1.46 Å
2.25 The minimum wavelength of the photon will correspond to maximum fre-
quency which will be determined by E = hvmax
λmin =
c
vmax
=
hc
hvmax
=
hc
E
=
2πc
E
=
2π × 197.3 fm − MeV
30 × 10−3 MeV
= 4.13 × 105
fm = 4.13 Å
2.26 λC =
hc
eV
h =
eV λC
c
=
1.6 × 10−19
× 80 × 103
× 0.15 × 10−10
3 × 108
= 6.4 × 10−34
J − s
2.27 λc =
hc
eV
h =
λceV
c
=
0.247 × 10−10
× 1.6 × 10−19
× 50,000
3 × 108
= 6.59 × 10−34
J − s
2.28 According to Mosley’s law
1
λ
= A(Z − 1)2
1
λI
= A(26 − 1)2
1
λCu
= A(29 − 1)2
λCu
λI
=
252
282
= 0.797 → λCu = 193 × 0.797 = 153.8 pm
2.29 λK − λC = 84 pm = 0.84 Å (1)
1,200
(28 − 1)2
−
12.4
V
= 0.84 (2)
where λC =
hc
eV
=
12.4
V
(V is in kV) (3)
Solving for V in (2), V = 15.4 kV
2.30 The Lα line is produced due to transition n = 3 → n = 2. For the n = 2 shell
the quantum numbers are l = 0 or l = 1 and j = l ± 1
2
, the energy states
being 2
S1/2, 2
P1/2, 2
P3/2. For n = 3 shell the energy states are 3
S1/2, 3
P1/2,
3
P3/2, 3
d3/2, 3
d5/2
The allowed transitions are
3
S1/2 → 2
P1/2, 3
S1/2 → 2
P3/2
3
P1/2 → 2
S1/2, 3
P3/2 → 2
S1/2

3
d → 2
P1/2, 3
d3/2 → 2
P3/2
3
d5/2 → 2
P3/2
In all there are seven allowed transitions.
2.31 Let the wavelength difference be Δλ when voltage V is applied.
1,200
(Z − 1)2
−
12.4
V
= Δλ (1)
1,200
(Z − 1)2
−
12.4
10
= Δλ (2)
1,200
(Z − 1)2
−
12.4
20
= 3Δλ (3)
Note that the first term on the LHS of (1) (Corresponding to the character-
istic X-rays) is unaffected due to the application of voltage. Eliminating Δλ
between (2) and (3), we find Z = 28.82 or 29. The target material is Copper.
2.32 λk =
1,200
(Z − 1)2
Å
Z1 = 1 +

1,200
2.4
1/2
= 23.36
Z2 = 1 +

1,200
1.8
1/2
= 26.82
The required elements have Z = 23, 24, 25 and 26
2.33 Bohr’s theory gives
ν =
mee4
8ε2
0h3

1
12
−
1
n2

(z − b)2
(1)
or
√
v =

mee4
8ε2
0h3

1
12
−
1
n2
1
2
(z − b) (2)
The factor within the square brackets is identified as a. Substitute me =
9.11 × 10−31
kg,
e = 1.6 × 10−19
C,
ε0 = 8.85 × 10−12
F/m and h = 6.626 × 10−34
J – s and put n = 2 to find a.
We get
a = 4.956 × 107
Hz1/2

2.3 Solutions 111
2.34 By Moseley’s law
1
λ
= A(Z1 − 1)2
= A(19 − 1)2
(1)
1
λ/4
= A(Z − 1)2
(2)
Dividing (2) by (1) and solving for Z, we get Z = 37
2.3.4 Spin and µ and Quantum Numbers – Stern–Gerlah’s
Experiment
2.35 The existence of electron spin and its value was provided by the Stern–Gerlah
experiment in which a beam of atoms is sent through an inhomogeneous mag-
netic field.
Schematic representation of the Stern–Gerlah experiment. to a force moment
tending to align the magnetic moment along the field direction, but also to
a deflecting force due to the difference in field strength at the two poles of
the particle. Depending on its orientation, the particle will be driven in the
direction of increasing or decreasing field strength. If atoms with all possible
orientations in the field are present, a sharp beam should be split up into 2J +1
components. In Fig. 2.2 the beam is shown to be split up into two components
corresponding to J = 1/2
Fig. 2.2 Schematic drawing
of Stern-Gerlah’s apparatus
2.36 (i) If l s, then there will be 2s +1 values of j; j = l +s, l +s −1 . . .l −s
If l s, then there will be 2l +1 values of j; j = s +l, s +l −1 . . . s −l
(ii) The spectroscopic notation for a term is 2S+1
LJ , s, p, d, f . . . refer to
l = 0, 1, 2, 3 . . . respectively.
Term L S J Possible values of J
2
S1/2 0 1/2 1/2 1/2
3
D2 2 1 2 3, 2, 1
5
P3 1 2 3 2, 1
(iii) Obviously the term 5
P3 cannot exist.

2.37 (a) By definition the magnetic moment of electron is given by the product of
the charge and the area A contained by the circular orbit.
μ = i A = −
eπr2
T
= −
ωeπr2
2π
=
−emeωr2
2me
= −
eL
2me
(b) μl = −
e
2mc
(L(L + 1))1/2

μs = −(2e/2mc) (S(S + 1))1/2

2.38 The principle of the Stern–Gerlah experiment is described in Problem 2.35.
While the atom is under the influence of inhomogeneous magnetic field the
constant force acting on the atom along y-direction perpendicular to the
straight line path OAF in the absence of the field, is a parabola (just like
an object thrown horizontally in a gravitational field). The equation to the
parabola is
y = kx2
(1)
where k is a constant, Fig 2.3. Let us focus on the atom which deviates upward.
After leaving the field at D, its path along DE is a staright line. It hits the
plate at E so that EF = s. When E D is extrapolated back, let it cut the line
OAF in C.
Taking the origin at O, Eq. (1) satisfies the relation at D,
h = kl2
(2)
Furthermore at D,

dy
dx

D
= 2K x|D = 2K.OA = 2K.l (3)

Dy
dx

D
=
AD
C A
=
h
C A
(4)
Combining (2), (3) and (4), we get
C A =
l
2
(5)
Now the time taken for the atom along the x-component is the same as for
along the y-component. Therefore
t =
l
ν
=

2h
a
1
2
(6)
or
h =
l2
a
2ν2
(7)

2.3 Solutions 113
From the geometry of the figure,
E F
C F
=
DA
C A
or
s
L + l
2
=
h
l/2
→ h =
s.l
2L + l
(8)
Eliminating h between (7) and (8), we find
a =
2sν2
l(2L + l)
(9)
Now the acceleration,
a =
F
m
=
μ
m

∂ B
∂y

(10)
Finally the separation between the images on the plate,
2s =
l(2L + l)
mν2
μ

∂ B
∂y

(11)
1/2 mν2
= 2kT
2s =
l(2L + l)μB(∂ B/∂y)
4kT
=
[0.6(2 × 1 + 0.6) × 9.27 × 10−24
× 20]
4 × 1.38 × 10−23 × 600
= 0.873 × 10−2
m = 8.73 mm
Fig. 2.3 Stern–Gerlah
experiment
2.39 H Na
1s 1s2
2s2
2p6
3s

Magnetic moment for both hydrogen and sodium is
1 Bohr magneton, μB =
e
2me
= 9.27 × 10−24
JT−1
2.40 From Problem 2.38, the distance of separation on the plate
2s =
l(2L + l)
mv2
μB

∂ B
∂y

Therefore, tan θ = s
L+l/2
= 2s
2L+l
=
lμB
∂ B
∂y

2E
=
lμB
∂ B
∂y

2×2kT
Substituting θ = 0.14◦
,l = 1.0 m, ∂ B
∂y
= 6 Tm−1
, k = 1.38 × 10−23
JK−1
And T = 400 K, we find μB = 8.99 × 10−24
J T−1
.
2.41 The total number of electrons is given by adding the numbers as superscripts
for each term. This number which is equal to the atomic number Z is found to
be 35. The transition elements have Z = 21−30, 39−48, 72−80, 104−112,
while the rare earths comprising the Lanthanide series have Z = 57 − 71
and actinides have Z = 89 − 100. Thus the element with Z = 35 does not
correspond to either a transition element or a rare earth element.
2.42 From Fig. 2.3 of Problem 2.38 the separation of the beams as they emerge
from the magnetic field is given by
2h = l2
a/v2
= (l2
μ/mv2
)(∂ B/∂y)
= (l2
/4kT )μ(∂ B/∂y)
Substituting l = 0.07 m, μ = 9.27 × 10−24
J T−1
(∂ B/∂y) = 5 Tmm−1
= 5,000 Tm−1
, k = 1.38 × 10−23
JK−1
, T = 1,250 K.
we find 2l = 3.29 × 10−3
m or 3.29 mm
2.43 (a) The magnetic moment for the silver atom is due to one unpaired electron
(b) In the 3
P0 state the atom has J = 0, therefore the magnetic moment is also
zero.
(c) The beam of neutral atoms with total angular momentum J is split into
2J + 1 components. 2J + 1 = 7, so J = 3
(d) Ratio of intensities,
I1
I2
= (2J1 + 1)/(2J2 + 1) =
2 × 1
2
+ 1
2 × 3
2
+ 1
=
1
2
2.44 Let an electron move in a circular orbit of radius r = 2
/me2
around a proton.
Assume that the z-component of the angular momentum is Lz = . Equating
Lz to the classical angular momentum
Lz = = mer2
ω (1)
An electron orbiting the proton with frequency v = ω
2π
constitutes a current
i = ωe
2π

2.3 Solutions 115
The magnetic moment μ0 produced is equal to this current multiplied by
the area enclosed.
μB = i A = ωe.
πr2
2π
(2)
Using (1) in (2)
μB =
e
2me
(3)
μB is known as Bohr magneton.
Electron with total angular momentum [ j( j + 1)]1/2
has a magnetic
moment μ = [ j( j + 1)]1/2
μB
The z-component of the magnetic moment is μJ = mJ μB
where mJ is the z-component of the angular momentum.
2.3.5 Spectroscopy
2.45 The principle quantum number n denotes the number of stationary states in
Bohr’s atom model. n = 1, 2, 3 . . .
l is called azimuthal or orbital angular quantum number. For a given value of
n,l takes the values 0, 1, 2 . . . n − 1
The quantum number ml, called the magnetic quantum number, takes the val-
ues −l, −l + 1, −l + 2, . . . , +l for a given pair of n and / values. This gives
the following scheme:
n 1 2 3
l 0 0 1 0 1 2
ml 0 0 −1 0 +1 0 −1 0 +1 −2 −1 0 +1 +2
n 4
l 0 1 2 3
ml 0 −1 0 +1 −2 −1 0 +1 +2 −3 −2 −1 0 +1 +2 +3
ms, the projection of electron spin along a specified axis can take on two values
±1/2.
Hence the total degeneracy is 2n2
. For n = 3, 2 × 32
= 18 electrons can be
accommodated.
2.46 According to Laporte rule, transitions via dipole radiation are forbidden
between atomic states with the same parity. This is because dipole moment
has odd parity and the integral
∞
−∞ ψ∗
f (dipole moment) ψidτ will vanish
between symmetric limits because the integrand will be odd when ψi and ψf
have the same parity. Now the parity of the state is determined by the factor
(−1)l
. Thus for the given terms the l-values and the parity are as below

Term S P D F
l 0 1 2 3
Parity = (−1)l
+1 −1 +1 −1
2.47 J = l + s = 0 + 1/2 = 1/2
F = I + J, I + J − 1, . . . I − J
= 2, 1, 0
2.48 The observed frequency (ω) of radiation from an atom that moves with the
velocity v at an angle θ to the line of sight is given by
ω = ω0(1 + (v/c) cos θ) (1)
where ω0 is the frequency that the atom radiates in its own frame of referenece.
The Doppler shift is then
Δω
ω0
=
ω − ω0
a0
=
v
c

cos θ (2)
As the radiating atoms are subject to random thermal motion, a variety of
Doppler shifts will be displayed. In equilibrium the Maxwellian distribution
gives the fraction dN
N
of atoms with x-component of velocity lying between vx
and vx + dvx
Fig. 2.4 Thermal broadening
due to random thermal
motion
dN
N
=
exp
*
−
vx
U
2
+
√
π
dvx
U
(3)
where u/
√
2 is the root-mean-square velocity for particles of mass M at tem-
perature T . Now
u =

2kT
M
1/2
(4)
where k = 1.38 × 10−23
J/K is Boltzmann’s constant.
Introducing the Doppler widths ΔωD and ΔλD in frequency and wavelength
ΔωD
ω0
=
ΔλD
λ0
=
U
c
=

2kT
Mc2
1/2
(5)

2.3 Solutions 117
Further from (2) d(Δω) = dω. The relative distribution of Doppler shift is
dN
N
=
exp

−

Δω
ΔωD
2
'
√
π
dω
ΔωD
(6)
Thus a Gaussian distribution is produced in the Doppler shift due to the
random thermal motion of the source (Fig. 2.4).
The intensity of radiation is
I(ω) =
exp

−

Δω
ΔωD
2
'
ΔωD
√
π
(7)
centered around the unshifted frequency ω0. The width of the distribution at
the frequencies where I(ω) falls to half the central intensity I(ω0) is known as
the half width
Doppler half width = 2(ln2)1/2
ΔωD
= 2ω0

2kT ln2
Mc2
1/2
(8)
Thermal broadening is most pronounced for light atoms such as hydrogen and
high temperatures, for example the Hα line (6,563 Å) has a Doppler width of
0.6 Å at 400 K.
2.49 Lande’ g-factor is
g = 1 + j( j + 1) + s(s + 1) − l(l + 1)/2 j( j + 1)
For the term 2
P3/2,l = 1, J = 3
2
, s = 1
2
and g = 4
3
For 2
S1/2,l = 0, j = 1
2
, s = 1
2
and g = 2
Fig. 2.5 Anamolous Zeeman
effect in an alkali atom. The
lines are not equidistant

The splitting of levels as in sodium is shown in Fig. 2.5. Transitions take place
with the selection rule
ΔM = 0, ±1.
2.50 Under the assumption of Russel–Saunders coupling, the ratios of the intervals
in a multiplet can be easily calculated as follows. The magnetic field produced
by L is proportional to [L(L + 1)]1/2
, and the component of S in the direction
of this field is [S(S + 1)]1/2
cos(L, S). The energy in the magnetic field is
W = W0 − BμB (1)
where μB is the component of the magnetic moment in the field direction and
W0 is the energy in the field-free case. From (1) the interaction energy is
μB B = A[L(L + 1)]1/2
[S(S + 1)]1/2
cos(L, S) (2)
where A is a constant. From Fig. 2.6 It follows that
cos(L, S) =
J(J + 1) − L(L + 1) − S(S + 1)
2
√
L(L + 1)
√
S(S + 1)
Consequently the interaction energy is A[J(J +1)−L(L +1)−S(S+1)]/2
As L and S are constant for a given multiplet term, the intervals between
successive multiplet components are in the ratio of the differences of the
corresponding J(J + 1) values. Now the difference between two successive
J(J + 1) values is
Fig. 2.6 Russel-Saunders
coupling
(J + 1)(J + 2) − J(J + 1) or 2(J + 1)
and therefore proportional to J + 1. This is known as Lande’s interval rule.
For the calcium triplet
(J + 2)/(J + 1) = 60 × 10−4
/30 × 10−4
= 2
whence J = 0. The three levels of increasing energy have J = 0, 1 and 2.
Now J = 0, 1 and 2 are produced from the combination of L and S. With
the spectroscopic notation 2S+1
LJ the terms for the three levels are 3
P0, 3
P1
and 3
P2.
2.51 ΔE = μB B
hΔv = hcΔλ/λ2
= μB B
B =
hcΔλ
μBλ2
=
(6.63 × 10−34
)(3 × 108
)(1.7 × 10−12
)
(9.17 × 10−24)(350 × 10−9)2
= 0.3 T

2.3 Solutions 119
2.52 ΔE = μB BΔm = μB B (because Δm = ±1)
= (9.27 × 10−24
)(1.0) = 9.27 × 10−24
J = 5.79 × 10−5
eV.
The splitting of levels by equal amount in the presence of magnetic field is
called normal Zeeman effect.
f = ΔE/h = 5.79 × 10−5
× 1.6 × 10−19
/6.625 × 10−34
= 1.398 × 1010
c/s
2.53 For the term 2
P1/2, l = 1, j = 1
2
, s = 1
2
and g = 2
3
. For the term 2
S1/2, l = 0,
j = 1
2
, s = 1
2
and g = 2. The energy levels and splitting of lines in sodium
are shown in Fig. 2.7.
2.54 The ground state energy is
E0 = hv =
hc
λ
= 6.63 × 10−34
× 3 × 1010
× 84,181/1.6 × 10−19
= 10.46 eV
The excitation lines E2 = 10.46 + 7.69 = 18.15 eV
The line 5461 Å is emitted when E2 is deexcited to a lower level E1 such
that
E2 − E1 =
1,241
λ(nm)
=
1,241
546.1
= 2.27 eV
Thus E1 = 18.15 − 2.27 = 15.88 eV
Fig. 2.7 Splitting of D1 lines
in magnetic field
Therefore the two levels involved in the emission of the 5,461 Å line are
18.15 eV and 15.88 eV
2.55 The 2s state of the hydrogen atom cannot decay by electric dipole radiation
because a 2s → 1s transition would violate the Δl = ±1 rule (Laporte
rule). In point of fact the 2s state is a metastable state with a long life time
which eventually decays to the 1s state by a mechanism, such as collision
with other gas molecules, which is much less probable than an electric dipole
transition.

2.56
n(E2)
n(E1)
= e−(E2−E1)/kT
=
1
10
T =
E2 − E1
kln 10
=
2.26
8.625 × 10−5 × 2.3
= 1.14 × 104
K
2.3.6 Molecules
2.57 The two modes of motion of a diatomic molecule are (i) rotation and (ii) vibra-
tion.
The first order rotational energy is 2
J(J + 1)/2I0, where I0 = M R2
0 is
the moment of inertia of the molecule about an axis perpendicular to the line
joining the nuclei; the energy being the same as for the rigid rotator. Clearly
the spacing between successive levels is unequal; it progressively increases
with the increasing value of J, where J = 0, 1, 2 . . . The spectrum called
band spectrum arises due to optical transitions between rotational levels. The
band spectrum is actually a line spectrum, but is thus called because the lines
are so closely spaced and unresolved with an ordinary spectrograph, and give
the appearance of a band.
The second mode consists of to and fro vibrations of the atoms about the
equilibrium position. The motion is described as simple harmonic motion.
The energy levels are given by En = ω (n + 1/2), where n = 0, 1, 2 . . .
and are equally spaced. However as J or n increases, the spacing between
levels becomes smaller than that predicted from the simple rigid rotator and
harmonic oscillator.
2.58 The rotational energy levels are given by
EJ = 2
J(J + 1)/2Io
where Io is the rotational inertia
ΔE = E1 − E0 = 2
/Io
If μ is the reduced mass,
I = μr2
=
mHmDr2
mH + mD
=
mH . 2mHr2
mH + 2mH
=

2
3

mHr2
(because mD ≈ 2mH)
ΔE =
32
2mHr2
=
3
2
.
(c MeV − fm)2
mHc2(0.075 × 10−9m)2
=
3
2
(197.3 × 10−15
MeV − m)2
938(0.075 × 10−9m)2
= 0.011 × 10−6
MeV
= 0.011 eV.
λ = 1,241/0.011 = 1.128 × 105
nm
= 0.113 mm
2.59 All nuclei of even A, with zero or non-zero spin obey Bose statistics and
all those of odd A obey Fermi statistics. The result has been crucial in

2.3 Solutions 121
deciding the model of the nucleus, that is discarding the electron–proton
hypothesis. Consider the nitrogen nucleus. The electron–proton hypothesis
implies 14 prorons+7 electrons. This means that it must have odd spin because
the total number of particles is odd (21) and Fermi statistics must be obeyed.
In the neutron–proton model the nitrogen nucleus has 7n + 7p = 14 parti-
cles (even). Therefore Bose statistics must be obeyed. If the electronic wave
function for the molecules is symmetric it was shown that the interchange of
nuclei produces a factor (−1)J
(J = rotational quantum number) in the total
wave function of the molecule. Thus, if the nuclei obey Bose statistics sym-
metric nuclear spin function must be combined with even J rotational states
and antisymmetric with odd J. Because of the statistical weight attached to
spin states, the intensity of even rotational lines will be (I + 1)/I as great
as that of neighboring odd rotational lines where I is the nuclear spin. For
Fermi statistics of the nuclei the spin and rotational states combine in a manner
opposite to that stated previously, the odd rotational lines being more intense
in the ratio (I + 1)/I. The experimental ratio (I + 1)/I = 2 for even to odd
lines, giving I = 1, is consistent with the neutron–proton model.
2.60 The vibrational energy level is
En =

n +
1
2

ω, n = 0, 1, 2 . . .
with ω =
√
(k/μ), k being the force constant and μ the reduced mass of the
oscillating atoms.
μ =
mcm0
mc + m0
=
12 × 16
12 + 16
= 6.857 amu
ω =

1908
6.857 × 1.67 × 10−27
1/2
= 4.082 × 1014
S−1
Number of molecules in state En is proportional to exp(−nω/kT ), k being
the Boltzmann constant and T the Kelvin temperature. The probability that
the molecule is in the first excited state is
P1 =
exp(−ω/kT )
∞
0 exp(−nω/kT )
= exp(−ω/kT )[1 − exp(−ω/kT )]
ω
kT
=
1.054 × 10−34
× 4.082 × 1014
1.38 × 10−23 × 1, 000
= 3.1177
Therefore, P1 = exp(−3.117)[1 − exp(−3.1177)]
= 0.042
2.61 The rotational energy state is given by
EJ = J
(J + 1)2
2I
, J = 0, 1, 2 . . .
The state with quantum number J is proportional to (2J + 1) exp(−EJ /kT )
The factor (2J + 1) arises from the J state.
N0/N1 = (1/3) exp

2
/IokT

μ = m1/2 = 0.5 mp
N0/N1 = (1/3) exp

2
/μr2
kT

= (1/3) exp

2
c2
/μc2
r2
kT

=
1
3
exp

197.3 × 10−15
2
0.5 × 938 ×

1.06 × 10−10
2
×

1.38 × 10−23/1.6 × 10−13

× 300
'
= 0.445
2.62 ω =
0
k
μ
μ =
mNm0
mN + m0
= 7.466 amu
ω =

1550
7.466 × 1.67 × 10−27
1/2
= 3.526 × 1014
s−1
E = ω =
1.055 × 10−34
× 3.526 × 1014
1.6 × 10−19
= 0.2325
λ =
1,241
0.2325
nm = 5,337 nm = 5.34 µm
This wavelength corresponds to Infrared region.
2.63 (i) λ =
c
ν
=
3 × 108
1.153 × 1011
= 2.6 × 10−3
m = 2.6 mm
It lies in the microwave part of electromagnetic spectrum.
(ii) E(eV ) =
1241
λ(nm)
=
1241
2.6 × 106(nm)
= 0.000477 eV
(iii) μ =
mcm0
mc + m0
=
12 × 16
12 + 16
= 6.857 u
(iv) E1 =
1(1 + 1)2
2μr2
=
2
μr2
E0 = 0
ΔE = E1 − E0 =
2
μr2
r =

2
c2
μc2ΔE
1/2
=

197.32
(MeV − fm)2
6.857 × 931.5 × 477 × 10−12
1/2
= 0.113 × 10−9
m
= 1.13 Å
2.64 EJ =
J(J + 1)2
2Io
Io = μr2
=

Mp
2

r2

2.3 Solutions 123
EJ = [J(J + 1)2
c2
]/(2)(0.5)(Mpc2
)r2
E2 = (2)(3)(197.3)2
(10−15
)2
/(940) × (10−10
)2
= 0.264 × 10−10
MeV
= 2.64 × 10−5
eV
2.65 ΔEJ =
J2
Io
=
J2
μr2
ΔEJ = hc/λ
∴ λ ∝ μ
λ1
λ2
=
0.00260
0.00272
=
μ1
μ2
(1)
μ1 =
16 × 12
16 + 12
; μ2 =
16x
16 + x
(2)
Using (2) in (1) and solving for x, we get x = 13.004. Hence the mass
number is 13.
2.66 En = ω

n +
1
2

=
0
k
μ

n +
1
2

4.5 =
1.054 × 10−34
1.6 × 10−19

573
0.5 × 1.67 × 10−27
1/2
n +
1
2

whence n = 7.75
Therefore the molecule would dissociate for n = 8.
2.3.7 Commutators
2.67 (a) Writing x = i
∂
∂p

eipα/

i
∂
∂p

e−ipα/

ψ|p|
= i eip α/

−
iα

e
ipα
ψ(p) + e−ipα/ ∂ψ(p)
∂p

= α +
i ∂ψ(p)
∂p
= α + x
(b) If A and B are Hermitian
(AB)†
= B†
A†
= B A
If the product is to be Hermitian then (AB)†
= AB i.e. AB = B A. Thus,
A and B must commute with each other.

2.68 (a) Let f = ei A
; then f †
=

ei A
†
= e−i A
Therefore f †
f = e−i A
ei A
= 1
Thus ei A
is unitary.
(b) (a) Momentum (b) Parity
2.69 (a) exp(iσx θ) = 1 + iσx θ + (iσx θ)2
/2! + (iσx θ)3
/3! + · · ·
= (1 − θ2
/2! + θ4
/4! . . .) + iσx (θ − θ3
/3! + θ5
/5!. . . .)
= cos θ + iσx sin θ
(where we have used the identity σ2
x = 1)
(b)

ϕ∗
(dψ/dx)dx = ϕ∗
ψ −

(dϕ∗
/dx)ψdx
But ϕ∗
ψ = 0
Hence

ϕ∗ dψ
dx

dx =

−dϕ∗
dx

ψdx =

−(dϕ/dx)†
ψdx
Therefore
d
dx
†
= −d/dx
2.70 (a) [x, Px ]ψ = x Px ψ − Px xψ
= x

−i
∂
∂x

ψ + i
∂
∂x
(xψ)
= −ix
∂ψ
∂x
+ ix
∂ψ
∂x
+ iψ
= iψ
∴ [x, Px ] = i
(b) [x2
, Px ]ψ = x2
(−i∂ ψ/∂x + i
∂
∂x
(x2
ψ)
= −ix2
∂ ψ/∂x + i x2
∂ψ/∂x + i(2x)ψ
= 2ixψ
Therefore [x2
, px ] = 2ix
2.71 By definition a transformation A is said to be linear if for any constant (possi-
bly complex) λ
A(λX) = λ A X
And if for any two vectors x and y
A(x + y) = Ax + Ay
If H is a hermitian operator
(x, Hλy) = (Hx, λy) = λ(Hx, y) = λ(x, Hy) = (x, λHy)
Or Hλy = λHy
for any y. Furthermore
(z, H(x + y)) = (Hz, x + y) = (Hz, x) + (Hz, y)
= (z, Hx) + (z, Hy) = (z, Hx + Hy)
∴ H(x + y) = Hx + Hy
2.72 Consider the equation
∂
∂x
(ψ∗
ψ) = ψ∗ ∂ψ
∂x
+ ψ
dψ∗
dx
(1)

2.3 Solutions 125
Integration of (1) yields

∂
∂x
(ψ∗
ψ)dτ +

ψ∗ ∂ψ
∂x
dτ +

ψ
dψ∗
dx
dτ (2)
The integral on the LHS vanishes because ψ∗
ψ vanishes for large values of
|x| if the particle is confined to some finite region. Thus

∂
∂x
(ψ∗
ψ)dxdydz =

|ψ∗
ψ|∞
−∞dydz = 0
Therefore (2) becomes

ψ∗

∂ψ
∂x

dτ = −

ψ

∂ψ∗
∂x

dτ (3)
Generalizing to all the three coordinates
(ψ, ∇ψ) = −(∇ψ, ψ) (4)
Hence
(ψ, i∇ψ) = (i∇ψ, ψ) (5)
where we have used, i∇ψ, ψ = −

i∇ ψ∗
ψ dτ.
This completes the proof that the momentum operator is hermitian.
2.73 Using the standard method explained in Chap. 1, define the eigen values
λ1 = 3 and λ2 = −1 for the matrix P and the eigen vectors 1
√
2
1
1

and
1
√
2
1
−1

. For the matrix Q, the eigen values are λ1 = 5 and λ2 = 1, the eigen
vectors being 1
√
2
1
1

and 1
√
2
1
−1

. Thus the eigen vectors for the commutat-
ing matrices are identical.
2.74 (a) A = αx + iβp
A†
= ax†
− iβp†
(b)[A, x] = α[x, x] + iβ[p, x]
= 0 + iβ(−i) = β
[A, A] = AA − AA = 0
[A, p] = α[x, p] + iβ[p, p]
= iα + 0 = iα
2.75 (a) As A satisfies a quadratic equation it can be represented by a 2 × 2 matrix.
Its eigen values are the roots of the quadratic equation
λ2
− 4λ + 3 = 0, λ1 = 1, λ2 = 3
(b) A is represented by the matrix
A =

1 0
0 3

The eigen value equation is

1 0
0 3

a
b

= λ

a
b

This gives a = 1, b = 0 for λ = 1 and a = 0, b = 1 for λ = 3
Hence the eigen states of A are 1
0

and 0
1

(c) As A = A†
, A is Hermitian and hence an observable.
2.76 (a) A general rule for commutators is
[A2
, B] = A[A, B] + [A, B]A
Here H = P2
/2μ
[H, X] = (1/2μ)[P2
, x] = (1/2μ)(P[P, x] + [P, x]P)
= (1/2μ)2p/i = p/iμ
Therefore [x, H] = ip/μ
(b) [[x, H], x] =

iPx
μ
, x

=

i
μ

[Px , x] =

i
μ

(−i) =
2
μ
2.77 [A2
, B] = A A B − B A A = A A B − A B A + A B A − B A A
= A[A, B] + [A, B]A
2.78 (σ.A)(σ.B) = (σx Ax + σy Ay + σz Az)(σx Bx + σy By + σz Bz)
= Ax Bx σ2
x + Ay Byσ2
y + Az Bzσ2
z + σx σy Ax By + σx σz Ax Bz
+ σyσx Ay Bx + σyσz Ay Bz + σzσx Az Bx + σzσy Az By
= A.B + iσz(Ax By − Ay Bx ) + iσx (Ay Bz − Az By)
+ iσy(Az Bx − Bz Ax )
= A.B + i[σ.(A × B)]z + i[σ.(A × B)]x + i[σ.(A × B)]y
= A.B + i σ.(A × B)
where we have used the identities in simplifying:
σ2
x = σ2
y = σ2
z = 1
and σyσx = −σx σy etc.
2.79 (a) σyμ = σμy† as can be seen from the matrix elements of σy. Therefore σy is
Hermitian. It is the matrix of a Hermitian operator whose eigen values are
real.
(b) The eigen values λ are found by setting
%
%
%
%
σ y11 −λ σ y11
σ y21 σ y22 −λ
%
%
%
% =
%
%
%
%
−λ −i
i −λ
%
%
%
% = 0
λ2
− 1 = 0, λ = ±1

2.3 Solutions 127
The eigen vector associated with λ1 = 1 is
|ψ1 =
2

n=1
Cn|n with − C1 − iC2 = 0, C2 = iC1,
|ψ1 =
1
√
2
|1 +
i
√
2
|2
The eigen vector associated with λ2 = −1 is
|ψ2 =
2

n=1
Cn|n with C1 − iC2 = 0, C2 = −iC1,
|ψ2 =
1
√
2
|1 −
i
√
2
|2
(c) The projector onto |ψi is Pi = |ψi ψi |.
Matrix of P1 =
1
2
− i
2
i
2
1
2
'
, matrix of P2 =
1
2
i
2
− i
2
1
2
'
P1
†
P2

0 0
0 0

= 0, P1 P1
†
+ P2 P2
†
= I
2.80 (i) σx
2
=

0 1
1 0

0 1
1 0

=

1 0
0 1

(ii) [σx , σy] =

0 1
1 0

0 −i
i 0

−

0 −i
i 0

0 1
1 0

=

i 0
0 −i

−

−i 0
0 i

=

2i 0
0 −2i

= 2i

1 0
0 −1

= 2iσz
2.81 Proof : A X = λ1 X (1)
BX = λ2 X (2)
where λ1 and λ2 are the eigen values belonging to the same state λ.
B AX = Bλ1 X = λ1 BX = λ1λ2 X (3)
ABX = Aλ2 X = λ2 AX = λ2λ1 X = λ1λ2 X (4)
Subtracting (3) from (4)
(A B − B A)X = 0
Therefore A B − B A = 0, because X = 0
Operate with B on A in (1) and with A and B in (2)
Or [A, B] = 0

2.82 ΔxΔpx∼/2
P =

2x
E =
p2
2m
+ 1/2 mω2
x2
=
2
8mx2
+ 1/2mω2
x2
The ground state energy is obtained by setting ∂E
∂x
= 0
∂E
∂x
= −
2
4mx3
+ mω2
x = 0
whence x2
=
2mω
∴ E = 1/4ω + 1/4ω = 1
2
ω
2.83 If E and p are to be measured simultaneously their operators must commute.
Now
H = −2
∇2
/2m + V and p = −i∇
[H, p] = [−2
∇2
/2m + V, −i∇]
= i3
∇2
∇/2m − iV ∇ − i3
∇∇2
/2m + i∇V
The first and the third term on the RHS get cancelled because ∇2
∇ = ∇∇2
.
Therefore
[H, P] = −i(V ∇ − ∇V )
If V = constant, the commutator vanishes. To put it differently energy
and momentum can be measured with arbitrary precision only for unbound
particles.
2.84 Consider the motion of a particle along x-direction.
The uncertainty Δx is defined as
(Δx)2
= (x− x )2
= x2
−2 x x + x 2
= x2
− x 2
(1)
Similarly
(ΔPx )2
= P2
x − Px 2
(2)

2.3 Solutions 129
The precise statement of the Heisenberg uncertainty principle is
ΔPx Δx ≥ /2
ΔPyΔy ≥ /2 (3)
ΔPzΔz ≥ /2
Consider the integral, a function of a real parameter λ
I(λ) =
∞
−∞
dx|(x− x )ψ + iλ(−i∂ψ/∂x− Px ψ|2
(4)
By definition, I(λ ≥ 0). Expanding (4)
I(λ) =
∞
−∞
dxψ∗
(x− x )2
ψ + λ
∞
−∞
dx

ψ∗ ∂ψ
∂x
+ ψ
∂ψ∗
∂x

(x− x )
+ λ2
2
∞
−∞

∂ψ∗
∂x

∂ψ
∂x
− iλ2
Px
∞
−∞
dx[ψ
∂ψ∗
∂x
+ λ2
Px 2
∞
−∞
dxψ∗
ψ (5)
The term in the second line can be written as
∞
−∞
dx
∂
∂x
(ψ∗
ψ)(x− x ) = [(x− x )ψ∗
ψ]∞
−∞ −
∞
−∞
dxψ∗
ψ = −1
because it is expected that ψ → 0. Sufficiently fast as x → ±∞ so that the
integrated term is zero. Similarly the third term can be re-written as
2
∞
−∞
dx

∂ψ∗
∂x

∂ψ
∂x

= 2

ψ∗ ∂ψ
∂x
∞
−∞
+
∞
−∞
dxψ∗
(−2
∂2
ψ/∂x2
) = P2
x
In term (4) rewrite
−i
∞
−∞
dx
∂ψ∗
∂x
ψ = −i
,
ψ∗
ψ
-∞
−∞
+
∞
−∞
dxψ∗
i
∂ψ
∂x
= − Px
So, the full term (4) becomes −2 Px 2
.
Collecting all the terms
I(λ) = (Δx)2
− λ + (ΔPx )2
λ2
≥ 0
Denoting I(λ) = aλ2
+ bλ + c, the condition I(λ ≥ 0) is satisfied if
b2
− 4ac ≥ 0.
Thus, 2
− 4(Δx)2
(ΔPx )2
≤ 0, and therefore
ΔPx Δx ≥ /2
2.85 ΔxΔp =
cΔP ≈ cp =
c
Δx
=
197.3MeV − fm
0.529 × 10−10 m
= 372.97 × 10−5
MeV = 3,730 eV
T = c2
p2
/mc2
= (3,730)2
/0.511 × 106
= 13.61 eV

This value is in agreement with 13.60 obtained from Bohr’s theory of hydro-
gen atom.
2.86 (i) ΔxΔPx = (ii) ΔEΔt =
ΔPx =

Δx
; cΔPx = cPx =
c
Δx
=
197.3 MeV − fm
10 fm
≈ 20 MeV/c
T =
P2
2M
=
c2
p2
2Mc2
=
202
(2)(940)
= 0.21 MeV

Chapter 3
Quantum Mechanics – II
Schrodinger’s equation
i
∂ψ
∂t
= −
2
2μ
∇2
ψ + V (r)ψ (3.1)
−
2
2μ
∇2
ψ + (V − E) ψ = 0 (Time independent equation) (3.2)
Probability density ρ = ψ∗
ψ (3.3)
Continuity equation
∂ρ
∂t
+ ∇. j = 0 (3.4)
where j is the probability current density.
Normalization of the wave function

all space
ρdτ =

all space
ψ∗
n ψndτ = 1 (3.5)
I =

ψmψndτ is called the overlapping integral.
Orthogonality
+∞
−∞
ψ∗
n (x)ψm(x)dx = 0, if m = n (3.6)
131

132 3 Quantum Mechanics – II
Table 3.1 Dynamic quantities and operators
Physical Quantity Operator
Position r R
Momentum P −i∇
Kinetic energy T −
2
2μ
∇2
Potential energy V V (r)
Angular momentum square L2
l(l + 1)2
z-component of angular momentum Lz −i
∂
∂φ
Expectation values of dynamical variables and operators
An arbitrary function of r has the expectation value
f (r)=

ψ∗
f (r)ψ dτ (3.7)
The expectation value of P
P =

ψ∗

i
∇ψ

dτ (3.8)
The expectation value of the kinetic energy
T=

ψ∗

−
2
2μ
∇2
ψ

dτ (3.9)
Pauli spin matrices
σx =

0 1
1 0

, σy =

0 −i
i 0

, σz =

1 0
0 −1

(3.10)
σ2
x = σ2
y = σ2
z = 1 (3.11a)
σx σy = iσz, σyσz = iσx , σzσx = iσy (3.11b)
These matrices are both Hermetian and unitary. Further, any two Pauli matrices
anticommute
σx σy + σyσx = 0, etc. (3.11c)

Commutators
AB − BA = [A, B] (3.12)
by definition.
Dirac’s Bra and Ket notation
A ket vector, or simply ket, is analogous to the wave function for a state. The symbol
|m denotes the ket vector that corresponds to the state m of the system. A bra
vector, or bra, is analogous to the complex conjugate of the wave function for a
state. The symbol n| denotes the bra vector that corresponds to the state n of the
system. Then

ψ∗
n ψm dτ = n|m (3.13)
And

ψ∗
n H ψm dτ = n|H|m (3.14)
Parity
Parity of a function can be positive or negative, and some functions may not have
any parity.
If ψ(−x, −y, −z) = +ψ(x, y, z) then ψ has positive or even parity.
If ψ(−x, −y, −z) = −ψ(x, y, z) then ψ has negative or odd parity.
(3.15)
Example of even parity is cos x. Example of odd parity is sin x.
For a function like ex
, parity cannot be defined. The parity due to orbital angular
momentum is determined by the function (−1)l
.
Laporte rule
An integral vanishes between, symmetric limits if the integrand has odd parity. Con-
sidering that the operator of the electric dipole moment has odd parity, the expecta-
tion value of the electric dipole moment has odd parity, the expectation value of the
electric dipole moment as well as the transition probability vanishes unless initial
and final state have different parity.
Even a more restrictive selection rule is
Δl = ±1 (3.16)

Table 3.2 Some selected eigen functions of hydrogen atom
State N L m u
1S 1 0 0 An e−x
2S 2 0 0 An e−x
(1 − x)
2P 2 1 0 Ane−x
x cos θ
2P 2 1 ±1 An
e−x
x sin θ e±iϕ
√
2
3S 3 0 0 Ane−x
1 − 2x + 2x2
3

3P 3 1 0 Ane−x 2
3
x(2 − x) cos θ
3P 3 1 ±1 Ane−x 1
√
3
x(2 − x) sin θe±iΦ
3d 3 2 0 Ane−x 1
2
√
3
x2
(3 cos2
θ − 1)
3d 3 2 ±1 Ane−x x2
√
3
sin θ cos θe±iϕ
3d 3 2 ±2 Ane−x 1
2
√
3
x2
sin2
θe±iϕ
where x = r/n a0; An = (1/
√
π)(1/na0)3/2
; a0 = 2
/me2
is the Bohr radius
Molecular spectra
Three types:
i. Electronic (Visible and ultraviolet)
ii. Vibrational (Near infrared)
iii. Rotational (Far infrared)
Because electron mass is much smaller than the nuclear mass, the three types of
motion can be treated separately. This is the Born–Oppenheimer approximation, in
which the complete ψ – function appears as the product of the wave functions of the
three types of motion, and the total energy as the sum of the energies of electronic
motion, of vibration, and of rotation.
ψ = ψel · ψv · ψrot
E = Eel + Evibr + Erot (3.17)
Eel : Evibr : Erot = 1 :

m/M : m
/M (3.18)
where m and M are the mass of electron and nucleus.
Thus Eel ≫ Evibr ≫ Erot.
The rotational energy
ER =
2
2I0
· J(J + 1) (3.19)
Permanent dipole moment is necessary, molecules with center of symmetry such
as C2H2 or O2 have no dipole moment and do not exhibit rotational spectrum.

Hetero-nuclear diatomic molecules such as CO and linear polyatomic molecule
such as HCN, do possess dipolemoment and so also rotational spectrum.
The selection rule is
ΔJ = 0, ±1. (3.20)
The occurrence of a rotational Raman spectrum depends on the change in polar-
izibility. Such a change can occur in symmetric top molecules. The selection rule is
ΔJ = 0, ±2. (3.21)
Required properties of the wave function
Ψ should be finite, single-valued and continuous. These requirements follow from
the interpretation of |ψ|2
as the probability density.
Ladder operators
Raising operator:
J+| jm =

( j + m + 1)( j − m)| j, m + 1
Lowering operator:
J−| jm =

( j − m + 1)( j + m)| j, m − 1 (3.23)
The Klein–Gordon equation for the relativistic free motion of a spinless particle
of rest mass m.
∇2
φ −
1
c2
∂2
ϕ
∂t2
=
m2
c2
ϕ
2
(3.24)
If the quantum of nuclear field is assumed to have mass m, then its range
R ∼

mc
(3.25)
The variation method
It consists of evaluating the integrals on the RHS of the inequality
E0 ≦

ψ∗
Hψ dτ

|ψ|2 dτ
(3.26)

with a trial function ψ that depends on a number of parameters, and varying these
parmeters until the expectation value of the energy is a minimum. This results in an
upper limit for the ground state energy of the system, which will be close if the form
of the trial function resembles that of the eigen function.
Table 3.3 Clebsch–Gordan coefficients (C.G.C)
1
2
×
1
2
J = 1 1 0 1
m1 m2 M = +1 0 0 −1
+1/2 +1/2 1
+1/2 −1/2
√
1/2
√
1/2
−1/2 +1/2
√
1/2 −
√
1/2
−1/2 −1/2 1
1 ×
1
2
J = 3/2 3/2 1/2 3/2 1/2 3/2
m1 m2 M = +3/2 +1/2 +1/2 −1/2 −1/2 −3/2
+1 +1/2 1
+1 −1/2
√
1/3
√
2/3
0 +1/2
√
2/3 −
√
1/3
0 −1/2
√
2/3
√
1/3
−1 +1/2
√
1/3 −
√
2/3
−1 −1/2 1
1 × 1
J = 2 2 1 2 1 0 2 1 2
m1 m2 M = +2 +1 +1 0 0 0 −1 −1 −2
+1 +1 1
+1 0
1
2
1
2
0 +1
1
2
−
1
2
+1 −1
1
6
1
2
1
3
0 0
2
3
0 −
1
3
−1 +1
1
6
−
1
2
1
3
0 −1
1
2
1
2
−1 0
1
2
−
1
2
−1 −1 1

3.2 Problems 137
The Born approximation
Here the entire potential energy of interaction between the colliding particles is
regarded as a perturbation. The approximation works well when the kinetic energy
of the colliding particles is large in comparision with the interaction energy. It there-
fore supplements the method of partial waves.
σ(θ) = | f (θ)|2
(3.27)
where
f (θ) = −K−1
∞
0
r sin Kr V (r)dr (3.28)
and
K = 2k sin
θ
2
, k = p. (3.29)
3.2 Problems
3.2.1 Wave Function
3.1 An electron is trapped in an infinitely deep potential well of width L = 106
fm.
Calculate the wavelength of photon emitted from the transition E4 → E3. (See
Problem 3.18).
3.2 Given ψ(x) =
π
α
− 1
4
exp −α2
x2
2

, calculate Var x
3.3 If ψ(x) = N
x2+a2 , calculate the normalization constant N.
3.4 Find the flux of particles represented by the wave function
ψ(x) = A eikx
+ Be−ikx
3.5 For Klein – Gordon equation obtain expressions for probability density and
current. Explain the significance of the result.
3.6 (a) Find the normalized wave functions for a particle of mass m and energy E
trapped in a square well of width 2a and depth V0 E.
(b) Sketch the first two wave functions in all the three regions. In what respect
do they differ from those for the infinite well depth.
3.7 The Thomas-Reich-Kuhn sum rule connects the complete set of eigen func-
tions and energies of a particle of mass m. Show that

2μ
2

k
(Ek − Es)|xsk|2
= 1

3.8 (a) State and explain Laporte rule for light emission.
(b) What are metastable states?
3.9 Show that the eigen values of a hermitian operator Q are real
3.10 The state of a free particle is described by the following wave function
(Fig. 3.1)
ψ(x) = 0 for x −3a
= c for − 3a x a
= 0 for x a
(a) Determine c using the normalization condition
(b) Find the probability of finding the particle in the interval [0, a]
Fig. 3.1 Uniform distribution
of ψ
3.11 In Problem 3.10,
(a) Compute x and σ2
(b) Calculate the momentum probability density.
3.12 Particle is described by the wavefunction
ψ = 0 x 0
=
√
2e−x/L
x ≥ 0
where L = 1 nm. Calculate the probability of finding the particle in the region
x ≥ 1 nm.
3.2.2 Schrodinger Equation
3.13 The radial Schrodinger equation, in atomic units, for an electron in a hydrogen
atom for which the orbital angular momentum quantum number, l = 0, is
((d2
/dr2
) + (2/r) + (2E))F(r) = 0,
where E is the total energy.

3.2 Problems 139
(a) Put F(r) = exp (−r/ν) y(r), where E = −1/(2ν2
), and show that
d2
y
dx2
=
2
ν
d
dr
−
ν
r
y
(b) Assuming that y(r) can be expanded as the series
y(r) =
∞
p=0
apr p+1
,
Show that the coefficients ap in the series satisfy the recurrence relation,
p(p + 1)ap =

2
ν

(p − ν)ap−1
(c) Solutions of the radial Schrodinger equation exist which are bounded for
all r provided that ν = n, where n is a positive integer. Show that the
un-normalized radial function for the n = 2 state is
F(r) = a0e−r/2
r(1 − r/2)
3.14 State Ehrenfest’s theorem. Show that
(a)
d x
dt
=
px
m
(b)
d px
dt
= −∂V/∂x
3.15 Consider the time-independent Schrodinger equation in three dimensions

−
2
2m

∇2
+ V (r)

ψ = Eψ
In spherical coordinates
∇2
=

1
r2

∂
∂r

r2 ∂
∂r

+

1
r2 sin θ

∂
∂θ

sin θ
∂
∂θ

+

1
r2 sin2
θ

∂2
∂ϕ2
(a) Write ψ(r, θ, ϕ) = ψr (r)Y(θ, ϕ) as a separable solution and split
Schrodinger’s equation into two independent differential equations, one
depending on r and the other depending on θ and ϕ.
(b) Further separate the angular equation into θ and ϕ parts
(c) Combine the angular part and the potential part of the radial equation and
write them as an effective potential Ve. Then make the substitution χ(r) =
rψr (r) and transform the radial equation into a form that resembles the
one-dimensional Schrodinger equation.
3.16 Consider a three-dimensional spherically symmetrical system. In this case,
Schrodinger’s equation can be decomposed into a radial equation and an angu-
lar equation. The angular equation is given by
−

1
sin θ

∂
∂θ

sin θ
∂
∂θ

+

1
sin2
θ

∂2
∂ϕ2

Y(θ, ϕ) = λY(θ, ϕ)
Solve the equation and, in the process, derive the quantum number m.

(a) Consider the case where m = 0. Make the change of variable μcos θ and
consider a series solution to the equation for (μ). Derive a recurrence rela-
tion for the coefficients of the series solution.
(b) Explain why the series solution should be cut off at some finite term, give
a mechanism for doing this and hence derive another quantum number l.
3.2.3 Potential Wells and Barriers
3.18 (a) The one-dimensional time-independent Schrodinger equation is

−
2
2m

d2
ψ(x)
dx2
+ U(x)ψ(x) = Eψ(x)
Give the meanings of the symbols in this equation.
(b) A particle of mass m is contained in a one-dimensional box of width a.
The potential energy U(x) is infinite at the walls of the box (x = 0 and
x = a) and zero in between (0 x a).
Solve the Schrodinger equation for this particle and hence show that the
normalized solutions have the form ψn(x) =
2
a
1
2
sin
nπx
a

, with energy
En = h2
n2
/8ma2
, where n is an integer (n 0).
(c) For the case n = 3, find the probability that the particle will be located in
the region a
3
x 2a
3
.
(d) Sketch the wave-functions and the corresponding probability density dis-
tributions for the cases n = 1, 2 and 3.
3.19 Deuteron is a loose system of neutron and proton each of mass M. Assuming
that the system can be described by a square well of depth V0 and width R,
show that to a good approximation
V0 R2
=
π
2
2

2
M

3.20 Show that the expectation value of the potential energy of deuteron described
by a square well of depth V0 and width R is given by
V = −V0 A2

R
2
−
sin 2kR
4k

where A is a constant.
3.21 Assuming that the radial wave function
U(r) = rψ(r) = C exp(−kr)
is valid for the deuteron from r = 0 to r = ∞ find the normalization
constant C.
Hence if k = 0.232 fm−1
find the probability that the neutron – proton separa-
tion in the deuteron exceeds 2 fm. Find also the average distance of interaction
for this wave function.
[Royal Holloway University of London 1999]

3.2 Problems 141
3.22 The small binding energy of the deuteron (2.2 MeV) implies that the maximum
of U(r) lies just inside the range R of the well. From this knowledge deduce
the value of V0 if R is approximately 1.5 fm.
3.23 Given that the normalized wave function
ψ =

1
r

α
2π
1/2
e−αr
(1/α = 4.3 fm) is a useful approximation to describe the ground state of the
deuteron, find the root mean square separation of the neutron and proton in
this nucleus.
[University of Durham 1972]
3.24 A particle of mass me trapped in an infinite depth well of width L = 1 nm.
Consider the transition from the excited state n = 2 to the ground state n = 1.
Calculate the wavelength of light emitted. In which region of electromagnetic
spectrum does it fall?
3.25 Consider a particle of mass m trapped in a potential well of finite depth V0
V (x) = V0, |x| a
= 0; |x| a
Discuss the solutions and eigen values for the class I and II solutions graphi-
cally.
3.26 Show that for deuteron, neutron and proton stay outside the range of nuclear
forces for 70% of the time. Take the binding energy of deuteron as 2.2 MeV.
3.27 Show that the results of the energy levels for infinite well follow from those
for the finite well.
3.28 Show that for deuteron excited states are not possible.
3.29 The small binding energy of the deuteron indicates that the maximum of U(r)
lies only just inside the range R of the square well potential. Use this informa-
tion to estimate the value of V0 if R is approximately 1.5 fm.
3.30 Consider a stream of particles with energy E travelling in one dimension from
x = −∞ to ∞. The particles have an average spacing of distance L. The
particle stream encounters a potential barrier at x = 0. The potential can be
written as
V (x) = 0 if x 0
= V if 0 x a
= 0 if x a
Suppose the particle energy is smaller than the potential barrier, i.e., Vb.
(a) For each of the three regions, write down Schrodinger’s equation and cal-
culate the wave-function ψ and its derivative dψ/dx.

Use the constants to represent the amplitude of the reflected and trans-
mitted particle streams respectively and take
k2
1 =
2mE
2
and k2
2 =
2m(Vb − E)
2
(b) At the boundaries to the potential barrier, ψ and dψ/dx must be continu-
ous. Equate the solutions that you have at x = 0 and x = a and manipulate
these equations to derive the following expression for the transmission
amplitude
τ =
4ik1k2e−ik1a
[(ik1 + k2)2e−k2a] − [(ik1 − k2)2ek2a]
(a) Show that the fraction of transmitted particles is given by Ftrans = τ∗
τ,
which when calculated evaluates to
Ftrans =

1 +
V2
b sinh2
(k2a)
4E(Vb − E)
'−1
(b) How would Ftrans vary if E Vb.
3.32 A particle is trapped in a one dimensional potential given by = kx2
/2. At a
time t = 0 the state of the particle is described by the wave function ψ =
C1ψ1 + C2ψ2, where ψ is the eigen function belonging to the eigen value En.
What is the expected value of the energy?
3.33 A particle is trapped in an infinitely deep square well of width a. Sud-
denly the walls are separated by infinite distance so that the particle becomes
free. What is the probability that the particle has momentum between p and
p + dp?
3.34 The alpha decay is explained as a quantum mechanical tunneling. Assuming
that the alpha particle energy is much smaller than the potential barrier the
alpha particle has to penetrate, the transmission coefficient is given by
T ≈ exp

−
2

b
a
[2m(U(r) − E)]1/2
dr

The integration limits a and b are determined as solutions to the equation
U(r) = E, where U(r) is the non-constant Coulomb’s potential energy. Cal-
culate the alpha transmission coefficient and the decay constant λ.
3.35 The one-dimensional square well shown in Fig. 3.2 rises to infinity at x = 0
and has range a and depth V1. Derive the condition for a spinless particle of
mass m to have (a) barely one bound state (b) two and only two bound states
in the well. Sketch the wave function of these two states inside and outside the
well and give their analytic expressions.

3.2 Problems 143
Fig. 3.2 Bound states in a
square well potential
3.36 In Problem 3.25 express the normalization constant A in terms of α, β and a.
3.37 A particle of mass m is trapped in a square well of width L and infinitely deep.
Its normalized wave function within the well for the nth state is
ψn =

2
L
1/2
sin
nπx
L

(a) Show that its mean position is L/2 and the variance is

L2
12

1 −
6
n2π2

(b) Show that these expectations are in agreement with the classical values
when n →∞.
3.38 The quantum mechanical Hamiltonian of a system has the form
H = (−2
/2m)∇2
+ ar2

1 −
5
6
sin2
θ cos2
ϕ

Find the energy eigen value of the two lowest lying stationary states.
3.39 (a) Write down the three-dimensional time-independent Scrodinger equation
in Cartesian co-ordinates. By separating the variables, ψ (x, y, z) =
X(x)Y(y)Z(z), solve this equation for a particle of mass m confined to
a rectangular box of sides a, b and c, with zero potential inside.
(b) Show that the particle has energy given by
E =

2
8m

n2
x
a2
+
n2
y
b2
+
n2
z
c2
'
3.40 In Problem 3.39, consider the special case of a cube a = b = c. Draw up a
table listing the first six energy levels, stating the degeneracy for each level.
3.41 A particle of mass m is moving in a region where there is a potential step at
x = 0 : V (x) = 0 for x 0 and V (x) = U0 (a positive constant) for x ≥ 0
(a) Determine ψ(x) separately for the regions x ≪ 0 and x ≫ 0 for the cases:
(i) U0 E
(ii) U0 E.
(b) Write down and justify briefly the boundary conditions that ψ(x) must sat-
isfy at the boundary between the two adjacent regions. Use these

conditions to sketch the form of ψ(x) in the region around x = 0 for
the cases (i) and (ii).
3.42 A steady stream of particles with energy E( V0) is incident on a potential
step of height V0 as shown in Fig. 3.3.
The wave functions in the two regions are given by
ψ1(x) = A0 exp(ik1x) + A exp(−ik1x)
ψ2(x) = B exp(ik2x)
Write down expressions for the quantities k1 and k2 in terms of E and V0.
Show that
A =

k1 − k2
k1 + k2

A0 and B =

2k1
k1 + k2

A0
and determine the reflection and transmission coefficients in terms of k1 and
k2.
If E = 4 V0/3 show that the reflection and transmission coefficients are 1/9
and 8/9 respectively.
Comment on why A2
+ B2
is not equal to 1.
Fig. 3.3 Potential step
3.43 (a) What boundary conditions do wave-functions obey?
A particle confined to a one-dimensional potential well has a wave-function
given by
ψ(x) = 0 for x −L/2;
ψ(x) = A cos

3πx
L

for −
L
2
≤ x ≤
L
2
;
ψ(x) = 0 for x
L
2
(b) Sketch the wave-function ψ(x).
(c) Calculate the normalization constant A.
(d) Calculate the probability of finding the particle in the interval −L
4
x
L
4
.
(e) By calculating d2
ψ/dx2
and writing the Schrodinger equation as

−
2
2m

d2
ψ
dx2

= Eψ.
show that the energy E corresponding to this wave-function is 9π2
2
2mL2 .
3.44 (a) Sketch the one-dimensional “top hat” potential (1) V = 0 for x 0; (2)
V = W = constant for 0 ≤ x ≤ L; (3) V = 0 for x L.
(b) Consider particles, of mass m and energy E W incident on this potential
barrier from the left (x 0). Including possible reflections from the barrier

3.2 Problems 145
boundaries, write down general expressions for the wavefunctions in these
regions and the form the time-independent Schrodinger equation takes in
each region. What ratio of wavefunction amplitudes is needed to determine
the transmission coefficient?
(c) Write down the boundary conditions for ψ and dψ/dx at x = 0 and x = L.
(d) A full algebraic solution for these boundary conditions is time consuming.
In the approximation for a tall or wide barrier, the transmission coefficient
T is given by
T = 16

E
W

1 −
E
W

e−2αL
, where α2
= 2m
W−E
2

Determine T for electrons of energy E = 2 eV, striking a potential of
value W = 5 eV and width L = 0.3 nm.
(e) Describe four examples where quantum mechanical tunneling is observed.
3.45 A particle of mass m moves in a 2-D potential well, V (x, y) = 0 for 0 x a
and 0 y a, with walls at x = 0, a and y = 0, a. Obtain the energy eigen
functions and eigen values.
3.46 A particle of mass m is trapped in a 3-D infinite potential well with sides of
length a each parallel to the x-, y-, z-axes. Obtain an expression for the number
of states N(N ≫ 1) with energy, say less than E.
3.47 A particle of mass m is trapped in a hollow sphere of radius R with impenetra-
ble walls. Obtain an expression for the force exerted on the walls of the sphere
by the particle in the ground state.
3.48 Starting from Schrodinger’s equation find the number of bound states for a par-
ticle of mass 2,200 electron mass in a square well potential of depth 70 MeV
and radius 1.42 × 10−13
cm.
[University of Glasgow 1959]
3.49 A beam of particles of momentum k1 are incident on a rectangular potential
well of depth V0 and width a. Show that the transmission amplitude is given by
τ =
4k1k2e−ik1a
(k2 + k1)2
e−ik2a − (k2 − k1)2
eik2a
where k2 =

2m(E − V0)
2
1/2
Show that τ ∗
= 1 when k2a = nπ. Further, show graphically the varia-
tion of T , the transmission coefficient as a function of E/V0, where E is the
incident particle energy.
3.50 (a) What are virtual particles? What are space-like and Time-like four momen-
tum vectors for real and virtual particles?
(b) Derive Klein – Gorden equation and deduce Yukawa’s potential.

3.2.4 Simple Harmonic Oscillator
3.51 Show that the wavefunction ψ0(x) = A exp(−x2
/2a2
) is a solution to the
time- independent Schrodinger equation for a simple harmonic oscillator
(SHO) potential.

−
2
2m

d2
ψ/dx2
+

1
2

mω0x2
ψ = Eψ
with energy E0 =
1
2

ω0, and determine a in terms of m and ω0.
The corresponding dimensionless form of this equation is
−d2
ψ/dR2
+ R2
ψ = εψ
where R = x/a and ε = E/E0.
Show that putting ψ(R) = AH (R) exp(−R2
/2) into this equation leads to
Hermite’s equation
d2
H
dR2
− 2R

dH
dR

+ (ε − 1) H = 0
H(R) is a polynomial of order n of the form anRn
+an−2 Rn−2
+an−4 Rn−4
+. . .
Deduce that ε is a simple function of n and that the energy levels are equally
spaced.
[Adapted from the University of London, Royal
Holloway and Bedford New College 2005]
3.52 Show that for a simple harmonic oscillator in the ground state the probability
for finding the particle in the classical forbidden region is approximately 16%
3.53 Determine the energy of a three dimensional harmonic oscillator.
3.54 Show that the zero point energy of a simple harmonic oscillator could not be
lower than ω/2 without violating the uncertainty principle.
3.55 Show that when n → ∞ the quantum mechanical simple harmonic oscillator
gives the same probability distribution as the classical one.
3.56 Derive the probability distribution for a classical simple harmonic oscillator
3.57 The wave function (unnormalized) for a particle moving in a one dimensional
potential well V (x) is given by ψ(x) = exp(−ax2
/2). If the potential is to
have minimum value at x = 0, determine (a) the eigen value (b) the poten-
tial V (x).
3.58 Show that for simple harmonic oscillator Δx.Δpx = (n + 1/2), and that this
is in agreement with the uncertainty principle.
3.59 In HCl gas, a number of absorption lines have been observed with the fol-
lowing wave numbers (in cm−1
): 83.03, 103.73, 124.30, 145.03, 165.51, and
185.86.
Are these vibrational or rotational transitions? (You may assume that tran-
sitions involve quantum numbers that change by only one unit). Explain your

3.2 Problems 147
reasoning briefly. (a) If the transitions are vibrational, estimate the spring con-
stant (in dyne/cm) (b) If the transitions are rotational, estimate the separation
between H and Cl nuclei. What J values do they correspond to, and what is
the moment of inertia of HCl (in g-cm2
)?
[Arizona State University 1996]
3.60 Determine the degeneracy of the energy levels of an isotropic harmonic oscil-
lator.
3.61 At time t = 0, particle in a harmonic oscillator potential V (x) = mω2
x2
2
has a
wavefunction
ψ(x, 0) =

1
√
2

[ψ0(x) + ψ1(x)]
where ψ0(x) and ψ1(x) are real ortho-normal eigen functions for the ground
and first-excited states of the oscillator. Show that the probability density
|ψ(x, t)|2
oscillates with angular frequency ω.
3.62 The quantum state of a harmonic oscillator has the eigen-function ψ(x, t) =

1
√
2

ψ0(x) exp

−
i E0t

+

1
√
3

ψ1(x) exp

−
i E1t

+

1
√
6

ψ2(x) exp

−
i E2t

where ψ0(x), ψ1(x) and ψ2(x) are real normalized eigen functions of the har-
monic oscillator with energy E0, E1 and E2 respectively. Find the expectation
value of the energy.
3.63 (a) Show that the wave-function ψ0(x) = A exp(−x2
/2a2
) with energy E =
ω/2 (where A and a are constants) is a solution for all values of x to the
one-dimensional time-independent Schrodinger equation (TISE) for the
simple harmonic oscillator (SHO) potential V (x) = mω2
x2
/2
(b) Sketch the function ψ1(x) = Bx exp(−x2
/2a2
)
(where B = constant), and show that it too is a solution of the TISE for
all values of x.
(c) Show that the corresponding energy E = (3/2)ω
(d) Determine the expectation value px of the momentum in state ψ1
(e) Briefly discuss the relevance of the SHO in describing the behavior of
diatomic molecules.
3.2.5 Hydrogen Atom
3.64 Find the expectation value of kinetic energy, potential energy, and total energy
of hydrogen atom in the ground state. Take ψ0 = e−r/a0
(π a3
0 )1/2 , where a0 = Bohr’s
radius

3.65 Show that (a) the electron density in the hydrogen atom is maximum at r = a0,
where a0 is the Bohr radius (b) the mean radius is 3a0/2
3.66 Refer to the hydrogen wave functions given in Table 3.2. Show that the func-
tions for 2p are normalized.
3.67 Show that the three 3d functions for H-atom are orthogonal to each other.
(Refer to Table 3.2)
3.68 What is the degree of degeneracy for n = 1, 2, 3 and 4 in hydrogen atom?
3.69 What is the parity of the 1s, 2p and 3d states of hydrogen atom.
3.70 Show that the 3d functions of hydrogen atom are spherically symmetric
3.71 In the ground state of hydrogen atom show that the probability (p) for the
electron to lie within a sphere of radius R is
P = 1 − exp (−2R/a0)

1 + 2R/a0 + 2R2
/a2
0

3.72 Locate the position of maximum and minimum electron density in the 2S orbit
(n = 2 and l = 0) of hydrogen atom
3.73 When a negative muon is captured by an atom of phosphorous (Z = 15) in
a high principal quantum number, it cascades down to lower state. When it
reaches inside the electron cloud it forms a hydrogen-like mesic atom with the
phosphorous nucleus.
(a) Calculate the wavelength of the photon for the transition 3d → 2p state.
(b) Calculate the mean lifetimes of this mesic atoms in the 3d-state, consider-
ing that the mean life of a hydrogen atom in the 3d state is 1.6 × 10−8
s.
(mass of muon = 106 MeV)
3.74 The momentum distribution of a particle in three dimensions is given by
ψ(p) = [1/(2π)3/2
]

e−p.r/
ψ(r)dτ. Take the ground state eigen function
ψ(r) =
$
πa3
0
− 1
2
exp (−r/a0)
Show that for an electron in the ground state of the hydrogen atom the
momentum probability distribution is given by
ψ|(p)|2
=
8
π2

a0
5

p2 +
a0
2
4
3.75 In Problem 3.74 (a) show that the most probable magnitude of the momentum
of the electron is /(
√
3 a0) and (b) its mean value is 8/3π a0, where a0 is
the Bohr radius.
3.76 Calculate the radius R inside which the probability for finding the electron in
the ground state of hydrogen atom is 50%.

3.2 Problems 149
3.2.6 Angular Momentum
3.77 Given that L = r × p, show that [Lx , Ly] = iLz
3.78 The spin wave function of two electrons is (x ↑ x ↓ –x ↓ x ↑)/
√
2. What is
the eigen value of S1.S2? S1 and S2 are spin operators of 1 and 2 electrons
3.79 Show that for proton – neutron system
σp.σn = −3 for singlet state
= 1 for triplet state
3.80 Write down an expression for the z-component of angular momentum, Lz, of a
particle moving in the (x, y) plane in terms of its linear momentum components
px and py.
Using the operator correspondence px = −i
∂
∂x
etc., show that
Lz = −i

x
∂
∂y
− y
∂
∂x

Hence show that Lz = −i
∂
∂ϕ
, where the coordinates (x,y) and (r, ϕ) are
related in the usual way.
Assuming that the wavefunction for this particle can be written in the form
ψ(r, ϕ) = R(r)Φ(ϕ) show that the z-component of angular momentum is
quantized with eigen value , where m is an integer.
3.81 Show that the operators Lx and Ly in the spherical polar coordinates are
given by
Lx
i
= sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ
Ly
i
= − cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ
3.82 Using the commutator [Lx , Ly] = i Lz, and its cyclic variants, prove that
total angular momentum squared and the individual components of angular
momentum commute, i.e [L2
, Lx ] = 0 etc.
3.83 Show that in the spherical polar coordinates
L2
(i)2
=
∂2
∂θ2
+

1
sin2
θ

∂2
∂ϕ2
+ cot θ
∂
∂θ
And show that in the expression for ∇2
in spherical polar coordinates the
angular terms are proportional to L2
.
3.84 (a) Obtain the angular momentum matrices for j = 1/2 particles
(b) Hence Obtain the matrix for J2
.

3.85 (a) Obtain the angular momentum matrices for j = 1
(b) Hence obtain the matrix for J2
.
3.86 Two angular momenta with j1 = 1 and j2 = 1/2 are vectorially added, obtain
the Clebsch – Gordan coefficients.
3.87 The wave function of a particle in a spherically symmetric potential is
Ψ (x, y, z) = C (xy + yz + zx)e−αr2
Show that the probability is zero for the angular momentum l = 0 and l = 1
and that it is unity for l = 2
3.88 Show that the states specified by the wave-functions
ψ1 = (x + iy) f (r)
ψ2 = zf (r)
ψ3 = (x − iy) f (r)
are eigen states of the z-component of angular momentum and obtain the cor-
responding eigen values.
[Adapted from the University of Manchester 1959]
3.89 The Schrodinger wave function for a stationary state of an atom is
ψ = Af (r) sin θ cos θeiϕ
where (r, θ, ϕ) are spherical polar coordinates. Find (a) the z component of the
angular momentum of the atom (b) the square of the total angular momentum
of the atom. (You may use the following transformations from Cartesian to
spherical polar coordinates
x
∂
∂y
− y
∂
∂x
= sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ
x
∂
∂z
− z
∂
∂x
= − cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ
y
∂
∂x
− x
∂
∂y
= −
∂
∂ϕ

3.90 The normalized Schrodinger wavefunctions for one of the stationary states of
the hydrogen atom is given in spherical polar coordinates, by
ψ(r, θ, ϕ) =

1
2a0
3/2
1
√
3
r
a0
exp

−
r
2a0

3
8π
1/2
sin θ exp (−iϕ)
(a) Find the value of the component of angular momentum along the z axis
(θ = 0)
(b) What is the parity of this wavefunction?
[a0 is the radius of the first Bohr orbit]
[Adapted from the University of New Castle 1964]

3.2 Problems 151
3.91 The normalized 2p eigen functions of hydrogen atom are
1
√
π
1
(2a0)3/2
e−r/2a0
r
2a0
sin θ eiΦ
,
1
√
π
1
(2a0)3/2
e−r/2a0
r
2a0
cos θ,
1
√
π
1
(2a0)3/2
e−r/2a0
r
2a0
sin θe−iΦ
, for m = +1, 0, −1 respectively.
Apply the raising operator L+ = Lx + iLy and lowering operator to show that
the states with m = ±2 do not exist.
3.92 How can nuclear spin be measured from the rotational spectra of diatomic
molecules?
3.93 An electron is described by the following angular wave function
u(θ, ϕ) =
1
4
15
π
sin2
θ cos 2ϕ
Re-express u in terms of spherical harmonics given below. Hence give the
probability that a measurement will yield the eigen value of L2
equal to 62
You may use the following:
Y20(θ, ϕ) =
5
16π

3 cos2
θ − 1

Y2±1(θ, ϕ) =
15
8π
sinθ cos θ exp (±iϕ)
Y2±2 (θ, ϕ) =
15
32π
sin2
θ exp (±2iϕ)
[University College, London]
3.94 Given that the complete wave function of a hydrogen-like atom in a particular
state is ψ(r, θ, ϕ) = Nr2
exp − Zr
3a0

sin2
θ e2iϕ
determine the eigen value of
Lz, the third component of the angular momentum operator.
3.95 Consider an electron in a state described by the wave function
ψ =
1
√
4π
(cosθ + sinθeiϕ
) f (r)
where
∞
0
| f (r)|2
r2
dr = 1
(a) Show that the possible values of Lz are + and zero
(b) Show that the probability for the occurrence of the Lz values in (a) is 2/3
and 1/3, respectively.
3.96 Show that (a) [Jz, J+] = J+ (b) J+|jm = Cjm + | j, m + 1 (c) [Jx , Jy]
= iJ z

3.2.7 Approximate Methods
3.97 Consider hydrogen atom with proton of finite size sphere with uniform
charge distribution and radius R. The potential is
V (r) = −
3e2
2R3
(R2
− r2
/3) for r R
= −e2
/r for r R
Calculate correction to first order for n = 1 and n = 2 with l = 0 states
[Adapted from University of Durham 1963]
3.98 A particle of mass m and charge q oscillating with frequency ω is subjected
to a uniform electric field E parallel to the direction of oscillation. Determine
the stationary energy levels.
3.99 Consider the Hermitian Hamiltonian H = H0 + H′
, where H′
is a small
perturbation. Assume that exact solutions H0|ψ = E0|ψ are known, two
of them, and that they are orthogonal and degenerate in energy. Work out to
first order in H′
, the energies of the perturbed levels in terms of the matrix
elements of H′
.
3.100 The helium atom has nuclear charge +2e surrounded by two electrons. The
Hamiltonian is
H =

−
2
2m

(∇2
1 + ∇2
2 ) − 2e2

1
r1
+
1
r2

+
e2
r12
where r1 and r2 are the position vectors of the two electrons with nucleus as
the origin, and r12 = |r1 − r2| is the distance between the two electrons. The
expectation value for the first two terms are evaluated in a straight forward
manner, the third term which is the interaction energy of the two electrons is
evaluated by taking the trial function as the product of two hydrogenic wave
functions for the ground state. The result is
H =
e2
Z2
a0
−
4e2
Z
a0
+
5e2
Z
8a0
=

e2
a0

Z2
−
27Z
8

Thus, the energy obtained by the trial function is
E(Z) =

−
e2
2a0

27Z
4
− 2Z2

Determine the ionization energy of the helium atom.
3.101 Consider the first-order change in the energy levels of a hydrogen atom due
to an external electric field of strength E directed along the z-axis. This phe-
nomenon is known as Stark effect.
(a) Show that the ground state (n = 1) of hydrogen atom has no first-order
effect.
(b) Show that two of the four degenerate levels for n = 2 are unaffected and
the other two are split up by an energy difference of 3eEa0.

3.2 Problems 153
3.102 A particle of mass m is trapped in a potential well which has the form,
V = 1/2 mω2
x2
. Use the variation method with the normalized trial function

1/
√
a

cos(πx/2a) in the limits −a x a, to find the best value of a.
3.103 In Problem 3.45, consider the perturbation W(x, y) = W0 for 0 x a/2
and 0 y a/2, and 0 elsewhere. Calculate the first order perturbation
energy.
3.2.8 Scattering (Phase-Shift Analysis)
3.104 A beam of particles of energy 2
k2
/2m, moving in the +z direction, is scat-
tered by a short-range central potential V (r). One looks for the stationary
solution of the Schrodinger equation which is of the asymptotic form,
ψ ≈ eikz
+ f (θ)eikr
/r
Derive the partial-wave decomposition
f (θ) = (2ik)−1
∞

l=0
(2l + 1) (exp(2iδl ) − 1)Pl(cos θ)
[Adapted from the University College, Dublin, Ireland, 1967]
3.105 In the case of α − He scattering the measured scattered intensity at 45◦
(laboratory coordinates) is twice the classical result. Indicate how the wave-
mechanical theory of collisions explains this experimental result.
[Adapted from the University of New Castle 1964]
3.106 In the analysis of scattering of particles of mass m and energy E from a fixed
centre with range a, the phase shift for the lth partial wave is given by
δl = sin−1

(iak)l
[(2l + 1)!(l!)]1/2

Show that the total cross-section at a given energy is approximately given by
σ =

2π2
mE

exp

−
2mEa2
2

[University of Cambridge, Tripos]
3.107 At what neutron lab energy will p-wave be important in n–p scattering?
3.108 1 MeV neutrons are scattered on a target. The angular distribution of the neu-
trons in the centre-of-mass is found to be isotropic and the total cross-section
is measured to be 0.1 b. Using the partial wave representation, calculate the
phase shifts of the partial waves involved.
3.109 Considering the scattering from a hard sphere of radius a such that only s-
and p-waves are involved, the potential being
V (r) = ∞ for r a
= 0 for r a.

show that σ(θ) = a2
*
1 − (ka)2
3
+ 2(ka)2
cosθ + · · ·
+
and σ = 4 π a2
[1 − (ka)2
/3]
3.110 Find the elastic and total cross-sections for a black sphere of radius R.
3.111 Ramsauer (1921) observed that monatomic gases such as argon is almost
completely transparent to electrons of 0.4 eV energy, although it strongly
scatters electrons which are slower as well as those which are faster. How is
this quantum mechanical peculiarity explained?
3.112 What conditions are necessary before the Schrodinger equation for the inter-
action of two nucleons can be reduced to the form
d2
U
dr2
+

2m
2

[E − V (r)]U = 0
where U(r) = rψ(r) and the other symbols have their usual meanings?
By solving this equation for a square-well potential V (r) for a neutron –
proton collision show that the neutron – proton scattering cross-section, as
calculated for high energies is about 3 barns compared with the experimental
value of 20 barns. What is the explanation of this discrepeancy and how has
this explanation been verified experimentally?
3.2.9 Scattering (Born Approximation)
3.113 In the case of scattering from a spherically symmetric charge distribution,
the form factor is given by
F(q2
) =
∞
0
ρ(r)
sin
qr

qr/
4π r2
dr
where ρ(r) is the normalized charge distribution.
(a) If the charge distribution of proton is approximated by ρ(r)= A exp(−r/a),
where A is a constant and a is some characteristic “radius” of the proton.
Show that the form factor is proportional to 1 + q2
q2
0
−2
where q0 is /a.
(b) If q2
0 = 0.71
GeV
c
2
, determine the characteristic radius of the proton.
3.114 The first Born approximation for the elastic scattering amplitude is
f = −

2μ
q2

V (r)eiq.r
d3
r
Show that for V (r) spherically symmetric it reduces to
f = −

2μ/q2

r sin(qr) V (r)dr

3.2 Problems 155
3.115 Given the scattering amplitude
f (θ) = (1/2ik)

(2l + 1)
,
e2iδl
− 1
-
Pl(cos θ)
Show that
Im f (0) = kσt /4π
3.116 Obtain the form factor F(q) for electron scattering from an extended nucleus
of radius R and charge Ze with constant charge density. Show that the minima
occur when the condition
tan qR = qR, is satisfied
3.117 In the Born’s approximation the scattering amplitude is given by
f (θ) = (−2μ/q2
)
∞
0
V (r) sin(qr) r dr
where μ is the reduced mass of the target-projectile system, and q is the
momentum transfer. Show that the form factor is given by the expression
F(q) = (4π/q)
∞
0
ρ(r) sin(qr)rdr
where ρ(r) is the charge density
3.118 Obtain the differential cross-section for scattering from the shielded Coulomb
potential for a point charge nucleus of the form
V = z1z2e2
exp(−r/r0)/r
where r0 is the shielding radius of the order of atomic dimension. Thence
deduce Rutherford’s scattering law.
3.119 Electrons with momentum 300 MeV/c are elastically scattered through an
angle of 12◦
by a nucleus of 64
Cu. If the charge distribution on the nucleus is
assumed to be that of a hard sphere, by what factor would the Mott scattering
be reduced?
3.120 An electron beam of momentum 200 MeV/c is elastically scattered through
an angle of 14◦
by a nucleus. It is observed that the differential cross-section
is reduced by 60% compared to that expected from a point charge nucleus.
Calculate the root mean square radius of the nucleus.
3.121 Assuming that the charge distribution in a nucleus is Gaussian, e−(r2/b2)
π3/2b3 then
show that the form factor is also Gaussian and that the mean square radius is
3b2
/2
3.122 In the Born approximation the scattering amplitude is given by
f (θ) = −
μ
2π2

V (r)eiq.r
d3
r
Show that for spherically symmetric potential it reduces to
f (θ) =

−
2μ
q2

r sin(qr)V (r)dr

3.123 Using the Born approximation, the amplitude of scattering by a spherically
symmetric potential V (r) with a momentum transfer q is given by
A =
∞
0

sin
qr

qr

'
V (r)4πr2
dr
Show that in the case of a Yukawa-type potential, this leads to an amplitude
proportional to (q2
+ m2
c2
)−1
.
3.3 Solutions
3.3.1 Wave Function
3.1 En =
n2
h2
8mL2
=
π2
n2
2
c2
2mc2 L2
=
π2
× (197.3 MeV − fm)2
n2
2x0.511(MeV) × (106fm)2
= 0.038 n2
eV
E1 = 0.038 eV, E2 = 0.152 eV, E3 = 0.342 eV, E4 = 0.608 eV
Δ E43 = E4 − E3 = 0.608 − 0.342 = 0.266 eV
λ =
1, 241
0.266
= 4,665 nm
3.2 ψ(x) = (π/α)−1/4
exp

−
α2
2
x2

Var x = x2
− x 2
The expectation value
x =
∞
−∞
ψ∗
x ψ dx = 0
because ψ and also ψ∗
are even functions while x is an odd function. There-
fore the integrand is an odd function
x2
=
π
α
−1/2
∞
−∞
x2
exp(−α2
x2
)dx
Put α2
x2
= y; dx = 1
/2 α
√
y
x2
=

πα5
−1/2
∞
0
y1/2
e−y
dy
But
∞
0 y1/2
e−y
dy = Γ(3/2) =
√
π/2
Var x = x2
= (4 α5
)−1/2

3.3 Solutions 157
3.3 Normalization condition is
∞
−∞
|ψ|2
dx = 1
N2
∞
−∞
(x2
+ a2
)−2
dx = 1
Put x = a tan θ; dx = sec2
θ dθ

2N2
α3
π/2
0
cos2
θ d θ = N2
π/2a3
= 1
Therefore N = 2a3
π
1/2
3.4 ψ = Aeikx
+ Be−ikx
The flux Jx =

2im
*
ψ∗ dψ
dx
− dψ∗
dx

ψ
+
=

2im

,
Ae−ikx
+ Beikx

ik

Aeikx
− Be−ikx

+ ik

Ae−ikx
− Beikx

Aeikx
+ Be−ikx
-
=

k
2m

,
A2
−B2
− ABe−2ikx
+ ABe2ikx
+ A2
−B2
+ ABe−2ikx
− ABe2ikx
-
=

k
m

A2
− B2

3.5 In natural units ( = c = 1) Klein – Gordon equation is
∇2
ϕ −
∂2
ϕ
dt2
− m2
ϕ = 0 (1)
The complex conjugate equation is
∇2
ϕ∗
−
∂2
ϕ∗
∂t2
− m2
ϕ∗
= 0 (2)
Multiplying (1) from left by ϕ∗
and (2) by ϕ and subtracting (1) from (2)
ϕ∇2
ϕ∗
− ϕ∗
∇2
ϕ − ϕ
∂2
ϕ∗
∂t2
− ϕ
∂2
ϕ∗
∂t2
+ ϕ∗ ∂2
ϕ
∂t2
= 0
∇.

ϕ∇ϕ∗
∇ϕ

−
∂
∂t

ϕ
∂ϕ∗
∂t
− ϕ∗ ∂ϕ
∂t

= 0
Changing the sign through out and multiplying by 1/2im
1
2im
∇ ·

ϕ∗
∇ϕ − ϕ∇ϕ∗

−
1
2im
∂
∂t

ϕ ∗
∂ϕ
∂t
− ϕ
∂ϕ∗
∂t

= 0
∇ ·

1
2im

ϕ∗
∇ϕ − ϕ∇ϕ∗

+
∂
∂t

i
2m

ϕ∗ ∂ϕ
∂t
− ϕ
∂ϕ∗
∂t

= 0
Or ∇ · J +
∂ρ
∂t
= 0

This is the continuity equation where the probability current J = 1
2im
(ϕ∗
∇ϕ−
ϕ∇ϕ∗
)
And probability density
ρ =
i
2m

ϕ∗ ∂ϕ
∂t
− ϕ
∂ϕ∗
∂t

For a force free particle the solution of the Klein – Gordan equation is ϕ =
A ei(p.x−Et)
The probability density is
ρ =
i
2m
,
A∗
e−i(p.x−Et)

(i AE) e−i(p.x−Et)
−

Ae−i(p.x−Et)

i A∗
E

e−i(p.x−Et)
-
=
i
2m
,
A∗
A (−i E) − A A∗
(i E)
-
=
|A|2
2m
[E + E] = E
|A|2
m
As E can have positive and negative values, the probability density could
then be negative
3.6 (a) Class I: Refer to Problem 3.25
ψ1 = Aeβx
(−∞ x −a)
ψ2 = D cos αx (−a x +a)
ψ3 = Ae−βx
(a x ∞)
Normalization implies that
−a
−∞
|ψ1|2
dx +
a
−a
|ψ2|2
dx +
∞
a
|ψ3|2
dx = 1
−a
−∞
A2
e2βx
dx +
a
−a
D2
cos2
αxdx +
∞
a
A2
e−2βx
dx = 1
A2
e−2βa
/2β + D2
[a + sin(2αa)/2α] + A2
e−2βa
/2β = 1
Or
A2
e−2βa
/β + D2
(a + sin(2αa)/2α) = 1 (1)
Boundary condition at x = a gives
D cos α a = a e−βa
(2)
Combining (1) and (2) gives
D =

a +
1
β
−1
A = eβa
cosαa

a +
1
β
−1

3.3 Solutions 159
(b) The difference between the wave functions in the infinite and finite poten-
tial wells is that in the former the wave function within the well terminates
at the potential well, while in the latter it penetrates the well.
Fig. 3.4 Wave functions in potential wells of infinite and finite depths
3.7 Consider the expression
[x, [H, x]] =
2
μ
(1)
The expectation value in the initial state s is
s
%
%
%
%
2
μ
%
%
%
% s =
2
μ
s [x, Hx − x H] |s (2)
Using the wave function ψs and expanding the commutator
2
μ
= s
%
%2x Hx − x2
H − Hx2
%
% s (3)
Further s|x Hx|s =

k
s|x|k k|x|s Ek =

|xks|2
Ek (4)
where the summation is to be taken over all the excited states of the atom.
Also s|Hx2
|s = s|x2
H|s
=

k
s|x|k k|x|s Es =

|xks|2
Es (5)
Using (4) and (5) in (3) we get
2
/2μ =

k
|xks|2
(Ek − Es)
3.8 (a) The wave functions of the hydrogen atom, or for that matter of any atom,
with a central potential, are of the form
u(r) = λ(r)Pm
l (cos θ)eimϕ
Exchange of r → −r implies r → r, θ → π −θ and ϕ → π +ϕ (parity
operation)

where Pm
l (cos θ) are the associated Legendre functions
Now, Pm
l (cos θ) = (1 − cos2
θ)m/2
dm
Pl (cos θ)/d cosm
θ
∴ Pm
l (cos(π − θ)) = (1 − cos2
θ)m/2
dm
Pl(− cos θ)/d(− cos θ)m
= (1 − cos2
θ)m/2
(−1)l
dm
Pl(cos θ)/d(− cos θ)m
= (1 − cos2
θ)m/2
(−1)l+m
dm
Pl(cos θ)/d cosm
θ = (−1)l+m
Pl
m
(cos θ)
Thus Pl
m
(cos(π − θ)) → (−1)l+m
Pl
m
(cos θ)
Further eim(ϕ+π)
= eimϕ
.eimπ
= eimϕ
.(cos mπ + i sin mπ)
= (−1)m
eimϕ
where m is an integer, positive or negative. So, under parity (p) operation,
the function overall F(r, θ, ϕ), goes as
P F(r, θ, ϕ) = P f (r)Pl
m
(cos θ)eimϕ
= f (r)Pl
m
(cos θ)eimϕ
(−1)l+m
(−1)m
= (−1)l+2m
F(r, θ, ϕ)
= (−1)l
F(r, θ, ϕ)
All the atomic functions with even values of l have even parity while
those with odd values of l have odd parity. Considering that an integral
vanishes between symmetrical limits if the integrand has odd parity, and
that the operator of the electric dipole moment has odd parity, the following
selection rule may be stated:- The expectation value of the electric dipole
moment, as well as the transition probability vanishes unless initial and
final state have different parity, that is linitial − lfinal = Δl = 0, 2, 4 . . .
This condition for the dipole radiation emission is known as Laportes’s
rule. Actually a more restrictive rule applies
Δ l = ±1
Note that even if the matrix element of electric dipole moment vanishes,
an atom will eventually go to the ground state by an alternative mechanism
such as magnetic dipole or electric quadrupole etc for which the transition
probability is much smaller than the dipole radiation.
(b) The 2s state of hydrogen can not decay to the 1s state via dipole radiation
because that would imply Δl = 0. Furthermore, there are no other electric
or magnetic moments to facilitate the transition. However, de-excitation
may occur in collision processes with other atoms. Even in perfect vacuum
transition may take place via two-photon emission, probability for which
is again very small compared to one-photon emission. The result is that
such a state is allowed to live for considerable time. Such states are known
as metastable states.

3.3 Solutions 161
3.9 (ψ, Qψ) = (ψ, qψ) = q(ψ, ψ)
(Qψ, ψ) = (qψ, ψ) = q∗
(ψ, ψ)
since Q is hermitian,
(ψ, Qψ) = (Qψ, ψ) and that q = q∗
That is, the eigen values are real. The converse of this theorem is also true,
namely, an operator whose eigen values are real, is hermitian.
3.10 (a) The normalization condition requires
∞
−∞
|ψ|2
dx =
a
−3a
|c|2
dx = 1 = 4a|c|2
Therefore c = 1/2
√
a
(b)
a
0 |ψ|2
dx =
α
0 c2
dx = 1/4
3.11 (a) The expectation values are
x =
∞
−∞
ψ∗
x ψ dx =
a
−3a
x
dx
4a
= −a
x2
=
∞
−∞
ψ∗
x2
ψ dx =
a
−3a
(1/4a) x2
dx =

7
3

a2
xσ2
= x2
− x 2
=

7
3

a2
− (−a)2
=
4
3
a2
(b) Momentum probability density is |ϕ(p)|2
ϕ(p) = (2π)−1/2
∞
−∞
dx ψ (x)e−ipx/
= (2π)−1/2
a
−3a
dxce−ipx/
=

ic
p

2π
1/2
⎡
⎣e
−
ipa
− e
3ipa

⎤
⎦
=

−
ic
p

2π
1/2
eipa/
⎡
⎣e
2ipa
− e
−2ipa

⎤
⎦
=

2c
p

2π
1
2 e
ipa
sin

2pa

Therefore |ϕ(p)|2
=
2πap2 sin2 2pa

3.12 First the wave function is normalized
N2
∞
0
ψ∗
ψ dx = 1
N2
∞
0
√
2e− x
L
2
dx = 1
N = 1/
√
L
The probability of finding the particle in the region x ≥ 1 nm is

1
L
∞
1
ψ∗
ψ dx =
∞
1

1
L
1
2
e− x
L
#2
dx =

2
L
∞
1
e−2x/L
dx
= −e−2x/L
%
%∞
1
= e−2
= 0.135
3.13

d2
dr2
+
2
r
+ 2E
#
F(r) = 0 (1)
(a) By using F(r) = exp(−r/v) y(r), and E = − 1
2ν2 , it is easily verified that
d2
y
dr2
=
2
v

d
dr
−
v
r

y (2)
(b) y(r) =
∞
p=0
apr p+1
(3)
dy
dr
=

ap(p + 1)r p
(4)
d2
y
dr2
=

ap p(p + 1)r p−1
(5)
Substitute (3), (4) and (5) in (2)
Σ ap p (p + 1)r p−1
=
2
v
Σ ap(p + 1)r p
− 2Σ apr p
Replace p by p − 1 in the RHS and simplify
Σ ap p(p + 1)r p−1
=
2
v
Σ ap−1(p − ν)r p−1
Comparing the coefficients of r p−1
on both sides
p(p + 1)ap =
2
v
(p − v)ap−1 (6)
(c) The series in (3) will terminate when ν = n where n is a positive integer.
Here n = 2
Using (3)
y(r) =
1

0
ap r p+1
= a0r + a1r2

3.3 Solutions 163
From the recurrence relation (6)
a1 = −
a0
2
Therefore, y(r) = a0r

1 − r
2

F(r) = a0e− r
2 r 1 −
r
2

The normalization constant is given to be 1/
√
2.
3.14 The statement that the variables in classical equations of motion can be
replaced by quantum mechanical expectation values is known as Ehrenfest’s
theorem. For simplicity we shall prove the theorem in one-dimension although
it can be adopted to three dimensions.
(a)
d x
dt
=
d
dt

ψ∗
x ψdx =

∂ψ∗
∂t
xψ + ψ∗
x
∂ψ
∂t

dx (1)
Now ψ satisfies Schrodinger’s one dimensional equation
i
∂ψ
∂t
= −
2
2m
∂2
ψ
∂x2
+ V (x)ψ (2)
− i
∂ψ∗
∂t
= −
2
2m
∂2
ψ∗
∂x2
+ V (x)ψ∗
(3)
Premultiply (2) by ψ∗
x and post multiply (3) by xψ and subtract the result-
ing equations
i

ψ∗
x
∂ψ
∂t
+
∂ψ∗
∂t
xψ

= −

2
2m

ψ∗
x
∂2
ψ
∂x2
−
∂2
ψ∗
∂x2
xψ

(4)
Using (4) in (1)
d x
dt
=
i
2m

ψ∗
x
d2
ψ
dx2
−
∂2
ψ∗
∂x2
xψ

dx (5)
The first integral can be evaluated by parts

ψ∗
x
∂2
ψ
∂x2
dx = ψ∗
x
dψ
dx
%
%
%
%
∞
0
−

(dψ/dx)

ψ∗
+ x
dψ∗
dx

dx (6)
The first term on RHS is zero at both the limits.
ψ∗
x
d2
ψ
dx2
dx = −

dψ
dx

ψ∗
+
xdψ∗
dx

dx
= −

ψ∗ dψ
dx
dx −

x

d
dx

d
dx

dx
(7)
Furthermore
−

∂2
ψ∗
∂x2
xψ dx = −ψx
dψ∗
dx
%
%
%
%
∞
0
−
dψ∗
dx

ψ + x
dψ
dx

dx

The first term on the RHS is zero at both the limits.
−

∂2
ψ∗
∂x2
xψdx = −

dψ∗
dx

ψ + x
dψ
dx

dx
= ψ
dψ∗
dx
dx +

x

dψ∗
dx

dψ
dx

dx (8)
Substituting (7) and (8) in (5), the terms underlined vanish together.
d x
dt
=

i
2m

−

ψ∗ ∂
∂x
dx +

ψ
dψ∗
dx
dx

=
1
2m

ψ∗

−i
∂
∂x

ψ dx +

ψ

i
∂
∂x

ψ∗
dx

(9)
Now the operator for Px is −i ∂
∂x
. The first term on RHS of (9) is the
average value of the momentum Px , the second term must represent the
average value of P∗
x . But px being real, P∗
x = Px . Therefore
d x
dt
=
1
m
Px (10)
Thus (10) is similar to classical equation x = p/m
Equation (10) can be interpreted by saying that if the “position” and
“momentum” vectors of a wave packet are regarded as the average or
expectation values of these quantities, then the classical and quantum
motions will agree.
(b)
d Px
dt
=
d
dt

ψ∗

−i
∂
∂x

ψdτ = −i
d
dt

ψ∗ ∂ψ
∂x
dτ
= −i

dψ∗
dt
∂ψ
∂x
dτ +

ψ∗ ∂
∂x
∂ψ
∂t
dτ (11)
Now i
∂ψ
∂t
= −

2
2m

∇2
ψ + V ψ
−
i∂ψ∗
∂t
= −

2
2m

∇2
ψ∗
+ V ψ∗
(12)
Using (12) in (11)
d
dt
Px = −

ψ∗ ∂
∂x

−

2
2m

∇2
ψ + V ψ

dτ
+

−
2
2m

∇2
ψ + V ψ

∂ψ
∂x
d τ

3.3 Solutions 165
Integrating by parts twice
d
dt
Px = −

ψ∗

∂
∂x
(V ψ) − V
∂ψ
∂x

dτ
= −

ψ∗ ∂V
∂x
ψdτ =
−∂V
∂x

These two examples support the correspondence principle as they show
that the wave packet moves like a classical particle provided the expecta-
tion value gives a good representation of the classical variable.
3.15 (a) Using the Laplacian in the time-independent Schrodinger equation
−
2
2m

1
r2
∂
∂r

r2 ∂
∂r

+
1
r2 sin θ

∂
∂θ

sin θ
∂
∂θ

+
1
r2 sin2
θ
∂2
∂ϕ2

ψ(r, θ, ϕ) + V (r)ψ(r, θ, ϕ) = Eψ(r, θ, ϕ) (1)
We solve this equation by method of separation of variables
Let ψ (r, θ, ϕ) = ψr (r) Y(θ, ϕ) (2)
Use (2) in (1) and multiply by

−2m
2 .r2

/ψr (r)Y(θ, ϕ) and rearrange
1
ψr
(r)
d
dr

r2
dψr (r)/dr

+
2mr2
2
[E − V (r)]
= −
1
Y
(θ, ϕ)

1
sin θ
∂
∂θ

sin θ
∂
∂θ
(sin θ∂Y (θ, ϕ) /∂θ) +
1
sin2
θ
∂2
Y (θ, ϕ) /∂ϕ2

(3)
It is assumed that V (r) depends only on r.
L.H.S. is a function of r only and R.H.S is a function of θ and ϕ only.
Then each side must be equal to a constant, say λ.
1
sin θ
∂
∂θ

sin θ
∂Y
∂θ

θ, ϕ

+
1
sin2
θ
∂2
Y
∂ϕ2

θ, ϕ

+ λY (θ, ϕ) = 0 (4)
The radial equation is
d
dr
r2 dψr (r)
dr
+
2mr2
2
[E − V (r) − λ] ψr (r) = 0 (5)
(b) The angular equation (4) can be further separated by substituting
Y(θ, ϕ) = f (θ)g(θ) (6)

Following the same procedure
g(ϕ)
sin θ
d
dθ
sin θ
d f (θ)
dθ
+
f (θ)
sin2
θ
d2
g(ϕ)
dϕ2
+ λf (θ)g(ϕ) = 0
sin θ
f (θ)
d
dθ

sin θ
d f (θ)
dθ

+ λ sin2
θ =
−1
g(ϕ)
d2
g(ϕ)
dϕ2
= m2
(7)
where m2
is a positive constant
d2
g
dϕ2
= −m2
ϕ (8)
gives the normalized function
g = (1/
√
2π) eimϕ
(9)
m is an integer since g(ϕ + 2π) = g(ϕ)
Dividing (6) by sin2
θ and multiplying by f , and rearranging
1
sin θ
d
dθ

sin θ
d f
dθ

+

λ −
m2
sin2
θ

f = 0 (10)
(c) The physically accepted solution of (10) is Legendre polynomials when
λ = l(l + 1) (11)
and l is an integer.
With the change of variable ψr (r) = χ(r)/r
The first term in (5) becomes
d
dr

r2 ψr
dr

=
d
dr

r2

−
χ
r2
+
1
r
dχ
dr

=
d
dr

r
dχ
dr
− χ

= r
d2
χ
dr2
+
dχ
dr
−
dχ
dr
= r
d2
χ
dr2
With the substitution of λ from (11), (5) becomes upon rearrangement

−
2
2m

d2
χ
dr2
+

V (r) +
l(l + 1)2
2mr2

χ = Eχ (12)
Thus, the radial motion is similar to one dimensional motion of a particle
in a potential
Ve = V (r) +
l(l + 1)2
2mr2
(13)
where Ve is the effective potential. The additional “potential energy” is
interpreted to arise physically from the angular momentum. A classical
particle that has angular momentum L about the axis through the origin
perpendicular to the plane of its path has the angular velocity ω = L/mr2
where its radial distance from the origin is r. An inward force mω2
r =
mL2
/ωr3
is required to keep the particle in the path. This “centripetal

3.3 Solutions 167
force” is supplied by the potential energy, and hence adds to the V (r)
which appears in (13) for the radial motion. This will have exactly the
form indicated in (13) if we put L =
√
l(l + 1)
3.16 −
*
1
sin θ
∂
∂θ

sin θ ∂
∂θ

+ 1
sin2
θ
∂2
∂ϕ2
+
Y (θ, ϕ) = λY(θ, ϕ)
We solve the equation by the method of separation of variables.
Let Y(θ, ϕ) = f (θ)g(ϕ) and multiply by sin2
θ
−g(ϕ) sin θ
∂
∂θ

sin θ
∂ f
∂θ

+ f (θ)
∂2
g
∂ϕ2
= λ sin2
θ fg
Divide through out by f (θ)g(ϕ) and separate the θ and ϕ variables.
1
f (θ)
sin θ
∂
∂θ

sin θ
∂ f
∂θ
+ λ sin2
θ

= −
1
g(ϕ)
∂2
g
∂ϕ2
= m2
(1)
LHS is a function of θ only and RHS function of ϕ only. The only way the
above equation can be satisfied is to equate each side to a constant, say −m2
,
where m2
is positive.
1
g(ϕ)
d2
g(ϕ)
dϕ2
= −m2
Therefore g(ϕ) = A eimϕ
We can now normalize g(ϕ) by requiring

g∗
(ϕ)g(ϕ)dϕ = 1
A2
2π
0
%
%eimϕ
%
%2
dϕ = 2π A2
= 1
Or A = (2π)−1/2
We shall now show that m is an integer
g(ϕ + 2π) = g(ϕ)
g(ϕ + 2π) = (2π)− 1
2 eim(ϕ+2π)
= (2π)− 1
2 eimϕ
.e2πmi
= g(ϕ)e2πmi
∴ e2πmi
= cos (2πm) + i sin(2πm)
= cos(2πm) = 1
Thus m is any integer, m = 0, ±1, ±2 . . .
3.17 (a) In Problem 3.16, going back to Eq. (1) and multiplying by f (θ) and divid-
ing by sin2
θ and putting μ = cos θ.


1
sin θ

∂
∂θ

sin θ
d f
dθ

+ λ f (θ) = 0 (2)

d
dθ

=

d
dμ

·

dμ
dθ

= − sin θ
d
dμ
Writing f (θ) = P(μ), Eq. (2) becomes
d
dμ

1 − μ2
dp
dμ

+ λP = 0
or

1 − μ2
d2
p
dμ2
− 2μ
dp
dμ
+ λp = 0 (3)
One can solve Eq. (3) by series method
Let P = Σ∞
k=1akμk
(4)
dp
dμ
=

k
akkμk−1
(5)
d2
P
dμ2
=

akk(k − 1)μk−2
(6)
Using (4), (5) and (6) in (3)

k(k − 1)akμk−2
−

k(k − 1)akμk
− 2

kakμk
+ λΣakμk
= 0
Equating equal powers of k
(k + 2)(k + 1)ak+2 − [k(k − 1) + 2k − λ] ak = 0
Or ak+2/ak = [k (k + 1) − λ] / (k + 1) (k + 2)
(b) If the infinite series is not terminated, it will diverge at μ = ±1, i.e. at
θ = 0 or θ = π. Because this should not happen the series needs to be
terminated which is possible only if λ = k(k + 1)
i.e. l(l + 1); l = 0, 1, 2 . . . Here l is known as the orbital angular momen-
tum quantum number. The resulting series P(μ) is then called Legendre
polynomial.
3.18 (a) The term − 2
d2
2mdx2 is the kinetic energy operator, U(x) is the potential
energy operator, ψ(x) is the eigen function and E is the eigen value.
(b) Put U(x) = 0 in the region 0 x a in the Schrodinger equation to
obtain

−
2
2m

d2
ψ(x)
dx2
= Eψ(x) (1)

3.3 Solutions 169
Or
d2
ψ(x)
dx2
+

2mE
2

ψ(x) = 0 (2)
Writing
α2
=
2mE
2
(3)
Equation (2) becomes
d2
ψ
dx2
+ α2
ψ = 0 (4)
which has the solution
ψ(x) = A sin αx + B cos αx (5)
where A and B are constants of integration. Take the origin at the left
corner, Fig 3.5.
Fig. 3.5 Square potential
well of infinite depth
Boundary condition:
ψ(0) = 0; ψ(a) = 0
The first one gives B = 0. We are left with
ψ = A sin αx (6)
The second one gives
αa = nπ, n = 1, 2, 3 . . . (7)
n = 0 is excluded as it would give a trivial solution.
Using the value of α in (6)
ψn(x) = Asin
nπx
a

(8)
This is an unnormalized solution. The constant A is determined from
normalization condition.

a
0
ψ∗
n (x)ψn(x)dx = 1
A2
α
0
sin2 nπx
a

dx = 1

A2
2

x − cos

2nπx
a
%
%
%
a
0
= A2
a = 1
Therefore, A =

2
a
1/2
(9)
The normalized wave function is
ψn(x) =

2
a
1
2
sin
nπx
a

(10)
Using the value of α from (7) in (3), the energy is
En =
n2
h2
8ma2
(11)
(c) probability p =
a
0 |ψ3(x)|2
dx
=
2a
3
a
3

2
a

sin2

3πx
a

dx =
1
3
(d) ψ(n) and probability density P(x) distributions for n = 1, 2 and 3 are
sketched in Fig 3.6
Fig. 3.6
3.19 The Schrodinger equation for the n – p system in the CMS is
∇2
ψ(r, θ, ϕ) +

2μ
2

[E − V (r)]ψ(r, θ, ϕ) = 0 (1)

3.3 Solutions 171
Fig. 3.7 Deuteron wave
function and energy
where μ is the reduced mass = M/2, M, being neutron of proton mass. With
the assumption of spherical symmetry, the angular derivatives in the Laplacian
vanish and the radial equation is
1
r2
d
dr

r2 d
dr

ψ(r) + (M/2
)[E − V (r)]ψ(r) = 0 (2)
With the change of variable
ψ(r) =
u(r)
r
(3)
d2
u
dr2
+

M
2

[E − V (r)]u = 0 (4)
The total energy = −W, where W = binding energy, is positive as the poten-
tial is positive
V0 = −V , where V0 is positive
Equation (4) then becomes
d2
u
dr2
+

M
2

(V0 − W) u = 0; r R (5)
d2
u
dr2
−
MWu
2
= 0; r R (6)
where R is the range of nuclear forces, Fig. 3.7.
Calling
M(V0 − W)
2
= k2
(7)
and
MW
2
= γ 2
(8)
(5) and (6) become
d2
u
dr2
+ k2
u = 0 (9)
d2
u
dr2
− γ 2
u = 0 (10)

The solutions are
u1(r) = A sin kr + B cos kr; r R (11)
u2(r) = Ce−γr
+ Deγr
; r R (12)
Boundary conditions: as r → 0, u1 → 0
and as r → ∞, u2 must be finite. This means that B = D = 0.
Therefore the physically accepted solutions are
u1 = A sin kr (13)
u2 = Ce−γr
(14)
At the boundary, r = R, u1 = u2 and their first derivatives

du1
dr

r=R
=

du2
dr

r=R
These lead to
A sin kR = Ce−γr
(15)
Ak cos kR = −γ Ce−γr
(16)
Dividing the two equations
k cot kR = −γ (17)
Or
cot kR = −
γ
k
(18)
Now V0 ≫ W, so cot kR is a small negative quantity. Therefore kR ≈
π/2 k2
R2
=
π
2
2
Or
M(V0 − W)R2
2
=
π2
4
Again neglecting W compared to V0
V0 R2
≈
π2
2
4M
3.20 The inside wave function is of the form u = A sin kr. Because V (r) = 0 for
r R, we need to consider contribution to V from within the well alone.
V =
R
0
u∗
(−V0)u dr = −V0 A2
R
0
sin2
krdr
=

−
V0 A2
2
R
0
(1 − cos 2kr)dr
= −V0 A2

R
2
−
sin 2kR
4k

3.3 Solutions 173
3.21 u = C e−kr
∞
0 |u|2
dr = c2
∞
0 e−2kr
= c2
2k
= 1
C =
√
2k
The probability that the neutron – proton separation in the deuteron exceeds
R is
P =
∞
R
|u|2
dr = 2k
∞
R
e−2kr
dr
= e−2kR
= e−(2×0.232×2)
≈ 0.4
Average distance of interaction
r =
∞
0
r|u2
|dr = 2k
∞
0
re−2kr
dr
=
1
2k
=
1
2 × 0.232
= 2.16 fm
3.22 The inside wave function u1 = A sin kr is maximum at r ≈ R. Therefore
kR = π/2
or k2
R2
= M(V0 − W)R2
/2
= π2
/4
V0 =
π2
2
c2
4Mc2 R2
+ W
Substituting c = 197.3 MeV − fm, Mc2
= 940 MeV, R = 1.5 fm and
W = 2.2 MeV, we find V0 ≈ 47 MeV
3.23 r2
=

ψ∗
r2
ψdτ
=
∞
0
r2
r2
α
2π

e−2αr
4πr2
dr
=
1
2α2
√
r2 =
1
√
2α
=
4.3 × 10−15
m
√
2
= 3.0 × 10−15
m = 3.0 fm
3.24 Referring to Problem 3.18, the energy of the nth level is
En =
n2
h2
8mL2
(1)
and
En+1 =
(n + 1)2
h2
8ML2
(2)
Therefore En+1 − En =
(2n + 1)h2
8mL2
(3)

The ground state corresponds to n = 1 and the first excited state to n =
2, m = 8me and L = 1 nm = 106
fm. Putting n = 1 in (3)
hv = E2 − E1 =
3h2
8mL2
= 3π2
2
c2
/16mec2
L2
= 3π2
(197.3)2
MeV2
− fm2
/(16 × 0.511 MeV)(106
)2
fm2
= 0.14 × 10−6
MeV = 0.14 eV
λ(nm) =
1,241
E(eV)
=
1,241
0.14
= 8864 nm
This corresponds to the microwave region of the electro-magnetic spectrum.
3.25 Consider a finite potential well. Take the origin at the centre of the well.
V (x) = V0; |x| a
= 0; |x| a
d2
ψ
dx2
+

2m
2

[E − V (r)] ψ = 0
Region 1 (E V0)
d2
ψ
dx2
−

2m
2

(V0 − E)ψ = 0 (1)
d2
ψ
dx2
− β2
ψ = 0 (2)
where β2
=

2m
2

(V0 − E) (3)
ψ1 = Aeβx
+ Be−βx
(4)
where A and B are constants of integration.
Since x is negative in region 1, and ψ1 has to remain finite we must set B = 0,
otherwise the wave function grows exponentially. The physically accepted
solution is
ψ1 = Aeβx
(5)
Region 2; (V = 0)
Fig. 3.8 Square potential
well of finite depth

3.3 Solutions 175
d2
ψ
dx2
+

2mE
2

ψ = 0
d2
ψ
dx2
+ α2
ψ = 0 (6)
with α2
=
2mE
2
(7)
ψ2 = C sin
Odd
αx + D cos
even
αx (8)
In this region either odd function must belong to a given value E or even
function, but not both,
Region 3; (E V0)
Solution will be identical to (4)
ψ3 = Aeβx
+ Be−βx
But physically accepted solution will be
ψ3 = Be−βx
(9)
because we must put A = 0 in this region where x takes positive values if the
wave function has to remain finite.
Class I (C = 0)
ψ2 = D cos αx (10)
Boundary conditions
ψ2(a) = ψ3(a) (11)
dψ2/dx|x=a = dψ3/dx|x=a (11a)
These lead to
D cos (αa) = B e−βa
(12)
− D α sin(αa) = −B β e−βa
(13)
Dividing (13) by (12)
α tan αa = β (14)
Class II (D = 0)
ψ2 = C sin(αx) (15)
Boundary conditions:
ψ2(−a) = ψ1(−a) (16)
dψ2/dx|x=−a = dψ1/dx|x=−a (17)
These lead to
C sin(−αa) = −C sin(αa) = Ae+βa
(18)
Cα cos(αa) = Aβeβa
(19)
α cot (αa) = −β (20)

Fig. 3.9 η − ξ curves for class I solutions. For explanation see the text
Note that from (15) and (2), α2
= −β2
, which is absurd because this implies
that α2
+ β2
= 0, that is 2mV 0/2
= 0, but V0 = 0. This simply means that
class I and class II solutions cannot coexist
Energy levels:
Class I: set ξ = αa; η = βa
where α and β are positive.
ξ tan ξ = η (21)
with ξ2
+ η2
= a2
(α2
+ β2
) = 2mV0a2
/2
= constant (22)
The energy levels are determined from the intersection of the curve ξ tan ξ
plotted against η with the circle of known radius 2mV 0a2
2
1/2
, in the first
quadrant since ξ and η are restricted to positive values.
The circles, Eq. (22), are drawn for V0a2
= 2
/2m, 42
/2m and 92
/2m
for curves 1, 2 and 3 respectively Fig 3.9. For the first two values there is only
one solution while for the third one there are two solutions.
For class II, energy levels are obtained from intersection of the same circles
with the curves of −ξ cot ξ in the first quadrant, Fig 3.10.
Curve (1) gives no solution while the other two yield one solution each.
Thus the three values of V0a2
in the increasing order give, one, two and three
energy levels, respectively. Note that for a given particle mass the energy levels
depend on the combination V0a2
. With the increasing depth and/or width of
the potential well, greater number of energy levels can be accommodated.
For ξ = 0 to π/2, that is V0a2
between 0 and π2
2
/8m there is just one
level of class I

3.3 Solutions 177
Fig. 3.10 η − ξ curves for class II solutions. For explanation see the text (After Leonard I. Schiff,
Quantum mechanics, McGraw-Hill 1955)
For V0a2
between 2π2
2
/8m and 4π2
2
/8m there is one energy level of
each class or two altogether. As V0a2
increases, energy levels appear succes-
sively first of one class and next of the other.
3.26 The probability for neutron and proton to be found outside the range of nuclear
forces (refer to Problem 3.21)
P =
∞
R
|ψ2|2
dτ =

(|u2(r)|/r)2
4πr2
dr
= 4πC2
∞
R
e−2γr
dr
P = 2πC2
e−2γ R/γ
(1)
By Eq. (15) of solution 3.19
A sin kR = Ce−γ R
or
C2
e−2γ R
= A2
sin2
kR ≈ A2
(2)
because, kR ≈ π/2
Therefore,
p =
2π A2
γ
(3)
We can now find the constant A from the normalization condition

R
0
|ψ1|2
dτ +
∞
R
|ψ2|2
dτ = 1
R
0
u2
1.4πr2
dr/r2
+
∞
R
u2
24πr2
dr/r2
= 1
A2
R
0
sin2
krdr + C2
∞
R
e−2γ r
dr = 1/4π
Integrating and using (2), we find
A2
=
γ
2π(γR + 1)
(4)
Using (4) in (3)
P =
1
γ R + 1
(5)
Now γ R =

MW
2
1/2
R =

Mc2
W
2c2
1/2
R
=

940 × 2.2
(197.3)2
1/2
× 2.1 = 0.48
where we have inserted Mc2
= 940 MeV /c2
,
W = 2.2MeV and R = 2.1 f m
Therefore p = 1
0.48+1
= 0.67
Thus neutron and proton stay outside the range of nuclear forces approxi-
mately 70% of time.
3.27 By Problem 3.25, for the finite well, for class I
α tan αa = β
with α =
(2mE)
1
2

; β =
[2m(V0 − E)]
1
2

As V0 → ∞, β → ∞ and αa = nπ/2 (n odd)
Therefore, α2
a2
= 2mEa2
2 = n2
π2
4
Or E = n2
π2
2
/8ma2
(n odd)
For class II
α cot αa = −β
As V0 → ∞, β → ∞ and αa = nπ/2 (n even)

3.3 Solutions 179
α2
a2
=
2mEa2
2
=
n2
π2
4
E =
n2
π2
2
8ma2
(n even)
Thus E =
n2
π2
2
8ma2
, n = 1, 2, 3 . . .
3.28 From problem 3.27,
α cot αa = −β = 0
The first solution is αa = π/2, for the ground state.
The second solution, αa = 3π/2, will correspond to the first excited state
(with l = 0). This will give
α2
a2
= 9π2
/4
Let the excited states be barely bound so that W = 0. Then,
α2
= 2mE/2
= 9π2
/4a2
E = V1 = 9V0
a value which is not possible. Thus, the physical reason why bound excited
states are not possible is that deuteron is a loose structure as the binding energy
(2.225 MeV) is small. The same conclusion is reached for higher excited states
including l = 1, 2 . . .
3.29 The inside wave function u1 = A sin kr is maximum at r ≈ R. Therefore
kR =
π
2
or k2
=
M(V0 − W)R2
2
=
π2
4
or V0 =

π2
2
c2
4Mc2 R2

+ W
Substituting c = 197.3 MeV − fm, Mc2
= 940 MeV, R = 1.5 fm and W =
2.2 Mev, we find V0 ≈ 47 MeV
3.30 Schrodinger’s equation in one dimension
(a) d2
ψ
dx2 +
2m
2

(E − V )ψ = 0
Region 1: (x 0)V = 0; d2
ψ
dx2 + k2
1ψ = 0
where k2
1 = 2mE
2 Solution: ψ1 = exp(ik1x) + A exp(−ik1x)
Region 2: (0 x a)V = Vb; d2
ψ
dx2 − k2
2ψ = 0
where k2
2 =
2m
2

(Vb − E)
Solution: ψ2 = B exp(k2x) + C exp(−k2x)

Region 3: (x a) V = 0
Solution: ψ3 = D exp(ik1x)
(b) Boundary conditions:
ψ1(0) = ψ2(0) → 1 + A = B + C (1)
dψ1
dx
%
%
%
%
x=0
=
dψ2
dx
%
%
%
%
x=0
→ ik1(1 − A) = k2(B − C) (2)
ψ2(a) = ψ3(a) → B exp(k2a) + C exp(−k2a) = D exp(ik1a) (3)
dψ2
dx
%
%
%
%
x=a
=
dψ3
dx
%
%
%
%
x=a
→ k2 (B exp(k2a) − Ck2 exp(−k2a))
= ik1 D exp(ik1a)
(4)
Eliminate A between (1) and (2) to get
B(k2 + ik1) − C(k2 − ik1) = 2ik1 (5)
Eliminate D between (3) and (4) to get
k2(B exp(k2a) − Ck2 exp(−k2a)) = ik1(B exp(k2a) + C exp(−k2a)) (6)
Solve (5) and (6) to get
B =
2ik1(k2 + ik1)
,
(k2 + ik1)2
− exp(2k2a)(k2 − ik1)2
- (7)
C =
2ik1(k2 − ik1)e2k2a
,
(k2 + ik1)2
− e2k2a (k2 − ik1)2
- (8)
Using the values of B and C in (3),
τ = D =
4ik1k2 exp(−ik1a)
(k2 + ik1)2 exp(−k2a) − (ik1 − k2)2 exp(k2a)
(9)
3.31 (a) Ftrans = τ∗
τ = |D|2
= 16
k2
1k2
2
(k2
1 + k2
2)2(e2k2a + e−2k2a) − 2(k4
2 − 6k2
2k2
1 + k4
1)
This expression simplifies to
Ftrans = T =
4k2
1 k2
2
(k2
1 + k2
2)2 sinh2(k2a) + 4k2
1k2
2
(10)
use k2
1 = 2mE/2
and k2
2 = 2m(Vb − E)/2
The reflection coefficient R is obtained by substituting (7) and (8) in (1)
to find the value of A. After similar algebraic manipulations we find
R = |A|2
=
(k2
1 + k2
2)2
sinh2
(k2a)
(k2
1 + k2
2)2 sinh2(k2a) + 4k2
1k2
2
(11)
Note that R + T = 1
(b) When E Vb, k2 becomes imaginary and
sinh (k2a) = i sin (k2a) (12)

3.3 Solutions 181
Using (11) in (9) and noting k2
1 + k2
2 = 2mV b
2
k2
2 =
2mVb
2
and k2
1k2
2 =

2m
2
2
E(Vb − E)
we find
T =
1
1 +
V2
b sin2
k2a
4E(E−Vb)
(13)
and
R =
1
1 + 4E(E−V0)
V2
0 sin2
k2a
(14)
A typical graph for T versus E
Vb
is shown in Fig. 3.12
Fig. 3.11 Transmission
through a rectangular
potential barrier
Fig. 3.12 Transmission as a
function of E/Vb
3.32 The form of potential corresponds to that of a linear Simple harmonic Oscil-
lator. The energy of the oscillator will be E1 = ω
2
and E2 = 3ω
2
.

E = ψ|H|ψ
= (C1ψ1 + C2ψ2)|H|(C1ψ1 + C2ψ2)
= C1ψ1 + C2ψ2|C1 Hψ1 + C2 Hψ2|
= C1ψ1 + C2ψ2|C1 E1ψ1 + C2 E2ψ2|
= C2
1 E1 + C2
2 E2
=
C2
1ω
2
+
C2
2 3ω
2
=
1
2
ω(C2
1 + 3C2
2 )
where ω =

k
m
1/2
3.33 The ground state is
ψ =

2
a
1/2
sin(πx/a)
The wave function corresponding to momentum p is
ψi = (2π)−1/2

k
Ckeikx
The probability that the particle has momentum between p and p+dp is given
by the value of
|Ck|2
, where Ck is the overlap integral
Ck = (2π)− 1
2

2
a
1/2 a
0
eikx
sin
πx
a

dx
Itegrating by parts twice,
Ck = (πa)
1
2

eika
+ 1

π2
− k2
a2
−1
The required probability is
|Ck|2
= πa

eika
+ 1

e−ika
+ 1

π2
− k2
a2
−2
= 4πa cos2

ka
2

π2
− k2
a2
−2
3.34 The transmission coefficient is given by
T = e−G
(1)
G =
2

b
a
[2m(U(r) − E)1/2
dr (2)
Put
U(r) =
zZe2
r
(3)
for the Coulomb potential energy between the alpha particle and the residual
nucleus at distance of separation r.

3.3 Solutions 183
G =
2

(2m)
1
2
b
a

zZe2
r
− E
1/2
dr
where z = 2
Now at distance b where the alpha energy with kinetic energy E, potential
energy = kinetic energy
E =
1
2
mv2
= zZe2
/b
G =

2

2mzZe2
1/2
b
a

1
r
−
1
b
1/2
dr
The integral is easily evaluated by the change of variable r = b cos2
θ
I =
√
b
1
cos−1 a
b

−
a
b
−
a2
b2
2
Finally
G =
2

2mz Z e2
b
1/2

cos−1

R
b

−
R
b
−
R2
b2
'
where a = R, the nuclear radius.
If vin is the velocity of the alpha particle inside the nucleus and R = a is
the nuclear radius then the decay constant λ = 1/τ ∼ (vin/R).e−G
3.35 (a) In Problem 3.19 the condition that a bound state be formed was obtained as
cot kR = −
γ
k
= −

W
V0 − W
1/2
where V0 is the potential depth and a is the width. Here the condition
would read
cot ka = −

W
V0 − W
1/2
where k2
= 2m(V0 − W)a2
/2
If we now make W = 0, the condition that only one bound is formed is
ka =
π
2
or V0 =
h2
32ma2
(b) The next solution is
ka =
3π
2
Here W1 = 0 for the first excited state
With the second solution we get
V1 =
9h2
32ma2
Note that in Problem 3.23 the reduced mass μ = M/2 while here μ = m.
The graphs are shown in Fig. 3.13.

Fig. 3.13
For (a) the inside and outside wave functions are as in the deuteron
Problem 3.19. For (b) the inside wave function is similar but the outside
function becomes constant (W1 = 0) and is a horizontal line.
3.36 (a) Class I: Refer to Problem 3.25
ψ1 = Aeβx
(−∞ x −a)
ψ2 = D cos ax(−a x +a)
ψ3 = A e−βx
(a x ∞)
Normalization implies that
−a
−∞
|ψ1|2
dx +
a
−a
|ψ2|2
dx +
∞
a
|ψ3|2
dx = 1
−a
−∞
A2
e2βx
dx +
a
−a
D2
cos2
αx dx +
∞
a
A2
e−2βx
dx = 1
A2
e−2βa
2β
+ D2

a +
sin(2αa)
2α

+
A2e−2βa
2β
= 1
Or
A2
e−2βa
/β + D2
(a + sin(2αa)/2α) = 1 (1)
Boundary condition at x = a gives
Dcos αa = ae−βa
(2)
Combining (1) and (2) gives
D =

a +
1
β
−1
A = eβa
cos αa

a +
1
β
−1

3.3 Solutions 185
3.37 (a) un =
2
L
1/2
sin
nπx
L

x =
L
0
u∗
n xundx =

2
L
L
0
x sin2 nπx
a

dx
=
L
2
+

L
4n2π2

(cos(2nπ) − 1)
The second term on the RHS vanishes for any integral value of n. Thus
x =
L
2
Var x = σ2
= (x− x )2
= x2
− x 2
= x2
−
L2
4
Now x2
=
L
0
u∗
n x2
undx =

2
L
L
0
x2
sin2 nπx
L

dx
=
L2
3
−
L2
2n2π2
σ2
= x2
− x 2
=
L2
3
−
L2
2n2π2
−
L2
4
=

L2
12

1 −
6
n2π2

For n → ∞, x = L
2
; σ2
→ L2
/12
(b) Classically the expected distribution is rectangular, that is flat.
The normalized function
f (x) =
1
L
x =

x f (x)dx =
L
0
xdx
L
=
L
2
σ2
= x2
− x 2
x2
=
L
0
x2
f (x)dx = L2
/3
∴ σ2
=
L2
3
−
L2
4
=
L2
12
Fig. 3.14
3.38 H =

−
2
2m

∇2
+ ar2

1 −
5
6
sin2
θ cos2
ϕ

(1)
In spherical coordinates x = r sin θ cos ϕ. Therefore

H =

−
2
2m

∇2
+ a

x2
+ y2
+ z2
−
5
6
x2

=

−
2
2m

∇2
+ a

x2
6
+ y2
+ z2

(2)
The Schrodinger’s equation is
Hψ(x, y, z) = Eψ(x, y, z) (3)
This equation can be solved by the method of separation of variables.
Let
ψ(x, y, z) = ψx ψyψz (4)
Hψ(x, y, z) = −
2
2m

∂2
∂x2
+
∂2
∂y2
+
∂2
∂z2

ψx ψyψz + a

x2
6
+ y2
+ z2

ψx ψyψz = Eψx ψyψz
−
2
2m
ψyψz
∂2
ψx
∂x2
−
2
2m
ψx ψz
∂2
ψy
∂x2
−
2
2m
ψx ψy
∂2
ψz
∂ X2
+
ax2
6
ψx ψyψz + ay2
ψx ψyψz + az2
ψx ψyψz = Eψx ψyψz
Dividing throughout by ψx ψyψz
−
2
2m
1
ψx
∂2
ψx
∂x2
−
2
2m
1
ψy
∂2
ψy
∂y2
−
2
2m
1
ψz
∂2
ψz
∂z2
+
ax2
6
+ ay2
+ az2
= E
(5)
−
2
2m
1
ψx
∂2
ψx
∂x2
+
ax2
6
= E1
a
6
=1
/2k1 (6)
−
2
2m
1
ψy
∂2
ψy
∂y2
+ ay2
= E2 a =1
/2k2 (7)
−
2
2m
1
ψz
∂2
ψz
∂z2
+ az2
= E3 a =1
/2k2 (8)
E1 = (n1 +1
/2)ω1; E2 = (n1 +1
/2)ω2; E3 = (n3 +1
/2)ω3
ω1 =
k1
m
=
a
3m
; ω2 = ω3 =
2a
m
E = E1 + E2 + E3 =

n1 +
1
2

ω1 + (n2 + n3 + 1)ω2
The lowest energy level corresponds to n1 = n2 = n3 = 0, with
E =
ω1
2
+ ω2 =

1
12
+
√
2
#

a
m

It is non-degenerate.
The next higher state is degenerate with n1 = 1, n2 = 0, n3 = 0;
E =
3
2

a
3m
+
2a
m
=

3
4
+
√
2
#

a
m
This is also non-degenerate.

3.3 Solutions 187
3.39 (a)

−
2
2m

∇2
+ V

ψ(x, y, z) = Eψ(x, y, z) (1)
Put V = 0

−
2
2m

∂2
∂x2
+
∂2
∂y2
+
∂2
∂z2

ψ(x, y, z) = Eψ(x, y, z)
Let
ψ (x, y, z) = X (x) Y (y) Z (x) (2)
Y Z
∂2
X
∂x2
+ Z X
∂2
Y
∂y2
+ XY
∂ Z
∂z2
= −

2mE
2

XY Z
Dividing throughout by XYZ
1
Y
∂2
Y
∂y2
+
1
Z
∂2
Z
∂z2
= −
1
X
∂2
X
∂x2
−
2mE
2
(3)
LHS is a function of y and z only while the RHS is a function of x only.
The only way (3) can be satisfied is that each side is equal to a constant, say –
α2
.
1
X
d2
X
dx2
+
2mE
2
− α2
= 0
∂2
X
dx2
+

2mE
2
− α2

X = 0
Or
∂2
X
∂x2
+ β2
X = 0
where
β2
=

2mE
2

− α2
(4)
X = A sin βx + B cos βx
Take the origin at the corner
Boundary condition: X = 0 when x = 0. This gives B = 0.
X = A sin βx
Further, X = 0 when x = a
Sinβa = 0 → βa = nx π
Or
β =
nx π
a
(nx = integer) (5)

Going back to (3)
1
Y
∂2
Y
∂y2
+
1
Z
∂2
Z
∂z2
= −α2
1
Z
∂2
Z
∂z2
= −
1
Y
∂2
Y
∂y2
− α2
(6)
Each side must be equal to a constant, say −γ 2
for the same argument as
before.
−
1
Y
∂2
Y
∂y2
− α2
= −γ 2
Or
−
1
Y
∂2
Y
∂y2
+

α2
− γ 2

= 0
Or
∂2
Y
∂y2
+ μ2
Y = 0
where μ2
= α2
− γ 2
(7)
Y = D sin μy
Y = 0 at y = b
This gives μ =
nyπ
b
(8)
Going back to (6)

1
Z

d2
Z
dz2
= −γ 2
This gives Z = F sin γz
where γ =
nzπ
c
(9)
∴ ψ ∼ sin
nx πx
a

sin
nyπy
b

sin
nzπz
c

(b) Combining (4), (5), (7), (8) and (9)
μ2
= α2
− γ 2
= (2mE/2
) − β2
− γ 2
Or
2mE
2
= μ2
+ β2
+ γ 2
=
nyπ
b
2
+
nx π
a
2
+
nzπ
c
2
Or
E =

h2
8m

n2
x
a2
+
n2
y
b2
+
n2
z
c2
#
(10)
3.40 For a = b = c
E = (h/8ma2
)

n2
x + n2
y + n2
z

(Equation 10 of Prob 3.39)

3.3 Solutions 189
None of the numbers nx , ny, or nz can be zero, otherwise ψ(x, y, z) itself
will vanish.
For an infinitely deep potential well En = h2
8ma2
,
n2
x + n2
y + n2
z
-
. The com-
bination nx = ny = nz = 0 is ruled out because the wave function will be
zero. Table 3.4 gives various energy levels along with the value of g, the degen-
eracy. The values of nx , ny and nz are such that n2
x +n2
y +n2
z = 8ma2
En/h2
=
constant for a given energy En. The energies of the excited states are expressed
in terms of the ground state energy E0 = h2
/8ma2
Table 3.4
nx ny nz g En
0 0 1 3-fold E0 = h2
/8ma2
0 1 0
1 0 0
0 1 1 3-fold 2E0
1 0 1
1 1 0
1 1 1 Non-degenerate 3E0
0 0 2 3-fold 4E0
0 2 0
2 0 0
0 1 2 6-fold 5E0
1 0 2
1 2 0
0 2 1
2 0 1
2 1 0
1 1 2 3-fold 6E0
1 2 1
2 1 1
3.41 (a) Case (i) U0 E, Region x ≪ 0
Putting V (x) = 0, Schrodinger’s equation is reduced to
d2
ψ
dx2
+

2mE
ℏ2

ψ = 0 (1)
ψ1 = A exp(ik1x) + B exp(−ik1x) (2)
where k2
1 =
2mE
ℏ2
(3)
ψ1 represents the incident wave moving from left to right (first term in (2))
plus the reflected wave (second term in (2)) moving from right to left
Region x ≫ 0 :
d2
ψ
dx2
+

2m(E − U0)
ℏ2

ψ = 0 (4)
which has the physical solution

ψ2 = C exp(ik2x) (5)
where
k2
2 =
2m(E − U0)
ℏ2
(6)
It represents the transmitted wave to the right with reduced amplitude.
Note that the second term is absent in (5) as there is no reflected wave in
the region x 0.
Case (ii), U0 E
Region x 0
ψ3 = A exp(ik1x) + B exp(−ik1x) (7)
Region x 0
d2
ψ
dx2
−
2mψ(U0 − E)
ℏ2
= 0
d2
ψ
dx2
− α2
ψ = 0
ψ4 = Ce−αx
+ Deαx
where α2
= 2m(U0−E)
2
ψ must be finite everywhere including at x = −∞. We therefore set
D = 0. The physically accepted solution is then
ψ4 = Ce−αx
(8)
(b) The continuity condition on the function and its derivative at x = 0 leads
to Eqs. (9) and (10).
ψ3(0) = ψ4(0)
A + B = C (9)
dψ3
dx
%
%
%
%
x=0
=
dψ4
dx
%
%
%
%
x=0
ik1(A − B) = −Cα (10)
Fig. 3.15 Case(i)

3.3 Solutions 191
Dividing (10) by (9) gives
ik(A − B)
A + B
= −α (11)
Diagrams for ψ at around x = 0
3.42 k1 =

2mE
ℏ2
1/2
; k2 =

2m(E − V0)
ℏ2
1/2
(1)
Boundary condition at x = 0:
ψ1(0) = ψ2(0)
dψ1
dx
%
%
%
%
x=0
=
dψ2(x)
dx
%
%
%
%
x=0
(2)
These lead to
A0 + A = B (3)
ik1(A0 − A) = ik2 B
Or
k1(A0 − A) = k2 B (4)
Solving (3) and (4)
A =

k1 − k2
k1 + k2

A0 (5)
B =
2k1 A0
k1 + k2
(6)
Reflection coefficient,
R =
|A|2
|A0|2
=
(k1 − k2)2
(k1 + k2)2
(7)
Transmission coefficient,
T =

k2
k1

|B|2
|A|2
=
4k1k2
(k1 + k2)2
(8)
Substituting the expressions for k1 and k2 from (1) and putting E = 4V0/3
we find that R = 1/9 and T = 8/9.
From (7) and (8) it is easily verified that
R + T = 1 (9)
Fig. 3.16 Graphs for
probability density

This is a direct result of the fact that the current density is constant for a
steady state.
Thus |A0|2
v1 = |A|2
v1 + |B|2
v2
where v1 = k1ℏ
m
and v2 = k2ℏ
m
A2
+ B2
= 1 because the sum of the intensities of the reflected intensity and
transmitted intensities does not add up to unity. What is true is relation (9)
which is relevant to current densities.
3.43 (a) The wave function must be finite, single-valued and continous. At the
boundary this is ensured by requiring the magnitude and the first derivative
be equal.
(b)
Fig. 3.17 Sketch of
ψ ∼ cos
3πx
L

(c)
L
2
− L
2
|ψ|2
dx = A2
L
2
− L
2
cos2 3πx
L
dx = 1
or
A2
L
2
− L
2
(1 + cos 6πx/L)dx = A2
L = 1
Therefore A = 1/
√
L
(d) P

−
L
4
x
L
4

= A2
L
4
L
4
cos2

3πx
L

dx
=

1
L
L/4
−L/4
1
/2(1 + cos

6πx
L

dx =
1
4
+
1
6π
= 0.303
(e)
d2
ψ
dx2
=
d2
dx2
A cos

3πx
L

= −9π2

A
L

cos

3πx
L

= −

9π2
L

ψ
Therefore,

−
ℏ2
2m

d2
ψ
dx2
=

9π2
ℏ2
2mL

ψ − Eψ
Or
E =
9π2
ℏ2
2mL

3.3 Solutions 193
3.44 (a)
Fig. 3.18 Penetration of a
rectangular barrier
(b) Region 1, x 0
d2
ψ
dx2
+ k2
ψ = 0
with k2
= 2mE
2
ψ1 = Aeikx
+ Be−ikx
Incident reflected at x = 0
Region 2, 0 x L
d2
ψ
dx2
− α2
ψ = 0
with α2
= 2m(W−E)
2
ψ2 = Ce−αx
+ Deαx
Region 3, x L
d2
ψ
dx2
+ k2
ψ = 0
with k2
= 2mE/2
ψ3 = Feikx
The second term is absent as there is no reflected wave coming from
right to left
The transmission coefficient T = |F|2
|A|2
(c) Boundary conditions
ψ1(0) = ψ2(0)
dψ1
dx
%
%
%
%
x=0
=
dψ2
dx
%
%
%
%
x=0
ψ2(L) = ψ3(L)
dψ2
dx
%
%
%
%
x=L
=
dψ3
dx
%
%
%
%
x=L
(d) T = 16
E
W

1 − E
W

e−2αL
α2
= 2m

W − E
ℏ2

→ α =

2mc2(W − E)
ℏc
=

(2 × 0.511 × (5 − 2) × 10−6
197.3 × 10−15
= 8.8748 × 109
m−1

Therefore 2αL = 2 × 8.8748 × 109
× 0.3 × 10−9
= 5.3249
T = 16

2
5

1 −
2
5

e−5.3249
= 0.0187
(e) Examples of quantum mechanical tunneling
(i) α-decay Observed α-energy may be ∼ 5 MeV although the Coulomb
barrier height is 20 or 30 MeV
(ii) Tunnel diode
(iii) Josephson effect In superconductivity electron emission in pairs
through insulator is possible via tunneling mechanism
(iv) Inversion spectral line in ammonia molecule. This arises due to tun-
neling through the potential barrier between two equilibrium posi-
tions of the nitrogen atom along the axis of the pyramid molecule
which is perpendicular to the plane of the hydrogen atoms. The oscil-
lation between the two equilibrium positions causes an intense spec-
tral line in the microwave region.
3.45 The wave function to the zeroeth order in infinitely deep 2-D potential well is
obtained by the method of separation of variables; the Schrodinger equation is
−
ℏ2
2m
∂2
ψ(x, y)
∂x2
−
ℏ2
2m
∂2
ψ(x, y)
∂y2
= Eψ(x, y)
Let ψ(x, y) = ψx ψy
−
ℏ2
2m
ψy
∂2
ψx
∂x2
−
ℏ2
2m
ψx
∂2
ψy
∂y2
= Eψx ψy
Divide through by ψx ψy
−
ℏ2
2m
1
ψx
∂2
ψx
∂x2
−
ℏ2
2m
1
ψy
∂2
ψy
∂y2
= E
−
ℏ2
2m
1
ψx
∂2
ψx
∂x2
− E =
ℏ2
2m
1
ψy
∂2
ψy
∂y2
= A = constant
∂2
ψx
∂x2
+ α2
ψx = 0
where α2
=
2m
2

(E + A)
ψx = C sin αx + D cos αx
ψx = 0 at x = 0
This gives D = 0
ψx = C sin αx
ψx = 0 at x = a
This gives αa = n1π or α = n1π
a
Thus ψx = C sin(n1πx/a)
Further
∂2
ψy
∂y2 =
2mAψy
2 = −β2
ψy
The negative sign on the RHS is necessary, otherwise the ψy will have an
exponential form which will be unphysical.
ψy = G sin βy

3.3 Solutions 195
When the boundary conditions are imposed,
β =
n2πy
a
ψy = G sin
n2πz
a

Thus ψ(x, y) = ψx ψy = K sin
n1πx
a

sin
n2πy
a

(K = constant)
and α2
+ β2
= 2mE/2
=
n1π
a
2
+
n2π
a
2
or
E =

ℏ2
π2
2ma2

n2
1 + n2
2

3.46 By Problem 3.39, E = h2
8ma2

n2
x + n2
y, n2
z

Therefore the number N of states whose energy is equal to or less than E is
given by the condition
n2
x + n2
y + n2
z ≤
8ma2
E
h2
The required number, N =

n2
x + n2
y + n2
z
1/2
, is numerically equal to the
volume in the first quadrant of a sphere of radius

8 m a2 E
h2
1/2
. Therefore
N =

1
8

·

4π
3

8ma2
E
ℏ2
3/2
=
2π
3

ma2
E
2ℏ2π2
3/2
3.47 Schrodinger’s radial equation for spherical symmetry and V = 0 is
d2
ψ(r)
dr2
+
2
r
dψ(r)
dr
+
2mE
ℏ2
ψ(r) = 0
Take the origin at the centre of the sphere. With the change of variable,
ψ =
u(r)
r
(1)
The above equation simplifies to
d2
u
dr2
+
2mEu
ℏ2
= 0
The solution is
u(r) = A sin kr + B cos kr
where k2
=
2mE
ℏ2
(2)
Boundary condition is: u(0) = 0, because ψ(r) must be finite at r = 0. This
gives B = 0
Therefore,
u(r) = A sin kr (3)
Further ψ(R) = u(r)
R
= 0
Sin kR = 0

or
kR = nπ → k =
nπ
R
(4)
Complete unnormalized solution is
u(r) = A sin
nπr
R

(5)
The normalization constant A is obtained from
R
0
|ψ(r)|2
· 4πr2
dr = 1 (6)
Using (1) and (5), we get
A =
1
√
2π R
(7)
The normalized solution is then
u(r) = (2π R)− 1
2 sin
nπr
R

(8)
From (2) and (4)
En =
π2
n2
ℏ2
2mR2
(9)
For ground state n = 1. Hence
E1 =
π2
ℏ2
2mR2
(10)
The force exerted by the particle on the walls is
F = −
∂V
∂ R
= −
∂ H
∂ R
= −
∂E1
∂ R
=
π2
ℏ2
mR3
The pressure exerted on the walls is
P =
F
4π R2
=
πℏ2
4mR5
3.48 The quantity π2
2
8m
= π2
2
c2
8mc2 = π2
(197.3)2
8×2,200 mcc2
=
π2
(197.3)2
8 × 2, 200 × 0.511
= 42.719 MeV − fm2
Now V0a2
= 70 × (1.42)2
= 141.148 MeV − fm2
It is seen that
π2
ℏ2
8m
V0a2

4π2
ℏ2
8m
(42.7 141 169)
From the results of Problem (3.25) there will be two energy levels, one belong-
ing to class I function and the other to class II function.
The particle of mass 2,200 me or 1,124 Mev/c2
is probably Λ-hyperon
(mass 1,116 MeV/c2
) which is sometimes trapped in a nucleus, to form a
hypernucleus before it decays(Chap. 10).

3.3 Solutions 197
Fig. 3.19 Class I and Class II wave functions
3.49 The analysis for the reflection and transmission of stream of particles from
the square well potential is similar to that for a barrier (Problem 3.30) except
that the potential Vb must be replaced by −V0 and in the region 2, k2 must be
replaced by ik2. Thus, from Eq. (9) of Problem 3.30, we get
τ =
4k1k2 exp(−ik1a)
(k2 + k1)2 exp(−ik2a) − (k2 − k1)2 exp(ik2a)
The fraction of transmitted particles when k2a = nπ is determined by the
imaginary exponential terms in the denominator.
e+inπ
= cos nπ ± i sin (nπ) = cos nπ
= 1; (n = 0, 2, 4·)
= −1; (n = 1, 3, 5 . . .)
Fig. 3.20 Transmission
coefficient T as a function of
the ratio E/Vo for attractive
square well potential
Therefore τ ∗
= 1
A typical graph for T as a function of E/V0 is shown in Fig. 3.20.
In general we get the transmission coefficient
T =
%
%
%
%
%
1 +
V 2
0 sin2
k2a
4E(E + V0)
%
%
%
%
%
−1
The transmission coefficient goes to zero at E = 0 because of the 1/E term
in the denominator. For E/V0 ≪ 1, narrow transmission bands occur when-

ever the condition k2a = nπ is satisfied, the width of the maxima becomes
broader as the electron’s incident energy increases. The occurrence of nearly
complete transparency to incident electrons in the atoms of noble gases is
known as the Ramsuer-Townsend effect.
The condition k2a = nπ implies that a = nλ/2, n = 1, 2 . . ., that is
whenever the barrier contains an integral number of half wavelengths leading
to complete transparency. Interference phenomenon of this type is analogous
to the transmission of light in optical layers.
3.50 (a) particles which can not be observed are called virtual particles
A 4-vector momemtum is p = (p, i E) so that
(4 − momentum)2
= (3 momentum
Space
)2
− (energy)2
time
The components p1,2,3 are said to be spacelike and the energy compo-
nent E, timelike. If q denotes the 4-momentum transfer in a reaction i.e.
q = p−p′
, where p, p′
are initial and 4-momenta, then
q2
0 is spacelike as in the scattering process
q2
0 is timelike as for (mass)2
of free particle
q2
= 0 is lightlike
(b) The relativistic relationship between total energy, momentum and mass for
the field quantum is
E2
− p2
c2
− m2
c4
= 0 (1)
We can now use the quantum mechanical operators
E → iℏ
∂
∂t
and p → −iℏ∇
To transform (1) into an operator equation
−ℏ2 ∂2
∂t2
+ ℏ2
∇2
− m2
c4
= 0
representing the force between nucleons by a potential ϕ(r, t) which may
be regarded as a field variable, we can write

∇2
−
1
c2
∂2
∂t2
−
m2
c2
ℏ2

ϕ = 0
as the wave equation describes the propagation of spinless particles in
free space.
The time independent part of the equation is

∇2
−
m2
c2
ℏ2

ϕ = 0 (2)
For m = 0, this equation is the same as that obeyed in electromagnetism,
for a point charge at the origin, the appropriate solution being
ϕ(r) =

e
4πε0

1
r
(3)
where ε0 is the permittivity.

3.3 Solutions 199
When m = 0, (2) can be written as

∂2
∂r2
+

2
r
∂
∂r
−
m2
c2
ℏ2

ϕ(r) = 0
or
1
r2
∂
∂r

r2 ∂ϕ
∂r

=
m2
c2
ϕ
ℏ2
(4)
For values of r 0 from a point source at the origin, r = 0. Integra-
tion gives
ϕ(r) =
ge
r
R
4πr
(5)
where R = ℏ/mc (6)
The quantity g plays the same role as charge in electrostatistics and mea-
sures the “strong nuclear charge”.
3.3.4 Simple Harmonic Oscillator
3.51 By substituting ψ(R) = AH(R) exp (−R2
/2) in the dimensionless form of
the equation and simplifying we easily get the Hermite’s equation
The problem is solved by the series method
H = ΣHn(R) = Σn=0,2,4an Rn
dH
dR
= annRn−1
d2
H
dR2
= Σn(n − 1)an Rn−2
Σn(n − 1)an Rn−2
− 2ΣannRn
+ (ε − 1)Σan Rn
= 0
Equating equal power of Rn
an+2 =
[2n − (ε − 1)] an
(n + 1)(n + 2)
If the series is to terminate for some value of n then
2n − (ε − 1) = 0 becuase an = 0. This gives ε = 2n + 1
Thus ε is a simple function of n
E = εE0 = (2n + 1)1
/2ℏω, n = 0, 2, 4, . . .
=1
/2ℏω, 3ℏω/2, 5ℏω/2, . . .
Thus energy levels are equally spaced.

3.52 u0 =
*
α
√
π
+
e−ξ2
/2
H0(ξ); ξ = αx
P = 1 −
a
−a
|u0|2
dx = 1 − 2
a
0
(α/
√
π)e−ξ2
/2
dx
= 1 −
2
√
π
aα
0
e−ξ2
dξ
E0 =1
/2ka2
=
ℏω
2
(n = 0)
Therefore a2
= ω
k
=

k
k
m
1/2
=
√
km
= 1
α2
Therefore α2
a2
= 1 or αa = 1
P = 1 −
2
√
π
1
0
e−ξ2
dξ
= 1 −
2
√
π

1 − ξ2
+
ξ4
2!
−
ξ6
3!
+
ξ8
4!

dξ
= 1 −
2
√
π

1 −
1
3
+
1
10
−
1
42
+
1
216
. . .

≈ 0.16
Therefore, p ≈ 16%
Fig. 3.21 Probability of the
particle found outside the
classical limits is shown
shaded
3.53 The potential is of the form V (r) = −V0 + γ 2
r2
(1)
Schrodinger’s radial equation is given by,
d2
u
dr2
=

l(l + 1)
r2
+
2μ
ℏ2(V (r) − E)

u (2)
Upon substituting (1) in (2), we obtain
d2
u
dr2
+

2μ
ℏ2(V0 + E − γ 2r2)
−
l(l + 1)
r2

u = 0 (3)
The quantity γ 2
can be expressed in terms of the classical oscillator fre-
quency
γ 2
=
μω2
2
(4)
For r → 0, (3) may be approximated to
d2
u
dr2
−
l(l + 1)u
r2
= 0
The solution of which is,
u(r) = a rl+1
+
b
r
with a and b as constants.

3.3 Solutions 201
The boundary condition that u/r be finite at r = 0 demands that b = 0.
Thus, ψ is proportional to rl
. The probability that a particle be in a spherical
shell of radii r and r + dr for small r, is proportional to r2l+2
dr. The larger l
is, the smaller is the probability that the particle be in the vicinity of the origin.
For the case of collision problems, there is a classical analogy: the larger the
orbital angular momentum the larger the impact parameter.
Thus u(r) ∼ rl+1
(r → 0)
For → ∞, we obtain, as an approximation to differential equation (3), as
d2
u
dr2
−
2μγ 2
r2
u
ℏ2
= 0
If we try a solution of the form,
u(r) = u0e−Br2
/2
the asymptotically valid solution is satisfied provided we change
B =
γ (2μ)
1
2
ℏ
=
μω
ℏ
Inorder to solve (3) for all r, we may first separate the asymptotic behaviour
by writing
u(r) = rl+1
eBr2
/2
V (r) (5)
Insert (5) in (3), and dividing by rl+1
e−Bν2
/2
We get
d2
ν
dr2
+
2dv
dr

l + 1
r

− Br

− Bv

2l + 3 −
2
ℏω
(V0 + E)

Define C = l + 3
2
4A = 2l + 3 −
2
ℏω
(V0 + E) (6)
d2
v
dr2
+
dv
dr

2C − 1
r

− 2Br

− 4ABv = 0 (7)
Set
Fig. 3.22 The parabolic
potential of the three
dimensional harmonic
oscillator

Br2
= x (8)
Substitute (8) in (7) and simplify, to obtain
xd2
v
dx2
+
dv
dx
(C − x) − Av = 0 (9)
which is the familiar confluent hyper geometric equation whose solution
which is regular at x = 0 is;
V (A, C, x) = α0

1 +
Ax
C
+
A(A + 1)x2
C(C + 1)2!
+
A(A + 1)(A + 2)x3
C(C + 1)(C + 2)3!
+ · · · .

V (A, C, Br2
) = α0

1 +
ABr2
C
+
A(A + 1)B2
r4
C(C + 1)2!
+
A(A + 1)(A + 2)B3
r6
C(C + 1)(C + 2)3!
+ · · ·

The asymptotic solution V (r) → 0, while r → ∞ implies that the series
must break off for finite powers of Br2
since α0 = 0. This means that A must
equal a negative integer −p; where p = 0, 1, 2, 3 . . .
Therefore −4p = 2l + 3 − 2
ω
(V0 + E)
Where we have used the definition of A (Eq. 6) from this we find, the energy
eigen values,
Ep,l = −V0 + ℏω

2p + l +
3
2

(p = 0, 1, 2, . . . .)
Setting n = 2p + l
En = −V0 + ω

n + 3
2

(which is different from one-dimensional harmonic
oscillator)
E0 = −V0 + 3ω
2
corresponds to ground state.
It is a single state (not degenerate)
since n = 0 can be formed only by the combination l = 0, p = 0.
3.54 When the oscillator is in the lowest energy state
H = V + T =

mω2
2

x2
+

1
2m

P2

Now, if a, b and c are three real numbers such that a + b = c, then
ab =
c2
4
−

a − b
2
2
or
ab ≤
c2
4
Apply this inequality to mω2
2

x2
,
1
2m

p2
and ω
2
Δx 2
= x2
− x 2
= x2
and x = 0

3.3 Solutions 203
Similarly Δpx 2
= p2

1
/2mω2
x2
·

1
2m

px 2
≤
1
4

ℏω
2
2
,
x2
p2
x
-1/2
≤
ℏ
2
or Δx.Δ px ≤
2
Now compare this result with the uncertainty principle
Δx · Δpx ≤
ℏ
2
We conclude that Δx.Δpx ≥
2
. Obviously the zero point energy could not
have been lower than ω
2
without violating the uncertainty principle.
3.55 The probability distribution for the quantum mechanical simple harmonic
oscillator (S.H.O) is
P(x) = |ψn|2
=
α exp(−ξ2
)H2
n (ξ)
√
π2nn!
(1)
ξ = αx; α4
= mk/ℏ2
Stirling approximation gives
n! → (2nπ)1/2
nn
e−n
(2)
Furthermore the asymptotic expression for Hermite function is
Hn(ξ)(for n → ∞) → 2n+1 (n/2e)
n
2
√
2 cos β
exp(nβ2
) cos
*
(2n +1
/2)β −
nπ
2
+
(3)
where sin β = ξ/
√
2n (4)
Using (2) and (3) in (1)
P(x) → 2α exp(−ξ2
) exp(2nβ2
)
cos2
,
2n + 1
2

β − nπ
2
-
π
√
2n cos β
But cos2
,
2n + 1
2

β − nπ
2
-
= 1
2
Therefore P(x) =
α exp(−ξ2
) exp(2nβ2
)
π
√
2n cos β
(5)
Fig. 3.23 Probability distribution of quantum mechanical oscillator and classical oscillator

Classically, E = ka2
2
=

n + 1
2

ω (quantum mechanically) ≈ nω(n → ∞)
Therefore a2
= 2nω
k
=
2n
k
k
m
1/2
= 2n
√
km
= 2n
α2
ω =
k
m
or
a =
√
2n
α
(6)
Sin β =
ξ
√
2n
=
αx
√
2n
=
x
a
Therefore
cos β =
(a2
− x2
)
1
2
a
(7)
P(x) =
exp(−ξ2
exp(2nβ2
))
π(a2 − x2)1/2
(8)
Now when n → ∞, sin β → β and
β → ξ/
√
2n, and exp(−ξ2
) exp(2nβ2
) → 1)
Therefore P(x) = 1
π
√
a2−x2
(classical)
3.56 One can expect the probability of finding the particle of mass m at distance x
from the equilibrium position to be inversely proportional to the velocity
P(x) =
A
v
(1)
where A =normalization constant. The equation for S.H.O. is
d2
x
dt2
+ ω2
x = 0
x = a sin ωt; (at t = 0, x = 0)
where a is the amplitude.
v =
dx
dt
= ω

a2 − x2 (2)
Using (2) in (1)
P(x) = A/ω

a2 − x2 (3)
We can find the normalization constant A.

P(x)dx =
a
−a
Adx
ω
√
a2 − x2
=
Aπ
ω
= 1
Therefore,
A =
ω
π
(4)
Using (4) in (3), the normalized distribution is

3.3 Solutions 205
P(x) =
1
π
√
a2 − x2
(5)
3.57 Schrodinger’s equation in one dimension is

−
ℏ2
2m

d2
ψ
dx2
+ V (x)ψ = Eψ (1)
Given
ψ = exp(−
1
2
ax2
) (2)
Differentiating twice,
we get
d2
ψ
dx2
exp(−
1
2
ax2
)(a2
x2
− a) (3)
Inserting (2) and (3) in (1), we get
V (x) = E +

ℏ2
2m

(a2
x2
− a) (4)
Minimum value of V (x) is determined from
dV
dx
=
ℏ2
a2
x
m
= 0
Minimum of V (x) occurs at x = 0
From (4) we find 0 = E − 2
a
2m
(a) Or the eigen value E = 2
a
2m
(b) V (x) = 2
a
2m
+ 2
2m

(a2
x2
− a) = 2
a2
x2
2m
3.58
V n=
En
2

P2
2m
n= H n − V n=1
/2 En
P2
n= mEn
Also x n= 0; P n= 0
(Δx)2
= x2
n − x2
n

ΔP2

= P2
n − Pn 2
= P2
n= mEn
But x2
n=
∞
−∞ u∗
n(x)x2
un(x)dx; ξ = αx

=
N2
n
α3
∞
−∞
H2
n (ξ)ξ2
e−ξ2
dξ
=

α
√
π2nn!

1
α3

(2n + 1)
2
2n
n!
√
π
=
1
α2

n +
1
2

=
ℏ
√
km

n +
1
2

∴ Δx.ΔP =
0
mEn
ℏ
√
km

n +
1
2

Now, ω =
k
m
∴ Δx.ΔPx =
0
ℏω

n +
1
2

ℏ
m
k

n +
1
2

= ℏ

n +
1
2

Thus, Δx.ΔP ≥
2
is in agreement with uncertainty principle.
3.59 The vibrational levels are equally spaced and so with the rule Δn = 1, the lines
should coincide. The rotational levels are progressively further spaced such
that the difference in the wave number of consecutive lines must be constant.
This can be seen as follows:
EJ =
J(J + 1)ℏ2
2I0
ℏc
λJ
= ΔE = EJ+1 − EJ =
[(J + 1) (J + 2) − J(J + 1)] ℏ2
2I0
=
(J + 1)ℏ2
2I0
Therefore 1
λJ+1
− 1
λJ
= Δ 1
λJ

α[(J + 2) − (J + 1)] = constant
103.73 − 83.03 = 20.70 cm−1
; 124.30 − 103.73 = 20.57 cm−1
;
145.03 − 124.30 = 20.73 cm−1
; 165.51 − 145.03
= 20.48 cm−1
; 185.86 − 165.51 = 20.35 cm−1
The data are consistent with a constant difference, the mean value being,
20.556 cm−1
. Thus the transitions are rotational.
EJ+1 EJ
EJ+2 − EJ+1
=
2(J + 1)
2(J + 2)
=
83.03
103.73
= 0.8
Therefore J = 3
The levels are characterized by J = 3, 4, 5, 6, 7, 8, 9

3.3 Solutions 207
ℏc

1
λj+1
−
1
λj

= ℏc × (20.556 cm−1
) =
ℏ2
I0
Moment of inertia I0 =
ℏ
4π2c × 20.556
=
6.63 × 10−27
erg − s−1
4π2 × 3 × 1010 cm − s−1 × 20.556 cm−1
= 2.727 × 10−40
g − cm2
I0 = μr2
μ =
m(H)m(Cl)
[m(H) + m(Cl)]
=
1 × 35 × 1.67
1 + 35
× 10−24
g
= 1.62 × 10−24
g
r =

I0
μ
1/2
=

2.727 × 10−40
1.62 × 10−24

= 1.3 × 10−8
cm = 1.3 Å
3.60 For the 3-D isotropic oscillator the energy levels are given by
EN = Ek + El + Em =

3
2
+ nk + nl + nm

ℏω
where ω is the angular frequency
N = nk + nl + nm = 0, 1, 2 . . .
For a given value of N, various possible combinations of nk, nl and nm are
given in Table 3.5, and the degeneracy indicated.
Table 3.5 Possible combinations of nk, nl and nm and degeneracy of energy levels
N nl nm nn Degeneracy (D)
0 0 0 0 Non-degenerate
1 1 0 0 Three fold (1 + 2)
0 1 0
0 0 1
2 1 1 0 Sixfold (1 + 2 + 3)
1 0 1
0 1 1
2 0 0
0 2 0
0 0 2
3 1 1 1 Tenfold (1 + 2 + 3 + 4)
1 2 0
1 0 2
2 1 0
2 0 1
0 2 1
0 1 2
3 0 0
0 3 0
0 0 3

It is seen from the last column of the table that the degeneracy D is given by
the sum of natural numbers, that is, = n(n + 1)/2, if we replace n by N + 1,
D = (N + 1)(N + 2)/2.
3.61 As the time evolves, the eigen function would be
ψ(x, t) =

n=0.1 Cnψn(x) exp(−i Ent/)
= C0ψ0(x) exp(−i E0t/) + C1ψ1(x) exp(−i E1t/)
The probability density
|ψ(x, t)|2
= C2
0 + C2
1 + C0C1ψ0(x)ψ1(x)[exp(i(E1 − E0)t/)
− exp(−i(E1 − E0)t/)]
= C2
0 + C2
1 + 2C0C1ψ0(x)ψ1(x) cos ω t
where we have used the energy difference E1 − E0 = ω. Thus the probability
density varies with the angular frequency.
3.62 E =

n=1,2,3
|Cn|2
En = C2
0 E0 + C2
1 E1 + C2
2 E2
=

1
2

·
ω
2
+

1
3

·
3ω
2
+

1
6

·
5ω
2
=
7ω
6
.
3.63 (a) ψ0(x) = A exp(−x2
/2a2
)
Differentiate twice and multiply by −2
/2m
−

2
2m

d2
ψ0
dx2
=

A2
2ma2

1 −
x2
a2

exp

−
x2
2a2

=

2
2ma2

ψ0 −

2
x2
2ma4

ψ0
or −

2
2m

d2
ψ0
dx2
+

2
x2
2ma4

ψ0 =

2
2ma2

ψ0
Compare the equation with the Schrodinger equation
E =
2
2ma2
=
ω
2
ω =

ma2
(1)
or a =

mω
1/2
Same relation is obtained by setting
V =
2
x2
2ma4
=
mω2
x2
2
(b) ψ1 = Bx exp − x2
2a2

Differentiate twice and multiply by − 2
2m
−
2
2m
d2
ψ1
dx2
=
B2
x3
exp x2
a

2ma4
+
3B2
exp − x2
2a2

2ma2

3.3 Solutions 209
Fig. 3.24 ψ1(x) for SHO
Substitute
Bx exp(−x2
/2a) = ψ1 and a =

mω
1/2
and rearrange to get
−

2
2m

d2
ψ1
dx2
+

mω2
2

ψ1 =

3ω
2

ψ1
(c) E = 3ω
2
(d) px =

ψ∗
1 −i∂ψ1
∂x

dx
= −iB2
∞
−∞ x exp −x2
a2

1 − x2
a2

dx
= zero (because integration over an odd function between symmetrical lim-
its is zero). This result is expected because half of the time the particle will
be pointing along positive direction and for the half of time in the negative
direction.
(e) The results of SHO are valuable for the analysis of vibrational spectra of
diatomic molecules, identification of unknown molecules, estimation of
force constants etc.
3.3.5 Hydrogen Atom
3.64 V = −e2
r
=

ψ∗
V ψdτ = − e2
πa3
0
∞

0
(exp −2r
a0

/r)4πr2
dr
where e = charge of electron
= −
4e2
a3
0
∞
0
exp

−
2r
a0

rdr = −
e2
a0
Therefore V = −e2
a0
T =

ψ∗

−
2
2m

∇2
ψ

dτ (1)
In polar coordinates (independent of θ and ϕ);

∇2
=
d2
dr2
+

2
r

d
dr
(2)
Inserting (2) in (1) and performing the integration we get
T =
2
2ma2
0
But a0 = 2
me2
Or 2
m
= a0e2
Therefore, T = e2
2a0
Also E = T + V = e2
2a0
− e2
a0
= −e2
/2a0
3.65 The normalized eigen function for the ground state of hydrogen atom is
ψ0 = 1/

πa3
0
1
2
e−r/a0
where a0 is the Bohr’s radius
(a) The probability that the electron will be formed in the volume element dτ
is
p(r)dr = |ψ0|2
dτ =

e
− 2r
a0
πa3
0
#
.4πr2
dr
= 4
a3
0

r2
e−2r/a0
dr
Maximum probability is found by setting dp
dr
= 0
d
dr

4r2
e
− 2r
a0
a3
0
#
=

8r
a3
0

e
− 2r
a0

1 −
r
a0

= 0
Therefore r = a0
(b) r =
∞
0
ψ∗
rψdτ =
1
π
a3
0
∞
0
r exp

−
2r
a0

.4πr2
dr
=

4
a3
0
∞

0
r3
exp

−
2r
a0

dr
Let
2r
a0
= x; dr =
a0dx
2
r =
a0
4
∞

0
x3
e−x
dx =
a0
4

3! =
3a0
2
3.66

u2
210 dτ = A2
2

e−2x
x2
cos2
θr2
sin θ dθ dr dϕ
=

1
π

1
2a0
3 ∞
0
exp

−
r
a0

r2
4a2
0

r2
dr
+1
−1
cos2
θ d(cos θ)
2π
0
dϕ
where we have put x = r/2a0. Put r/a0 = y, dr = a0dy

3.3 Solutions 211
u2
210 dτ =

1
8πa3
0
∞
0
y4
e−y

dy
4a2
0

a5
0

cos3
θ
3
1
−1
(2π)
=

1
24

× 4! = 1
Similarly u2
21±1dτ = 1
π(2a0)3
e−2x
2
x2
sin2
θ e±iϕ
e∓iϕ
r2
sin θdθdϕdr
=

1
8πa3
0
∞
0

e
− r
a0
2
#
r4
dr
4a2
0
+1
−1
(1 − cos2
θ) d(cos θ)
2π
0
dϕ
=

a5
0
64πa5
0
∞
0
y4
e−y
dy

4
3

(2 π)
=

1
192

(4!)(8) = 1
3.67

u21±1 u210 dr =
A2
2
√
2

e−x
x cos θ e−x
x sin θ e+ϕ
r2
sin θ dθdϕdr
The integral
2π
0 e±iϕ
dϕ = 0
Therefore u21±1 and u210 are orthogonal.
Further the integral

u∗
211 U21−1dτ involves the integral
2π
0
e−iϕ
e−iϕ
dϕ or
2π
0
e−2iϕ
dϕ = 0
So, the functions u211 and u21−1 are also orthogonal.
3.68 The degree of degenerating is given by 2n2
. So for n = 1, degenerary is 2, for
n = 2 it is 8, for n = 3, it is 18 and for n = 4, it is 32.
3.69 Parity of the state is determined by the factor (−1)l
. For 1s, l = 0, parity= +1,
for 2p, l = 1, parity = − 1 and for 3d, l = 2, parity = +1.
3.70 To show that the probability density of the 3d state is independent of the polar
angle θ. We form
u∗
u = u∗
(3, 2, 0)u(3, 2, 0) + 2 u∗
(3, 2, 1)u(3, 2, 1) + 2 u∗
(3, 2, 2)
u(3, 2, 2)
The factor 2 takes care of m values ±1 and ±2, as in Table 3.2. Inserting the
functions the azimuth part, eiϕ
or e−iϕ
drop off when we multiply with the
complex conjugate,
i.e.

eiϕ
∗
eiϕ
= 1 or (e−iϕ
)∗
(e−iϕ
) = 1
u∗
u = A2
3e−2x
x4

1
18

(3 cos2
θ − 1)2
+

2
3

sin2
θ cos2
θ +

1
6

sin4
θ

Writing sin2
θ = 1−cos2
θ and simplifying we get u∗
u = 2
9
A2
3e−2x
x4
which
is independent of both θ and ϕ. Therefore the 3d functions are spherically
symmetrical or isotropic.
3.71 ψ100 =

πa3
0
− 1
2
exp − r
a0

The probability p of finding the electron within a sphere of radius R is
P =
R
0
|ψ100|2
.4πr2
dr =

4π
πa3
0
R
0
r2
exp

−
2r
a0

dr
Set 2r
a0
= x; dr =
a0
2

dx
P =

4
a3
0

·

a2
0
4

·
a0
2

x2
e−x
dx = 1
/2

x2
e−x
dx
Integrating by parts
P = 1
/2[−x2
e−x
+ 2

xe−x
dx]
= 1
/2

−x2
e−x
+ 2

−x e−x
+

e−x
dx

= 1
/2
,
−x2
e−x
− 2xe−x
− 2e−x
-2R/a0
0
= 1
/2

−

2R
a0
2
exp

−
2R
a0

− 2

2R
a0

exp

−
2R
a0

−2 exp

−
2R
a0

+ 2

P = 1 − e
− 2R
a0

1 +
2R
a0
+
2R2
a2
0

3.72 The hydrogen wave function for n = 2 orbit is
ψ200 =

1
4

2πa3
0
− 1
2

2 −
r
a0

e−r/2a0
The probability of finding the electron at a distance r from the nucleus
P = |ψ200|2
· 4πr2
=
1
8
r2
a3
0

2 −
r
a0
2
exp

−
r
a0

Maxima are obtained from the condition dp/dr = 0.
The maxima occur at r = (3 −
√
3)a0 and r = (3 +
√
3)a0 while minimum
occur at r = 0, 2a0 and ∞ (Fig. 3.25) (the minima are found from the condi-
tion p = 0).
3.73 (a) hν = 13.6 Z2 mμ
me
1
22 − 1
32

Put mμ = 106 MeV (instead of 106.7 MeV for muon)
Z = 15 and me = 0.511 MeV

3.3 Solutions 213
Fig. 3.25 Probability
distribution of electron in
n = 2 orbit of H-atom
hν = 0.088 MeV
λ = 1241
8.8×104 nm = 0.0141 nm
(b) The transition probability per unit time is
A ∝ ω3
|rkk′ |2
For hydrogen-like atoms, such as the meisc atom
|rkk′ | ∝
1
z
, ω ∝ mμ Z2
, so that
A ∝ m3
μ Z4
The mean life time of the mesic atom in the 3d state is
τμ =

AH
Aμ

τH =

me
mμ
3
τH /Z4
=

0.511
106
3

1.6 × 10−8

1
154

= 3.5 × 10−20
s.
3.74 Take k along the z-axis so that p.r/ = k.r = kr cos θ. Write
dτ = r2
drd(cos θ)dϕ
ψ(p) =
1
(2πℏ)
3
2
∞
r=0
π
θ=0
2π
ϕ=0
e−ikr cos θ

πa3
0
− 1
2
exp

−
r
a0

r2
drd (cos θ) dϕ
(1)
2π
0
dϕ = 2π (2)
+1
−1
e−ikr cos θ
d (cos θ) =

2
kr

sin kr (3)
With the aid of (2) and (3), (1) becomes
ψ(p) =
√
2
πk
#
(ℏa0)− 3
2
∞
0
r sin kr exp

−
r
a0

dr (4)
Now
∞
0 r sin kr e−br
= −∂
∂b
∞
0 sin kr e−br

= −
∂
∂b

k
b2 + k2

=
2kb
(b2 + k2)2
Therefore the integral in (4) is evaluated as

2k
a0
3
1
a2
0

+ k2

(5)
Using the result (5) in (4), putting k = p/, and rearranging, we get
ψ(p) =

2
√
2
π
#
ℏ
a0
5
2
4
p2
+

ℏ
a0
2
'2
or
|ψ(p)|2
=
8
π2
(ℏ/a0)5
[p2 + (ℏ/a0)2
]4
(6)
3.75 (a) |ψ(p)|2
=
8
π2

ℏ
a0
5
.4 πp2
4
p2
+

ℏ
a0
2
'4
(1)
Maximize (1)
d
dp
|ψ(p)|2
= 0
This gives Pmost probable = /
√
3 a0
(b) p =
∞
0
ψ∗
p pψp.4πp2
dp
=

32
π

ℏ
a0
5 ∞
0
p3
dp
4
p2
+

ℏ
a0
2
'4
The integral I1 is easily evaluated by the change of variable
p =
a0

tan θ. Then
I1 =

1
8

a0
ℏ
4
π/2
0
sin3
2θdθ =

1
12

a0
ℏ
4
Thus p = 8ℏ
3πa0
3.76 By Problem 3.71, the probability that
P

r
a0

= 1 − exp

−
2r
a0

1 +
2r
a0
+
2r2
a2
0

Put p r
a0

= 0.5 and solve the above equation numerically (see Chap. 1). We
get r = 1.337a0, with an error of 2 parts in 105
.

3.3 Solutions 215
3.77 L =
%
%
%
%
%
%
i j k
x y z
px py pz
%
%
%
%
%
%
= i(ypz − zpy) − j(xpz − zpx ) + k(xpy − ypx )
= i Lx + j Ly + kLz
Thus, Lx = ypz − zpy, Ly = zpx − xpz, Lz = xpy − ypx (1)
[Lx , Ly] = Lx Ly − Ly Lx
= (ypz − zpy)(zpx − xpz) − (zpx − xpz)(ypz − zpy)
= ypz zpx − ypz xpz − zpyzpx + zpy xpz − zpx ypz + zpx zpy
+ xpz ypz − xpzzpy (2)
But [px , py] = [x, py] = 0, etc (3)
(2) becomes
[Lx , Ly] = ypx pzz − yxp2
z − z2
py px + xpyzpy − ypx zpz + z2
px py
+ yxp2
z − xpy pz z = [z, pz](xpy − ypz) (4)
But [z, pz] = [z, −i
∂
∂z
] = −i

z,
∂
∂z

(5)
Further,

z,
∂
∂z

= −1 (6)
So
[z, pz ] = i (7)
Combining (1), (4) and (7) we get
[Lx , Ly] = iLz (8)
3.78 Given spin state is a singlet state, that is S = 0
S1 + S2 = S
Form scalar product by itself
S1 · S1 + S1 · S2 + S2 · S1 + S2 · S2 = S · S
S1
2
+ 2 S1 · S2 + S2
2
= S2
= 0
Now, S1
2
= S2
2
= (1/2)(1/2 + 1) = 3/4
Therefore S1 · S2 = −(3/4) 2
3.79 For the n – p system
Sp + Sn = S
and Sp
2
= Sn
2
= s(s + 1) with s = 1/2
(i) For singlet state, S = 0

∴ (Sp + Sn) · (Sp + Sn) = S · S
Sp
2
+ Sn
2
+ 2 Sp · Sn = S2
= 0
1/2(1/2 + 1) + 1/2(1/2 + 1) + 2 Sp · Sn = 0
Or Sp · Sn = −3/4. Or σp · σn = −3
(ii) For triplet state S = 1
3/4 + 3/4 + 2 Sp · Sn = 1(1 + 1)
∴ Sp · Sn = 1/4
But Sp = 1/2σp and Sn = 1/2 σn
∴ σp · σn = 1
3.80 From the definition of angular momentum
L = r × p, we can write
L =
%
%
%
%
%
%
i j k
x y z
px py pz
%
%
%
%
%
%
= i(ypz − zpy) + j(zpx − xpz)
+ k(xpy − ypx )
= iLx + jLy + kLz
Fig. 3.26 Cartesian and polar
coordinates
Lx = ypz − zpy = −i

y
∂
∂z
− z
∂
∂y

Ly = zpx − xpz = −i

z
∂
∂x
− x
∂
∂z

(1)
Lz = xpy − ypx = −i

x
∂
∂y
− y
∂
∂x

If θ is the polar angle, ϕ the azimuthal angle and r the radial distance,
(Fig. 3.26). Then

3.3 Solutions 217
x = r sin θ cos ϕ
y = r sin θ sin ϕ (2)
z = r cos θ
r2
= x2
+ y2
+ z2
(3)
tan2
θ =
x2
+ y2
z2
(4)
tan ϕ =
y
x
(5)
Differentiating (3), (4) and (5) partially with respect to x
∂r
∂x
= sin θ cos ϕ;
∂r
∂y
= sin θ sin ϕ;
∂r
∂z
= cos θ (6)
∂θ
∂x
=

1
r

cos θ cos ϕ;
∂θ
∂y
=

1
r

cos θ sin ϕ;
∂θ
∂z
= −
sin θ
r
(7)
∂ϕ
∂x
= −

1
r

cosecθ sin ϕ;
∂ϕ
∂y
=
cos ϕ
r sin θ
;
∂ϕ
∂z
= 0 (8)
Lzψ(r, θϕ) = −i

x∂ψ
∂y
−
y∂ψ
∂x

= −i

x

∂ψ
∂r

·

∂r
∂y

+

∂ψ
∂θ

·

∂θ
∂y

+

∂ψ
∂ϕ

·

∂ϕ
∂y

−y

∂ψ
∂r

·

∂r
∂x

+

∂ψ
∂θ

·

∂θ
∂x

+

∂ψ
∂ϕ

·

∂ϕ
∂x

Lzψ(r, θ, ϕ)
= −i

∂ψ
∂r

x
∂r
∂y
− y
∂r
∂x

+
∂ψ
∂θ

x
∂θ
∂y
− y
∂θ
∂x

+
∂ψ
∂ϕ

x
∂ϕ
∂y
− y
∂ϕ
∂x

(9)
Substituting (2), (6), (7) and (8) in (9) and simplifying, the first two terms drop
off and the third one reduces to ∂ψ/∂ϕ, yielding
Lz = −i
∂
∂ϕ
(10)
In Problem 3.15 it was shown that the Schrodinger equation was separated
into radial (r) and angular parts (θ and ϕ). The angular part was shown to be
separated into θ and ϕ components. The solution to ϕ was shown to be
g(ϕ) =
1
√
2π
eimϕ
where m is an integer.

Now Lz g(ϕ) = −i∂g
∂ϕ
= mg(ϕ)
Thus the z-component of angular momentum is quantized with eigen
value .
3.81 One can do similar calculations for Lx and Ly as in Problem 3.80 and obtain
Lx
i
= sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ
Ly
i
= − cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ
3.82 Using L2
= L2
x + L2
y + L2
z , the commutator with total angular momentum
squared can be evaluated
[L2
, Lz] =
,
L2
x + L2
y + L2
z , Lz
-
=
,
L2
x + L2
y, Lz
-
= Lx [Lx , Lz] + [Lx , Lz] Lx + Ly
,
Ly, Lz
-
+
,
Ly, Lz
-
Ly (1)
= −iLx Ly − iLy Lx + iLy Lx + iLx Ly = 0
Similarly
,
L2
, Lx
-
= [L2
, Ly] = [L2
, L] = 0
3.83 L2
= L2
x + L2
y + L2
z
Using the expressions for Lx , Ly and Lz from problem (3.81)
L2
(i)2
=

sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ

sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ

+

− cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ

− cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ

+

−
∂
∂ϕ

−
∂
∂ϕ

= sin2
ϕ
∂2
∂θ2
+ cot2
θ cos2
ϕ
∂2
∂ϕ2
− sin ϕ cos ϕ cosec2
θ
∂
∂ϕ
+ cos2
ϕ cot θ
∂
∂θ
+ cos2
ϕ
∂2
∂θ2
+ cot2
θ sin2
ϕ
∂2
∂ϕ2
+ sin ϕ cos ϕcosec2
θ
∂
∂ϕ
+ sin2
ϕ cot θ
∂
∂θ
+
∂2
∂ϕ2
The cross terms get cancelled and the expression is reduced to
∂2
∂θ2
+
1
sin2
θ
∂2
∂ϕ2
+ cot θ
∂
∂θ
∇2
ψ =
1
r2
∂
∂r

r2 ∂ψ
∂r

+
1
r2 sin θ
∂
∂θ

sin θ
∂ψ
∂θ

+
1
r2 sin2
θ
∂2
ψ
∂ϕ2

3.3 Solutions 219
Apart from the factor 1/r2
, the angular part is seen to be
∂2
ψ
∂θ2
+ cot θ
∂ψ
∂θ
+
1
sin2
∂2
ψ
∂ϕ2
3.84 (a) The (i, j)th matrix element of an operator O is defined by
Oij = i|o| j (1)
For j = 1/2, m = 1/2 and −1/2. The two states are
|1 = |
1
2
,
1
2
and |2 = |
1
2
, −
1
2
(2)
With the notation | j, m
Now
j′
m′
|Jz|jm = mδjj ′ δmm′ (3)
Thus
(Jz)11 = 1|Jz|1 =
1
2
(4)
(Jz)22 = 2|Jz|2 = −
1
2
(5)
(Jz)12 = 1|Jz|2 =
5
1
2
,
1
2
|Jz|
1
2
,
1
2
6
= 0 (6)
because of (3).
Similarly
(Jz)21 = 0 (7)
Therefore
Jz =

2

1 0
0 −1

(8)
For Jx and Jy, we use the relations
Jx = 1
/2(J+ + J−) and Jy = −

1
2i

(J+ − J−)
j, m|Jx | j, m = j, m|
1
2
(J+ + J−)| j, m
= 1
/2[( j + m + 1)( j − m)]1/2
j, m′
| j, m + 1
+ 1
/2[( j − m + 1)( j + m)]1/2
j, m′
| j, m − 1
= 1
/2[( j + m + 1)( j − m)]1/2
δm′,m+1
+ 1
/2[( j − m + 1)( j + m)]
1
/2
δm′,m−1
That is the matrix element is zero unless m′
= m + 1 or m′
= m − 1.
The first delta factor survives

(Jx )12 = 1|Jx |2 =
5
1
2
,
1
2
|Jx |
1
2
, −
1
2
6
=
1
2

1
2
−
1
2
+ 1

1
2
+
1
2
1/2
= 1
/2
(Jx )21 =
5
1
2
, −
1
2
|J|
1
2
,
1
2
6
= 1
/2
because the second delta factor survives
(Jx )11 = 1|Jx |1 =
71
2
, 1
2
|Jx |1
2
, 1
2
8
= 0 because of delta factors.
Similarly, (Jx )22 = 0
Thus Jx =
2

0 1
1 0

; Jy =
2

0 −i
i 0

Jz =

2

1 0
0 −1

(9)
These three matrices are known as Pauli matrices.
(b) J2
= J2
x + J2
y + J2
z
Using the matrices given in (6), squaring them and adding we get
J2
= 3
/42

1 0
0 1

3.85 (a) For j = 1, m = 1, 0 and −1, the three base states are denoted by
|1 , |2 and |3 . In the | j, m notation |1 = |1, 1 , |2 =
|1, 0 , |3 = |1, −1
Jz|1 = m|1 = |1
Jz|2 = 0.|2 = 0
Jz|3 = −|3
(Jz)11 = 1|Jz|1 = 1, 1|Jz|1, 1 =
(Jz)22 = 2|Jz|2 = 1, 0|Jz|1, 0 = 0
(Jz)33 = 3|Jz|3 = 1, −1|Jz|1, −1 = −
(Jz)12 = (Jz)21 = (Jz)13 = (Jz)31 = (Jz)23 = (Jz)32 = 0
because of δ – factor δmm′
Jz =
⎛
⎝
1 0 0
0 0 0
0 0 −1
⎞
⎠
For the calculation of Jx and Jy, we need to work out J+ and J−.
J+|1 = 0
J+|2 = [( j − m)( j + m + 1)]1/2
|1

3.3 Solutions 221
= [(1 − 0)(1 + 0 + 1)]1/2
|1 =
√
2|1
J+
%
%
%3 = [[1 − (−1)](1 − 1 + 1)]
1
2
%
%
% 2 =
√
2 |2
J−|1 = [( j + m)( j − m + 1)]1/2
|2
= [(1 + 1)(1 − 1 + 1)]1/2
|2 =
√
2 |2
J−|2 =
√
2 |3
J−|3 = 0
Matrices for Jx and Jy:
(Jx )11 = 1|Jx |1 = 1
/2 1|J+ + J−|1
= 1
/2 1|J+|1 +
1
2
1|J−|1
= 1
/2[( j − m)( j + m + 1)]1/2
δm′m+1
+ 1
/2[( j + m)( j − m + 1)]1/2
δm′,m−1=0
Similarly; (Jx )22 = (Jx )33 = 0
(Jx )12 = 1|Jx |2 = 1/2 1|J+ + J−|2 = 1/2 1, 1|J+ + J−|1, 0
= 1
/2[( j − m)( j + m + 1)]1/2
δm′m+1 + 1/2[( j + m)( j − m + 1)]1/2
δm′m−1
The second delta is zero
∴ (Jx )12 =
1
2
[(1 + 0)(1 + 0 + 1)]1/2
=

√
2
Similarly, (Jx )21 = (Jx )23 = (Jx )32 =
√
2
By a similar procedure the matrix elements of Jy can be found out. Thus
Jx =

√
2
⎛
⎝
0 1 0
1 0 1
0 1 0
⎞
⎠ ; Jy =

√
2
⎛
⎝
0 −i 0
i 0 −i
0 i 0
⎞
⎠ ; Jz =
⎛
⎝
1 0 0
0 0 0
0 0 −1
⎞
⎠
(b) For the matrix elements of J we can use the relation
j′
m′
|J2
|jm = j( j + 1)2
δj′j δm′m
Thus (J2
)11 = (J2
)22 = (J2
)33 = 1(1 + 1)2
= 22
(J2
)12 = (J2
)21 = (J2
)13 = (J2
)31 = (J2
)23 = (J2
)32 = 0
J2
= 22
⎛
⎝
1 0 0
0 1 0
0 0 1
⎞
⎠
Alternatively, J2
= J2
x + J2
y + J2
z
Using the matrices which have been derived the same result is obtained.
3.86 With the addition of j1 = 1 and j2 = 1/2, one can get J = 3/2 or 1/2. In the
(J, M) notation in all one gets 6 states
ψ

3
2
,
3
2

, ψ

3
2
,
1
2

, ψ

3
2
, −
1
2

, ψ

3
2
, −
3
2

and ψ

1
2
,
1
2

, ψ

1
2
, −
1
2

Clearly the states ψ
3
2
, 3
2

and ψ
3
2
, −3
2

can be formed in only one way
ψ

3
2
,
3
2

= ϕ(1, 1)ϕ

1
2
,
1
2

(1)
ψ

3
2
, −
3
2

= ϕ(1, −1)ϕ

1
2
, −
1
2

(2)
We now use the ladder operators J+ and J− to generate the second and the
third states.
J+ϕ( j, m) = [( j − m)( j + m + 1)]
1
2 ϕ( j, m + 1) (3)
J−ϕ( j, m) = [( j + m)( j − m + 1)]
1
2 ϕ( j, m − 1) (4)
Applying (4) to (1) on both sides
J−ψ

3
2
,
3
2

=

3
2
+
3
2

3
2
−
3
2
+ 1
1/2
=
√
3ψ

3
2
,
1
2

= J−ϕ(1, 1)ϕ

1
2
,
1
2

= ϕ(1, 1)J−

1
2
,
1
2

+ ϕ

1
2
,
1
2

J−ϕ(1, 1)
= ϕ(1, 1)ϕ

1
2
, −
1
2

+ ϕ

1
2
,
1
2

√
2ϕ(1, 0)
Thus ψ

3
2
,
1
2

=
2
3
ϕ(1, 0)ϕ

1
2
,
1
2

+
1
3
ϕ(1, 1)ϕ

1
2
, −
1
2

(5)
Similarly, applying J+ operator given by (3) to the state ψ
3
2
, −3
2

we
obtain
ψ

3
2
, −
1
2

=
2
3
ϕ(1, 0)ϕ

1
2
, −
1
2

+
1
3
ϕ(1, −1)ϕ

1
2
,
1
2

(6)
The J = 1/2 state can be obtained by making it as a linear combination
ψ

1
2
,
1
2

= aϕ(1, 1)ϕ

1
2
, −
1
2

+ bϕ(1, 0)ϕ

1
2
,
1
2

(7)
For normalization reason,
a2
+ b2
= 1 (8)
We can obtain one other relation by making (7) orthogonal to (5)
a
1
3
+ b
2
3
= 0
Or a = −
√
2b (9)
Same result is obtained by applying J+ operator to (7). J+ψ(1/2, 1/2) = 0
Solving (8) and (9), =
2
3
, b = −
1
3
(10)

3.3 Solutions 223
Thus
ψ

1
2
,
1
2

=
2
3
ϕ(1, 1)ϕ

1
2
, −
1
2

−
1
3
ϕ(1, 0)ϕ

1
2
,
1
2

(11)
Similarly
ψ

1
2
, −
1
2

=
1
3
ϕ(1, 0)ϕ

1
2
, −
1
2

−
2
3
ϕ(1, −1)ϕ

1
2
,
1
2

(12)
The coefficients appearing in (1), (2), (5), (6), (11) and (12) are known as
Clebsch – Gordon coefficients. These are displayed in Table 3.3.
3.87 In spherical coordinates
x = r sin θ cos ϕ ; y = r sin θ sin ϕ ; z = r cos θ (1)
So that xy + yz + zx = r2
sin2
θ sin ϕ cos ϕ + r2
sin θ cos θ sin ϕ
+ r2
sin θ cos θ cos ϕ (2)
The spherical harmonics are
Y00 =

1
4π
1/2
; Y10 =

3
4π
1/2
cos θ ; Y1±1 = ∓

3
8π
1/2
sin θ e±ϕ
Y20 =

5
16π
1
2
(3 cos2
θ − 1); Y2±1 = ∓

15
8π
1
2
sin θ cos θ e±iϕ
;
Y2±2 =

15
32π
1/2
sin2
θ e±2iϕ
(3)
Expressing (2) in terms of (3),
sin2
θ sin ϕ cos ϕ = 1
/2 sin2
θ sin 2ϕ =
Y22 − Y2−2
4i

32π
15
1/2
Similarly sin θ cos θ sin ϕ =
8π
15
1/2 (Y21−Y2−1)
2i
sin θ cos θ cos ϕ =

8π
15
1
2
(Y21 − Y2−1)/2
Hence, xy + yz + zx = r2
8π
15
1
2
[(Y22 − Y2−1)/i + (Y21 + Y2−1)/2i
+(Y21 − Y2−1)/2]
The above expression does not contain Y00 corresponding to l = 0, nor
Y10 and Y1±1 corresponding to l = 1. All the terms belong to l = 2, and the
probability for finding l = 2 and therefore L2
= l(l + 1) = 62
is unity.
3.88 Lz = −i
∂
∂ϕ
x = r sin θ cos ϕ ; y = r sin θ sin ϕ ; z = r cos θ
ψ1 = (x + iy) f (r) = r sin θ(cos ϕ + i sin ϕ) f (r)
= (cos ϕ + i sin ϕ) sin θ F(r)

where F(r) = r f (r)
Lzψ1 = −i
∂
∂ϕ
(cos ϕ + i sin ϕ) sin θ F(r)
= −i(− sin ϕ + i cos ϕ) sin θ F(r)
= (cos ϕ + i sin ϕ) sin θ F(r)
= ψ1
Thus ψ1 is the eigen state and the eigen value is
ψ2 = zf (r) = r cos θ f (r)
Lzψ2 = −i
∂
∂ϕ
(r cos θ f (r)) = 0
The eigen value is zero.
ψ3 = (x − iy) f (r) = r sin θ(cos ϕ − i sin ϕ) f (r)
= (cos ϕ − i sin ϕ) sin θ F(r)
Lzψ3 = −i
∂
∂ϕ
(cos ϕ − i sin ϕ) sin θ F(r)
= −i(− sin ϕ − i cos ϕ) sin θ F(r)
= i (sin ϕ + i cos ϕ) sin θ F(r)
= −(cos ϕ − i sin ϕ) sin θ F(r) = −ψ3
Thus ψ3 is an eigen state and the eigen value is −.
3.89 (a) Lz = −i
∂
∂ϕ
Lzψ = −i
∂
∂ϕ
Af (r) sin θ cos θ eiϕ
= (i) (−i A f (r) sin θ cos θ eiϕ
)
= A f (r) sin θ cos θ eiϕ
= ψ
Therefore, the z-component of the angular momentum is .
(b) L2
= −2
.
∂2
∂θ2 + cot θ ∂
∂θ
+ 1
sin2
θ

∂2
∂ϕ2
/
Expressions for Lz and L2
are derived in Problems 3.80 and 3.83.
L2
ψ = −2

∂2
∂θ2
+ cot θ
∂
∂θ
+

1
sin2
θ

·
∂2
∂ϕ2

= −2
Af (r)eiϕ

−4 sin θ cos θ + cot θ (cos2
θ − sin2
θ) −
sin θ cos θ
sin2
θ

= 62
= 6 2
ψ
Thus L2
= 6 2
But L2
= l(l + 1). Therefore l = 2

3.3 Solutions 225
3.90 With reference to the Table 3.2, the function ψ(r, θ, ϕ) is the 2p function for
the hydrogen atom.
(a) Lz = −i ∂
∂ϕ
Applying Lz to the wavefunction
Lzψ(r, θ, ϕ) = −i
∂ψ(r, θ, ϕ)
∂ϕ
= (−i)(−i)ψ(r, θ, ϕ)
= −ψ(r, θ, ϕ)
Therefore, the value of Lz is −
(b) As it is a p-state, l = 1 and the parity is (−1)l
= (−1)l
= −1, that is an
odd parity.
3.91 Lx = i

sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ

(1)
Ly = i

− cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ

(2)
L+ = Lx + i Ly = i

sin ϕ
∂
∂θ
+ cot θ cos ϕ
∂
∂ϕ

−

− cos ϕ
∂
∂θ
+ cot θ sin ϕ
∂
∂ϕ

(3)
Apply (3) to the m = +1 state which is proportional to sin θ eiϕ
L+(sin θ eiϕ
) = i(sin ϕ cos θ + i cot θ cos ϕ)eiϕ
− (− cos ϕ cos θi + cot θ sin ϕ sin θ)eiϕ
= i(sin ϕ cos θ − cot θ sin ϕ sin θ)eiϕ
− (cos θ cos ϕ − cos ϕ cos θ)eiϕ
= 0
Thus the state with m = 2 does not exist. Similarly by applying L− to the
state with m = −1, it can be shown that the m = −2 state does not exist.
3.92 Particles with even spin (0, 2, 4 . . .) obey Bose statistics and those with odd
spin (1/2, 3/2, 5/2 . . .) obey Fermi – Dirac statistics.
Consider a diatomic molecule with identical nuclei. The total wave function
may be written as
ψ = ψelecζvibρrotσnuc
Let p be an operator which exchanges the space and spin coordinates.
Now pψelec = ±ψelec
It is known from molecular spectroscopy, that for the ground state it is pos-
itive. Furthermore, Pζvib = +ζvib, because ζvib depends only on the distance
of separation of nuclei.
Now ρ = Pl
m
(cos θ)eimϕ
, where θ is the polar angle and ϕ the azimuth
angle; ρ is represented by the associated Legendre function.

Exchange of x → −x, y → −y, z → −z implies θ → π − θ and
ϕ → π + ϕ, so that
P1
m
(cos θ) → (−1)l+m
and eimϕ
→ (−1)m
eimϕ
, Thus Pρ = (−1)2m
(−1)l
= (−1)l
ρ
where m is an integer. Thus ρ is symmetrical for even l and antisymmetrical
for odd l.
First consider zero nuclear spin. The total wave function ψ is antisymmet-
rical for odd l and symmetrical for even l. As the nuclei must obey either
Fermi or Bose statistics, either only the l = odd states must exist or only the
l = even states must exist. It turns out that for nuclei with zero spin only the
even rotational states exist and odd rotational states are missing.
Next consider the case of non-zero spin. A nucleus of total angular momen-
tum I can have a component M in any prescribed direction taking 2I+1 values
in all (I, I − 1, . . . − I), that is 2I + 1 states exist. For the two identical nuclei
(2I + 1)2
wave functions of the form ψM1(A)ψM2(B) can be constructed. If
the two nuclei are identical, these simple products must be replaced by lin-
ear combination of those products which are symmetric or antisymmetric for
interchange of nuclei. If M1 = M2, the products themselves are (2I + 1)
symmetric wave functions, the remaining (2I + 1)2
− (2I + 1) = 2I(2I + 1)
functions with M1 = M2 have the form ψM1(A)ψM2(B) and ψM2(A)ψM1(B).
Each such pair can be replaced by one symmetric and one antisymmetric
wave function of the form ψM1(A)ψM2(B) ± ψM2(A)ψM1(B). Thus, half of
2I(2I + 1) functions, that is I(2I + 1) are symmetric and an equal num-
ber antisymmetric. Therefore, total number of symmetric wave functions
= (2I + 1) + I(2I + 1) = (2I + 1)(I + 1). Total number of antisymmetric
wave functions = I(2I + 1). Therefore, the ratio of the number of symmetric
and antisymmetric functions is (I + 1)/I.
From the previous discussion it was shown that for the symmetric elec-
tronic wave function of the molecule the interchange of nuclei produces a
factor (−1)l
in the molecular wave function. Thus, for nuclei obeying Bose
statistics symmetric nuclear spin functions must combine with even l. Because
of the statistical weight attached to spin states, the intensity of even rota-
tional lines will be (I + 1)/I as great as that of neighboring odd rotational
lines.
For nuclei obeying Fermi statistics, the spin and rotational states combine
in a manner opposite to the previously described and the odd rotational lines
are more intense in the ratio (I + 1)/I.
Thus, by determining which lines are more intense, even or odd, the nuclear
statistics is determined and by measuring the ratio of intensities of adjacent
lines the nuclear spin is obtained. The reason for comparing the intensity of
neighboring lines is that the intensity of rotational lines varies according to
the occupation number of rotational states governed by the Boltzmann distri-
bution.

3.3 Solutions 227
3.93
u(θ, ϕ) = 1
/4
15
π
sin 2θ cos 2ϕ
= 1
/4
15
π
sin2
θ

(e2iϕ
+ e−2iϕ
)
2

(1)
But Y2+2(θ, ϕ) =
15
32π
sin2
θe2iϕ
(2)
Y2−2(θ, ϕ) =
15
32π
sin2
θe−2iϕ
(3)
Adding (2) and (3)
Y2+2(θ, ϕ) + Y2−2(θ, ϕ) =
15
32π
sin2
θ(e2iϕ
+ e−2iϕ
) (4)
Dividing (1) by (4) and simplifying we get
u(θ, ϕ) =
1
√
2
(Y2+2(θ, ϕ) + Y2−2(θ, ϕ))
We know that
L2
Ylm = 2
l(l + 1)Ylm
So L2
u(θ, ϕ) = L2
[Y22(θ,ϕ)+Y2−2(θ,ϕ)]
√
2
= 2(2 + 1)2 [Y22(θ,ϕ)+Y2−2(θ,ϕ)]
√
2
= 62
u(θ, ϕ)
Thus the eigen value of L2
is 62
3.94 The wave function is identified as ψ322
LZ ψ = −i
∂ψ
∂ϕ
= 2ψ
Thus the eigen value of LZ is 2.
3.95 (a) Using the values, Y10 =
$
3
4π
cos θ and Y1,±1 = ∓
$
3
8π
sin θ exp(±iϕ), we
can write
ψ =
1
3
−
√
2Y11 + Y10

f (r)
Hence the possible values of LZ are +, 0

(b) First we show that the wavefunction is normalized.

|ψ|2
dr =
1
4π
∞
0
|g(r)|2
r2
dr
π
0
dθ
2π
0
(1 + cos ϕ sin 2θ) sin θdϕ
=
1
2
π
0
sin θdθ = 1
The probability for the occurrence of Lz = is

1
3
−
√
2

Y11
'2
dΩ = (2/3)

(3/8π) sin2
θ.2π sin θdθ
= (1/2)
+1
−1
(1 − cos2
θ)d cos θ = 2/3
The probability for the occurrence of Lz = 0 is

1
3
Y10
'2
dΩ =
+1
−1

3
4π

.2π cos2
θd cos θ = 1/3
3.96 (a) [Jz, J+] = Jz J+ − J+ Jz = Jz(Jx + iJ y) − (Jx + iJ y)Jz
= Jz Jx − Jx Jz + i(Jz Jy − Jy Jz)
= [Jz, Jx ] + i[Jz, Jy] = iJy − iiJx
= iJy + Jx = (Jx + iJ y) = J+
= J+, in units of .
(b) From (a), Jz J+ = J+ Jz + J+
Jz J+|j m = J+ Jz|jm +J+|jm
= J+m|jm +J+|jm
= (m + 1)J+|jm
J+|jm is nothing but | j, m + 1 apart from a possible normalization
constant. Thus
J+|jm = Cjm+ | j, m + 1
Given a state |jm , the state | j, m + 1 must exist unless Cjm+ van-
ishes for that particular m. Since j is the maximum value of m by definition.
There can not be a state | j, j + 1 , i.e. Cjj+ must vanish. J+ is known as
the ladder operator. Similarly, J− lowers m by one unit.
(c) J+ = Jx + iJ y
J− = Jx − iJ y
Therefore, Jx =
1
2
(J+ + J−) =
⎛
⎝
0 1/
√
2 0
1/
√
2 0 1/
√
2
0 1/
√
2 0
⎞
⎠
Jy =
1
2i
(J+ − J−) =
⎛
⎝
0 −i/
√
2 0
i/
√
2 0 −i/
√
2
0 i/
√
2 0
⎞
⎠
[Jx , Jy] = Jx Jy − Jy Jx = i
⎛
⎝
1 0 0
0 0 0
0 0 −1
⎞
⎠ = iJ z

3.3 Solutions 229
3.97 ΔE = ψ|δU|ψ
δU = U (interaction energy of electron with point charge nucleus)
δU = e2
/r −
3e2
2R

R2
−
r2
3

for r ≤ R
= 0 for r ≥ R
(a) First we consider n = 1 state
ΔE =

e2
πa3
0
R
0
exp

−
2r
a0

1
r
−
3
2R
+ 1
/2
r2
R3

4πr2
dr
=
2e2
a3
0
R
0
e−2r/a0

2r −
3r2
R
+
r4
R3

dr
Now R = 10−13
cm ≪ 10−8
cm = a0, the factor exp −2r
a0

≈ 1
ΔE =

2e2
a3
0

R2
− R2
+
R2
5

=
4
5

e2
2a0

R
a0
2
= (0.8)(13.6)

10−13
0.53 × 10−8
2
= 3.87 × 10−9
eV
(b) n = 2
ψ200 =

1
8πa3
0
1
2

2 −
r
a0

exp

−
r
2a0

ΔE =

e2
8πa3
0
R
0
exp

−
r
a0

2 −
r
a0
2
1
r
−
3
2R
+ 1
/2
r2
R3

4πr2
dr
Here also exp − r
a0

∼ 1, for reasons indicated in (a)
When the remaining factors are integrated we get
ΔE =

e2
2a0

R2
a2
0

2
5
−
1
6
R
a0
+
3
140
R2
a2
0

≈
2
5
.

e2
2a0

R
a0
2
as
R
a0
1

= 1
/2 × 3.87 × 10−9
= 1.93 × 10−9
eV
where we used the result of (a)
3.98 Schrodinger’s equation in the presence of electric field is

−
2
2m

d2
dx2
+ 1
/2m ω2
x2
+ qEx

ψn = En x (1)
Now, 1
/2mω2
x2
+ qEx = 1
/2mω2
*
x2
+ 2qEx
mω2
+

= 1
/2 m ω2
[(x + qE/mω2
)2
− q2
E2
/m2
ω4
]
= 1
/2 mω2

x +
qE
mω2
2
−
q2
E2
2mω2
(2)
Put X = x + qE
mω2 ; then we can write
d
dx
=
d
dX
and
d2
dx2
=
d2
dX2

−
2
2m
d2
dX2
+
1
2
mω2
X2

ψn(X) =

En +
q2
E2
2mω2

ψn(X) (3)
Left hand side of (3) is the familiar Hamiltonian for the Simple harmonic
oscillator. The modified eigen values are then given by the right hand side
En +
q2
E2
2mω2
=

n +
1
2

ω
or
En =

n +
1
2

ω − q2
E2
/2mω2
(4)
3.99 The matrix of H′
is
(H′
) =

H′
11 H′
12
H′
21 H′
22

, with H′
12 = H′∗
21
The matrix of H is
H =

E0 + H′
11 H′
12
H′
21 E0 + H′
22

%
%
%
%
E0 + H′
11 − E H′
12
H′
21 E0 + H′
22 − E
%
%
%
% = 0
E1 = 1
/2

2E0 + H′
11 + H′
22

+ 1
/2
*
H′
11 − H′
22
2
+ 4
%
%H′
12
%
%2
+1/2
E2 = 1
/2

2E0 + H′
11 + H′
22

− 1
/2
*
H′
11 − H′
22
2
+ 4
%
%H′
12
%
%2
+1/2
These are the energy levels of a two state system with Hamiltonian H =
H0 + H′
. The perturbation theory requires finding the eigen values of H′
and
adding them to E0, which gives an exact result.
3.100 The nuclear charge seen by the electron is Z and not 2. This is because of
screening the effective charge is reduced.
The smallest value of E(Z) = − e2
2a0
27
4
Z − 2Z2

must be determined
which is done by setting

3.3 Solutions 231
∂E
∂ Z
= 0; this yields Z =
27
16
E

27
16

= −

2e2
2a0

27
16
2
= −(2)(13.6)

27
16
2
= −77.45 eV
The estimated value of ionization energy 77.4 eV may be compared to the
74.4 eV derived by perturbation theory and an experimental value of 78.6 eV.
In general the variation method gives better accuracy then the perturbation
theory.
3.101 (a) The unperturbed Hamiltonian for a hydrogen atom is
H0 = −
2
2μ
∇2
−
e2
r
(1)
where μ is the reduced mass.
H′
is the extra energy of the nucleus and electron due to external field
and is
H′
= −eEz = eEr cos θ (2)
where the polar axis is in the direction of positive z.
Now, the perturbation in (2) is an odd operator since it changes sign
when the coordinates are reflected through the origin. Thus, the only non-
vanishing matrix elements are those for unperturbed states that have oppo-
site parities. In particular all diagonal elements of H′
of hydrogen atom
wave functions are zero. This shows that a non-degenerate state, like the
ground state (n = 1) has no first-order Stark effect.
(b) The first excited state (n = 2) of hydrogen atom is fourfold degener-
ate, the quantum numbers n, l and m have the values (2,0,0), (2,1,1),
(2,1,0), (2,1,−1). The first one (2S) has even parity while the remaining
three(2P) states have odd parity of the degenerate states only |2, 0, 0
and |2, 1, 0 are mixed by the perturbation, but |2, 1, 1 and (2, 1, −1
are not and do not exhibit the Stark effect. It remains to solve the secular
equation,
%
%
%
%
e|E| 2, 0, 0|z|2, 0, 0 −λ e|E| 2, 0, 0|z|2, 1, 0
e|E| 2, 1, 0|z|2, 0, 0 e|E| 2, 1, 0|z|2, 1, 0 −λ
%
%
%
% = 0
Because the conservation of parity the diagonal elements 2, 0, 0|z|2,
0, 0 and 2, 1, 0|z|2, 1, 0 vanish. Hence the first-order change in
energy λ = ±e|E 2, 0, 0|z|2, 1, 0
Thus, only one matrix element needs to be evaluated, using the unper-
turbed eigen functions explicitly. (see Table 3.2)

2S(m = 0); ψ2s(0) = (4π)− 1
2

1
2a
3
2
2 −
r
a

exp −
r
a

2P(m = 0); ψ2p(0) = (4π)− 1
2

1
2a
3
2 r
a

exp −
r
2a

cos θ
We can calculate
2, 0, 0|z|2, 1, 0 = 2, 0, 0|r cos θ|2, 1, 0
=

1
4π

1
2a
3
1
a
∞
0
r4
2 −
r
a

exp −
r
a

dr
π
0
cos2
θ sin θdθ
2π
0
dϕ
= −3a
Thus, the linear Stark effect splits the degenerate m = 0 level into two
components, with the shift
ΔE = ±3 ae |E|
The corresponding eigen functions are 1
√
2
(ψs(0) ∓ ψp(0))
The two components being mixed in equal proportion (Fig. 3.27).
Fig. 3.27 Stark effect in
Hydrogen
3.102 E =
+a
−a

1
√
a

cos
πx
2a

−
2
2m

d2
dx2
+ 1/2mω2
x2

1
√
a
cos
πx
2a

dx
=
π2
2
8ma2
+
mω2
a4
10
+
8a5
π2

1 −
6
π2

The best approximation to the ground-state wave function is obtained by
setting ∂E
∂α
= 0. This gives
a =

3π2
2
5m2 ω2(π2 − 3)
1/4
3.103 The unperturbed wave function is
ψ0
= k sin
n1πx
a

sin(n2π y/a); H′
= W0
E =

π2
2ma2

n2
1 + n2
2

3.3 Solutions 233
First order correction is
ΔE =

ψ∗
0 H′
ψdτ

ψ∗
0 ψdτ

ψ∗
0 H′
ψdτ = K2
W
a
2
0
sin2 n1πx
a

dx
a/2
0
sin2 n2πy
a

dy
= K2
W

x
2
−

a
4n1π

sin

2n1πx
a
a/2
0

y
2
−

a
4n2π

sin

2n2πy
a
a/2
0
=
K2
Wa2
16
a
0
ψ∗
0 ψ0dτ = K2
a
0
a
0
sin2
n1πx
a
sin2
n2πy
a
dx dy =
k2
a2
4
Therefore ΔE =
K2
W0a2
16
/
K2
a2
4
=
W0
4
3.3.8 Scattering (Phase Shift Analysis)
3.104 Let the total wave function be
ψ = ψi + ψs (1)
where ψi represents the incident wave and ψs the scattered wave.
In the absence of potential, the incident plane wave
ψi = Aeikz
= eikz
(2)
where we have dropped off A to choose unit amplitude.
Assume
ψs =
f (θ)eikr
r
(3)
which ensures inverse square r dependence of the scattered wave from the
scattering centre.
σ(θ) = | f (θ)|2
(4)
f(θ) being the scattering amplitude.
We can write (1)
ψ = eikrcosθ
+
f (θ)eikr
r
(5)
or
f θ = re−ikr
(ψ − eikrcosθ
) (6)
Lt r → ∞
The azimuth angle ϕ has been omitted in f (θ) as the scattering is assumed
to have azimuthal symmetry. In the absence of potential ψi is the most
general solution of the wave equation.
∇2
ψi + k2
ψi = 0 (7)

Now ψi can be expanded as a sum of partial waves
ψi = eikr cos θ
=
∞
l=0
Al jl (kr)pl(cos θ) (8)
where jl(kr) are the spherical Bessel functions and pl(cos θ) are the Legen-
dre polynomials of degree l. For r → ∞, jl (kr) ≈ 1
kr
sin

kr − πl
2

. The Al
are some constants which can be evaluated as follows.
Multiply both sides of (8) by Pl (cos θ) sin θdθ and integrate. Put cos θ = t
Al jl(kr)2/(2l + 1) =
+1
−1
eikrt
pl(t)(d)t
where we have used the orthonormal property of Legendre polynomials.
Integrating the RHS by parts
(1/ikr)
,
eikrt
pl (t)
-+1
−1
− (1/ikr)

eikrt
pl
′
(t)dt
where prime (′
) means differentiation with respect to t. The second term is of
the order of 1/r2
which can be neglected. Therefore

2
2l + 1

Al jl(kr) ≈

1
ikr

,
eikr
− e−ikr
(−1)l
-
(9)
where we have used pl(1) = 1 and pl(−1) = (−1)l
Also, using the identity
eiπl/2
= il
(10)
(9) becomes

2
2l + 1

Al jl(kr) ≈
*
2il
kr
+ *
ei(kr− πl
2 ) − e−i(kr− πl
2 )
+
2i
=
2il
sin

kr − πl
2

kr
Thus
Al jl(kr) =
(2l + 1)il
sin

kr − πl
2

kr
(11)
Similarly, we can expand the total wave function into components
ψ(r, θ) =
∞
l=0
Bl Rl (r)pl(cos θ)
=

r→∞

Bk
kr

sin

kr −
πl
2
+ δl

pl(cos θ)
where Bl are arbitrary coefficients and δl is the phase-shift of the lth wave.
From (6)
f (θ) = re−ikr

Bl

1
kr
sin

kr −
πl
2
+ δl

pl(cos θ)

−
Σ il
(2l + 1)
kr
sin

kr −
πl
2

pl(cos θ)

3.3 Solutions 235
or eikr
f (θ) =

Bl pl(cos θ)
*
ei(kr− πl
2
+δl ) − e−i(kr− πl
2
+δl )
+
2ik
−

il
(2l + 1)pl(cos θ)
*
ei(kr− πl
2 ) − e−i(kr− πl
2 )
+
2ik
Equating coefficients of e−ikr
0 = −

1
2ik

Bl pl (cos θ)

e−i(−πl/2+δl )
+

il
(2l + 1)Pl(cos θ)
2ik
eiπl/2

Therefore
Bl = il
(2l + 1)eiδl
(12)
Equating coefficients of eikr
, and using the value of Bl
f (θ) =

1
2ik
*
il
(2l + 1)eiδl
pl(cos θ)ei(− πl
2
+δl )
−

il
(2l + 1)pl (cos θ)e− πl
2
+
=

1
2ik

il
(2l + 1)Pl(cos θ)e−iπl/2
,
e2iδl
− 1
-
(13)
Using (10), formula (12) becomes
f (θ) =
1
2ik

(2l + 1)(e2iδl
− 1)Pl(cos θ)
The above method is called the method of partial wave analysis. The sum-
mation over various integral values of l means physically summing over var-
ious values of angular momenta associated with various partial waves. The
quantity δl is understood to be the phase shift when the potential is present. At
low energies only a few l values would be adequate to describe the scattering.
3.105 The differential cross-section for the scattering of identical particles of spin
s is given by
σ(θ)∗
= | f (θ∗
)|2
+ | f (π − θ∗
)|2
+

(−1)2s
2s + 1

2Re[ f (θ∗
) f ∗
(π − θ∗
)] (1)
where f is assumed to be independent of the azimuth angle ϕ. The angles refer
to the CM-system. The first two terms on RHS are given by the Rutherford
scattering, one for the scattered particle and the other for the target particle.
In the CMS the identical particles are oppositely directed and the detector
cannot tell one from the other. The third term on the RHS is due to quantum
mechanical interference and does not occur in the classical formula. Now for
alpha-alpha scattering s = 0 and (1) reduces to
σ(θ∗
) = | f (θ∗
)|2
+ | f (π − θ∗
)|2
+ 2Re[ f (θ∗
) f ∗
(π − θ∗
)].
Furthermore if the scattering at θ∗
= 90◦
is considered then obviously f (π −
θ∗
) = f (θ∗
) and the lab angle θ = 45◦
. In that case classically σL(45◦
) =
2| f (90◦
)|2
CM while quantum mechanically

σL (45◦
) = 4| f (90◦
)|CM .
Thus quantum mechanics explains the experimental result
3.106 σ =

4π
k2

l=0
(2l + 1) sin2
δl (1)
By problem
sin δl =
(iak)l
√
(2l + 1)l!
(2)
Therefore,
sin2
δl =
(−a2
k2
)l
(2l + 1)l!
(3)
Using (3) in (1)
σ =

4π
k2

l=0
(−a2
k2
)l
l!
Summing over infinite number of terms for the summation and writing
k2
=
2mE
2
,
σ =

2π2
mE

exp(−a2
k2
)
=
2π2
mE
exp

−
2mEa2
2
#
3.107 Let b be the impact parameter. In the c-system
bPcm = l =
where we have set l = 1 for the p-wave scattering
ECM =
P2
CM
2μ
=
P2
CM
M
= 2
/Mb2
(Since the reduced mass μ = M/2, where
M is the mass of neutron or proton)
ELab = 2ECM =
22
Mb2
=
22
c2
Mc2
b2
Inserting c = 197.3 MeV.fm, Mc2
= 940 MeV and b = 2 fm, we find
ELab = 20.6 MeV. Thus up to 20 MeV Lab energy, s-waves (l = 0) alone
are important
3.108 Only s-waves (l = 0) are expected to be involved as the scattering is
isotropic.
σ =
4π sin2
δ0
k2
Now k2
2
= p2
= 2mE
sin2
δ0 = 2mEσ
4π2 = 2mc2
Eσ
4π2c2
Inserting mc2
= 940 MeV; E = 1.0 MeV,
σ = 0.1 b = 10−25
cm2
= 10 fm2
, c = 197.3 MeV − fm
sin2
δ0 = 0.03845
δ0 = ±11.3◦

3.3 Solutions 237
3.109 By Problem 3.104
σ(θ) =
1
k2
,
sin2
δ0 + 6 sin δ0 sin δ1 cos(δ1 − δ0) cos θ + 9 sin2
δ1 cos2
θ
-
(1)
We assume that at low energies δ1 ≪ δ0. Now in the scattering with a hard
sphere
tan δl = −
(ka)2l+1
(2l + 1)(1.1.3.5 . . . 2l − 1)2
δ0 (H.sphere) = −ka, for all ka
And δ1 (H.sphere) = −(ka)3
3
, for ka ≪ 1
Neglecting higher powers of δ′
s, we can write (1)
σ(θ) =
1
k2

δ0 −
δ3
0
3!
2
+ 6δ0δ1 cos δ
'
=
1
k2

δ2
0 −
δ4
0
3
+ 6δ0δ1 cos δ

=
1
k2

k2
a2
−
k4
a4
3
+ 6(−ka)

−
k3
a3
3

cos θ

= a2

1 −
k2
a2
3
+ 2k2
a2
cos θ

σ =

dσ
dΩ

.2π sin θ dθ = 2π
+1
−1
a2

1 −
k2
a2
3
+ 2k2
a2
cos θ

d cos θ
= 4πa2

1 −
(ka)2
3

3.110 A spherical nucleus of radius R will be totally absorbing, or appear “black”
when the angular momentum l R/λ. In that case ηl = 0 in the reaction
and scattering formulae.
σr = π-
λ2

l
(2l + 1)(1 − |ηl|2
)
σs = π-
λ2

l
(2l + 1)|1 − ηl |2
(| ηl 0)
Putting ηl = 0
σr = σs = π -
λ2
R/λ
l=0
(2l + 1)
The summation can be carried out by using the formula for arithmetic
progression
S = na +
n(n − 1)d
2
Here a = 1, d = 2, n = R/-
λ

R/λ
l=0
(2l + 1) = (R/-
λ)2
∴ σr = σs = π-
λ2

R
-
λ2

= π R2
The total cross-section
σt = σr + σs = 2π R2
which is twice the geometrical cross-section
3.111 The potential which an electron sees as it approaches an atom of a monatomic
gas can be qualitatively represented by a square well. Slow particles are con-
sidered.
V (r) = −V0; r ≤ R
= 0; r R
corresponding to an attractive potential. Scattering of slow particles for
which kR 1, is determined by the equation

∇2
+ k2
−
2μV
2

ψ2 = 0 (inside the well) (1)
with k2
= 2μE/2
, and the wave number k = p/
Outside the well the equation is
(∇2
+ k2
)ψ1 = 0 (2)
Further writing
k2
1 = k2
+ k2
0
where k2
0 = 2μV
2
and V = −V0
The solutions are found to be
ψ2 = A sin k1r (3)
ψ1 = B sin(kr + δ0) (4)
ψ1(r) is the asymptotic solution at large distances with the boundary condi-
tion
ψ1(0) = 0
Matching the solutions (3) and (4) at r = R both in amplitude and first
derivative,
A sin k1 R = B sin(kR + δ0) (5)
Ak1 cos k1 R = Bk cos(kR + δ0) (6)
Dividing one equation by the other, and setting k1cot k1 R = 1
D
, and with
simple algebraic manipulations we get
tan δ0 = (kD − tan kR)(1 + kD tan kR)−1
(7)
The phase shift δ0 determined from (7) is a multivalued function but we are
only interested in the principle value lying within the interval −π
2
≤ δ0 ≤ π
2
.
For small values of the energy of the relative motion

3.3 Solutions 239
tan kR ≈ kR +
(kR)3
3
+ · · ·
We therefore have
tan δ0 ≈
k
*
D − R − (kR)3
3k
+
1 + k2 DR
If the inequalities kR 1 and k2
DR 1 are satisfied we can still
further simplify the expression for tan δ0.
tan δ0 ≈ k(D − R) = kR

tan k1 R
k1 R
− 1

(8)
The total cross-section is then
σ =
4π
k2
sin2
δ0 ≈ 4π(D − R)2
= 4π R2

1 −
tan k1 R
k1 R
2
(9)
It follows that if the condition
tan k1 R = k1 R (10)
is satisfied, the phase shift and the scattering cross-section both vanish. This
phenomenon is known as the Ramsauer-Townsend effect. The field of the
inert gas atoms decreases appreciably faster with distance than the field of
any other atom, so that to a first approximation, we can replace this field by a
rectangular spherical well with sharply defined range and use Equation (10)
to evaluate the cross-section for slow electrons.
Physically, the Ramasuer – Townsend effect is explained as the diffraction
of the electron around the rare-gas atom, in which the wave function inside
the atom is distorted in such a way that it fits on smoothly to an undistorted
function outside.
Here the partial wave wave with l = 0 has exactly a half cycle more of
oscillation inside the atomic potential then the wave in the force-free field,
and the wavelength of the electron is large enough in comparision with R so
that higher l phase-shifts are negligible.
3.112 In order that the Schrodinger equation is reduced to the given form is that the
potential V (r) does not depend on time. From Problem 3.104 the total wave
function
ψ = Σil
(2l + 1)

eiδl
kr

sin

kr −
1
2
πl + δl

pl(cos θ)
For slow neutrons only the first term (l = 0) in the summation is important.
As p0(cos θ) = 1
ψ =
exp(iδ0)
kr
sin(kr + δ0)
u = ψr =
exp(iδ0
k
sin(kr + δ0) = const. sin(kr + δ0)
We assume that the wave function inside the well is identical with that
in the deuteron problem. This is justifiable since the total energy inside the
potential well is raised by little over 2 MeV corresponding to the

binding energy W of the deuteron, which is much smaller than the well depth
(∼25 MeV). In the deuteron problem the outside function Ce−γr
, where
γ =

MW /2, is matched with the inside function A sin kr. Here we
match the functions sin(kr + δ0) and Ce−γr
at r = R, both in magnitude
and first derivative.
This gives us k cot(kR+δ0) = −γ . Further, R ≈ 0. This is also reasonable
since for the square well the main features of the deuteron problem remain
unaltered by narrowing the well width and deepening the well. It follows that
sin2
δ0 = k2
/k2
+ γ 2
)
But the s-wave cross-section is given by
σ = 4π sin2
δ0/k2
= 4π/(k2
+ γ 2
)
Substituting k2
= ME/2
and γ 2
= MW /2
σ =
4π2
M
1
W + E
(1)
where M is proton or neutron mass, W is the deuteron binding energy
(2.225 MeV), and E is the lab kinetic energy.
Formula (1) agrees well with experiment at relatively higher energies (say
5–10 MeV) but fails badly at very low energies. For E ≪ W, for example,
(1) predicts σ = 2 barns which is far from the experimental value of 20
barns. Wigner pointed out that in n−p scattering the spins of the colliding
nucleons could be either parallel or antiparallel. Formula (1) holds for the
parallel case because the analogy is made with the deuteron problem which
has parallel spins. Now for random orientations of spins:
σ =
3
4
σt +
1
4
σs (2)
where σt and σs are the cross-sections for the triplet and singlet scattering,
the factors 3
4
and 1
4
being the statistical weights. In (1), W is the binding
energy of the n−p system for the triplet state. Corresponding to the singlet
state the quality Ws is introduced, although it is a virtual state.
σ =
3π2
M(E + W)
+
π2
M(E + Ws)
(3)
Ws takes a value of 70 keV if agreement is to reach with the experiments.
Agreement at higher energies is preserved because for E ≫ W or Ws, (3)
reduces to (1).
3.113 (a) F(q) ≈
∞
0
ρ(r) sin(qr/)4πr2
qr/
dr
ρ(r) = A exp(−r/a)
F(q) ≈ 4π A
∞
0
r exp(−r/a)[sin(qr/)/(q/)]dr

3.3 Solutions 241
Put α = 1/a and β = q/
F(q) ≈ 4π A
∞
0
r e−αr sin βr
β
dr
I = −
∂
∂α
∞
0
e−αr
sin βrdr
=
−1
β
∂
∂α
β
α2 + β2
=
1
β
2αβ
(α2 + β2)2
=
2α
(α2 + β2)2
=
2
α3

1 +
β2
α2
−2
= 2a3
/(1 + q2
a2
/2
)
F(q) = 8π A a3
/

1 + q2
/q2
o

, where qo = /a
thus F(q) ≈ 1/ 1 + q2
q2
0
2
(b) The characteristic radius
a =

qo
=
c
qoc
=
197.3 MeV − fm
0.71 × 1,000 MeV
= 0.278 fm
3.114 f (θ) = − μ
2π2

V (r)eiq.r
d3
r
= −
μ
2π2
∞
r=0
π
θ=0
2π
ϕ=0
V (r)eiqr cos θ
r2
sin θdθdϕdr
= −
μ
2π2
∞
0
V (r)r2
dr
+1
−1
eiqr cos θ
d(cos θ
2π
0
dϕ
= −
2μ
2

V (r)r2
dr
qr

eiqr
− e−iqr
2i

= −
2μ
q2

r sin(qr)V (r)dr
3.115 From the partial wave analysis of scattering the scattering amplitude
f (θ) =
1
k
Σl(2l + 1)(ηl exp(2iδl) − 1)/2i)pl (cos θ).
For elastic scattering without absorption ηl = 1, and
f (θ) =
1
k
Σl(2l + 1)
,
exp(2iδl) − 1)/2i
-
pl (cos θ)
=
1
k
Σl (2l + 1) exp(iδl) sin δl pl(cos θ).
Now for θ = 0, pl(cos θ) = pl(1) = 1 for any value of l, and exp(iδl) =
cos δl + i sin δl. Therefore the imaginary part of the forward scattering
amplitude
Im f (0) =
1
k

l
(2l + 1) sin2
δl.

But the total cross-section is given by
σt =
4π
k2
(2l + 1) sin2
δl .
It follows that Im f (0) = kσt /4π. The last equation is known as the opti-
cal theorem.
3.116 V (r) =

−
Ze2
2R
#
3 −
r2
R2

; 0 r R (1)
= −
Ze2
e−ar
r
; R r ∞ (2)
Inside the nucleus the electron sees the potential as given by (1) corre-
sponding to constant charge distribution, while outside it sees the shielded
potential given by (2). The scattering amplitude is given by
f (θ) = −(2μ/q2
)
∞
0
V (r)r sin(qr)dr
= 2μ
Ze2
q2

1
2R
R
0

3 −
r2
R2

r sin(qr)dr +
∞
R
sin(qr)e−ar
dr

(3)
The first integral is easily evaluated and the second integral can be written
as
∞
R
sin(qr)e−ar
dr =
∞
0
sin(qr)e−ar
dr −
R
0
sin(qr)e−ar
dr (4)
=
q
q2 + a2
−
R
0
sin(qr)e−ar
dr (5)
(Lim a → 0) =
1
q
−
R
0
sin(qr)dr =
1
q
cos(qr)
We finally obtain
f (θ) =

−
2μZe2
q22
#
3
q2 R2

sin(qR)
qR
− cos qR

σ(θ)finite size = σ(θ)point charge|F(q)|2
where the form factor is identified as
F(q) =

3
q2 R2

sin(qR)
qR
− cos(qR)

The angular distribution no longer decreases smoothly but exhibits sharp
maxima and minima reminiscent of optical diffraction pattern from objects
with sharp edges. The minima occur whenever the condition tan qR = qR,
is satisfied. This feature is in contrst with the angular distribution from a
smoothly varying charge distribution, such as Gaussian, Yakawa, Wood-
Saxon or exponential, wherein the charge varies smoothly and the maxima

3.3 Solutions 243
and minima are smeared out, just as in the case of optical diffraction from a
diffuse boundary of objects characterized by a slow varying refractive index.
3.117 f (θ) = −(2μ/q2
)
∞
0 V (r) sin(qr) r dr
Integrate by parts
∞
0
V (r) sin(qr)r dr = V (r)

1
q2
sin qr −
r
q
cos qr
∞
0
−
∞
0
dV
dr

1
q2
sin qr −
r
q
cos qr

dr
The first term on the right hand side vanishes at both limits because V (∞) =
0, Therefore:
∞
0
V (r) sin(qr)r dr = −
1
q2
∞
0
dV
dr
sin qrdr +
1
q
∞
0
dV
dr
r cos qrdr
Evaluate the second integral by parts
1
q
∞
0
dV
dr
r cos(qr)dr =
1
q

dV
dr

r
q
sin qr +
cos qr
q2
∞
0
−
1
q
∞
0

r
q
sin qr +
cos qr
q2

d2
V
dr2
dr
Now the term 1
q2

r
dV
dr

sin qr
%
%∞
0
vanishes at both the limits because it is
expected that (dV/dr)r=∞ = 0.
Integrating by parts again

1
q3

cos qr
d2
V
dr2
dr =

1
q3

cos qr
dV
dr
%
%
%
%
∞
0
+

1
q2
∞
0

dV
dr

sin qrdr

1
q
∞
0

dV
dr

r cos qr dr =

1
q3

dV
dr

cos qr
%
%
%
%
∞
0
−

1
q2
∞
0

d2
V
dr2

r sin qr dr −

1
q3

dV
dr

cos qr
%
%
%
%
∞
0
−

1
q2
∞
0

dV
dr

sin qr dr.
The first and third terms on the right hand side get cancelled
∞
0
V (r) sin(qr)r dr = −

1
q2

d2
V
dr2
+
2
r
dV
dr

sin(qr)r dr.
Now for spherically symmetric potential
∇2
V =
d2
V
dr2
+

2
r

dV
dr
.
Furthermore by Poisson’s equation:

∇2
V = −4π Ze2
ρ
where Ze is the nuclear charge and ρ is the charge density.
∴ f (θ) = −8πμ

Ze2
q32
# ∞
0
ρ(r) sin(qr) r dr
=

2μZe2
q22
#
4π
q
∞
0
ρ(r)sin (qr) r dr
The quantity 4π
q
∞
0 ρ(r) sin(qr)r dr is known as the form factor.
3.118 f (θ) =

−
2μ
q2
∞
0
V (r) sin(qr)r dr (1)
Substituting,
V (r) =
z1z2e2
r
e−ar
(2)
Where a = 1/ro, (1) becomes
f (θ) = −

2μz1z2e2
q2
∞
0
e−ar
sin(qr)dr
=
−2μz1z2e2
q2
q
q2 + a2
=
−2μz1z2e2
2

q2 + 1/r2
0
(3)
But the momentum transfer
q = 2k sin

θ
2

(4)
The differential cross-section
σ(θ) = | f (θ)|2
=
4μ2
z2
1z2
2e4
4

4k2 sin2
(θ/2) + 1/r2
0
2
(5)
The general angular distribution of scattered particles is reminiscent of
Rutherford scattering. However for θ θ0, where
sin(θo/2) ≈ 1/2kro (6)
the curve does not rise indefinitely but tends to flatten out because when
qro ≪ 1, the angular dependence of σ(θ) is damped out resulting in the
flattening of the curve. The angle θo may be considered as the limiting angle
below which the Rutherford scattering is inoperative because of the shielding
of the atomic nucleus by the electron cloud.
Rutherford scattering is derived from (5) by letting ro → ∞, in which case
the scattering would occur from a bare nucleus. The screening potential (2)
now reduces to Coulomb potential. Furthermore, writing k = p = μv, (5)
becomes

3.3 Solutions 245
σ(θ) =
1
4

z1z2e2
μv2
2
1
sin4
θ
2
(Rutherford scattering formula)
3.119 By Problem 3.116
F(q2
) =
3
q2 R2

sin qR
qR
− cos qR

(1)
R = ro A1/3
= 1.3 × (64)1/3
= 5.2 fm
q = 2po sin(θ/2)
qR = 2cpo R sin(θ/2)/c = 2 × 300 ×
5.2 × sin 6◦
197.3
= 1.653 radians (2)
sin qR = 0.9966, cos qR = −0.0819 (3)
Inserting (2) and (3) in (1), we find F(q) = 0.75, F2
≈ 0.57. Thus Mott’s
scattering is reduced by 57%.
3.120 F(q2
) = 1 − q2
62 r2
+ · · ·
q = 2po sin(θ/2) = 2 × 200 × (sin 7◦
) MeV/c = 48.75 MeV/c
r2
=
62
q2
[1 − F(q2
)]
= 6 ×
(197.3)2
(48.75)2
(1 − 0.6) fm2
= 39.3
∴ Root mean square radius = 6.27 fm
3.121 F(q) = (4π/q)
∞
0
ρ(r) sin(qr)r dr
= (4π/π3/2
b3
q)
∞
0
e−r2
/b2
sin(qr)r dr
= (−4/π1/2
b3
q)
∂
∂q
∞
0
e−r2
/b2
cos(qr) dr
= (−4/π1/2
b3
q)
∂
∂q

1
2
(πb2
)1/2
e−b2
q2
/4

F(q) = exp(−b2
q2
/4)
r2
=
∞
0 r2
ρ(r)4πr2
dr

ρ(r)4πr2dr
=
∞
0 r4
er2
/b2
dr
∞
0 r2e−r2/b2
dr
where we have put ρ(r) = (1/π3/2
b3
)e−r2
/b2
dr
With the change of variable r2
/b2
= x, we get
r2
=
b2
∞
0 x3/2
e−x
dx
∞
0 x1/2e−x dx
=
Γ
5
2

b2
Γ
3
2
=
3b2
2

3.122 f (θ) = −
μ
2π2

V (r)eiqr
d3
r
= −
μ
2π2
∞
0
V (r)r2
dr
+1
−1
eiqr cos θ
d(cos θ)
2π
0
dϕ
= −
μ
2π2

V (r)r2

(eiqr
− e−iqr
)
iqr

2πdr
= −

2μ
q2
∞
0
V (r)r sin(qr)dr
3.123 A =
∞
0
sin(qr/)
(qr/)
V (r)4πr2
dr
Substitute V (r) ∼ e−r/R
r
where R = /mc
A ∼
∞
0
e−r/R sin(qr/)
q
dr
Put 1/R = a and q/ = b
A ∼
∞
0
e−ar sin(br)
b
dr
=
1
b

b
a2 + b2

=
1
a2 + b2
=
1
1
R2 + q2
2
∼
1
2
R2 + q2
=
1
q2 + m2c2

Chapter 4
Thermodynamics and Statistical Physics
Kinetic theory of gases
Pressure
p =
1
3
ρ ν2
(4.1)
Root-mean-square velocity
νrms =

3P/ρ (4.2)
νrms =

3kT/m =

3RT/M (4.3)
Average speed
ν =
8kT
πm
=
8RT
π M
(4.4)
Most probable speed
νp =
2kT
m
=
2RT
M
(4.5)
where m is the mass of the molecule, M is the molar weight, ρ the gas density,
k = 1.38 × 10−23
/K, the Boltzmann constant, R = 8.31 J/mol-K, is the universal
gas constant, and K is the Kelvin (absolute) temperature.
νp : ν : νrms ::
√
2 :

8/π :
√
3 (4.6)
247

248 4 Thermodynamics and Statistical Physics
The Maxwell distribution
N(ν)dν = 4π
m
2πkT
3/2
ν2
e−mν2
/2kT
dv (4.7)
Fig. 4.1 The Maxwell distribution
Flux
∅ =
1
4
n ν (number of molecules striking unit area per second) (4.8)
where n is the number of molecules per unit volume.
Mean free path (M.F.P)
λ =
1
√
2πnσ2
(4.9)
where n is the number of molecules per unit volume and σ is the diameter of the
molecule.
Collision frequency
f =
ν
λ
(4.10)
Viscosity of gas (η)
η =
1
3
ρλ ν (4.11)
Thermal conductivity (K)
K = ηCν (4.12)
where Cν is the specific heat at constant volume.

Coefficient of diffusion (D)
D =
η
ρ
(4.13)
Clausius Clepeyron equation
dP
dT
=
L
T (ν2 − ν1)
(4.14)
where ν1 and ν2 are the initial and final specific volumes (volume per unit mass) and
L is the latent heat.
Vander Waal’s equation
P +
a
V 2

(V − b) = RT (for one mole of gas) (4.15)
The Stefan-Boltzmann law
E = σ T4
(4.16)
If a blackbody at absolute temperature T be surrounded by another blackbody at
absolute temperature T0, the amount of energy E lost per second per square metre
of the former is
E = σ(T4
− T 4
0 ) (4.17)
where σ = 5.67 × 10−8
W/m2
.K4
is known as Stefan-Boltzmann constant.
Maxwell’s thermodynamic relations
First relation:

∂S
∂V

T
=

∂ P
∂T

V
(4.18)
Second relation:

∂S
∂ P

T
= −

∂V
∂T

P
(4.19)
Third relation:

∂T
∂V

S
= −

∂ P
∂S

V
(4.20)
Fourth relation:

∂T
∂ P

S
=

∂V
∂S

P
(4.21)

Thermodynamical potentials
(i) Internal energy (U)
(ii) Free energy (F)
(iii) Gibb’s function (G)
(iv) Enthalpy (H)
H = U + PV (4.22)

∂U
∂T

V
= CV (4.23)

∂ H
∂S

P
= T (4.24)

∂ H
∂ P

S
= V (4.25)

∂ H
∂T

P
= CP (4.26)
The Joule-Kelvin effect
ΔT =
,
T
∂V
∂T

P
− V
-
ΔP
CP
(4.27)
Black body radiation
Pradiatio =
u
3
(4.28)
where u is the radiation density
Wein’s displacement law
λm T = 0.29 cm - K (4.29)
Planck’s radiation law
uνdν =
8πhν3
dν
c3(ehν/kT − 1)
(4.30)
uλdλ =
8πhc
λ5
.
1
(ehc/λkT − 1)
(4.31)
σ =
2
15
.
π5
k4
h3c2
(4.32)
ΔS =
ΔQ
T
(4.33)
ΔS = kln(ΔW) (4.34)

4.2 Problems 251
where W is the number of accessible states.
Probability for finding a particle in the nth state at temperature T
P(n, T ) =
e−En/kT
Σ∞
n=0e−En /kT
(4.35)
Stirling’s approximation
n! =
√
2πn nn
e−n
(4.36)
4.2 Problems
4.2.1 Kinetic Theory of Gases
4.1 Derive the formula for the velocity distribution of gas molecules of mass m at
Kelvin temperature T .
4.2 Assuming that low energy neutrons are in thermal equilibrium with the sur-
roundings without absorption and that the Maxwellian distribution for veloci-
ties is valid, deduce their energy distribution.
4.3 In Problem 4.1 show that the average speed of gas molecule ν =
√
8kT/πm.
4.4 Show that for Maxwellian distribution of velocities of gas molecules, the root
mean square of speed ν2
1/2
= (3kT/m)1/2
4.5 (a) Show that in Problem 4.1 the most probable speed of the gas molecules
νp = (2kT/m)1/2
(b) Show that the ratio νp : ν : ν2
1/2
::
√
2 :
√
8/π :
√
3
4.6 Estimate the rms velocity of hydrogen molecules at NT P and at 127◦
C
[Sri Venkateswara University 2001]
4.7 Find the rms speed for molecules of a gas with density of 0.3 g/l of a pressure
of 300 mm of mercury.
[Nagarjuna University 2004]
4.8 The Maxwell’s distribution for velocities of molecules is given by N(ν)dν =
2π N(m/2πkT )3/2
ν2
exp(−mν2
/2kT )dν
Calculate the value of 1/ν
4.9 The Maxwell’s distribution of velocities is given in Problem 4.8. Show that the
probability distribution of molecular velocities in terms of the most probable
velocity between α and α + dα is given by

N(α)dα =
4N
√
π
α2
e−α2
dα
where, α = ν/νp and νp = (2kT/m)1/2
.
4.10 Calculate the fraction of the oxygen molecule with velocities between 199 m/s
and 201 m/s at 27◦
C
4.11 Assuming that the hydrogen molecules have a root-mean-square speed of
1,270 m/s at 300 K, calculate the rms at 600 K.
4.12 Clausius had assumed that all molecules move with velocity v with respect
to the container. Under this assumption show that the mean relative velocity
νrel of one molecule with another is given by νrel = 4ν/3.
4.13 Estimate the temperature at which the root-mean-square of nitrogen molecule
in earth’s atmosphere equals the escape velocity from earth’s gravitational
field. Take the mass of nitrogen molecule = 23.24 amu, and radius of
earth = 6,400 km.
4.14 Calculate the fraction of gas molecules which have the mean-free-path in the
range λ to 2λ.
4.15 If ρ is the density, ν the mean speed and λ the mean free path of the
gas molecules, then show that the coefficient of viscosity is given by η = 1
3
ρ
ν λ.
4.16 At STP, the rms velocity of the molecules of a gas is 105
cm/s. The molecular
density is 3 × 1025
m−3
and the diameter (σ) of the molecule 2.5 × 10−10
m.
Find the mean-free-path and the collision frequency.
4.17 When a gas expands adiabatically its volume is doubled while its Kelvin tem-
perature is decreased by a factor of 1.32. Calculate the number of degrees of
freedom for the gas molecules.
4.18 What is the temperature at which an ideal gas whose molecules have an aver-
age kinetic energy of 1 eV?
4.19 (a) If γ is the ratio of the specific heats and n is the degrees of freedom then
show that for a perfect gas
γ = 1 + 2/n
(b) Calculate γ for monatomic and diatomic molecules without vibration.
4.20 If K is the thermal conductivity, η the coefficient of viscosity, Cν the specific
heat at constant volume and γ the ratio of specific heats then show that for the
general case of any molecule
K
ηCν
=
1
4
(9γ − 5)

4.2 Problems 253
4.2.2 Maxwell’s Thermodynamic Relations
4.21 Obtain Maxwell’s Thermodynamic Relations
(a)

∂s
∂V

T
=

∂ P
∂T

V
(b)

∂s
∂ P

T
= −

∂V
∂T

P
4.22 Obtain Maxwell’s thermodynamic relation.

∂T
∂V

S
= −

∂p
∂S

V
4.23 Obtain Maxwell’s thermodynamic relation.

∂T
∂ P

S
=

∂V
∂S

P
4.24 Using Maxwell’s thermodynamic relations deduce Clausius Clapeyron equa-
tion

∂p
∂T

saturation
=
L
T (ν2 − ν1)
where p refers to the saturation vapor pressure, L is the latent heat, T the
temperature, ν1 and ν2 are the specific volumes (volume per unit mass) of the
liquid and vapor, respectively.
4.25 Calculate the latent heat of vaporization of water from the following data:
T = 373.2 K, ν1 = 1 cm3
, ν2 = 1, 674 cm3
, dp/dT = 2.71 cm of mercury
K−1
4.26 Using the thermodynamic relation

∂s
∂V

T
=

∂p
∂T

V
,
derive the Stefan-Boltzmann law of radiation.
4.27 Use the thermodynamic relations to show that for an ideal gas
CP − CV = R.
4.28 For an imperfect gas, Vander Waal’s equation is obeyed
p +
a
V 2

(V − b) = RT
with the approximation b/V ≪ 1, show that
CP − CV
∼
= R

1 +
2a
RT V

4.29 If E is the isothermal bulk modulus, α the coefficient of volume expansion
then show that
CP − CV = T Eα2
V

4.30 Obtain the following T ds equation
T ds = CV dT + T αET dV
where ET = −V
∂ P
∂V

T
is the isothermal elasticity and α = 1
V
∂V
∂T

P
is the volume coefficient of expansion, S is the entropy and T the Kelvin
temperature.
4.31 Obtain the equation
T ds = CpdT − T V αdp
4.32 Obtain the equation
T ds = CV

∂T
∂ P

V
dP + CP

∂T
∂V

P
dV
4.33 Obtain the formula for the Joule–Thompson effect
ΔT =
[T (∂V/∂T )P − V ]ΔP
CP
4.34 (a) Show that for a perfect gas governed by the equation of state PV = RT
the Joule-Thompson effect does not take place.
(b) Show that for an imperfect gas governed by the equation of state P +
a
V 2

(V − b) = RT , the Joule-Thompson effect is given by
ΔT =
1
CP

2a
RT
− b

ΔP.
4.35 Explain graphically the condition for realizing cooling in the Joule-Thompson
effect using the concept of the inversion temperature.
4.36 Prove that for any substance the ratio of the adiabatic and isothermal elastici-
ties is equal to the ratio of the two specific heats.
4.37 Prove that the ratio of the adiabatic to the isobaric pressure coefficient of
expansion is 1/(1 − γ ).
4.38 Show that the ratio of the adiabatic to the isochoric pressure coefficient is
γ/(γ − 1).
4.39 If U is the internal energy then show that for an ideal gas (∂U/∂V )T = 0.
4.40 Find the change in boiling point when the pressure on water at 100◦
C is
increased by 2 atmospheres. (L = 540 Calg−1
, volume of 1 g of steam =
1,677 cc)
4.41 If 1 g of water freezes into ice, the change in its specific volume is 0.091 cc
Calculate the pressure required to be applied to freeze 10 g of water at −1◦
C.
[Sri Venkateswara University 1999]

4.2 Problems 255
4.42 Calculate the change of melting point of naphthalene per atmospheric change
of pressure, given melting point = 80◦
C, latent heat = 35.5 cal/g, density of
solid = 1.145 g/cc and density of liquid = 0.981 g/cc
[University of Calcutta]
4.43 The total energy of blackbody radiation in a cavity of volume V at temperature
T is given by u = aV T 4
, where a = 4σ/c is a constant.
(a) Obtain an expression for the entropy S in terms of T, V and a.
(b) Using the expression for the free energy F, show that the pressure P = 1
3
u.
4.44 Given that the specific heat of Copper is 387 J/kg K−1
, calculate the atomic
mass of Copper in amu using Dulong Petit law.
4.45 Calculate the ratio of the number of molecules in the lowest two rotational
states in a gas of H2 at 50 K (take inter atomic distance = 1.05 A◦
)
[University of Cambridge, Tripos 2004]
4.46 Consider a photon gas in equilibrium contained in a cubical box of volume
V = a3
. Calculate the number of allowed normal modes of frequency ω in the
interval dω.
4.47 Show that for very large numbers, the Stirling’s approximation gives
n! ∼
=
√
2πn nn
e−n
4.48 Show that the rotational level with the highest population is given by
J max(pop) =
√
I0kT

−
1
2
4.49 Assuming that the moment of inertia of the H2 molecule is 4.64×10−48
kg-m2
,
find the relative population of the J = 0, 1, 2 and 3 rotational states at 400 K.
4.50 In Problem 4.49, at what temperature would the population for the rotational
states J = 2 and J = 3 be equal.
4.51 Calculate the relative numbers of hydrogen atoms in the chromosphere with
the principal quantum numbers n = 1, 2, 3 and 4 at temperature 6,000 K.
4.52 Calculate the probability that an allowed state is occupied if it lies above the
Fermi level by kT , by 5kT , by 10 kT .
4.53 If n is the number of conduction electrons per unit volume and m the electron
mass then show that the Fermi energy is given by the expression
EF =
h2
8m

3n
π
2/3

4.54 The probability for occupying the Fermi level PF = 1/2. If the probability for
occupying a level ΔE above EF is P+ and that for a level ΔE below EF is P−,
then show that for ΔE
kT
≪ 1, PF is the mean of P+ and P−
4.55 Find the number of ways in which two particles can be distributed in six states
if
(a) the particles are distinguishable
(b) the particles are indistinguishable and obey Bose-Einstein statistics
(c) the particles are indistinguishable and only one particle can occupy any
one state.
4.56 From observations on the intensities of lines in the optical spectrum of nitro-
gen in a flame the population of various vibrationally excited molecules rela-
tive to the ground state is found as follows:
v 0 1 2 3
Nv/N0 1.000 0.210 0.043 0.009
Show that the gas is in thermodynamic equilibrium in the flame and calcu-
late the temperature of the gas (θv = 3,350 K)
4.57 How much heat (in eV) must be added to a system at 27◦
C for the number of
accessible states to increase by a factor of 108
?
4.58 The counting rate of Alpha particles from a certain radioactive source shows
a normal distribution with a mean value of 104
per second and a standard
deviation of 100 per second. What percentage of counts will have values
(a) between 9,900 and 10,100
(b) between 9,800 and 10,200
(c) between 9,700 and 10,300
4.59 A system has non-degenerate energy levels with energy E =

n + 1
2

ω,
where ω = 8.625×10−5
eV, and n = 0, 1, 2, 3 . . . Calculate the probability
that the system is in the n = 10 state if it is in contact with a heat bath at room
temperature (T = 300 K). What will be the probability for the limiting cases
of very low temperature and very high temperature?
4.60 Derive Boltzmann’s formula for the probability of atoms in thermal equilib-
rium occupying a state E at absolute temperature T .
4.2.4 Blackbody Radiation
4.61 A wire of length 1 m and radius 1 mm is heated via an electric current to pro-
duce 1 kW of radiant power. Treating the wire as a perfect blackbody and
ignoring any end effects, calculate the temperature of the wire.

4.2 Problems 257
4.62 When the sun is directly overhead, the thermal energy incident on the earth is
1.4 kWm−2
. Assuming that the sun behaves like a perfect blackbody of radius
7 × 105
km, which is 1.5 × 108
km from the earth show that the total intensity
of radiation emitted from the sun is 6.4 × 107
Wm−2
and hence estimate the
sun’s temperature.
4.63 If u is the energy density of radiation then show that the radiation pressure is
given by Prad = u/3.
4.64 If the temperature difference between the source and surroundings is small
then show that the Stefan’s law reduces to Newton’s law of cooling.
4.65 The pressure inside the sun is estimated to be of the order of 400 million atmo-
spheres. Estimate the temperature corresponding to such a pressure assuming
it to result from the radiation.
4.66 The mass of the sun is 2 × 1030
Kg, its radius 7 × 108
m and its effective
surface temperature 5,700 K.
(a) Calculate the mass of the sun lost per second by radiation.
(b) Calculate the time necessary for the mass of the sun to diminish by 1%.
4.67 Compare the rate of fall of temperature of two solid spheres of the same
material and similar surfaces, where the radius of one surface is four times
of the other and when the Kelvin temperature of the large sphere is twice that
of the small one (Assume that the temperature of the spheres is so high that
absorption from the surroundings may be ignored).
4.68 A cavity radiator has its maximum spectral radiance at a wavelength of 1.0 µm
in the infrared region of the spectrum. The temperature of the body is now
increased so that the radiant intensity of the body is doubled.
(a) What is the new temperature?
(b) At what wavelength will the spectral radiance have its maximum value?
(Wien’s constant b = 2.897 × 10−3
m-K)
4.69 In the quantum theory of blackbody radiation Planck assumed that the oscil-
lators are allowed to have energy, 0, ε, 2ε . . . Show that the mean energy of
the oscillator is ε̄ = ε/[exp(ε/kT ) − 1] where ε = hν
4.70 Planck’s formula for the blackbody radiation is
uλdλ =
8πhc
λ5
1
ehc/λkT − 1
dλ
(a) Show that for long wavelengths and high temperatures it reduces to
Rayleigh-Jeans law.
(b) Show that for short wavelengths it reduces to Wien’s distribution law
4.71 Starting from Planck’s formula for blackbody radiation deduce Wien’s dis-
placement law and calculate Wien’s constant b, assuming the values of h, c
and k.

4.72 Using Planck’s formula for blackbody radiation show that Stefan’s constant
σ =
2
15
π5
k4
h3c2
= 5.67 × 10−8
W.m−2
.K−4
4.73 A blackbody has its cavity of cubical shape. Determine the number of modes
of vibration per unit volume in the wavelength region 4,990–5,010 A◦
.
4.74 A cavity kept at 4,000 K has a circular aperture 5.0 mm diameter. Calculate (a)
the power radiated in the visible region (0.4–0.7 µm) from the aperture (b) the
number of photons emitted per second in the visible region
4.75 Planck’s formula for the black body radiation is
uλdλ =
8πhc
λ5
1
ehc/λkT − 1
dλ
Express this formula in terms of frequency.
4.76 Estimate the temperature TE of the earth, assuming that it is in radiation
equilibrium with the sun (assume the radius of sun Rs = 7 × 108
m, the
earth-sun distance r = 1.5 × 1011
m, the temperature of solar surface Ts =
5,800 K)
4.77 Calculate the solar constant, that is the radiation power received by 1 m2
of earth’s surface. (Assume the sun’s radius Rs = 7 × 108
m, the earth-
sun distance r = 1.5 × 1011
m, the earth’s radius RE = 6.4 × 106
m,
sun’s surface temperature, Ts = 5,800 K and Stefan-Boltzmann constant
σ = 5.7 × 10−8 W
m2 − K4
).
4.78 A nuclear bomb at the instant of explosion may be approximated to a black-
body of radius 0.3 m with a surface temperature of 107
K. Show that the bomb
emits a power of 6.4 × 1020
W.
4.3 Solutions
4.1 Consider a two-body collision between two similar gas molecules of initial
velocity ν1 and ν2. After the collision, let the final velocities be ν3 and ν4.
The probability for the occurrence of such a collision will be proportional to
the number of molecules per unit volume having these velocities, that is to
the product f (ν1) f (ν2). Thus the number of each collisions per unit volume
per unit time is c f (ν1) f (ν2) where c is a constant. Similarly, the number of
inverse collisions per unit volume per unit time is c′
f (ν3) f (ν4) where c′
is
also a constant. Since the gas is in equilibrium and the velocity distribution is
unchanged by collisions, these two rates must be equal. Further in the centre

4.3 Solutions 259
of mass these two collisions appear to be equivalent so that c′
= c. We can
then write
f (ν1) f (ν2) = f (ν3) f (ν4)
or ln f (ν1) + ln f (ν2) = ln f (ν3) + ln f (ν4) (1)
Since kinetic energy is conserved
ν2
1 + ν2
2 = ν2
3 + ν2
4 (2)
Equations (1) and (2) are satisfied if
ln f (ν) ∝ ν2
(3)
or f (ν) = A exp(−αν2
) (4)
where A and α are constants. The negative sign is essential to ensure that no
molecule can have infinite energy.
Let N(ν)dν be the number of molecules per unit volume with speeds ν to
+dν, irrespective of direction. As the velocity distribution is assumed to be
spherically symmetrical, N(ν)dν is equal to the number of velocity vectors
whose tips end up in the volume of the shell defined by the radii ν and +dν,
so that
N(ν)dν = 4πν2
f (ν)dν (5)
Using (4) in (5)
N(ν)dν = 4π Aν2
exp(−αν2
) (6)
We can now determine A and α. If N is the total number of molecules per
unit volume,
N =
∞
0
N(ν)dν (7)
Using (6) in (7)
N = 4π A
∞
0
ν2
exp(−αν2
) dν = 4π A(1/4)(π/α3
)1/2
or N = A(π/α)3/2
(8)
If E is the total kinetic energy of the molecules per unit volume
E =
1
2
m
∞
0
ν2
N(ν)dν =
4π Am
2
∞
0
ν4
exp(−αν2
)dν
or E = (3m A/4)(π3
/α5
)1/2
(9)
where gamma functions have been used for the evaluation of the two integrals.
Further,

E = 3NkT/2 (10)
Combining (8), (9) and (10)
α =
m
2kT
(11)
and A = N(α/π)3/2
= N(m/2πkT )3/2
(12)
Using (11) and (12) in (5)
N(ν)dν = 4π N(m/2πkT )3/2
ν2
exp(−mν2
/2kT )dν
4.2 N(ν)dν = 4π N(m/2πkT )3/2
ν2
exp(−mν2
/2kT )dν (1)
Put E = 1
2
mν2
, dE = mνdν (2)
Use (2) in (1) and simplify to obtain
N(E)dE =
2π N E1/2
(πkT )3/2
exp

−
E
kT

dE
4.3 The average speed
ν =
∞
0 νN(ν)dν
N
= 4π
m
2πkT
3/2
∞
0
ν3
exp(−mν2
/2kT )dν (1)
where we have used the Maxwellian distribution
Put α =
m
2kT
(2)
so that
∞
0
ν3
e−αν2
dν =
1
2α2
(3)
ν =

8kT
πm
1/2
=
8RT
M
(4)
where m is the mass of the molecule, M is the molecular weight and R the gas
constant.
4.4 ν2
=
∞
0 ν2
N(ν)dν
N
= 4π
m
2πkT
3/2
∞
0
ν4
exp(−mν2
/2kT )dν
with α =
m
2kT
and x = αν2
; dx = 2ανdν
The integral, I =
∞
0
ν4
e−αν2
dν =
1
2α5/2
∞
0
x3/2
e−x
dx =
3
√
π
8α5/2
Therefore, ν2
= 4π
m
2πkT
3/2 3
√
π
8
m
2kT
5/2
=
3kT
m
ν2
1/2
= (3kT/m)1/2

4.3 Solutions 261
4.5 (a) νp is found by maximizing the Maxwellian distribution.
d
dν
[ν2
exp(−mν2
/2kT )] = 0
exp(−mν2
/kT )[2ν − mν3
/kT ] = 0
whence ν = νp = (2kT/m)1/2
(b) νp : ν : ν2
1/2
:: (2kT/m)1/2
: (8kT/πm)1/2
: (3kT/m)1/2
=
√
2 :

8/π :
√
3
4.6 ν2
1/2
=

3kT
m
1/2
=

3 × 1.38 × 10−23
× 273
1.67 × 10−27
1/2
= 2,601 m/s at N.T.P
ν2
1/2
=

3 × 1.38 × 10−23
× 400
1.67 × 10−27
1/2
= 3,149 m/s at 127◦
C.
4.7 ν2
1/2
=

3p
ρ
1/2
=

3 × (300/760) × 1.013 × 105
0.3
1/2
= 632 m/s
4.8
1
ν
=
1
N
∞
0
1
ν
N(ν) dν
=
1
N
∞
0
1
ν
.4π N
m
2πkT
3/2
v2
exp(−mν2
/2kT ) dν
Set mν2
/2kT = x; vdν = kT dx/m

1
ν
= (2m/πkT )1/2
∞

0
exp(−x) dx = (2m/πkT )1/2
4.9 N(ν)dν = 4π N(m/2πkT )3/2
ν2
exp(−mv2
/2kT )dν (1)
νp = (2kT/m)1/2
(2)
Let ν/νp = α; dν = νpdα (3)
Use (2) and (3) in (1)
N(α)dα =
4N
√
π
α2
exp(−α2
)dα
4.10 Fraction
f =
N(ν)dv
N
= 4π
* m
2πkT
+3/2
ν2
exp(−mν2
/2kT )dν
ν =
199 + 201
2
= 200 m/s
dν = 201 − 199 = 2 m/s

f = 4π

32 × 1.67 × 10−27
2π × 1.38 × 10−23 × 300
3/2
(200)2
exp

−
32 × 1.67 × 10−27
× 2002
2 × 1.38 × 10−23 × 300

× (2)
= 2.29 × 10−3
4.11 ν2
1/2
=

3kT
m
1/2
νrms(600 K) = [νrms(300 K)](600/300)1/2
= 1270 ×
√
2 = 1,796 m/s
4.12 Relative velocity νrel of one molecule and another making an angle θ is
νrel = (ν2
+ ν2
− 2(ν)(ν) cos θ)1/2
= 2ν sin(θ/2)
Now, all the direction of velocities v are equally probable. The probability
f (θ) that v lies within an element of solid angle between θ and θ +dθ is given
by
f (θ) = 2πsin θdθ/4π =
1
2
sin θdθ
νrel is obtained by integrating over f (θ) in the angular interval 0 to π.
νrel =
π
0
νrel f (θ) =
π
0
2ν sin

θ
2

1
2
sin θdθ

= 2ν
π
0
sin2

θ
2

cos
θ
2
dθ = 4ν
π
0
sin2 θ
2
d

sin
θ
2

= 4ν/3
4.13 νe = (2gR)1/2
; νrms = (3kT/m)1/2
νrms = νe
T =
2mgR
3k
=
2 × (2 × 23.24 × 10−27
)(9.8)(6.37 × 106
)
3 × 1.38 × 10−23
= 1.4 × 105
K
4.14 Fraction of gas molecules that do not undergo collisions after path length x
is exp(−x/λ). Therefore the fraction of molecules that has free path values
between λ to 2λ is
f = exp(−λ/λ) − exp(−2λ/λ)
= exp(−1) − exp(−2)
= 0.37 − 0.14 = 0.23
4.15 Consider a volume element dV = 2πr2
sin θdθdr located on a layer at a
height z = r cos θ. If mu is the momentum of a molecule at the XY-plane
at z = 0, then its value at dV will be mu +
d
dz
mu

r cos θ (Fig. 4.2). At an
identical layer below the reference plane dA, the momentum would be

4.3 Solutions 263
mu −

d
dz
mu

r cos θ
Let dn be the number of molecules with velocity between ν and ν + dν
per unit volume. The number of molecules with velocity ν and ν + dν in the
volume element dν is dndν. Molecules within the volume element undergo
collisions and are scattered in various directions.
Fig. 4.2 Transport of
momentum of gas molecules
Number of collisions that occur in dV in time dt will be 1
2
ν
λ
dt. The fac-
tor 1
2
is introduced to avoid counting each collision twice, since the collision
between molecules 1 and 2 and that between 2 and 1 is same.
Each collision results in two new paths for the scattered molecules. Hence
the number of molecules that are scattered in various directions from this vol-
ume element dV in time dt will be 2 × 1
2
ν
λ
dt × dndV or ν
λ
dtdndV .
Now the solid angle subtended by dA of the reference plane at dV is
dA cos θ/r2
.
Assuming the scattering to be isotropic the number of molecules moving
downward toward dA is
ν
λ
dtdndV dA cos θ
4πr2
or
νdtdn(2πr2
sin θdθdr)dA cos θ
λ.4πr2
or
νdtdndA sin θ cos θ
2λ
Transport of momentum downward from molecules in the upper hemi-
sphere through dA in time dt is

P− =
dAdt
2λ
∞
0
νdn
∞
0
e−r/λ
dr
π/2
0
sin θ cos θ

mu + r cos θ
dmu
dz

dθ
The factor e−r/λ
is included to ensure that the molecule in traversing the
distance r toward dA does not get scattered and prevented from reaching dA.
Similarly, transport of momentum upward, from molecules in the lower
hemisphere through dA in time dt is
P+ =
dAdt
2λ
∞
0
νdn
∞
0
e−r/λ
dr
π/2
0
sin θ cos θ

mu − r cos θ
dmu
dz

dθ
Hence net momentum transfer to the reference plane through an area dA in
time dt is
P = P− − P+ =
dAdt
λ
mdu
dz
∞
0
νdn
∞
0
re−r/λ
dr
π/2
0
cos2
θ sin θdθ
=
dAdt
λ
m
du
dz
n ν
λ2
3
=
m
3
dAdt
du
dz
λn ν
(the first integral gives n ν , the second one λ2
and the third one a factor
1/3)
Momentum transported per second is force
F =
λ
3
dAn ν m
du
dz
The viscous force is
ηdA
du
dz
=
λ
3
dAn ν m
du
dz
or η =
1
3
mn ν λ =
1
3
ρ ν λ
where mn = ρ = density of molecules.
4.16 λ =
1
√
2πnσ2
=
1
√
2π × 3 × 1025 × (2.5 × 10−10)2
= 1.2 × 10−7
m
f =
ν
λ
=
1, 000
1.2 × 10−7
= 8.33 × 109
s−1
4.17 T1V
γ −1
1 = T2V
γ −1
2 ,

V2
V1
γ −1
=
T1
T2
or 2γ −1
= 1.32, γ = 1.4
Number of degrees of freedom,
f =
2
γ − 1
=
2
1.4 − 1
= 5

4.3 Solutions 265
4.18 1eV = kT
T =
1eV
k
=
1.6 × 10−19
J
1.38 × 10−23 J/K
= 11,594 K
4.19 (a) For a perfect gas at temperature T , the kinetic energy from translation
motion
1
2
m ν2
x +
1
2
m ν2
y +
1
2
m ν2
z =
3
2
RT
N0
(1)
where R is the gas constant and N0 is Avagadro’s number. The energy of
the 3 degrees of freedom of translation is therefore on the average equal to
3
2
RT/N0 for each molecule. Using this result together with the principle of
the equipartition of energy, it is concluded that in a system at temperature T
each degree of freedom contributes, 1
2
R
N0
T to the total energy.
If each molecule has n degrees of freedom, the total internal energy U of a
gram-molecule of a perfect gas at temperature T ,
U =
1
2
nRT (2)
The molecular heat at constant volume C is equal to
∂U
∂T

ν
, and is therefore
given by
Cν =
1
2
nR (3)
For a perfect gas
Cp − Cν = R (4)
Therefore Cp = Cν + R =
(n + 2)R
2
(5)
and γ =
Cp
Cν
= 1 +
2
n
(6)
(b) For monatomic molecule n = 3, for translation (rotation and vibration are
absent), γ = 1.667.
For diatomic molecule n = 5 (3 from translation and only 2 from rotation
as the rotation about an axis joining the centres of atoms does not contribute)
and γ = 1.4
If vibration is included then n = 7 and γ = 1.286
4.20 According to Chapman and Enskog
K =
η
m

5
2
dEt
dT
+
dE
′
dT
'
(1)
where Et is the translational energy and E
′
the energy of other types.
If β denotes the number of degrees of freedom of the molecule due to causes
other than translation, the total number of degrees of freedom of the molecule
will be 3 + β.

From the law of equipartition of energy we have
dEt
dT
=
3
2
k;
dE
′
dT
=
β
2
k (2)
Hence,
K
η
=

5
2
.
3
2
+
β
2

k
m
(3)
We can express the result in terms of Cν and γ . From the law of equiparti-
tion of energy
Cν =
(3 + β)
2
.
k
m
; Cp =
(5 + β)
2
.
k
m
whence γ =
Cp
Cν
= 1 +
2
3 + β
or β =
5 − 3γ
γ − 1
(4)
Furthermore
Cν =
k
m(γ − 1)
(5)
K
ηCν
=
1
4
(9γ − 5)
4.3.2 Maxwell’s Thermodynamic Relations
4.21 Let f (x, y) = 0 (1)
d f =

∂ f
∂x

y
dx +

∂ f
∂y

x
dy = 0 (2)
Equation of state can be written as f (P, V, T ) = 0. By first law of thermody-
namics
dQ = dU + dW (3)
By second law of thermodynamics
dQ = T ds (4)
for infinitesimal reversible process
dW = pdV (5)

4.3 Solutions 267
Therefore,
dU = T ds − PdV (6)
where U is the internal energy, Q the heat absorbed, W the work done by the
system, S the entropy, P the pressure and T the Kelvin temperature.
Let the independent variables be called x and y. Then
U = U(x, y); V = V (x, y); S = S(x, y) (7)
Now,
d f =

∂ f
∂x

y
dx +

∂ f
∂y

x
dy (8)
Therefore
dU =

∂U
∂x

y
dx +

∂U
∂y

x
dy (9)
dV =

∂V
∂x

y
dx +

∂V
∂y

x
dy (10)
dS =

∂S
∂x

y
dx +

∂S
∂y

x
dy (11)
Eliminating internal energy U and substituting (9), (10) and (11) in (6)

∂U
∂x

y
dx +

∂U
∂y

x
dy = T

∂S
∂x

y
dx +

∂S
∂y

x
dy
'
−P

∂V
∂x

y
dx +

∂V
∂y

x
dy
'
(12)
Equating the coefficients of dx and dy

∂U
∂x

y
= T

∂S
∂x

y
− P

∂V
∂x

y
(13)

∂U
∂y

x
= T

∂S
∂y

x
− P

∂V
∂y

x
(14)
Differentiating (13) with respect to y with x fixed, and differentiating (14)
with respect to x with y fixed

1
∂
∂y

∂U
∂x

y
2
x
=

∂T
∂y

x

∂S
∂x

y
+ T
1
∂
∂y

∂S
∂x

y
2
x
(15)
−

∂ P
∂y

x

∂V
∂x

y
− P
1
∂
∂y

∂V
∂x

y
2
x

∂
∂x

∂U
∂y

x

y
=

∂T
∂x

y

∂S
∂y

x
+ T

∂
∂x

∂S
∂y

x

y
(16)
−

∂ P
∂x

y

∂V
∂y

x
− P

∂
∂x

∂V
∂y

x

y
Since the order of differentiation is immaterial, dU being a perfect differ-
ential, the left hand sides of (15) and (16) are equal. Further, since dS and dV
are perfect differentials.
1
∂
∂y

∂S
∂x

y
2
x
=

∂
∂x

∂S
∂y

x

y
(17)
and
1
∂
∂y

∂V
∂x

y
2
x
=

∂
∂x

∂V
∂y

x

y
(18)
Using (15), (16), (17), and (18),

∂ P
∂x

y

∂V
∂y

x
−

∂ P
∂y

x

∂V
∂x

y
=

∂T
∂x

y

∂S
∂y

x
−

∂T
∂y

x

∂S
∂x

y
(19)
Equation (19) can be written in the form of determinants
%
%
%
%
%
%
%
%

∂ P
∂x

y

∂ P
∂y

x

∂V
∂x

y

∂V
∂y

x
%
%
%
%
%
%
%
%
=
%
%
%
%
%
%
%
%

∂T
∂x

y

∂T
∂y

x

∂S
∂x

y

∂S
∂y

x
%
%
%
%
%
%
%
%
(20)
(a) Let the temperature and volume be independent variables. Put x = T and
y = V in (20). Then

∂T
∂x

y
=

∂V
∂y

x
= 1;

∂T
∂y

x
=

∂V
∂x

y
= 0
Since T and V are independent, we find

∂S
∂V

T
=

∂ P
∂T

V
(21)

4.3 Solutions 269
(b) Let the temperature and pressure be independent variables. Put x = T and
y = P in (20).

∂T
∂x

y
=

∂ P
∂y

x
= 1;

∂T
∂y

x
=

∂ P
∂x

y
= 0

∂S
∂ P

T
= −

∂V
∂T

P
(22)
4.22 In Problem 4.21 let the entropy and volume be independent variables. Put
x = s and y = V in Eq. (19)

∂S
∂x

y
=

∂V
∂y

x
= 1;

∂S
∂y

x
=

∂V
∂x

y
= 0

∂T
∂V

T
= −

∂ P
∂S

V
(23)
4.23 Let the entropy and pressure be independent variables. Put x = s and y = p
in Eq. (19) of Problem 4.21.

∂S
∂x

y
=

∂ P
∂y

x
= 1
Therefore,

∂T
∂p

s
=

∂V
∂S

p
(24)
4.24 Consider Maxwell’s relation (21) of Problem 4.21

∂S
∂V

T
=

∂ P
∂T

V
(1)
Multiply both sides by T,
T

∂S
∂V

T
= T

∂ P
∂T

V
(2)
or

∂ Q
∂V

T
= T

∂ P
∂T

V
(3)
which means that the latent heat of isothermal expansion is equal to the prod-
uct of the absolute temperature and the rate of increase of pressure with tem-
perature at constant volume. Apply (3) to the phase transition of a substance.
Consider a vessel containing a liquid in equilibrium with its vapor. The pres-
sure is due to the saturated vapor pressure which is a function of temperature
only and is independent of the volume of liquid and vapor present. If the vessel
is allowed to expand at constant temperature the vapor pressure would remain
constant. However, some liquid of mass δm would evaporate to fill the extra
space with vapor. If L is the latent heat absorbed per unit mass,

δQ = Ldm (4)
If ν1 and ν2 are the specific volumes (volumes per unit mass) of the liquid and
vapor respectively
δν = (ν2 − ν1)dm (5)
L
ν2 − ν1
= T

∂ P
∂T

V
(6)
Here, various thermodynamic quantities refer to a mixture of the liquid and
vapor in equilibrium. In this case

∂ P
∂T

V
=

∂V
∂T

sat
since the pressure is due to the saturated vapor and is therefore independent of
V , being only a function of T . Thus (6) can be written as

∂ P
∂T

sat
=
L
T (ν2 − ν1)
(Clapeyron’s equation) (7)
4.25 L = T (ν2 − ν1)
dP
dT
= 373.2(1,674 − 1) ×

2.71
76

× 1.013 × 106
= 2.255 × 1010
erg g−1
= 2.255 J/g
=
2.255
4.18
= 539.5 cal/g
4.26

∂S
∂V

T
=

∂ P
∂T

V
(1)
Substitute
dS =
dU + PdV
T
(2)
in (1)

∂U
∂V

T
= T

∂ P
∂T

V
− P (3)
If u is the energy density and P the total pressure,
∂U
∂V

= u and the total
pressure P = u/3, since the radiation is diffuse. Hence (3) reduces to

4.3 Solutions 271
u =
T
3
∂u
∂T
−
u
3
or
du
u
+ 4
dT
T
= 0
Integrating,
ln u = 4 ln T + ln a = ln aT 4
where ln a is the constant of integration. Thus,
u = aT 4
4.27 S = f (T, V )
where T and V are independent variables.
dS =

∂S
∂T

V
dT +

∂S
∂V

T
dV

∂S
∂T

p
=

∂S
∂T

V
+

∂S
∂V

T

∂V
∂T

P
Multiplying out by T and re-arranging
T

∂S
∂T

p
− T

∂S
∂T

V
= T

∂S
∂V

T

∂V
∂T

P
Now,
T

∂S
∂T

p
= Cp; T

∂S
∂T

ν
= Cν
and from Maxwell’s relation,

∂S
∂V

T
=

∂ P
∂T

ν
Therefore,
Cp − Cν = T

∂ P
∂T

V

∂V
∂T

P
(1)
For one mole of a perfect gas, PV = RT . Therefore

∂ P
∂T

V
=
R
V
and

∂V
∂T

P
=
R
P
It follows that
Cp − Cν = RT
4.28 P +
a
V 2

(V − b) = RT (1)
Neglecting b in comparison with V ,
P =
RT
V
−
a
V 2
(2)


∂ P
∂T

V
=
R
V
(3)
Re-writing (1)
PV +
a
V
= RT
Differentiating V with respect to T , keeping P fixed
P

∂V
∂T

P
−
a
V 2

∂V
∂T

P
= R
or

∂V
∂T

P
=
R
P − a/V 2
(4)
Now,
Cp − Cν = T

∂ P
∂T

V

∂V
∂T

P
(5)
(By Problem 4.27)
Cp − Cν =
R2
T
V (P − a/V 2)
= R
(P + a/V 2
)
(P − a/V 2)
≈ R(1 + 2a/PV2
)
= R

1 +
2a
RT V

4.29 If f (x, y, z) = 0, then it can be shown that

∂x
∂y

z

∂y
∂z

x

∂z
∂x

y
= −1 (1)
Thus, if f (P, V, T ) = 0

∂ P
∂V

T

∂V
∂T

P

∂T
∂ P

V
= −1 (1)
or

∂ P
∂T

V
= −

∂ P
∂V

T

∂V
∂T

P
(2)
and

∂V
∂T

P
= −

∂ P
∂T

V

∂V
∂ P

T
(3)
But
CP − CV = T

∂ P
∂T

V

∂V
∂T

P
(4)

4.3 Solutions 273
Use (2) and (3) in (4)
CP − CV = −T

∂ P
∂V

T

∂V
∂T
2
P
(5)
CP − CV = −T

∂V
∂ P

T

∂ P
∂T
2
V
(6)
Equation (5) can be written in terms of the bulk modulus E at constant tem-
perature and the coefficient of volume expansion ∝.
E = −

∂ P
∂V/V

; α =
1
V

∂V
∂T

(7)
Cp − Cν = T Eα2
V (8)
4.30 Taking T and V as independent variables
S = f (T, V )
dS =

∂S
∂T

V
dT + T

∂S
∂V

T
dV
Multiplying by T ,
T dS = T

∂S
∂T

V
dT + T

∂S
∂V

T
dV
= CV dT + T

∂S
∂V

T
dV
But

∂S
∂V

T
=

∂ P
∂T

ν
∴ T dS = CV dT + T

∂ P
∂T

V
dV
Also,

∂ P
∂T

V
= −

∂ P
∂V

∂V
∂T

P
∴ T dS = CV dT − T

∂ P
∂V

∂V
∂T

P
dV
Introducing relations α = 1
V
(∂V/∂T )P and ET = −V (∂ P/∂V )T for volume
coefficient of expansion and isothermal elasticity
T dS = CV dT + T αET dV

4.31 Taking T and P as independent variables
S = f (T, P)
dS =

∂S
∂T

P
dT +

∂S
∂ P

T
dP
or T dS = T

∂S
∂T

P
dT + T

∂S
∂ P

T
dP
= CP dT + T

∂S
∂ P

T
dP
or T dS = CPdT − T

∂V
∂T

P
dP
= CP dT − T V αdP
4.32 Taking P and V as independent variables,
S = f (P, V )
dS =

∂S
∂ P

V
dP +

∂S
∂V

P
dV
T dS = T

∂S
∂ P

V
dP + T

∂S
∂V

P
dV
= T

∂S
∂T

V

∂T
∂ P

V
dP + T

∂S
∂T

P

∂T
∂V

P
dV
= CV

∂T
∂ P

V
dP + CP

∂T
∂V

P
dV
4.33 In the Joule–Thompson effect heat does not enter the expanding gas, that is
ΔQ = 0. The net work done by the external forces on a unit mass of the gas
is (P1V1 − P2V2), where P1 and P2 refer to higher and lower pressure across
the plug respectively.
ΔW = P1V1 − P2V2
If the internal energy of unit mass is U1 and U2 before and after the gas
passes through the plug
ΔU = U1 − U2
By the first law of Thermodynamics
ΔQ = 0 = ΔW + ΔU
or U2 − U1 = P1V1 − P2V2
or Δ(U + PV ) = 0
or ΔH = 0

4.3 Solutions 275
where H is the enthalpy
∴ T ΔS + V ΔP = 0
But by Problem 4.31
T ΔS = CPΔT − T

∂V
∂T

P
ΔP
∴ CP ΔT +

V − T

∂V
∂T

P

ΔP = 0
or ΔT =
,
T
∂V
∂T

P
− V
-
ΔP
CP
4.34 (a) For perfect gas
PV = RT
P

∂V
∂T

P
= R
T

∂V
∂T

P
=
T R
P
= V
or T

∂V
∂T

P
− V = 0
∴ ΔT = 0 by Problem 4.31
(b) For imperfect gas
P +
a
V 2

(V − b) = RT
or PV = RT −
a
V
+ bP +
ab
V 2
P

∂V
∂T

P
= R +
a
V 2

∂V
∂T

P
−
2ab
V 3

∂V
∂T

P
Re-arranging

∂V
∂T

P
=
R
P − a
V 2 + 2ab
V 3
=
R
RT
V −b
− 2a
V2

1 − b
V

Multiplying both numerator and denominator of RHS by (V − b)/R
T

∂V
∂T

P
= (V − b)

1 −
2a
RT V3
(V − b)2
−1
= (V − b)

1 +
2a
RT V3
(V − b)2

= (V − b) +
2a
RT V3
(V − b)3
T

∂V
∂T

P
− V =
2a
RT
− b (∴ b ≪ V )
Using this in the expression for Joule–Thompson effect (Problem 4.31),
ΔT =
1
Cp

2a
RT
− b

ΔP
4.35 The equation of state for an imperfect gas is
p +
a
V 2

(V − b) = RT
It can be shown that
ΔT =
1
Cp

2a
RT
− b

Δp
If T 2a/bR, ΔT/Δp is positive and there will be cooling.
If T 2a/bR, ΔT/Δp will be negative and the gas is heated on undergo-
ing Joule–Kelvin expansion.
If T = 2a/bR, ΔT/Δp = 0, there is neither heating nor cooling.
The temperature given by Ti = 2a
bR
is called the temperature of inversion
since on passing through this temperature the Joule–Kelvin effect changes its
sign. Figure 4.3 shows the required curve.
Fig. 4.3 Joule-Thompson
effect
4.36 By definition
ET = −V

∂ P
∂V

T
; ES = −V

∂ P
∂V

S

4.3 Solutions 277
ES
ET
=
(∂ P/∂V )S
(∂ P/∂V )T
=
(∂ P/∂V )S

∂T
∂V

S
(∂ P/∂S)T

∂S
∂V

T
=
(∂T/∂V )S

∂S
∂ P

T
(∂T/∂ P)S

∂S
∂V

T
=
(∂ P/∂S)V (∂V/∂T )P
(∂V/∂S)P(∂ P/∂T )V
from the relations given in Problems 4.21 and 4.22
∴
ES
ET
=
(∂S/∂T )P
(∂S/∂T )V
=
(∂ Q/∂T )P
(∂ Q/∂T )V
=
CP
CV
= γ
4.37
(∂V/∂T )S
(∂V/∂T )P
=
1
(∂T/∂V )S(∂V/∂T )P
=
1
−

∂ P
∂S

V

∂V
∂T

P
where we have used Eq. (23) of Problem 4.22.
Writing

∂ P
∂S

V
=

∂ P
∂T

V

∂T
∂S

V
=
(∂ P/∂T )V
(∂S/∂T )V
(∂V/∂T )S
(∂V/∂T )P
=
(∂S/∂T )V
(∂ P/∂T )V (∂V/∂T )P
=
(∂S/∂T )V
−(CP − CV )/T
(by Eq. (4.1) of Problem 4.27
T (∂S/∂T )V
−(CP − CV )
=
CV
−(CP − CV )
=
1
1 − γ
4.38
(∂ P/∂T )S
(∂ P/∂T )V
=
1
(∂T/∂ P)S(∂ P/∂T )V
=
1

∂V
∂S

P

∂ P
∂T

V
=
1

∂V
∂T

P

∂T
∂S

P

∂ P
∂T

V
=
(∂S/∂T )P

∂V
∂T

P
(∂ P/∂T )V
=
T (∂S/∂T )P
(CP − CV )
=
CP
(CP − CV )
=
γ
γ − 1
where we have used Eq. (4.24) of Problem 4.22 and the relation
CP = T

∂S
∂T

P

4.39 By Maxwell’s first equation

∂S
∂V

T
=

∂ P
∂T

V
(1)
dS =
dU + PdV
T
(2)
using (2) in (1)

∂U
∂V

T
= T

∂ P
∂T

− P
For perfect gases,
P =
RT
V

∂U
∂V

T
=
RT
V
− P = 0
Thus, temperature remaining constant, the internal energy of an ideal gas
is independent of the volume.
4.40
dP
dT
=
L
T (ν2 − ν1)
dT =
T
L
(ν2 − ν1)dP
=
373(1677 − 1)(2 × 106
)
546 × 4.2 × 107
= 55.1◦
C
4.41 ν1 = 1 cm3
; v2 =
1
0.091
= 10.981 cm3
dP =
LdT
T (ν2 − ν1)
=
80 × 4.2 × 107
× 1
(−1 + 273)(10.981 − 1.0)
= 1.238 ×
106
dynes
cm2
= 1.24 atm
P2 = P1 + dP = 1.0 + 1.24 = 2.24 atm
4.42 ν1 =
1
ρ1
=
1
1.145
= 0.873 cm3
/g
ν2 =
1
ρ2
=
1
0.981
= 1.019 cm3
/g
dT =
T (ν2 − ν1)dP
L
=
(80 + 273)(1.019 − 0.873)(1.0 × 106
)
35.5 × 4.2 × 107
= 0.0346◦
C

4.3 Solutions 279
4.43 (a) Use the relation
dU = T ds − PdV (1)
Here,
dV = 0(∵ V = constant) and
U = aV T4
(2)
dU = 4aV T 3
dT = T ds

ds
dT

V
= 4aV T 2
Integrating S = 4
3
aT3
V
(b) F = U − T S = aV T 4
−
4
3
aT 4
V = −
1
3
aV T 4
p = −

∂ F
∂V

T
=
1
3
aT 4
=
1
3
u
4.44 According to Dulong-Petit’s law the molar specific heats of all substances,
with a few exceptions like carbon, have values close to 6 cal/mol◦
C−1
. The
specific heat of Cu is 387
kgK−1 = 0.387J
gK−1 = 0.0926cal/gK−1
. Therefore, the atomic
mass of Cu = 6
0.0926
= 64.79 amu.
4.45 Probability for the rotational state to be found with quantum number J is given
by the Boltzmann’s law.
P(E) ∝ (2J + 1) exp[−J(J + 1)2
/2I0kT
where I0 is the moment of inertia of the molecule, k is Boltzmann’s constant,
and T the Kelvin temperature. The two lowest states have J = 0 and J = 1
I0 = M(r/2)2
+ M(r/2)2
=
1
2
Mr2
, where M = 938 MeV/c2
2I0 = Mr2
= 938 × (1.05 × 10−10
)2
/c2
c = 197.3 MeV − 10−15
m
kT = 1.38 × 10−23
×
50
1.6 × 10−13
= 43.125 × 10−10
2
2I0kT
=
2
c2
Mc2r2kT
=
(197.3)2
× 10−30
938 × (1.05 × 10−10)2 × 43.125 × 10−10
= 0.8728
For J = 1,
J(J + 1)2
2kT
= 1 × (1 + 1) × 0.8728 = 1.7457
For J = 0, P(E0) ∝ 1.0
For J = 1, P(E1) ∝ (2 × 1 + 1) exp(−1.7457) = 0.52
∴ P(E0) : P(E1) :: 1 : 0.52

4.46 For stationary waves, in the x-direction
kx a = nx π
or nx = kx a/π
dnx = (a/π)dkx
Similar expressions are obtained for y and z directions.
dn = dnx dnydnz
= (a/π)3
d3
k
However only the first octant of number space is physically meaningful.
Therefore
dn = (1/8)(a/π)3
d3
k
Taking into account the two possible polarizations
dn =
2V
(2π)3
d3
k =
2V
8π3
.4πk2
dk
But k =
ω
c
; dk = dω/c
∴ dn =
V ω2
dω
π2c3
4.47 n! = n(n − 1)(n − 2) . . . (4)(3)(2)
Take the natural logarithm of n!
ln n! = ln 2 + ln 3 + ln 4 + · · · + ln(n − 2) + ln(n − 1) + ln n
= Σn
n=1 ln n
=
n
1
ln n dn
= n ln n − n + 1
≈ n ln n − n
where we have neglected 1 for n ≫ 1
4.48 p(E) = (2J + 1)e−J(J+1)2
/2ikT
The maximum value of p(E) is found by setting dp(E)/dJ = 0

2 −
(2J + 1)2
2
2I0kT

e−J(J+1)2
/2I0kT
= 0
Since the exponential factor will be zero only for J = ∞,

2 −
(2J + 1)2
2
2I0kT

= 0
Solving for J, we get
Jmax =
√
I0kT

−
1
2

4.3 Solutions 281
4.49 p(EJ ) = (2 j + 1)e
−J(J+1)2
2I0kT
The factor
2
2I0k
=
(1.055 × 10−34
)2
2 × 4.64 × 10−48 × 1.38 × 10−23 J
= 86.9
p(E0) = 1
p(E1) = 3e−2×86.9/400
= 1.942
p(E2) = 5e−6×86.9/400
= 1.358
p(E3) = 7e−12×86.9/400
= 0.516
4.50 p(E2) = 5e−6×86.9/T
= 5e−521.4/T
(1)
p(E3) = 7e−12×86.9/T
= 7e−1042.8/T
Equating p(E2) and p(E3) and solving for T , we find T = 1,549 K
4.51 For Boltzmann statistics p(E) ∝ e−E/kT
Therefore,
p(En)
p(E1)
= e−(En−E1)/kT
In hydrogen atom, if the ground state energy E1 = 0, then E2 = 10.2,
E3 = 12.09 and E4 = 12.75 eV
The factor kT = 8.625 × 10−5
× 6, 000 = 0.5175
P(E2)/P(E1) = e−10.2/0.5175
= 2.75 × 10−9
P(E3)/P(E1) = e−12.09/0.5175
= 1.4 × 10−10
P(E4)/P(E1) = e−12.75/0.5175
= 1.99 × 10−11
Thus P(E1) : P(E2) : P(E3) :: 1 : 2.8 × 10−9
: 1.4 × 10−10
: 2.0 × 10−11
This then means that the hydrogen atoms in the chromospheres are predomi-
nantly in the ground state.
4.52 p(E) =
1
e(E−EF )/kT + 1
For E − EF = kT, p(E) =
1
e + 1
= 0.269
For E − EF = 5kT, p(E) =
1
e5 + 1
= 6.69 × 10−3
For E − EF = 10kT, p(E) =
1
e10 + 1
= 4.54 × 10−5
4.53 For the conduction electrons, the number of states per unit volume with energy
in the range E and E+dE, can be written as n(E)dE where n(E) is the density
of states. Now, for a free electron gas
n(E) =
8
√
2πm3/2
h3
E1/2

Let P(E) be the probability function which gives the probability of the state
at the energy E to be occupied. At T = 0 all states below a certain energy are
filled (P = 1) and all states above that energy are vacant (P = 0). The highest
occupied state under the given conditions is called the Fermi energy.
The product of the density n(E) of available states and the probability P(E)
that those states are occupied, gives the density of occupied states n0(E);
that is
n0(E) = n(E)P(E)
The total number of occupied states per unit volume is given by
n =
EF
0
n0(E)dE
=
8
√
2πm3/2
h3
EF
0
E1/2
d(E)
=
8
√
2πm3/2
h3
.
2
3
E
3/2
F
or EF =
h2
8m

3n
π
2/3
4.54 P+ =
1
e(E−EF )/kT + 1
=
1
eΔ/kT + 1
≈
1
2 + Δ/kT
=
1
2
(1 − Δ/2kT )
P− =
1
2
(1 + Δ/2kT )
∴
P+ + P−
2
= 1/2 = PF
4.55 (a) For n states, the number of ways is N = n2
. Therefore, for n = 6 states
N = 36
(b) For n states the number of ways is N = n2
−(n−1)orn2
−n+1. Therefore,
for n = 6, N = 31
(c) For n states, N = n2
− n + 1 − n or n2
− 2n + 1. Therefore for n = 6,
N = 25
4.56 If the gas is in equilibrium, the number of particles in a vibrational state is
Nν = N0 exp

−
hν
kT

= N0 exp

−
θ
T

.
The ratios, N0/N1 = 4.7619, N1/N2 = 4.8837, N2/N3 = 4.7778, are seen
to be constant at 4.8078. Thus the ratio Nν/Nν+1 is constant equal to 4.81,
showing the gas to be in equilibrium at a temperature
T = 3, 350/(ln 4.81) ≈ 2, 130 K
4.57 ΔS = k ln(ΔW)
But ΔS = ΔQ/T

4.3 Solutions 283
or ΔQ = T ΔS = kT ln ΔW
= (1.38 × 10−23
)(300) ln 108
= 7.626 × 10−20
J = 0.477 eV
4.58 The Gaussian (normal) distribution is
f (x) =
1
σ
√
2π
e−(x−μ)2
/2σ2
where μ is the mean and σ is the standard deviation. The probability is found
from
(a) P(μ − σ x μ + σ) =
μ+σ
μ−σ
f (x)dx
Letting z = x−μ
σ
P(−1 z 1) =
1

−1
φ(z)dz
= 2
1
0
φ(z)dz (from symmetry)
= 2 × 0.3413 = 0.6826 (from tables)
or 68.26%(shown shaded under the curve, Fig 4.4)
Fig. 4.4
(b) Similarly
P(μ − 2σ) x μ + 2σ) = 0.9544 or 95.44%
(c) P(μ − 3σ) x (μ + 3σ) = 0.9973 or 99.73%
4.59 P(n, T ) =
e−
(n+ 1
2 )ω
kT
Σ∞
n=0e−
(n+ 1
2 )ω
kT
=
e−(n+ 1
2 )ω/kT
e− 1
2 ω/kT
Σ∞
n=1enω/kT

=
e−nω/kT
e−ω/kT
1 − e−ω/kT
=
e−nω/kT
1
eω/kT −1
= e−nω/kT

eω/kT
− 1

Substitute n = 10,
ω
k
=
8.625 × 10−5
(1.38 × 10−23/1.6 × 10−19)
= 1.0
P(10, 300) = 3.2 × 10−3
In the limit T → 0, the state n = 0 alone is populated so that n = 10 state is
unpopulated.
In the limit T → ∞, probability for n = 10 again goes to zero, as higher
states which are numerous, are likely to be populated.
4.60 Consider a collection of N molecules of a large number of energy states,
E1, E2, E3 etc such that there are N1 molecules in state E1, N2 in E2 and
so on. The nature of energy is immaterial. The number of ways in which N
molecules can be accommodated in various states is given by
W =
N!
N1!N2! . . .
(1)
The underlying idea is that the state of the system would be state if W is a
maximum.
Taking logs on both sides and applying Stirling’s approximation ln W =
N ln N − N − ΣNi ln Ni + ΣNi
= N ln N − ΣNi ln Ni (2)
because ΣNi = N (3)
ΣNi Ei = E (4)
If the system is in a state of maximum thermodynamic probability, the varia-
tion of W with respect to change in Ni is zero, that is
ΣδNi = 0 (5)
ΣEi δNi = 0 (6)
Σ(1 + ln Ni )δNi = 0 (7)
We now use the Lagrange method of undetermined multipliers. Multiplying
(5) by α and (6) by β and adding to (7), we get
Σ{(1 + ln Ni ) + α + βEi }δNi = 0 (8)
Therefore
ln Ni + 1 + α + βEi = 0 (9)
or Ni = Ce−βEi
(10)

4.3 Solutions 285
Where C = constant which can be determined as follows.
ΣNi = N = CΣe−βEi
(11)
or C =
N
Σe−βEi
(12)
Ni =
Ne−βEi
Σe−βEi
(13)
The denominator in (13)
Z = Σe−βEi
(14)
Is known as the partition function. It can be shown that the quantity
β =
1
kT
(15)
where k is the Boltzmann constant and T is the absolute temperature.
α =
N
Z
(16)
4.61 Electric power = power radiated
W = σ T 4
A
A = 2πrl = 2π × 10−3
× 1.0 = 6.283 × 10−3
m2
T =

W
σ A
1/4
=

1,000
5.67 × 10−8 × 6.283 × 10−3
1/4
= 1,294 K
4.62 The Solar constant S is the heat energy received by 1 m2
of earth’s surface per
second. If R is the radius of the sun and r the earth-sun distance, then the total
intensity of radiation emitted from the sun will be σ T 4
W m−2
and from the
sun’s surface σ T 4
.4π R2
. The radiation received per second per m2
of earth’s
surface will be
S = σ T4
.
4π R2
4πr2
Solving,
σ T 4
= S.
r2
R2
= 1,400

1.5 × 108
7 × 105
2
= 6.43 × 107
W m−2
T =

6.43 × 107
σ
1/4
=

6.43 × 107
5.67 × 10−8
1/4
= 5,800 K
4.63 Using the analogy between radiation (photon gas) and gas molecules, the pho-
tons move in a cavity at random in all directions, rebounding elastically from
the walls of the cavity. The pressure exerted by an ideal photon gas is

p =
1
3
ρ ν2

where ρ is the mass density. In the case of photon gas, the speed of all photons
is identical being equal to c. Furthermore, from Einstein’s relation
u = ρc2
where u is the energy density. Replacing ν2
by c2
prad =
1
3
ρc2
=
u
3
4.64 Let T and T0 be the Kelvin temperatures of the body and the surroundings.
Then, by Stefan–Boltzmann law, the rate of loss of heat per unit area of the
body is
dQ
dt
= σ(T4
− T 4
0 )
= σ(T − T0)(T + T0)(T 2
+ T2
0 )
If (T − T0) be small, (T ≈ T0), and
dQ
dt
= σ(T − T0) × 4T 3
0
Since T0 is constant,
dT
dt
∝ (T − T0); (Newton’s law of cooling).
4.65 The energy density u and pressure p of radiation are related by
p =
u
3
Furthermore, u = 4σ T 4
/c
Eliminating u,
T =

3cp
4σ
1/4
=

3 × 3 × 108
× 4 × 108
× 1.013 × 105
4 × 5.67 × 10−8
1/4
= 2 × 107
K
4.66 (a) Power, P = σ AT 4
= 4π R2
σ T 4
= 4π(7 × 108
)2
(5.67 × 10−8
)(5,700)4
= 3.68 × 1026
W
Mass lost per second, m = P/c2
=
3.68 × 1026
(3 × 108)2
= 4.1 × 109
kg/s
(b) Time taken for the mass of sun (M) to decrease by 1% is
t =
M
100
×
1
m
=
2 × 1030
100
×
1
4.1 × 109
= 4.88 × 1018
s
=
4.88 × 1018
3.15 × 107
= 1.55 × 1011
years

4.3 Solutions 287
4.67 Power radiated, P = σ AT 4
= 4π R2
σ T 4
P2
P1
=
R2
2
R2
1
.
T 4
2
T4
1
=
(4R1)2
R2
1
.
(2T1)4
T4
1
= 256
Furthermore,
P2
P1
=
dQ2/dt
dQ1/dt
=
m2s(dT/dt)2
m1s(dT/dt)1
where s is the specific heat
But m2 ∝ R3
2 and m1 ∝ R3
1
∴
(dT/dt)2
(dT/dt)1
=
P2
P1
.
R3
1
R3
2
=
256
43
= 4
4.68 (a) λm.T = b
T =
b
λm
=
2.897 × 10−3
1 × 10−6
= 2,897 K
P2
P1
=
T 4
2
T 4
1
= 2
New temperature, T2 = T1 × 21/4
= 2,897 × 1.189 = 3,445K
(b) The wavelength at which the radiation has maximum intensity
λm =
2.897 × 10−3
3445
= 0.84 × 10−6
m = 0.84 µm
4.69 The mean value ∈ is determined from;
∈ =
Σ∞
n=0n ∈ e−βn∈
Σ∞
n=0e−βn∈
= −
d
dβ
ln
∞

n=0
e−βn∈
= −
d
dβ
ln

1 + e−β∈
+ e−2β∈
+ · · ·

= −
d
dβ
ln
1
1 − e−β∈
where we have used the formula for the sum of terms of an infinite geometric
series.
∈ =
∈ e−β∈
1 − e−β∈
=
∈
eβ∈ − 1
(β = 1/kT )
4.70 (a)
uλdλ =
8πhc
λ5
.
1
ehc/λkT − 1
dλ (Planck’s formula) (1)
For long wavelengths (low frequencies) and high temperatures the ratio
hc
λkT
≪ 1 so that we can expand the exponential in (1) and retain only the
first two terms
uλdλ =
8πhc
λ5[(1 + hc/λkT + . . .) − 1]
=
8πkT
λ4
dλ
writing λ = c
υ
; dλ = − c
υ2 dν

uν =
8πν2
c3
kT (Rayleigh-Jeans law)
(b) If hν/kT ≫ 1 i.e hc/λkT ≫ 1 then we can ignore 1 in the denominator
in comparison with the exponential term in Planck’s formula
uλdλ = c1e−c2/λkT
dλ (Wien’s distribution law)
where the constants, c1 = 8πhc and c2 = hc
4.71 uλdλ =
8πhc
λ5
.
1
ehc/λkT − 1
dλ (Planck’s formula)
The wavelength λm corresponding to the maximum of the distribution curve
is obtained from the condition

duλ
dλ

λ=λm
= 0
Differentiating and writing hc/kT λm = β, gives
e−β
+
β
5
− 1 = 0
This is a transcental equation and has the solution
β = 4.9651, so that
λm T =
hc
4.9651k
= b = constant.
Thus, the constant
b =
6.626068 × 10−34
× 2.99792 × 108
4.9651 × 1.38065 × 10−23
= 2.8978 × 10−3
m-K
a value which is in excellent agreement with the experiment.
4.72 By definition
u =

uνdν = aT 4
(1)
Inserting Planck’s formula in (1)
u = aT4
=
8πh
c3
∞
0
ν3
dν
ehν/kT − 1
=
8πk4
T4
h3c3
∞
0
x3
dx
ex − 1
where x = hν/kT
a =
8πk4
h3c3
∞
0
x3
(e−x
+ e−2x
+ . . . e−rx
+ . . .)
Now,
∞
0 x3
e−rx
dx = 6
r4 , and Σ∞
r=1
1
r4 = π2
90
a =
48πk4
h3c3
.
π4
90
=
8
15
π5
k4
h3c3
∴ σ =
ac
4
=
2
15
π5
k4
h3c2

4.3 Solutions 289
=
2
15
(3.14159)5
(1.38065)4
× 10−92
(6.626068 × 10−34)3(2.99792 × 108)2
= 5.67 × 10−8
W-m−2
-K−4
a value which is in excellent agreement with the experiment.
4.73 Number of modes per m3
in the frequency interval dν is
N =
8πν2
dν
c3
But,
ν =
c
λ
; dν = −
dλ
λ2
; λ =
4,990 + 5,010
2
= 5,000 A0
dλ = 5,010 − 4,990 = 20 A0
∴ N =
8πdλ
λ4
=
8π × 20 × 10−10
(5 × 10−7)4
= 8.038 × 1017
/m3
4.74 (a)
(1)
P = AEλdλ =
8πhcAdλ
λ5(ehc/λkT − 1)
(2)
Mean wavelength λ = 0.55 µm = 5.5 × 10−7
m.
dλ = (0.7 − 0.4) µm = 3 × 10−7
m
A = πr2
= π(2.5 × 10−3
)2
= 1.96 × 10−5
m2
hc
λkT
=
(6.63 × 10−34
)(3 × 108
)
(5.5 × 10−7)(1.38 × 10−23)(4,000)
= 6.55
Using the above values in (2) we find
P = AEλdλ = 0.84 × 10−6
W = 0.84 µW.
(b) hν =
hc
λ
=
6.63 × 10−34
× 3 × 108
5.5 × 10−7
= 3.616 × 10−19
Number of photons emitted per second
n =
P
hν
= 0.84 × 10−6
/3.616 × 10−19
= 2.32 × 1012
/s
4.75 uλdλ =
8πhc
λ5
1
ehc/λkT − 1
dλ (1)
Put λ = c/ν (2)
and dλ = −
c
ν2

dν (3)
in the RHS of (1) and simplify
uνdν =
8πhν3
c3(ehν/kT − 1)
dν (4)

The negative sign in (3) is omitted because as λ increases v decreases.
4.76 Power radiated from the sun = σ × (surface area) × T 4
s
Ps = σ4π R2
s T4
s
Power received by the earth,
PE =
π R2
e
4πr2
.Ps
The factor π R2
e represents the effective (projected) area of the earth on
which the sun’s radiation is incident at a distance r from the sun. The factor
4πr2
is the surface area of a sphere scooped with the centre on the sun. Thus
π R2
e /4πr2
is the fraction of the radiation intercepted by the earth’s surface
area.
Now power radiated by earth,
PE = σ4π R2
E T 4
E
For radiation equilibrium, power radiated by the earth=power received by
the earth.
σ4π R2
E T 4
E = σ4π R2
s T 4
s .
π R2
E
4πr2
or TE = Ts

Rs
2r
1/2
= 5,800

7 × 108
2 × 1.5 × 1011
1/2
= 280 K = 7◦
C
Note that the calculations are approximate in that the earth and sun are not
black bodies and that the contribution of heat from the interior of the earth has
not been taken into account.
4.77 Power radiated by the sun, Ps = σ4π R2
s T4
s
Power received by 1 m2
of earth’s surface,
S =
σ4π R2
s T4
s
4πr2
=
(5.7 × 10−8
)(7 × 108
)2
(5,800)4
(1.5 × 1011)2
= 1,400 W/m2
4.78 P = 4πr2
σ T 4
= 4π(0.3)2
(5.67 × 10−8
)(107
)4
= 6.4 × 1020
W

Chapter 5
Solid State Physics
Crystal Structure
There are seven crystal systems Cubic, Tetragonal, Orhtorhombic, monoclinic, tri-
clinic, Rhombohedral, Hexagonal. They are distinguished by the axial lengths and
axial angles. The lengths are taken as a, b and c. In the cubic system a = b = c and
the angle between any two axes is a right angle.
Bragg’s equation
2d sin θ = nλ (5.1)
where d is the distance between parallel diffraction planes, θ is the angle between
the incident beam and the diffraction plane and n is the order of diffraction. The
distance d can be related to the lattice parameters of the crystal cell. In the simple
cubic cell the distance between (100) planes is “a”, the lattice parameter. The dis-
tance between parallel (110) planes passing through lattice points is a/
√
2; for (111),
it is a/
√
3. In general distance between parallel planes of indices (hkl) in terms of
the parameter “a” for the cubic system is
d2
=
a2
h2 + k2 + l2
(5.2)
Electrical properties of crystals
The principal attractive force between ions of opposite sign is an electrostatic force.
The repulsive force arises from the interaction of the electron clouds surrounding an
atom. This force arises because of the exclusion principle and is not electrostatic in
nature. Empirically this is represented by b/rn
, where b and n are constants, and r
is the anion-cation distance.
291

292 5 Solid State Physics
The total energy of the lattice is the sum of the attractive and repulsive energies
U = −N A
e2
r
+
b
rn
(5.3)
where N is the number of molecules and A is known as the Madelung constant.
For equilibrium
dU
dr
= 0 (5.4)
The force is
−
dU
dr
= 0 (5.5)
The current density
J = i/A (5.6)
where A is the cross-section of the conductor.
The drift speed
vd = j/ne (5.7)
where n is the number of conduction electrons per unit volume. The resistivity is
given by
R = ρL/A (5.8)
The conductivity is given by
σ = 1/ρ (5.9)
vd = eEτ/m (5.10)
where E is the electric field and τ is the mean time between collisions.
ρ = me/ne2
τ (5.11)
τ = meσ/ne2
(5.12)
The mean free path
λ = τ v (5.13)
Hall effect
If a thin strip of material carrying a constant current is placed in a magnetic field
B perpendicular to the strip a potential difference appears across the strip. This is
known as Hall effect.
E = jB/qn (5.14)

Metals, insulators and semiconductors
Materials are distinguished by the extent to which the valence and conduction bands
are filled by electrons. The bands in solids may be filled, partially or empty. A good
conductor has a conduction band that is approximately half filled or the conduction
band overlaps the next higher band. In this case it is very easy for the valence elec-
tron to be raised to a higher energy level under the application of electric field and
provide electrical conduction.
In an insulator the valency band is completely filled and the energy gap (Eg) with
the conduction gap is large (∼ 5 eV).
In the case of semiconductors the valence band is completely filled, like an insu-
lator. However, the conduction band is empty, so that at room temperature some
of the electrons acquire sufficient energy to be found in the conduction band. Fur-
thermore, the electrons leave behind unfilled “holes” into which other electrons in
the valence band can move in the electrical conduction regime. The excitation of
electron into these holes has the net effect of positive charge carriers aiding the
electrical conduction. Such semiconductors are known as intrinsic semiconductors.
However, with the introduction of certain impurities into a material in a controlled
way, a procedure known as doping conduction is dramatically increased. Such doped
semiconductors are known as extrinsic semiconductors, on which are based numer-
ous semiconductor devices. If the majority charge carriers are electrons, the material
is called an n-type semiconductor and if the holes are the majority charge carrier the
material is called a p-type semiconductor.
The Fermi energy EF lies in the middle of the energy gap.
The mobility of charge carriers is defined as
μ = vd/E (5.15)
The conductivity σ has two contributions, one from the electrons and the other
from the holes.
σ = nneμn + npeμp (5.16)
n = σ/eμ (5.17)
τ = μm/e (5.18)
Superconductivity
Some materials when cooled below a certain temperature, called critical temperature
(Tc), have zero resistance. The material is said to be a superconductor. Tc varies from
one superconductor to another.
When a superconductor is placed in a magnetic field, Tc decreases with the
increasing B. When B is increased beyond a critical magnetic field Bc, the super-
conductivity will not take place no matter how low the temperature.
Tc(B) = Tc0(1–B/Bc)1/2
(5.19)
where Tc0 is the critical temperature with zero magnetic field, and B is the applied
field.

According to the BCS theory superconductivity is due to a weak binding of two
electrons of equal and opposite momenta and spin to form the so-called a Cooper
pair which behaves as a single particle, a Boson.
An energy Eg, called the superconducting energy gap, is required to break the
Cooper pair. At T = 0,
Eg = 3.53 kTc (5.20)
where k is the Boltzmann constant.
5.2 Problems
5.2.1 Crystal Structure
5.1 Show that π/6 of the available volume is occupied by hard spheres in contact
in a simple cubic arrangement.
5.2 Show that
√
3π/8 of the available volume is occupied by hard spheres in con-
tact in a body-centered cubic arrangement.
5.3 Calculate the separations of the sets of planes which produce strong x-ray
diffractions beams at angles 4◦
and 8◦
in the first order, given that the x-ray
wavelength is 0.1 nm.
5.4 At what angle will a diffracted beam emerge from the (111) planes of a face
centered cubic crystal of unit cell length 0.4 nm? Assume diffraction occurs in
the first order and that the x-ray wavelength is 0.3 nm.
5.5 An x-ray beam of wavelength 0.16 nm is incident on a set of planes of a certain
crystal. The first Bragg reflection is observed for an incidence angle of 36◦
.
What is the plane separation? Will there be any higher order reflections?
5.6 In the historical experiment of Davisson and Germer electrons of 54 eV at nor-
mal incidence on a crystal showed a peak at reflection angle θr = 400
. At what
energy neutrons would also show a peak at θr = 400
for the same order.
5.7 Write down the atomic radii r in terms of the lattice constant a, for (a) Simple
cubic structure (b) FCC structure (c) BCC structure (d) Diamond structure.
5.2.2 Crystal Properties
5.8 Show that the Madelung constant for a one-dimensional array of ions of alter-
nating sign with equal distance between successive ions is equal to 2 ln 2.
5.9 Write down the first five terms for the Madelung constant corresponding to the
NaCl crystal.

5.2 Problems 295
5.10 The energy of interaction of two atoms a distance r apart can be written as:
E(r) = −
a
r
+
b
r7
where a and b are constants.
(a) Show that for the particles to be in equilibrium, r = ro = (7 b/a)1/6
(b) In stable equilibrium, show that the energy of attraction is seven times
that of the repulsion in contrast to the forces of attraction and repulsion
being equal.
5.11 In Problem 5.10, if the two atoms are pulled apart, show that they will separate
most easily when r = (28 b/a)1/6
.
5.12 Let the interaction energy between two atoms be given by:
E(r) = −
A
r2
+
B
r8
If the atoms form a stable molecule with an inter-nuclear distance of 0.4 nm
and a dissociation energy of 3 eV, calculate A and B.
5.13 Lead is a fcc with lattice constant 4.94 Å. Lead melts when the average ampli-
tude of its atomic vibrations is 0.46 Å. Assuming that for lead the Young’s
modulus is 1.6 × 1010
N/m2
, find the melting point of lead.
5.2.3 Metals
5.14 Take the Fermi energy of silver to be 5.52 eV.
(a) Find the corresponding velocity of conduction electron.
(b) If the resistivity of silver at room temperature is 1.62 × 10−8
Ωm estimate
the average time between collisions.
(c) Determine the mean free path. Assume the number of conduction electrons
as 5.86 × 1028
m−3
.
5.15 Find the drift velocity of electron subjected to an electric field of 20 Vm−1
,
given that the inter-collision time is 10−14
s.
5.16 Aluminum is trivalent with atomic weight 27 and density 2.7 g/cm3
, while the
mean collision time between electrons is 4 × 10−14
s. Calculate the current
flowing through an aluminum wire 20 m long and 2 mm2
cross-sectional area
when a potential of 3 V is applied to its ends.
5.17 Given that the conductivity of sodium is 2.17 × 107
Ω−1
m−1
, calculate:
(a) The inter-collision time at 300 K, and
(b) The drift velocity in a field of 200 Vm−1
.
5.18 Given that the inter-collision time in copper is 2.3 × 10−14
s, calculate its
thermal conductivity at 300 K. Assume the Wiedemann-Franz constant is
CWF = 2.31 × 10−8
W ΩK−2
.

5.19 The resistivity of a certain material is 1.72 × 10−8
Ωm whilst the Hall coeffi-
cient is −0.55 × 10−10
m3
C
−1
. Deduce:
(a) The electrical conductivity (σ)
(b) Mobility (μ)
(c) The inter-collision time (τ)
(d) Electron density (n)
5.20 In a Hall effect experiment on zinc, a potential of 4.5 µV is developed across
a foil of thickness 0.02 mm when a current of 1.5 A is passed in a direction
perpendicular to a magnetic field of 2.0 T . Calculate:
(a) The Hall coefficient for zinc
(b) The electron density
5.21 The density of states function for electrons in a metal is given by:
Z(E)dE = 13.6 × 1027
E1/2
dE
Calculate the Fermi level at a temperature few degrees above absolute zero
for copper which has 8.5 × 1028
electrons per cubic metre.
5.22 Using the results of Problem 5.21, calculate the velocity of electrons at the
Fermi level in copper.
5.23 For silver (A = 108), the resistivity is 1.5 × 10−8
Ωm at 0 ◦
C density is
10.5 × 103
kg/m3
and Fermi energy EF = 5.5 eV. Assuming that each atom
contributes one electron for conduction, find the ratio of the mean free path λ
to the interatomic spacing d.
5.24 Calculate the average amplitude of the vibrations of aluminum atoms at 500 K,
given that the force constant K = 20 N/m.
5.25 The Fermi energy in gold is 5.54 eV (a) calculate the average energy of the free
electrons in gold at 0 ◦
K. (b) Find the corresponding speed of free electrons (c)
What temperature is necessary for the average kinetic energy of gas molecules
to posses this value?
5.26 The density of copper is 8.94 g/cm3
and its atomic weight is 63.5 per mole,
the effective mass of electron being 1.01. Calculate the Fermi energy in copper
assuming that each atom gives one electron.
5.27 Find the probability of occupancy of a state of energy (a) 0.05 eV above
the Fermi energy (b) 0.05 eV below the Fermi energy (c) equal to the Fermi
energy. Assume a temperature of 300 K.
5.28 What is the probability at 400 K that a state at the bottom of the conduction
band is occupied in silicon. Given that Eg = 1.1 eV
5.29 The Debye temperature θ for iron is known to be 360 K. Calculate νm, the
maximum frequency.

5.2 Problems 297
5.30 Einstein’s model of solids gives the expression for the specific heat
Cv = 3N0k

θE
T
2
eθE /T
(eθE /T − 1)2
where θE = hνE /k.
The factor θE is called the characteristic temperature. Show that (a) at high
temperatures Dulong Petit law is reproduced. (b) But at very low temperatures
the T3
law is not given.
5.31 Debye’s model of solids gives the expression for specific heat
Cv = 9N0k
1
x3
x
0
ξ4
eξ
(eξ − 1)2
dξ
where ξ = hν/kT, x = hνm/kT and θD = hνm/k is the Debye’s charac-
teristic temperature. Show that (a) at high temperatures Debye’s model gives
Dulong Petit law (b) at low temperatures it gives Cv ∝ T 3
in agreement with
the experiment.
5.32 For a free electron gas in a metal, the number of states per unit volume with
energies from E to E + dE is given by
n(E)dE =
2π
h3
(2m)3/2
E1/2
dE
Show that the total energy = 3NEmax/5.
5.33 Assuming that the conduction electrons in a cube of a metal on edge 1 cm
behave as a free quantized gas, calculate the number of states that are available
in the energy interval 4.00–4.01 eV, per unit volume.
5.34 Calculate the Fermi energy for silver given that the number of conduction
electrons per unit volume is 5.86 × 1028
m−3
.
5.35 Calculate for silver the energy at which the probability that a conduction
electron state will be occupied is 90%. Assume EF = 5.52 eV for silver and
temperature T = 800 K.
5.2.4 Semiconductors
5.36 An LED is constructed from a Pn junction based on a certain semi-conducting
material with energy gap of 1.55 eV. What is the wavelength of the emit-
ted light?
5.37 Suppose that the Fermi level in a semiconductor lies more than a few kT below
the bottom of the conduction band and more than a few kT above the top of
the valence band, then show that the product of the number of free electrons
and the number of free holes per cm3
is given by

nenh = 2.33 × 1031
T 3
e−Eg/kT
where Eg is the gap width
5.38 The effective mass m∗
of an electron or hole in a band is defined by
1
m∗
=
1
2
·
d2
E
dk2
where k is the wave number (k = 2π/λ). For a free electron show that
m∗
= m.
5.39 After adding an impurity atom that donates an extra electron to the conduction
band of silicon (μn = 0.13 m2
/Vs), the conductivity of the doped silicon is
measured as 1.08 (Ωm−1
). Determine the doped ratio (density of silicon is
2,420 kg/m3
).
5.40 Estimate the ratio of the electron densities in the conduction bands of silicon
(Eg = 1.14 eV) and germanium (Eg = 0.7 eV) at 400 K.
5.41 Show that at the room temperature (300 K) the electron densities in the con-
duction bands of the insulator carbon (Eg = 5.33 eV) and the semiconductor
like germanium (Eg = 0.7 eV) is extremely small.
5.42 A current of 8×10−11
A flows through a silicon p −n junction at temperature
27 ◦
C. Calculate the current for a forward bias of 0.5 V.
5.43 Calculate the depletion layer width for a pn junction with zero bias in ger-
manium, given that the impurity concentrations are Na = 1 × 1023
m−3
and
Nd = 2 × 1022
m−3
, respectively at T = 300 K, ∈r = 16 and contact potential
difference V0 = 0.8 V.
5.44 Consider the Shockley equation for the diode
I = I0 exp[(eV /kT) − 1]
Show that the slope resistance re of the I − V curve at a particular d.c bias
is given to a good approximation, at room temperature (T = 300 K) by the
expression re = 26
I
Ω (forward bias) where I is in milliampere, and that for
the reverse bias re tends to infinity.
5.45 Given that a piece of n-type silicon contains 8 × 1021
m−3
phosphorus impu-
rity atoms, calculate the carrier concentrations at room temperature. It may be
assumed that the intrinsic electron concentration in silicon at room tempera-
ture is 1.6 × 1016
m−3
.
5.2.5 Superconductor
5.46 It is required to break up a Cooper pair in lead which has the energy gap of
2.73 eV. What is the maximum wavelength of photon which will accomplish
the job?

5.3 Solutions 299
5.47 Given that the maximum wavelength of photon to break up Cooper pair in tin
is 1.08 × 10−3
m, calculate the energy gap.
5.48 A Josephson junction consists of two super conductors separated by a very
thin insulating layer. When a DC voltage is applied across the junction an
AC current is produced, a phenomenon called Josephson effect. Calculate the
frequency of the AC current produced when a DC voltage of 1.5 µV is applied.
5.49 Use the BCS theory to calculate the energy gap for indium whose critical
temperature Tc = 3.4 K.
5.50 For lead superconductivity ensues at 7.19 K, when there is a zero applied mag-
netic field. When the magnetic field of 0.074 T is applied at temperature 2.0 K
superconductivity will stop. Find the magnetic field that should be applied so
that superconductivity will not occur at any temperature?
5.51 An ac current of frequency 1 GHz is observed through a Josephson junction.
Calculate the applied dc voltage.
5.52 If a superconducting Quantum Interference Device which consists of a 3 mm
ring can measure 1/5,000 of a fluxon, calculate the magnetic field that can be
detected (1 fluxon, φ0 = h/2e = 2.0678 × 10−15
T m2
, is the smallest unit
of flux).
5.3 Solutions
5.1 The cross-section of the portions of four spheres each of radius r touching each
other and lying in a cell of edge a is shown in Fig. 5.1. The volume of each
sphere lying within the cell is 1
4
× 4
3
πr3
or π
3
r3
. Volume of four spheres lying
within the cell is 4π
3
r3
. Volume of the cell is a3
or (2r)3
. Therefore, the available
volume occupied by hard spheres in the simple cubic structure is 4πr3
/3
(2r)3 or π
6
.
Fig. 5.1 Hard spheres in the
simple structure

5.2 Volume of the unit cell = a3
. Since there are two atoms per unit cell, 8 × 1/8
for the corner atoms and 1 × 1 for the centre atom,
Volume = 2 ×
4
3
πr3
Since the body diagonal atoms touch one another,
4r = a
√
3
Volume of atoms in terms of a is
2 ×
4
3
πr3
= 2 ×
4
3
π[a
√
3/4]3
=
√
3πa3
/8
Or the fraction of the volume occupied by the body-centred cubic structure is
√
3π/8.
5.3
2d sin θ = nλ
d1 =
1.λ
2 sin θ
=
0.1
2 sin 4◦
= 0.717 nm
d2 =
0.1
2 sin 8◦
= 0.359 nm
5.4 nλ =
2a
(h2 + k2 + l2)1/2
sin θ =
2 × 0.4
(12 + 12 + 12)1/2
sin θ
sin θ =
0.3
√
3
0.8
= 0.6495
θ = 40.5◦
5.5 2d sin θ = nλ
d =
1.λ
2 sin θ
=
0.16
2 sin 30◦
= 0.136 nm
For n = 2,
sin θ =
2 × 0.16
2 × 0.136
= 1.176
a value which is not possible. Thus higher order reflections are not possible.
5.6 The de Broglie wavelength for electrons is calculated from
λ =
150
V
=
150
54
= 1.66 Å
Bragg’s equation will be satisfied for neutrons of the same wavelength.
λ =
0.286
√
E
Å

5.3 Solutions 301
where E is in eV,
E =

0.286
λ
2
=

0.286
1.66
2
= 0.0297 eV.
5.7 (a) r = a/2
(b) r =
√
2 a/4
(c) r =
√
3 a/4
(d)
√
3 a/8
5.8 Consider an infinite line of ions of alternating sign, as in Fig. 5.2. Let a nega-
tive ion be a reference ion and let a be the distance between adjacent ions. By
definition the Madelung Constant ∝ is given by:
Fig. 5.2 Infinite line of ions
of alternating sign
α
a
=

j
(±)
rj
(1)
where rj is the distance of the jth ion from the reference ion and a is the
nearest neighbor distance. Thus:
α
a
= 2

1
a
−
1
2a
+
1
3a
−
1
4a
+ · · ·

Or, α = 2

1 −
1
2
+
1
3
−
1
4
+ · · ·

(2)
The factor 2 occurs because there are two ions, one to the right and one to
the left, at equal distances rj . We sum the series by the expansion:
ln(1 + x) = x −
x2
2
+
x3
3
−
x4
4
+ · · · (3)
Putting x = 1, the RHS in (3) is identified as In 2. Thus ∝ = 2 ln 2.

5.9 The lattice of the NaCl structure which is face-centered is represented in
Fig. 5.3. The shortest inter ionic distance is represented by r. A given sodium
ion is surrounded by 6Cl−
ions at a distance r, 12Na+
ions at a distance
r
√
2, 8Cl−
ions at a distance r
√
3, 6Na+
√
4, 24Cl−
√
5, etc. The coulomb energy of of this ion in the field of
all other ions is therefore
Ec = −
e2
r

6
√
1
−
12
√
2
+
8
√
3
−
6
√
4
+
24
√
5
− · · ·

where e is the charge per ion. Series of this sort which consists of pure numbers
depends on the crystal structure and is known as the Madelung constant.
Fig. 5.3 The lattice of the
NaCl structure
5.10 Er = −
a
r
+
b
r7
(a) For equilibrium, Er must be minimum, so that
dE
dr
= 0
a
r2
−
7b
r8
= 0
Or, r = ro = (7b/a)1/6
(b) Energy of attraction, EA =
−a
ro
Energy of repulsion, ER =
b
r7
0
|EA|
|ER|
=
a
b
r6
o =
a
b
·
7b
a
= 7
Force, F = −
dV
dr
Attractive force FA =
a
r2
o
(r = ro)
Repulsive force FR =
−7b
r8
o
=
−7b
r2
o
·
a
7b
= −
a
r2
o
Thus the two forces are equal in magnitude.

5.3 Solutions 303
5.11 Force, F = −
dV
dr
=
a
r2
−
7b
r8
The particles will separate most easily when the force between them is a min-
imum, that is when
dF
dr
= 0. This gives:
dF
dr
= −
2a
r3
+
56b
r9
= 0
r =

28b
a
1/6
5.12 The inter-nuclear distance is found from
dE
dr
= 0
2A
r3
o
−
8B
r9
o
= 0 → r6
o =
4B
A
(1)
The dissociation energy D is formed from −D = E(ro)
−D = −
A
r2
o
+
B
r8
o
= −
A
r2
o
+
A
4r2
o
= −
3A
4r2
o
where we have used (1).
A =
4Dr2
o
3
=
4
3
× 3 × 1.6 × 10−19
× (0.4 × 10−9
)2
= 1.02 × 10−37
B =
A
4
r6
0 =
1.02 × 10−37
4
× (0.4 × 10−9
)6
= 1.04 × 10−91
5.13 A=

2kT
K
1/2
The force constant K = Ya0 = 1.6 × 1010
× 4.94 × 10−10
= 7.9 N/m2
T =
K
2k
A 2
=
7.9 × (0.46 × 10−10
)2
2 × 1.38 × 10−23
= 606 K = 333◦
C
5.3.3 Metals
5.14 (a) vF =

2EF
m
1/2
=

2 × 5.52 × 1.6 × 10−19
9.11 × 10−31
1/2
= 1.39 × 106
m/s
(b) τ =
m
ne2ρ
=
9.11 × 10−31
(5.86 × 1028)(1.6 × 10−19)2(1.62 × 10−8)
= 3.7 × 10−14
s

(c) λ = vFτ = (1.39 × 106
)(3.7 × 10−14
)
= 5.14 × 10−8
m
5.15 vD =
e
m
ετ
=
(1.6 × 10−19
)(20)(10−14
)
9.11 × 10−31
= 0.0351 m/s
= 3.51 cm/s
Note that the drift velocities are much smaller than the average thermal
velocities which are of the order of 105
m/s.
,
νT = (3kT/me)1/2
-
5.16 Current,
i =
V
R
(1)
R =
ρl
A
(2)
where the resistivity,
ρ =
me
ne2τ
(3)
n =
Nod
A
× 3 × 104
(4) (4)
where n is the number of electrons per m3
, No being Avagardro’s number, A
the atomic weight and d the density, the factor 3 is for the trivalency.
n = 6.02 × 1023
×
2.7
27
× 3 × 104
= 1.806 × 1027
ρ =
9.11 × 10−31
1.806 × 1027 × (1.6 × 10−19)2 × 4 × 10−14
= 4.92 × 10−9
R =
4.92 × 10−9
× 20
2 × 10−6
= 0.0492 Ω
i =
3
0.0492
= 61 Å
5.17 (a) τ =
mσ
ne2
Assuming that one conduction electron will be available for each sodium
atom,
n =
Noρ
A
=
6.02 × 1023
× 0.97
23
cm−3
= 2.539 × 1028
m−3
τ =
9.11 × 10−31
× 2.17 × 107
2.539 × 1028 × (1.6 × 10−19)2
= 3.04 × 10−14
s
(b) vD =
e
m
ετ
=
(1.6 × 10−19
)(200)(3.04 × 10−14
)
9.11 × 10−31
= 1.07 m/s

5.3 Solutions 305
5.18 σ =
ne2
τ
m
n =
6.02 × 1023
× 8.88 × 106
63.57
= 8.38 × 1028
m−3
σ =
8.38 × 1028
× (1.6 × 10−19
)2
× 2.3 × 10−14
9.11 × 10−31
= 5.422 × 107
Ω−1
m−1
K = σCWFT = (5.422 × 107
)(2.31 × 10−8
)(300) = 376 Wm−1
K−2
5.19 (a) σ =
1
ρ
=
1
1.72 × 10−8
= 5.8 × 107
Ω−1
m−1
(b) μ = RH σ = (0.55 × 10−10
)(5.8 × 107
) = 0.0032 m2
V−1
s
−1
(c) τ =
μm
e
=
(0.0032)(9.11 × 10−31
)
1.6 × 10−19
= 1.82 × 10−14
s
(d) n =
σ
eμ
=
5.8 × 107
(1.6 × 10−19)(0.0032)
= 1.13 × 1029
m−3
5.20 (a) RH =
VHt
i B
=
(4.5 × 10−6
)(2 × 10−5
)
(1.5)(2)
= 0.3 × 10−10
m3
C
−1
(b) n =
1
RHe
=
1
(0.3 × 10−10)(1.6 × 10−19)
= 2.08 × 1029
m−3
5.21 Integrating n(E) dE from zero to EF:
13.6 × 1027
EF

0
E1/2
dE = 8.5 × 1028
E
3/2
F = 9.375
Or EF = 4.445 eV
5.22 v =
2E
m
=
0
2 × 4.445 × 1.6 × 10−19
9.11 × 10−31
= 1.25 × 106
m/s
5.23 vF =

2EF
mc2
1/2
c =

2 × 5.5
0.511 × 106
1/2
(3 × 108
) = 1.39 × 106
m/s
Since each atom contributes one electron, the density of electrons is equal to
that of atoms.
n =
6.02 × 1026
× 10.5 × 103
108
= 5.85 × 1028
e/m3
Each atom occupies approximately a volume d3
. Therefore
d =

1
5.85 × 1028
1/3
= 2.576 × 10−10
m

Now, ρ =
meVF
e2nλ
or λ =
meVF
ρe2n
=
(9.11 × 10−31
)(1.39 × 106
)
(1.5 × 10−8)(1.6 × 10−19)2(5.85 × 1028)
= 9.02 × 10−8
m
∴
λ
d
=
9.02 × 10−8
2.58 × 10−10
= 350
5.24 A=

2kT
K
1/2
=

2 × 1.38 × 10−23
× 500
20
1/2
=2.63×10−11
m=0.26 Å
5.25 (a) E=
3
5
EF =
3 × 5.54
5
= 3.32 eV
(b) v = c

2E
Mc2
1/2
= 3 × 108

2 × 3.32
0.511 × 106
1/2
= 1.08 × 106
m/s
(c)
3
2
kT =
3
5
EF = 3.32 eV = 3.32 × 1.6 × 10−19
J
T =
2
3
×
3.32 × 1.6 × 10−19
1.38 × 10−23
= 2.56 × 104
K
5.26 EF =
h2
2m∗

3N
8πV
2/3
where m∗
is the effective mass.
Density of Cu atoms =
N0ρ
A
=
6.02 × 1023
× 8.94
63.5
= 8.475 × 1022
atoms/cm3
= 8.475 × 1028
atoms/m3
= 8.475 × 1028
e/m3
(∵ each atom gives one electron)
EF =
(6.625 × 10−34
)2
2 × 1.01 × 9.11 × 10−31

3
8π
× 8.475 × 1028
2/3
= 11.157 × 10−19
J = 6.97 eV
5.27 The probability is given by Fermi-Dirac distribution p(E) =
1
e(E−EF)/kT + 1
(a) K = 1.38 × 10−23
J = 8.625 × 10−5
eV K−1
E − EF
kT
=
0.05
(8.625 × 10−5)(300)
= 1.932
p(E) =
1
e1.932 + 1
= 0.126

5.3 Solutions 307
(b) p(E) =
1
e−1.932 + 1
= 0.873
(c) p(E) =
1
e0 + 1
= 0.5
5.28 Assuming that the Fermi energy is to be at the middle of the gap between the
conduction and valence bands, E − EF = 1/2 Eg
p(E) =
1
e(E−EF)/kT +1
=
1
eEg/2kT + 1
The factor Eg/2kT =
1.1
2 × 8.625 × 10−5 × 400
= 15.942
p(E) ≈ e−15.942
= 8.4 × 10−6
5.29 The Debye temperature θ is
θ =
hνm
k
νm =
k
h
θ =
1.38 × 10−23
× 360
6.625 × 10−34
= 7.5 × 1012
Hz
5.30 (a) At high temperatures T θE, in the denominator (eθE/T
− 1)2
≈ θ2
E/T 2
,
and in the numerator eθE/T
→ 1, so that Cv → 3N0k = 3R, the Dulong –
Petit’s value
(b) When the temperature is very low T θE, and in the bracket of the
denominator, 1 is negligible in comparison with the exponential term.
Therefore, Cv → 3R(θE/T )2
e−θE/T
. Thus the specific heat goes to zero
as T → 0. However, the experimentally observed specific heats at low
temperatures decrease more gradually than the exponential decrease sug-
gested by Einstein’s formula.
5.31 Cv =
9R
x3
x
0
ξ4
eξ
(eξ − 1)2
dξ (1)
This equation may be integrated by parts,
x
0
ξ4
eξ
(eξ − 1)2
dξ = −

ξ4 d
dξ

1
eξ − 1

dξ
= −ξ4 1
eξ − 1
+

1
eξ − 1
dξ4
dξ
dξ
= −ξ4 1
eξ − 1
+ 4

ξ3
eξ − 1
dξ
Thus (1) becomes
Cv = 9R

4
x3
x
0
ξ3
eξ − 1
dξ −
x
ex − 1

(2)

(a) At high temperatures, θD T , or x 1, and the exponential can be
expanded to give
Cv = 9R

4
3
− 1

= 3R (Dulong Petit’s law)
(b) At very low temperatures T θD x 1, (2) can be approximated to
Cv = 9R
4
x3
∞
0
ξ3
dξ
eξ − 1
=
12
5
π4

T
θD
3
where the value of the integral is π4
/15. Thus, Cv ∝ T 3
5.32 If there are N free electrons in the metal there will be N/2 occupied quantum
states at the absolute zero of temperature in accordance with the Fermi Dirac
statistics. In Fermi-Dirac statistics at absolute zero, kinetic energy is not zero
as would be required if the Boltzmann statistics were assumed.
As N(E)dE gives the number of states per unit volume, in a crystal of volume
V , the number of electrons in the range from E to E + dE is
2V ·
2π
h3
(2m)3/2
E1/2
dE (1)
The total energy of these electrons would be
Etotal =
Emax
0
4πV
h3
(2m)3/2
E3/2
dE =
4πV (2m)
3
2
h3
·
2
5
E5/2
max (2)
But,
Emax =
h2
8m

3N
πV
2/3
(3)
Combining (2) and (3),
Etotal =
3
5
N Emax (4)
or per electron 3Emax/5. The quantity Emax = EF, the Fermi energy
5.33 The density of states n(E) (the number of states per unit volume of the solid
in the unit energy interval) is given by
n(E) =
8
√
2πm3/2
h3
E1/2
=
(8
√
2π)(9.11 × 10−31
)3/2
(6.63 × 10−34)3
(4 × 1.6 × 10−19
)1/2
= 8.478 × 1046
m−3
J−1
= 1.356 × 1028
m−3
eV−1
Number of states N that lie in the range E = 4.00eV to E = 4.01eV, for
volume, V = a3

5.3 Solutions 309
N = n(E)ΔEa3
= 1.356 × 1028
× 0.01 × (10−2
)3
= 1.356 × 1020
5.34 EF =
h2
8m

3n
π
2/3
=
(6.63 × 10−34
)2
(8)(9.11 × 10−31)

3 × 5.86 × 1028
π
2/3
= 8.827 × 10−19
J = 5.517 eV
5.35 P(E) =
1
eΔE/kT + 1
= 0.9
Substituting kT = 5.52 × 10−5
× 800 = 0.04416 eV
Solving for ΔE, we get ΔE = E − EF = −2.2 × 0.04416 = −0.097
Therefore, E = 5.52 − 0.10 = 5.42 eV
5.36 λ =
1241
1.55
= 800 nm
5.37 The number of electrons and holes per unit volume are given by
ne = 2

2πm
kT
h2
3/2
e(EF−Eg)/kT
(1)
and nh = 2

2πm
kT
h2
3/2
e−EF/kT
(2)
Multiplying (1) and (2), one can write
nenh = 4

mc2
k
2π2c2
3
T 3
e−Eg/kT
(3)
= 2.34 × 1031
T3
e−Eg/kT
cm−6
where we have substituted the values of the constants.
5.38 p = k (1)
E = p2
/2m = k2
2
/2m (2)
1
m∗
=
1
2
d2
E
dk2
(3)
Using (2) in Eq. (3)
1
m∗
=
1
2
d2
dk2

k2
2
2m

=
22
2m2
=
1
m
∴ m∗
= m

5.39 The number of silicon atoms/m3
n =
N0d
A
=
6.02 × 1026
× 2,420
28
= 0.52 × 1029
Let x be the fraction of impurity atom (donor). The general expression for the
conductivity is
σ = nneμn + npeμp
where nn and np are the densities of the negative and positive charge carriers.
Because nn ≫ np,
σ ∼
= nneμn = xnpeμn
x =
σ
npeμn
=
1.08
0.52 × 1029 × 1.6 × 10−19 × 0.13
=
9.985
1010
or 1 part in 109
.
Note that in normal silicon the conductivity is of the order of 10−4
(Ω− m)−1
.
A small fraction of doping (10−9
) has dramatically increased the value by four
orders of magnitude.
5.40 ne = (4.83 × 1021
)T 3/2
e−Eg/2kT
e/m3
nGe
nSi
= e(ESi−EGe)/2kT
kT =
1.38 × 10−23
× 400
1.6 × 10−19
= 0.0345 eV
nGe
nSi
= e(1.14−0.7)/(2×0.0345)
= 588
5.41 kT =
1.38 × 10−23
× 300
1.6 × 10−19
= 0.0259 eV
nC
nGe
= e−(EGe−EC )/2K T
= e−(5.33−07)/0.052
= e−89
≈ 2.2 × 10−39
5.42 I = I0[exp(eV/kT ) − 1]
where I0 is the forward bias saturation current.
I = 8×10−11

exp

0.5 × 1.6 × 10−19
1.38 × 10−23 × 300

− 1

= 19.7×10−3
A = 19.7 mA
5.43 W =

2ǫ0ǫr
e
(V0 − Vb)

1
Na
+
1
Nd
1/2
where ǫr is the relative permittivity, Vb is the bias voltage applied to the junc-
tion (here Vb = 0), Na and Nd are carrier concentrations in n-type and p-type
respectively.
W =

2 × 8.85 × 10−12
× 16 × 0.8
1.6 × 10−19

1
1 × 1023
+
1
2 × 1022
1/2
= 0.29 µm

5.3 Solutions 311
5.44 I = I0[exp(eV/kT ) − 1] (1)
1
re
=
dI
dV
=
eI0
kT
exp(eV/kT ) (2)
But exp (eV/kT) ≫ 1. Therefore
1
re
=
eI
kT
or re =
kT
eI
=
1.38 × 10−23
× 300
1.6 × 10−19 I
=
25.875 × 10−3
I
If I is in milliamp.
re ≈
26
I
(forward bias) (3)
For the reversed bias we note from (2)
1
re
=
dI
dV
=
e
kT
I0 exp (eV/kT ) = 0
For V ≤ −4kT/e,re → ∞. (reverse bias) (4)
5.45 For a semiconductor in equilibrium the product of n(= Nd) and p(= Na) is
equal to n2
i , the square of the intrinsic concentration.
n × p = n2
i
p =
n2
i
n
=
(1.6 × 1016
)2
8 × 1021
= 3.2 × 1010
m−3
5.46 λ =
1241
E(eV)
nm =
1, 241
2.73 × 10−3
= 4.546 × 105
nm
5.47 Eg =
1241
λ(nm)
=
1, 241
1.08 × 106
= 1.15 × 10−3
eV
5.48 f =
2eV
h
=
2(1.602 × 10−19
)(1.5 × 10−6
)
6.626 × 10−34
= 7.253×108
Hz = 0.7253 GHz
5.49 Eg = 3.53kTc = (3.53)
(1.38 × 10−23
)
1.6 × 10−19
(3.4) = 1.035 × 10−3
eV
5.50 Tc(B) = Tc

1 −
B
Bc
1/2
2.0 = 7.19

1 −
0.074
Bc
1/2
Solving for Bc, we get Bc = 0.079 T .

5.51 f = 2eV/h → V = h f/2e
V = (6.625 × 10−34
)(109
)/2 × (1.6 × 10−19
)
= 2.07 × 10−6
V
= 2.07 µV
5.52 ΔB =
ϕ
A
=

1
5, 000

ϕ0/π(3 × 10−3
)2
Substituting φ0 = 2.0678 × 10−15
Tm2
, we get
ΔB = 2.93 × 10−14
T

Chapter 6
Special Theory of Relativity
Inertial frame
Laws of mechanics take the same form (invariant) in all inertial frames. An iner-
tial frame of reference is the one which moves with constant relative velocity in
which Newton’s laws of motion are valid. The principle that all inertial frames are
equivalent for the description of nature is called the principle of relativity.
Galilean Transformations
Reference frame S′
moves along x-axis with velocity ν relative to S. Spatial coor-
dinates x, y, z are measured in S and x′
, y′
, z′
in S′
and time t and t′
in S and S′
respectively. For simplicity, x and x′
axes coincide. At the beginning (t = 0), S and
S′
coincide. After time t, S′
would have moved through a distance νt. The Galilean
transformations are given by the set of relations.
x′
= x − νt (6.1)
y′
= y (6.2)
z′
= z (6.3)
t′
= t (6.4)
In Galilean relativity time is absolute.
Fig. 6.1 Reference frames S
and S′
313

314 6 Special Theory of Relativity
Transformation of velocities
Differentiating (6.1) with respect to time and noting that t′
= t and ν is constant
dx′
dt
=

dx
dt

− ν (6.5)
U′
= U − ν
or
V = U′
+ ν (6.6)
Invariance of Newton’s second law of motion
In S, force is given by
F = ma =
md2
x
dt2
= m
d
dt

dx
dt

= m
du
dt
In S′
, force is given by
F′
=
md2
x
dt2
=
md2
x
dt2
=
md2
x
dt2
− 0 = F (6.7)
Galilean Relativity fails for Electromagnetism as evidenced by the negative result
of Michelson–Morley experiment to measure earth’s velocity in the hypothetical
medium of ether.
Einstein took the view that the principle of Relative is correct but time is not
absolute but only relative.
Postulates of special Theory of relativity
(1) The laws of physics apply equally well in all inertial frames of reference, that
is no preferred system exists (the principle of relativity)
(2) The speed of light in free space has the same value c(= 3 × 108
ms−1
) in all
inertial frames (the principle of constancy of light)
Lorentz Transformations
x′
= γ (x − νt) (6.8)
y′
= y (6.9)
z′
= z (6.10)
t′
= γ t −
νx
c2

(6.11)

Inverse transformations
x = γ (x′
+ νt′
) (6.12)
y = y′
(6.13)
z = z′
(6.14)
t = γ

t′
+
νx′
c2

(6.15)
with
γ =
1

(1 − β2)
=
1
√
(1 − ν2/c2)
(6.16)
and
β =
ν
c
(6.17)
Transformation matrix
The Lorentz transformations (6.8), (6.9), (6.10), and (6.11) can be condensed in the
matrix form
X′
= ΛX (6.18)
where X =
⎡
⎢
⎢
⎣
x1
x2
x3
x4
⎤
⎥
⎥
⎦ and X′
=
⎡
⎢
⎢
⎣
x1
′
x2
′
x3
′
x4
′
⎤
⎥
⎥
⎦ (6.19)
are the column vectors with components
x1 = x, x2 = y, x3 = z, x4 = τ = ict (6.20)
x′
1 = x′
, x′
2 = y′
, x′
3 = z′
, x′
4 = τ′
= ict′
(6.21)
with i =
√
− 1, and Λ is an orthogonal matrix
Λ =
⎡
⎢
⎢
⎣
γ 0 0 iβγ
0 1 0 0
0 0 1 0
−iβγ 0 0 γ
⎤
⎥
⎥
⎦ (6.22)

The inverse transformations (6.12), (6.13), (6.14), and (6.15) are immediately writ-
ten with the aid of inverse matrix
Λ−1
= Λ̃
Λ−1
=
⎡
⎢
⎢
⎣
γ 0 0 − iβγ
0 1 0 0
0 0 1 0
iβγ 0 0 γ
⎤
⎥
⎥
⎦ (6.23)
Four vectors
If s =
$
Σμx2
μ = invariant, μ = 1, 2, 3, 4 (6.24)
under Lorentz transformation, then s is said to be a four vector, s can be
positive or negative or zero. Examples of Four vectors are
X = (x1, x2, x3, ict) (6.25)
cP = (cPx , cPy, cPz, i E) (6.26)
In (6.25) the first three space components of X define the ordinary three-
dimensional position vector x and the fourth, a time component ict. The four-
momentum in (6.26) has the first three components of ordinary momentum and
E is the total energy of the particle. The four-vectors have the properties which are
similar to those of ordinary vectors. Thus, the scalar product of two four-vectors,
A.B = A1 B1 + A2 B2 + A3 B3 + A4 B4
Consequences of Lorentz transformations
Time Dilation
Δt′
= γ Δt (6.27)
Rule: Every clock appears to go at its fastest rate when it is at rest relative to
the observer. If it moves relative to the observer with velocity ν, its rates appears
slowed down by the factor

1 − ν2/c2. No distinction need be made between the
stationary and moving frame. Each observer will think that the other observer’s
clock has slowed down. What matters is the only the relative motion.
The Lorentz contraction
l′
= l

1 − β2 (6.28)

Rule: Every rigid body appears to be longest when at rest relative to the observer.
When it is moving relative to the observer it appears contracted in the direction of
its relative motion by the factor

1 − ν2/c2, while its dimensions perpendicular to
the direction of motion are unaffected.
Addition of velocities
β =
β1 + β2
1 + β1β2
(6.29)
Mass, energy and momentum
m = m0γ =
m0

1 − ν2/c2
(6.30)
where m is the effective mass and m0 is the rest mass.
The rest mass energy
E0 = m0c2
(6.31)
The total energy of a free particle is
E = T + m0c2
= mc2
= m0γ c2
(6.32)
where T is the kinetic energy.
The momentum p is given by
P = mν = m0γβc (6.33)
E2
= c2
p2
+ m2
0c4
(6.34)
cp = βE (6.35)
c2
p2
= T 2
+ 2T m0c2
(6.36)
Lorentz transformations of momentum and energy
cpx′ = γ (cpx − βE) (6.37)
cpy′ = cpy (6.38)
cpz = cpz (6.39)
E′
= γ (E − βcpx ) (6.40)

cpx = γ (cp′
x + βE′
) (6.41)
cpy = cp′
y (6.42)
cpz = cp′
z (6.43)
E = γ (E
′
− βcp′
x∗ ) (6.44)
E′2
− c2
p′2
= E2
− c2
p2
= m2
0 = Invariant (6.45)
γ ′
= γ γ0(1 − ββ0 cos θ0) (6.46)
γ0 = γ γ ′
(1 + ββ∗
cos θ∗
) (6.47)
where zeros refer to the particle’s velocity, Lorentz factor and the angle in the
S-system while primes refer to the corresponding quantities in the S′
system.
Transformation of angles
tan θ′
=
sin θ
γ cos θ − β
β0
(6.48)
tan θ =
sin θ′
γc(cos θ′ − β/β′)
(6.49)
Optical Doppler effect
ν′
= γ ν(1 − β cos θ) (6.50)
ν = γ ν′
(1 + β cos θ∗
) (6.51)
where ν is the frequency in the S-system and ν′
is the frequency in the S′
-system,
θ and θ′
are the corresponding angles, β is the source velocity and γ is the corre-
sponding Lorentz factor.
Threshold for particle production
Consider the reaction
m1 + m2 → m3 + m4 + M (6.52)
T (threshold) =
1
2m2
[(m3 + m4 + M)2
− (m1 + m2)2
]
T (threshold) =
(Sum of final masses)2
− (sum of initial masses)2
2 × mass of target particle
(6.53)

6.2 Problems 319
6.2 Problems
6.2.1 Lorentz Transformations
6.1 In the inertial system S, an event is observed to take place at point A on the
x-axis and 10−6
S later another event takes place at point B, 900 m further
down. Find the magnitude and direction of the velocity of S′
with respect to S
in which these two events appear simultaneous.
6.2 Show that the Lorentz-transformations connecting the S′
and S systems may
be expressed as
x1
′
= x1 cosh α−ct sinh α
x2
′
= x2
x3
′
= x3
t′
= t cosh α − (x1t sinh α)/c
where tanh α = ν/c. Also show that the Lorentz transformations correspond
to a rotation through an angle iα in four-dimensional space.
6.3 A pion moving along x-axis with β = 0.8 in the lab system decays by emitting
a muon with β′
= 0.268 along the incident direction (x′
-axis) in the rest
system of pion. Find the velocity of the muon (magnitude and direction) in the
lab system.
6.4 In Problem 6.3, the muon is emitted along the y′
-axis. Find the velocity of
muon in the lab frame
6.5 In Problem 6.3, the muon is emitted along the positive y-axis (i.e. perpendic-
ular to the incidental direction of pion in the lab frame). Find the speed of
muon in the lab frame and the direction of emission in the rest frame of pion.
Assume βc = 0.2
6.6 Show that Maxwell’s equations for the propagation of electromagnetic waves
are Lorentz invariant.
6.7 A neutral K meson decays in flight via K0
→ π+
π−
. If the negative pion is
produced at rest, calculate the kinetic energy of the positive pion.
[Mass of K0
is 498 MeV/c2
; that of π±
is 140 MeV/c2
]
6.8 A pion travelling with speed ν = |ν| in the laboratory decays via π → μ + ν.
If the neutrino emerges at right angles to ν, find an expression for the angle θ
at which the muon emerges.
6.9 Determine the speed of the Lorentz transformation in the x-direction for which
the velocity in the frame S of a particle is u = (c/
√
2, c/
√
2) and the velocity
in frame S′
is seen as
u′
= (−c/
√
2, c/
√
2).
6.10 A particle decays into two particles of mass m1 and m2 with a release of energy
Q. Calculate relativistically the energy carried by the decay products in the
rest frame of the decaying particle.

6.11 A π-meson with a kinetic energy of 140 MeV decays in flight into μ-meson
and a neutrino. Calculate the maximum energy which (a) the μ-meson (b) the
neutrino may have in the Laboratory system (Mass of π-meson = 140 MeV/c2
,
mass of μ-meson = 106 MeV/c, mass of neutrino = 0)
[University of Bristol 1968]
6.12 A positron of energy E+ , and momentum p+ and an electron, energy E−,
momentum p− are produced in a pair creation process
(a) What is the velocity of their CMS?
(b) What is the energy of either particle in the CMS?
6.13 A particle of mass m collides elastically with another identical particle at rest.
Show that for a relativistic collision
tan θ tan ϕ = 2/(γ + 1)
where θ, ϕ are the angles of the out-going particles with respect to the direc-
tion of the incident particle and γ is the Lorentz factor before the collision.
Also, show that θ + ϕ ≤ π/2 where the equal sign is valid in the classical
limit
6.14 A K+
meson at rest decays into a π+
meson and π0
meson. The π+
meson
decays into a μ meson and a neutrino. What is the maximum energy of the
final μ meson? What is its minimum energy?
(mK = 493.5 MeV/c2
, mπ+ = 139.5 MeV/c2
, mπ0 = 135 MeV/c2
,
mμ = 106 MeV/c2
, mν = 0)
6.15 An unstable particle decays in its flight into three charged pions (mass
140 MeV/c2
). The tracks recorded are shown in Fig. 6.2, the event being
coplanar. The kinetic energies and the emission angles are
T1 = 190 MeV, T2 = 321 MeV, T3 = 58 MeV
θ1 = 22.4◦
, θ2 = 12.25◦
Estimate the mass of the primary particle and identify it. In what direction was
it moving?
Fig. 6.2 Decay of a kaon into
three poins
6.2.2 Length, Time, Velocity
6.16 If a rod travels with a speed ν = 0.8 c along its length, how much does it
shrink?

6.2 Problems 321
6.17 If a rod is to appear shrunk by half along its direction of motion, at what speed
should it travel?
6.18 Assuming that the rest radius of earth is 6,400 km and its orbital speed about
the sun is 30 km−1
, how much does earth’s diameter appear to be shortened to
an observer on the sun, due to earth’s orbital motion?
6.19 The mean life-time of muons at rest is found to be about 2.2 × 10−6
s, while
the mean life time in a burst of cosmic rays is found to be 1.5 × 10−5
s. What
is the speed of these cosmic ray muons?
6.20 A beam of muons travels with a speed of v = 0.6 c. Their mean life-time
as observed in the laboratory is found to be 2.9 × 10−6
s. What is the mean
life-time of muons when they decay at rest?
6.21 (a) If the mean proper life-time of muons is 2.2 × 10−6
s, what average dis-
tance would they travel in vacuum before decaying in the reference frame
in which its velocity is measured as 0.6 c?
(b) Compare this distance with the distance the muon sees while travelling.
6.22 With what constant velocity must a person travel from the centre to the edge
of our galaxy so that the trip may last 40 years (proper time)? Assume that the
radius of the galaxy is 3 × 104
light years?
6.23 A pion is produced in a high energy collision of a primary cosmic ray particle
in the earth’s atmosphere 1 km above the sea level. It proceeds vertically down
at a speed of 0.99 c and decays in its rest frame 2.5×10−8
s after its production.
At what altitude above the sea level is it observed from the ground to decay?
6.24 One cosmic particle approaches the earth along its axis with a velocity of
0.9 c toward the North Pole and another one with a velocity of 0.5 c toward
the South Pole. Find the relative speed of approach of one particle with respect
to another.
6.25 A 100 MeV electron moves along the axis of an evacuated tube of length 4 m
fixed to the laboratory frame. What length of the tube would be measured by
the observer moving with the electron?
6.26 A man has a mass of 100 kg on the earth. When he is in the space-craft, an
observer from the earth registers his mass as 101 kg. Determine the speed of
the space-craft.
6.27 At the time a space ship moving with speed ν = 0.5 c passes a space station
located near Mars, a radio signal is sent from the station to earth. This signal
is received on earth 1,125 s later.
How long does the spaceship take to reach the earth according to the observers
on earth?
6.28 In Problem 6.27, what is the duration according to the crew of the spaceship?

6.29 A spaceship is moving away from earth with speed ν = 0.6 c. When the ship
is at a distance d = 5 × 108
km from earth. A radio signal is sent to the ship
by the observers on earth.
How long does the signal take to reach the ship as measured by the scientist
on earth?
6.30 In Problem 6.29, how long does the signal take to reach the ship as measured
by the crew of the spaceship?
6.31 If the mean track length of 100 MeV π mesons is 4.88 m up to the point of
decay, calculate their mean lifetime.
6.32 A beam of π+
mesons of energy 1 GeV has an intensity of 106
particles per sec
at the beginning of a 10 m flight path. Calculate the intensity of the neutrino
flux at the end of the flight path (mass of π meson = 139 MeV/c2
, lifetime =
2.56 × 10−8
s)
6.33 A π+
meson at rest decays into a μ+
meson and a neutrino in 2.5 × 10−8
s.
Assuming that the π+
meson has kinetic energy equal to its rest energy. What
distance would the meson travel before decaying as seen by an observer at rest?
6.34 Beams of high-energy muon neutrinos can be obtained by generating intense
beams of π+
mesons and allowing them to decay while in flight. What frac-
tion of the π+
mesons in a beam of momentum 200 GeV/c will decay while
travelling a distance of 300 m?
At the end of the decay path (an evacuated tunnel) the beam is a mixture of
π+
-mesons, muons and neutrinos. What distinguishes these particles in their
interactions with matter, and how is a neutrino beam free of contamination by
π-mesons and muons obtained? [π+
-meson mean-life: 2.6 × 10−8
s]
6.35 A beam of 140 MeV kinetic energy π+
mesons through three counters A, B,
C spaced 10 m apart. If 1,000 pions pass through counter A and 470 in B. (a)
how many pions are expected to be recorded in C? (b) Find the mean life time
of pions (Take mass of pion as 140 MeV/c2
)
6.36 A moving object heads toward a stationary one with a velocity αc. At what
velocity βc would an observer have to move so that in his frame of reference
the objects would have equal and opposite velocities?
6.37 A beam of identical unstable particles flying at a speed βc is sent through
two counters separated by a distance L. It is observed that N1 particles are
recorded at the first counter and N2 at the second counter, the reduction being
solely due to the decay of the particles in flight.
(i) Show that the lifetime of the particles at rest is given by
τ =
L
ln N1
N2

(

γ 2 − 1 c

6.2 Problems 323
where the Lorentz factor γ is defined as usual
(ii) Hence determine the lifetime of muons at rest, knowing that when trav-
elling at a speed c
√
8/3 through the apparatus described above (with L =
200 m) N1 and N2 were measured to be 10,000 and 8,983, respectively.
[adapted from University of London, Royal Holloway and Bedford New
College]
6.38 A particle X at rest is a sphere of rest-mass m and radius r and has a proper
lifetime τ. If the particle is moving with speed
√
3
2
c with respect to the lab
frame (c is the speed of light):
(a) Determine the total energy of the particle in the lab frame
(b) The average distance the particle travels in the lab before decaying
(c) Sketch the shape and dimensions of the particle when viewed perpendicular
to its motion in the lab frame, include an arrow to indicate its direction of
motion on your sketch.
[adapted from the University of London Royal Holloway and Bedford New
College 2006]
6.39 The Lorentz velocity transformation is ν′
= ν−u
1−uν/c2 , where ν′
and ν are the
velocities of an object parallel to u as measured in two inertial frames with
relative velocity u. Show that a photon moving at c, the speed of light will
have the same speed in all frames of reference.
6.2.3 Mass, Momentum, Energy
6.40 The mean life-time of muons at rest is 2.2 × 10−6
s. The observed mean life-
time of muons as measured in the laboratory is 6.6 × 10−6
s. Find
(a) The effective mass of a muon at this speed when its rest mass is 207 me
(b) its kinetic energy (c) its momentum
6.41 Calculate the energy that can be obtained from complete annihilation of 1 g of
mass.
6.42 What is the speed of a proton whose kinetic energy equals its rest energy?
Does the result depend on the mass of proton?
6.43 What is the speed of a particle when accelerated to 1.0 GeV when the
particle is (a) proton (b) electron
6.44 (a) Calculate the energy needed to break up the 12
C nucleus into its con-
stituents. The rest masses in amu are:
12
C 12.000000; p 1.007825; n 1.008665; α 4.002603
(b) If 12
C nucleus is to break up into 3 alphas. Calculate the energy that is
released.

(c) If the alphas are to further break into neutrons and protons, then show that
the overall energy needed is identical with the results in (a)
6.45 The kinetic energy and the momentum of a particle deduced from mea-
surements on its track in nuclear photographic emulsions are 250 MeV and
368 MeV/c, respectively. Determine the mass of the particle in terms of elec-
tron mass and identify it.
[University of Durham]
6.46 What is the rest mass energy of an electron (me = 9.1 × 10−31
kg)?
6.47 What potential difference is required to accelerate an electron from rest to
velocity 0.6 c?
6.48 At what velocity does the relativistic kinetic energy differ from the classical
energy by
(a) 1% (b) 10%?
6.49 Prove that if ν/c ≪ 1, the kinetic energy of a particle will be much less
than its rest energy. Further show that the relativistic expression reduces to the
classical one for small velocities.
6.50 Find the effective mass of a photon for
(a) λ = 5,000Å (visible region) (b) λ = 1Å (X-ray region)
6.51 Show that 1 amu = 931.5 MeV/c2
6.52 Estimate the energy that is released in the explosion of a fission bomb con-
taining 5.0 kg of fissionable material
6.53 A proton moving with a velocity βc collides with a stationary electron of mass
m and knocks it off at an angle θ with the incident direction. Show that the
energy imparted to the electron is approximately
T = 2mc2
β2
cos2
θ/(1 − β2
cos2
θ)
6.54 A positive pion (mπ = 273 me) decays into a muon (mμ = 207 me) and
a neutrino (mν = 0) at rest. Calculate the energy carried by the muon and
neutrino, given that mec2
= 0.511 MeV
6.55 A body of rest mass m travelling initially at a speed of β = 0.6 makes a com-
pletely inelastic collision with an identical body initially at rest. Find (a) the
speed of the resulting body (b) its rest mass in terms of m.
6.56 A neutral particle is observed to decay into a kaon and a pion. They are pro-
duced in the opposite direction, each of them with momentum 861 MeV/c.
Calculate the mass of the neutral particle and identify it. (Mass of kaon is
494 MeV/c2
, Mass of pion is 140 MeV/c2
).
6.57 A pion at rest decays via π → μ + ν. Find the speed of the muon in terms of
the masses involved.

6.2 Problems 325
6.58 A particle A decays at rest via A → B + C. Find the total energy of B in
terms of the masses of A, B and C.
6.59 Calculate the maximum energy of the positron emitted in kaon decay at rest
K+
→ e+
+ π0
+ γe.
6.60 Consider a symmetric elastic collision between a particle of mass m and
kinetic energy T and a particle of the same mass at rest. Relativistically, show
that the cosine of the angle between the two particles after the collision is
T/(T + 4 mc2
)
6.61 An electron has kinetic energy equal to its rest energy. Show that the energy
of a photon which has the same momentum as this electron is given by
Eγ =
√
3E0, where E0 = mec2
6.62 Consider the decay of muon at rest. If the energy released is divided equally
among the final leptons, then show that the angle between paths of any two
leptons is approximately 120◦
(neglect the mass of leptons compared to the
mass of muon mass).
6.63 If a proton of 109
eV collides with a stationary electron and knock it off at
3◦
with respect to the incident direction, what is the energy acquired by the
electron?
6.64 Calculate (a) the mass of the pion in terms of the mass of the electron, given
that the kinetic energy of the muon from the pion decay at rest is 4.12 MeV
and (b) the maximum energy of electron (in MeV) from the decay of muon
at rest (mass of muon is 206.9 me). The mass of the electron is equivalent to
0.511 MeV)
6.65 Antiprotons are captured at rest in deuterium giving rise to the reaction.
p−
+ d → n + π0
Find the total energy of the π0
. The rest energies for p−
, d, n, π0
are 938.2,
1875.5, 939.5 and 135.0 MeV respectively.
6.66 As a result of a nuclear interaction a K∗+
particle is created which decays to a
K meson and a π−
meson with rest masses equal to 966 me and 273 me respec-
tively. From the curvature of the resulting tracks in a magnetic field, it is con-
cluded that the momentum of the secondary K and π mesons are 394 MeV/c
and 254 MeV/c respectively, their initial directions of motion being inclined
to one another at 154◦
. Calculate the rest mass of the K∗
particle
[Bristol 1964, 1966]
6.67 A proton of kinetic energy 940 MeV makes an elastic collision with a station-
ary proton in such a way that after collision, the protons are travelling at equal
angles on either side of the incident proton. Calculate the angle between the
directions of motion of the protons.
[Liverpool 1963]

6.68 A proton of momentum p large compared with its rest mass M, collides with
a proton inside a target nucleus with Fermi momentum pf. Find the available
kinetic energy in the collision, as compared with that for a free-nucleon target,
when p and pf are (a) parallel (b) anti parallel (c) orthogonal.
6.69 An antiproton of momentum 5 GeV/c suffers a scattering. The angles of the
recoil proton and scattered antiproton are found to be 82◦
and 2◦
30′
with
respect to the incident direction. Show that the event is consistent with an
elastic scattering of an antiproton with a free proton.
6.70 Show that if E is the ultra-relativistic laboratory energy of electrons incident
on a nucleus of mass M, the nucleus will acquire kinetic energy
EN = (E2
/Mc2
)(1 − cos θ)/(1 + E(1 − cos θ)/Mc2
)
where θ is the scattering angle.
6.71 A particle of mass M ≫ me scatters elastically from an electron. If the inci-
dent particle’s momentum is p and the scattered electron’s relativistic energy
is E and φ is the angle the electron makes with the incident particle, show that
M = P[{[E + me]/[E − me]} cos2
φ − 1]]1/2
6.72 A neutrino of energy 2 GeV collides with an electron. Calculate the maximum
momentum transfer to the electron.
6.73 A particle of mass m1 collides elastically target particle of mass m2 at rela-
tivistic energy. Show that the maximum angle at which m1 is scattered in the
lab system is dependent only on the masses of particles provided m1 m2
6.74 Show that if energy ν( mec2
) and momentum q are transferred to a free sta-
tionary electron the four-momentum transfer squared is given by q2
= −2meν
6.75 A photon of energy E travelling in the +x direction collides elastically with
an electron of mass m moving in the opposite direction. After the collision,
the photon travels back along the –x direction with the same energy E.
(a) Use the conservation of energy and momentum to demonstrate that the
initial and final electron momenta are equal and opposite and of magni-
tude E/c.
(b) Hence show that the electron speed is given by
v/c = (1 + (mc
2
/E)2
)−1/2
[adapted from the University of Manchester 2008]
6.2.4 Invariance Principle
6.76 Use the invariance of scalar product of two four-vectors under Lorentz trans-
formation to obtain the expression for Compton scattering wavelength shift.
6.77 Show that for a high energy electron scattering at an angle θ, the value of
the squared four-momentum transfer is given approximately by Q2
= 2E2

6.2 Problems 327
(1 − cos θ)/c2
, where E is the total initial electron’s energy in the lab system.
State when this approximation is justified.
6.78 A neutral unstable particle decays into π+
and π−
, each of which has a
momentum 530 MeV/c. The angle between the two pions is 90◦
. Calculate
the mass of the unstable particle.
6.79 If a particle of mass M decays in flight into m1 and another m2; m1 has
momentum p1 and total energy E1, where as m2 has momentum p2 and total
energy E2. p1 and p2 make an angle θ. Show that
E1 E2 − p1 p2 cos θ = invariant = 1
2
[M2
− m1
2
− m2
2
]
6.80 The Mandelstam variables s, t, and u are defined for the reaction A + B →
C + D, by
s = (PA + PB)2
/c2
, t = (PA − PC)2
/c2
, u = (PA − PD)2
/c2
where PA, PB, PC, PD are the relevant energy-momentum four vectors. Show
that
s + t + u =

mj
2
( j = A, B, C, D)
6.81 In Problem 6.80 show for the elastic scattering t = −2p2
(1−cos θ)/c2
where
p = |p|.p is the center of mass momentum of particle and θ is its scattering
angle in the CMS.
6.82 A neutral pion undergoes radioactive decay into two γ -rays. Obtain the
expression for the laboratory angle between the direction of the γ -rays,
and find the minimum value for the angle when the pion energy is 10 GeV
(mπ = 0.14 GeV)
6.83 A bubble chamber event was identified in the reaction
p−
+ p → π+
+ π−
+ ω0
The total energy available was 2.29 GeV while the kinetic energy of the resid-
ual particles was 1.22 GeV. What is the rest energy of ω0
in MeV?
6.84 A particle of rest mass m1 and velocity v1 collides with a particle of mass
m2 at rest after which the two particles coalesce. Show that the mass M
and velocity v of the composite particle are related by M2
= m1
2
+ m2
2
+
2m1m2/

1 − v2/c2
6.85 Show that for the decay in flight of a Λ-hyperon into a proton and a pion with
Laboratory momenta Pp and Pπ respectively, the Q value can be calculated
from
Q = (mp
2
+ mπ
2
+ 2Ep Eπ − 2Pp Pπ cos θ)1/2
− (mp + mπ )
where θ is the angle between Pp and Pπ in the Laboratory system and E is
the total relativistic energy
[University of Dublin 1967]
6.86 Two particles are moving with relativistic velocities in directions at right
angles, they have momenta p1 and p2 and total energies E1 and E2. If they

are the sole products of disintegration of a heavier object, what was the rest
mass, velocity and direction of motion?
6.87 A V-type of event is observed in a bubble chamber. The curvature measure-
ments on the two tracks show that their momenta are p+ = 1.670 GeV/c and
p− = 0.408 GeV/c. The angle contained between the two tracks is θ = 15◦
. It
is obviously due to the decay of a neutral unstable particle. It is suspected that
it is due to the decay (a) K0
→ π+
+ π−
or (b) Λ → p + π−
. Identify the
neutral particle.
6.88 Derive the formula in Problem 6.103 using the invariance of (ΣE)2
− |Σp|2
6.89 Calculate the maximum four momentum transfer to proton in the decay of
neutron at rest.
6.2.5 Transformation of Angles and Doppler Effect
6.90 Find the Doppler shift in wavelength of H line at 6,563 Å emitted by a star
receding with a relative velocity of 3 × 106
ms−1
.
6.91 Certain radiation of a distant nebula appears to have a wavelength 656 nm
instead of 434 nm as observed in the laboratory. (a) if the nebula is mov-
ing in the line of sight of the observer, what is its speed? (b) Is the nebula
approaching or receding?
6.92 Show that for slow speeds, the Doppler shift can be approximated as
Δλ/λ = v/c
where Δλ is the change in wavelength.
6.93 A physicist was arrested for going over the railway level crossing on a motor-
cycle when the lights were red. When he was produced before the magistrate
the physicist declared that he was not guilty as red lights (λ = 670 nm)
appeared green (λ = 525 nm) due to Doppler Effect. At what speed he was
travelling for the explanation to be valid? Do you think such a speed is
feasible?
6.94 A spaceship is receding from earth at a speed of 0.21 c. A light from the
spaceship appears as yellow (λ = 589.3 nm) to an observer on earth. What
would be its color as seen by the passenger of the spaceship?
6.95 Find the wavelength shift in the Doppler effect for the sodium line 589 nm
emitted by a source moving in a circle with a constant speed 0.05 c observed
by a person fixed at the center of the circle.
6.96 A neutrino of energy E0 and negligible mass collides with a stationary elec-
tron. Find an expression for the laboratory angle of emission of the electron
in terms of its recoil energy E and calculate its value when E0 = 2 GeV and
E = 0.5 GeV

6.2 Problems 329
6.97 Assume the decay K0
→ π+
+π−
. Calculate the mass of the primary particle
if the momentum of each of the secondary particles is 3 × 108
eV and the
angle between the tracks is 70◦
6.98 Neutral pions of fixed energy decay in flight into two γ -rays. Show that the
velocity of pion is given by
β = (Emax − Emin)/(Emax + Emin)
where E is the γ -ray energy in the laboratory
6.99 In Problem 6.98 show that the rest mass energy of π0
is given by mc2
=
2(Emax Emin)1/2
6.100 In Problem 6.98 show that the energy distribution of γ -rays in the laboratory
is uniform under the assumption that γ -rays are emitted isotropically in the
rest system of π0
6.101 In Problem 6.98 show that the angular distribution of γ -rays in the laboratory
is given by
I(θ) = 1/4πγ 2
(1 − β cos θ)2
6.102 In Problem 6.98 show that the locus of the tip of the momentum vector is an
ellipse
6.103 In Problem 6.98 show that in a given decay the angle φ between two γ -rays
is given by
sin(φ/2) = mc2
/2(E1 E2)1/2
6.104 In Problem 6.98 show that the minimum angle between the two γ -rays is
given by
φmin = 2mc2
/Eπ
6.105 In Problem 6.98 find an expression for the disparity D (the ratio of energies)
of the γ -rays and show that D 3 in half the decays and D 7 in one
quarter of them
6.106 In the interaction π−
+ p → K∗
(890) + Y0
∗
(1, 800) at pion momentum
10 GeV/c , K∗
is produced at an angle θ in the lab system. Calculate the
maximum value θm, given mπ = 0.140 GeV/c2
and mp = 0.940 GeV/c2
6.107 A particle of mass m1 travelling with a velocity v = βc collides elastically
with the particle m2 at rest. The scattering angles of m1 in the LS and CMS
are θ and θ∗
. Show that
(a) γc = (γ + ν)/(1 + 2γ ν + ν2
)1/2
(b) γ ∗
= (γ + 1/γ )/
√
(1 + 2γ/ν + 1/ν2
)
(c) tan θ = sin θ∗
/γc(cos θ∗
+ βc/β∗
)
(d) tan θ∗
= sin θ/γc(cos θ − βc/β∗
)
where βc is the CMS velocity, β∗
c is the velocity of m1 in CMS,
γc = (1 − βc
2
)−1/2
, γ ∗
= (1 − β∗2
)−1/2
, ν = m2/m1

6.108 A linear accelerator produces a beam of excited carbon atoms of kinetic
energy 120 MeV. Light emitted on de-excitation is viewed at right angles
to the beam and has a wavelength λ′
. If λ is the wavelength emitted by a
stationary atom, what is the value of (λ′
− λ)/λ? (Take the rest energies of
both protons and neutrons to be 109
eV)
[University of Manchester 1970]
6.109 A certain spectral line of a star has natural frequency of 5 × 1017
c/s. If the
star is approaching the earth at 300 km/s, what would be the fractional change
of frequency?
6.110 A neutral pion (mass 135 MeV/c2
) travelling with speed β = v/c = 0.8
decays into two photons at right angle to the line of flight. Find the angle
between the two photons as observed in the lab system.
6.111 An observer O sights light coming to him by an object X at 45◦
to its path as
in the diagram. If the corresponding angle of emission of light in the frame
of reference of the object is 60◦
, calculate the velocity of the object.
Fig. 6.3 Aberration of light
6.112 In an inertial frame S a rod of proper length L is at rest and at an angle θ
with respect to the x-axis with relativistic speed relative to S
(a) Show that the product tan θ. Lx is independent of which frame it is evalu-
ated in:
tan θ.Lx = tan θ′
.Lx
′
where Lx and Lx
′
are the projections of the rod length onto the x-axis in
frames S and S′
, respectively.
(b) In frame S the rod is at an angle θ = 30◦
knowing that an observer in
frame S′
measures the rod to be at an angle θ′
= 45◦
with respect to the
x-axis, determine the speed at which the rod is moving with respect to the
observer.
[adapted from University of London 2004 Royal Holloway]
6.2.6 Threshold of Particle Production
6.113 Show that the threshold energy for the production of a proton–antiproton pair
in the collision of a proton with hydrogen target (P + P → P + P + P + P−
)
is 6 Mc2
, where M is the mass of a proton or antiproton.

6.2 Problems 331
6.114 A positron–electron pair production can occur in the interaction of a gamma
ray with electron, via γ + e−
→ e−
+ e+
+ e−
. Determine the threshold.
6.115 Find the threshold energy for the pion production in the reaction N + N →
N + N + π, given
MN c2
= 940 MeV and mπ c2
= 140 MeV
6.116 Show that a Fermi energy of 25 MeV lowers the threshold incident kinetic
energy for antiproton production by proton incident on nucleus to 4.3 GeV.
6.117 Find the threshold energy of the reaction
γ + p → K+∗
+ Λ
The laboratory proton is at rest. The following rest energies may be assumed,
for proton 940 MeV for K∗
890 MeV, for Λ 1,110 MeV
6.118 Find the threshold energy for the production of two pions by a pion incident
on a hydrogen target. Assume the rest masses of the pion and proton are 273
me and 1,837 me, where the rest energy of the electron mec2
= 0.51 MeV
6.119 Calculate the threshold energy of the following reaction π−
+ p → K0
+ Λ
The masses for π−
and p, K0
and Λ are, 140, 938, 498, 1,115 MeV respec-
tively
6.120 Show that the threshold kinetic energy in the Laboratory for the production
of n pions in the collision of protons with a hydrogen target is given by
T = 2nmπ (1 + nmπ /4mp)
where mπ and mp are respectively the pion and proton masses.
6.121 A gamma ray interacts with a stationary proton and produces a neutral pion
according to the scheme
γ + p → p + π0
Calculate the threshold energy given Mp = 940 MeV and Mπ = 135 MeV
6.122 Calculate the threshold energy of the reaction
π−
+ p → Ξ−
+ K+
+ K0
The masses for π−
, p, Ξ−
, K+
, K0
are respectively 140, 938, 1,321, 494,
498, MeV
6.123 Attempts have been made to produce a hypothetical new particle, the W+
,
using the reaction ν + p → p + μ−
+ W+
where stationary protons are
bombarded with neutrinos. If a neutrino energy of 5 GeV is not high enough
for this reaction to proceed, estimate a lower limit for the mass of the W+
(mass of proton 938 MeV/c2
, mass of muon 106 MeV/c2
, neutrino has zero
rest mass)

6.124 For the reaction p + p → p + Λ + K+
calculate the threshold energy and
the invariant mass of the system at threshold energy. The rest energies of the
p, Λ and K+
are respectively 938, 1,115 and 494 MeV
6.125 The Ω−
has been produced in the reaction
K−
+ p → K0
+ K+
+ Ω−
What is the minimum momentum of the K−
in the Laboratory for this reac-
tion to proceed assuming that the target proton is at rest in the laboratory?
Assume that a Ω−
is produced in the above reaction with this momentum
of K−
. What is the probability that Ω−
will travel 3 cm in the Lab before
decaying? You may ignore any likelihood of the Ω−
interacting. The rest
energies of K+
, K−
, K0
, P, Ω−
are 494, 494, 498, 938, 1,675 MeV, respec-
tively. Lifetime of Ω−
= 1.3 × 10−10
s.
6.126 Assuming that the nucleons in the nucleus behave as particles moving inde-
pendently and contained within a hard-walled box of volume (4/3)π R3
where
R is the nuclear radius, calculate the maximum Fermi momentum for a pro-
ton in 29Cu63
the nuclear radius being 5.17 fm.
A proton beam of kinetic energy 100 MeV (momentum 570.4 MeV/c) is
incident on a target 29Cu63
. Would you expect any pions to be produced by
the reaction p + p → d + π from protons within the nucleus (neglect the
binding energy of nucleons in the copper, i.e assume a head-on collision with
a freely moving proton having maximum Fermi momentum and calculate the
total energy in the CMS of the p + p collision). The binding energy of the
deuteron is 2.2 MeV
6.3 Solutions
6.1 t1 = γ (t1
′
+ vx1
′
/c2
)
t2 = γ (t2
′
+ vx2
′
/c2
)
t2 − t1 = γ (t2
′
− t1
′
) + γ v(x2
′
− x1
′
)/c2
= 0 + γ ν(x2
′
− x1
′
)/c2
10−6
= γβ × 900/c
γβ = 10−6
× 3 × 108
/900 = 1/3
β = 0.316. The velocity of S′
is 0.316 c with respect to S along the positive
direction.

6.3 Solutions 333
6.2 The transformation matrix is
Λ =
⎡
⎢
⎢
⎣
γ 0 0 iβγ
0 1 0 0
0 0 1 0
−iβγ 0 0 γ
⎤
⎥
⎥
⎦
Set γ = cosh α and β = tanh α, so that γβ = sinh α, the transformation
matrix becomes
Λ =
⎡
⎢
⎢
⎣
cosh α 0 0 i sinh α
0 1 0 0
0 0 1 0
− i sinh α 0 0 cosh α
⎤
⎥
⎥
⎦ .
Since we can write i sinh α = siniα and cosh α = cosiα, the matrix
Λ corresponds to a rotation through an angle iα in four-dimensional space,
Further the transformation equations can be obtained from
⎡
⎢
⎢
⎣
x′
1
x′
2
x′
3
ict′
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
cosh α 0 0 i sinh α
0 1 0 0
0 0 1 0
−i sinh α 0 0 cosh α
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
x1
x2
x3
ict
⎤
⎥
⎥
⎦ .
and the inverse transformation equations from
⎡
⎢
⎢
⎣
x1
x2
x3
ict
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
cosh α 0 0 −i sinh α
0 1 0 0
0 0 1 0
i sinh α 0 0 cosh α
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
x′
1
x′
2
x′
3
ict′
⎤
⎥
⎥
⎦ .
6.3 β∗
= 0.268 and βc = 0.8
γc = 1/(1 − βc
2
)1/2
= 1/(1 − 0.82
)1/2
= 1/0.6 = 1.667
γ ∗
= 1/(1 − β∗2
)1/2
= 1/(1 − 0.2682
)1/2
= 1.038
γ = γcγ ∗
(1 + βcβ∗
cos θ∗
), β = (γ 2
− 1)1/2
/γ
tan θ = sin θ∗
/γc(cos θ∗
+ βc/β∗
)
θ∗
= 0
γ = (1.667 × 1.038)(1 + 0.8 × 0.268) = 1.4248
β = 0.712
tan θ = 0 (Because θ∗
= 0)
Therefore, θ = 0

6.4 θ∗
= 90◦
γ = γcγ ∗
= 1.667 × 1.038 = 1.73
β = (1.732
− 1)1/2
/1.73 = 0.816
tan θ = sin θ∗
/(γcβc/β∗
) = 1/(1.667×0.8/0.268) = 0.2
θ = 11.3◦
6.5 θ = 90◦
γ ∗
= γcγ (1 − βcβ cos θ) = γcγ
γ = γ ∗
/γc
γc = 1/[1 − (0.2)2
]1/2
= 1.02
γ ∗
= 1/[1 − (0.268)2
]1/2
= 1.038
γ = 1.038/1.02 = 1.0176
β = (γ 2
− 1)1/2
/γ = (1.01762
− 1)1/2
/1.0176 = 0.185
tan θ∗
= sin θ/γc(cos θ − βc/β∗
)
= −β∗
/βcγc (Because θ = 90◦
)
βc = βπ = 0.2
tan θ∗
= −0.268/1.02 × 0.2 = 1.3137
θ∗
= 127◦
6.6 The scalar wave equation for the propagation of electromagnetic waves deriv-
able from Maxwell’s equations is:
(∂2
/∂x2
+ ∂2
/∂y2
+ ∂2
/∂z2
− (1/c2
)∂2
/∂t2
)ϕ(x, y, z, t) = 0 (1)
for the S system. We are required to show that in S′
system, the equation has
the form:
∂2
∂x′2
+
∂2
∂y′2
+
∂2
∂z′2
−

1
c2

∂2
∂t′2

ϕ(x′
, y′
, z′
, t′
) = 0
The Lorentz transformations are:
x′
= γ (x − βct) (2)
y′
= y (3)
z′
= z (4)
t′
= γ (t − βx′
c) (5)
Assume that we have propagation along x-axis so that the wave func-
tion will depend only on x and t. Now the function ϕ(x′
, y′
, z′
, t′
) = 0
is obtained from ϕ(x, y, z, t) = 0 by a substitution of variables. We have
ϕ(x, t) = ϕ(x′
, t′
). Then,
dϕ = (∂ϕ/∂x) dx + (∂ϕ/∂t) dt = (∂ϕ/∂x′
) dx′
+ (∂ϕ/∂t′
)dt′
(6)
Differentiating (2) and (5),
dx′
= γ (dx − βcdt) (7)
dt′
= γ (dt − βc dt) (8)
Substituting (7) and (8) in (6) and equating the coefficients of dx and dt:
(∂ϕ/∂x) dx + (∂ϕ/∂t) dt = (γ ∂ϕ/∂x′
− γβc∂ϕ/∂t′
) dx +
(γ ∂ϕ/∂t′
− γβc∂ϕ/∂x′
) dt

6.3 Solutions 335
∂ϕ/∂x = γ (∂ϕ/∂x′
− βc∂ϕ/∂t′
) (9)
∂ϕ/∂t = γ (∂ϕ/∂t′
− βc∂ϕ/∂x′
) (10)
∂2
ϕ
∂x2
= γ 2 ∂2
ϕ
∂x′2
+ (γ 2
β2
c2
)
∂2
cϕ
∂t′2
− (2γβc)
∂2
ϕ′
∂x′∂t′
(11)
∂2
ϕ
∂t2
=
γ 2
∂2
ϕ
∂t′2
+ (γ 2
β2
c2
)
∂2
ϕ
∂x′2
− (2γβc)
∂2
ϕ′
∂x′∂t′
(12)
Dividing (12) through c2
and subtracting the resulting equation from (11)
∂2
ϕ
∂x2 −
1
c2
∂2
ϕ
∂t2
= (γ 2
− γ 2
β2
)
∂2
ϕ
∂x′2
−

1
c2

(γ 2
− γ 2
β2
)
∂2
ϕ
∂t′2
=
∂2
ϕ
∂x′2
−

1
c2

∂2
ϕ
∂t′2
since γ 2
− γ 2
β2
= γ 2
(1 − β2
) = 1
Similarly, the Klein–Gordon equation
(∇2
− (1/c2
)∂2
/∂t2
+ m2
c2
/2
) = 0 is Lorentz invariant.
6.7 The only way π−
is emitted at rest in the lab system is when it is emitted at
θ1
∗
= 180◦
in the CMS (rest frame of K◦
) with with the same speed as K◦
in
the lab system. In that case π−
will be emitted at θ∗
2 = 0◦
in the CMS.
The energy released Q = 498 − 2 × 140 = 218 MeV
As the product particles are identical, each pion carries half of the enrgy,
109 MeV
γ ∗
= 1 + T ∗
/mπ = 1 + 109/140 = 1.778
From the above discussion
γc = γ ∗
, βc = β∗
γ = γ ∗
γc(1 + β∗
βc) = γ ∗2
(1 + β∗2
)
= γ ∗2
(1 + (γ ∗2
− 1)/γ ∗2
) = 2γ ∗2
− 1
= 2 × 1.7782
− 1 = 5.3266
T = (γ − 1)mπ = (5.3266 − 1) × 140
= 605.7 MeV
6.8 tan θν
∗
= sin θν/γc(cos θν − βc/βν
∗
) = −1/γcβc (1)
(Because θν = 90◦
and β∗
ν = 1). Here βc is the velocity of the pion.
It follows that
sin θν
∗
= 1/γc and cos θ∗
ν = −βc (2)
In the CMS (the system in which the pion is at rest)
θμ
∗
= π − θν
∗
, because the muon and neutrino must fly in the opposite direc-
tion to conserve momentum.
tan θμ
∗
= tan(π − θν
∗
) = − tan θν
∗
= −1/γcβc

or
tan θμ
∗
= 1/βcγc (3)
tan θμ = sin θμ
∗
/γc(cos θμ
∗
+ βc/βμ
∗
) = 1/γc
2
βc(1 + 1/β∗
μ) (4)
(Since θμ
∗
= π − θν
∗
) and we have used (2)
But β∗
μ = (mπ
2
− mμ
2
)/(mπ
2
+ mμ
2
) (5)
Substituting (5) in (4) and simplifying
tan θμ = (mπ
2
− m2
μ)/2γ 2
cβcm2
π
6.9 The z-component of velocity is zero; hence the particle must be moving in the
xy =plane. Further, the y-component of velocity is unchanged. This implies
that the Lorentz transformation is to be made along x-axis
cPx = γ (cP′
x + βE′
) (1)
c2
mβx γ0 = γ (c2
mβ′
x γ ′
+ mβγ ′
c2
) (2)
βx = 1/21/2
, γ0 = 21/2
, β′
x = −1/21/2
, γ ′
= 21/2
, γ = 1/(1 − β2
)1/2
(3)
Using (4) in (1) and simplifying we get β = 2 × 21/2
/3
6.10 Energy conservation gives
T1 + T2 = Q (1)
Momentum conservation gives
P1 + P2 = 0
or p1
2
= p2
2
T1
2
+ 2T1 m1 = T2
2
+ 2T2 m2 (2)
Solving (1) and (2)
T1 = Q(Q + 2m2)/2(m1 + m2 + Q); T2 = Q(Q + 2m1)/2(m1 + m2 + Q)
6.11 γπ = 1 + Tπ /mπ = 1 + 140/140 = 2
βπ = (γπ
2
− 1)1/2
/γπ = (22
− 1)1/2
/2 = 0.866
By Problem 6.54 , Tμ
∗
= 4.0 MeV therefore
γμ
∗
= 1 + Tμ
∗
/mμ = 1 + (4/106) = 1.038
βμ
∗
= (1.037772
− 1)1/2
/1.0377 = 0.267
γμ = γ γμ
∗
(1 + βπ βμ
∗
cos θ∗
)
γμ(max) = γπ γμ
∗
(1 + βπ βμ
∗
) = 2 × 1.038(1 + 0.866 × 0.267) = 2.556
(Because θ∗
= 0)
Tμ(max) = (γμ(max) − 1)mμ = 165 MeV
Using the formula for optical Doppler effect
Tν(max) = γπ Tν
∗
(1 + βπ ) = 2 × 29.5(1 + 0.866) = 110 MeV
6.12 βc = |p+ + p−|/(E+ + E−)
Using the invariance principle
(total energy)2
− (total momentum)2
= invariant
(E+ + E−)2
− |p+ + p−|2
= (E1
∗
+ E2
∗
)2
− |p1
∗
+ p2
∗
|2
But E1
∗
= E2
∗
since the particles have equal masses. Also by definition of
center of mass, |p1
∗
+ p2
∗
| = 0
Therefore, E1
∗2
= E2
∗2
= 1
4
[E+ + E−)2
− (p+
2
+ p−
2
+ 2 p+ p− cos θ)]
where θ is the angle between e+
− e−
pair

6.3 Solutions 337
E1
∗
= E2
∗
= (1/4)[(E+
2
− p+
2
+ E−
2
− p−
2
+ 2(E+ E− − p+ p− cos θ)]
E1
∗
= E2
∗
= (1/4)[m2
+ m2
+ 2(E+ E− − p+ p− cos θ)]
= (1/2)(m2
+ E+ E− − p+ p− cos θ)
or E1
∗
= E2
∗
= [1/2(m2
+ E+ E− − p+ p− cos θ)]1/2
6.13 tan θ = sin θ∗
/γc(cos θ∗
+ βc/β1
∗
) = (1/γc) tan θ∗
/2 (1)
(Because βc = β1
∗
)
Also
tan ϕ = (1/γc) tan θ∗
/2 (2)
Where θ∗
and ϕ∗
are the corresponding angles in the CM system
Multiply (1) and (2)
tan θ tan ϕ = (1/γc
2
). tan θ∗
/2. tan ϕ∗
/2
But ϕ∗
= π − θ∗
and for m1 = m2, γc = ((γ + 1)/2)1/2
tan θ tan ϕ = 2/(γ + 1)
In the classical limit, γ → 1 and tan θ tan ϕ = 1
But since tan(θ + ϕ) = (tan θ + tan ϕ)/(1 − tan θ tan ϕ)
tan(ϕ + θ) → ∞
i.e. θ + ϕ = π/2
For γ 1, tan θ tan ϕ 1. For tan(θ + ϕ) to be finite (θ + ϕ) π/2
Hence, for γ 1; θ + ϕ π/2
6.14 The total energy carried by π+
in the rest frame of K+
can be calculated from
Eπ+ = (mK
2
+ mπ
2
− mν
2
)/2mk = 265 MeV (1)
γc = γ ∗
π+ = 265/139.5 = 1.9
and βc = βπ
∗
= (γ ∗2
π+ − 1)1/2
/γ ∗
π+
In the rest frame of pion, the total energy of muon is obtained again by (1)
Eμ+
2
= (mμ+
2
+ mπ+
2
− 0)/2mπ+ = 110 MeV
γμ
∗
= 110/106 = 1.0377
βμ
∗
= (γ ∗2
μ − 1)1/2
/γμ
∗
= 0.267
The Lorentz factor γμ for the muon in the LS is obtained from γμ = γcγ ∗
μ(1 +
β∗
μβc cos θ∗
μ) where θ∗
μ is the emission angle of the muon in the rest frame of
pion. Put θ∗
μ = 0 to obtain γμ(max) and θ∗
μ = 180◦
to obtain γμ(min). The
maximum and minimum kinetic energies are 150 and 55.5 MeV.
6.15 First we find the momenta of three pions by using the formula
P1 = (E2
1 −m2
)1/2
etc, where the total energy, E1 = T1 +m, m = 140 MeV is
the pion mass. We need to find the total momentum of the three particles. Take
the direction of the middle particle as x-axis. Calculate these components as
below:
T1(MeV) = 190, E1 = 330(MeV), P1 = 299 MeV/c,
p1(x) = 276.4 MeV/c, p1(y) = −114 MeV/c
T2 = 321, E2 = 461, P2 = 439, p2(x) = 439, p2(y) = 0
T3 = 58, E3 = 198, P3 = 140, p3(x) = 137, p3(y) = 227

E = 989 MeV,

p(x) = 852.4 MeV/c

P(y) = −84 MeV/c

Fig. 6.4 Decay of a charged
unstable particle into three
pions
Therefore, P = [

p(x)2
+ (

p(y)2
]1/2
= 856.4 MeV/c
The mass of the particle is given by
M =
*
E
2
−
%
%
p
%
%2
+1/2
= [(989)2
− (856.4)2
]1/2
= 494.7 MeV
It is a K meson
We can find the direction of K meson by calculating the resultant momenta
of the three pions and its orientation with respect to one of the pions.
By the vector addition of p2 and p3 we find the resultant P23 = 576.6 MeV/c
inclined at angle α = 2.84◦
above p2 as in the Fig. 6.4. The angle inclined
between P23 and p1 is ϕ = 2.84 + 22.4 = 25.24◦
.
When p23 is combined with vector p we find that the resultant is inclined at
an angle of 16.7◦
above p1.
6.16 L = γ L0 = (1 − β2
)1/2
L0 = (1 − 0.82
)1/2
L0
= 0.6L0
ΔL = L0 − L = 0.4L0
6.17 L = L0/γ = L0/2
γ = 2 → β = (γ 2
− 1)1/2
/γ = (22
− 1)1/2
/2 = 0.866
v = βc = 0.866 × 3 × 108
= 2.448 × 108
ms−1
6.18 β = v/c = 30 km s−1
/3 × 105
km s−1
= 10−4
1/γ = (1 − β2
)1/2
= (1 − 1/2 × β2
)
ΔL = L0 − L = L0 − L0/γ = L0 − L0(1−β2
)1/2
= 1
2
L0β2
= 1/2×6,400×
10−8
km = 3.2 cm
Thus the earth appears to be shrunk by 3.2 cm.
6.19 τ = γ τ0
1.5 × 10−5
= 2.2 × 10−6
γ
γ = 6.818
β = (γ2
− 1)1/2
/γ = 0.9892
v = 0.9890c

6.3 Solutions 339
6.20 τ0 = τ/γ = τ(1 − β2
)1/2
= 2.9 × 10−6
× (1 − 0.62
)1/2
= 2.32 × 10−6
s
6.21 (a) Time, τ = τ0/(1 − β2
)1/2
= 2.2 × 10−6
/(1 − 0.62
)1/2
= 2.75 × 10−6
s
Distance d = vτ = 0.6 × 3 × 108
× 2.75 × 10−6
= 495 m
(b) d0 = vτ0 = 0.6 × 3 × 108
× 2.2 × 10−6
= 396 m.
6.22 d = vt = vγ t0 = 3 × 104
c
γβ = 3 × 104
/40 = 750
β/(1 − β2
)1/2
= 750
β = 0.99999956
6.23 d = vγ t0 = βγ ct0 = 0.99 × 3 × 108
× 2.5 × 10−8
/(1 − 0.992
)1/2
= 373 m.
It is therefore observed at an altitude of 1,000 − 373 = 627 m above the sea
level.
6.24 β = (β1 + β2)/(1 + β1β2) = (0.9 + 0.9)/(1 + 0.9 × 0.9) = 0.994475
6.25 E = T + m0c2
= 100 + 0.51 = 100.51 MeV
γ = E/m0c2
= 100.51/0.51 = 197
L = L0/γ = 4/197 = 0.02 m = 2 cm
6.26 m = γ m0
γ = m/m0 = 101/100 = 1.01
β = (γ 2
− 1)1/2
/γ = 0.14
6.27 If the space station is located at a distance d from the earth then d is fixed by
the time taken by the radio signal to reach the earth is
d = ct
As observed from the earth, at t1 = 0 the spaceship was at a distance d
approaching with speed 0.5 c. It will arrive at time
t1 = d/βc = ct/βc = 1,125/0.5 = 2,250 s
6.28 The time t2 recorded in the spaceship related to t1 is shortened by γ , the
Lorentz factor.
t2 = t1/γ = t(1 − β2
)1/2
= t(1 − 0.52
)1/2
= 0.866 t
= 0.866 × 2,250 = 1,948 s
6.29 Let system S be attached to the ground and S′
to the spaceship.
Let t1 be the time when the radio signal reaches the ship. In that time the
signal traveled a distance
d1 = ct1
At time t1 = 0, the ship was at a distance d.
At time t1 it is now at a distance
d2 = d + vt1 = d + βct1
Now d1 = d2

So ct1 = d + βct1
Solving for t1,
t1 = d/c(1 − β)
For β = 0.6, γ = 1.25
t1 = 5 × 108
× 103
/3 × 108
× 0.4 = 4167 s
6.30 From Lorentz transformations we get
Δt′
= γ Δt = γ d1/c = 1.25 × 5 × 108
× 103
/3 × 108
= 2,083 s
6.31 τ = d/vγ = d/γβc = d/c(γ 2
− 1)1/2
(1)
γ = E/m = 1 + (T/m) = 1 + (100/140) = 1.714
D = 4.88 m, c = 3 × 108
ms−1
Using these values in (1), we get τ = 1.17 × 10−8
s
6.32 I = I0 e−t/τ
= I0 e−γ d/cβτ
γ = 1/0.14 = 7.143, d = 10 m, c = 3 × 108
ms−1
β = (1 − 1/7.1432
)1/2
= 0.99, τ = 2.56 × 10−8
s
I0 = 106
Using the above values, we find I = 83
6.33 (γ − 1)M = M
Or γ = 2
β = (γ 2
− 1)1/2
/γ = (22
− 1)1/2
/2 =
√
3/2
The dilated time T = γ T0 = 2 × 2.5 × 10−8
= 5 × 10−8
s
The distance traveled before decaying is
d = vT = β cT =
√
3/2 × 3 × 108
× 5 × 10−8
= 13 m
6.34 Time t = d/v = 300/3 × 108
= 1.0 × 10−6
s
As v ≈ c at ultrarelativistic velocity
The proper lifetime is dilated
τ = τ0γ = τ0 E/m = 2.6 × 10−8
× (200 × 103
+ 140)/140
= 3.71 × 10−3
s
Fraction f of pions decaying is given by the radioactive law
f = 1 − exp(−T/τ)
= 1 − exp(−0.0269)
= 0.027
The pions and muons are subsequently stopped in thick walls of steel and
concrete, pions through their nuclear interactions and muons through absorp-
tion by ionization. The neutrinos being stable, neutral and weakly interacting
will survive.
6.35 Assuming that the pions decay exponentially (the law of radioactivity), then
after time t they travel a distance d, with velocity v = βc so that t = d/βc
and their mean lifetime is lengthened by the Lorentz factor γ .
(a) The intensity at counter B will be
IB = IA exp (−d/γβcτ) (1)

6.3 Solutions 341
and at the counter C
Ic = IA exp (−2d/γβcτ) (2)
∴ Ic =
I2
B
IA
=
(470)2
1000
= 221
(b) Take logarithm on both sides of (1) and simplify.
τ =
d
γβcln(IA/IB)
(3)
γ = 1 + T/mc2
= 1 + 140/140 = 2.0
β = (γ 2
− 1)1/2
/γ = 0.866
d = 10 m; c = 3 × 108
m/s
ln(IA/IB) = ln(1000/470) = 0.755
Substituting the above values in (3),
τ = 2.55 × 10−8
s
The accepted value is 2.6 × 10−8
s
6.36 The stationary object will appear to move with velocity −βc toward the
observer. The object moving with velocity αc toward the stationary object
would appear to have velocity
(αc − βc)/(1 − αβ), as seen by the observer. If these two velocities are to be
equal then (αc − βc)/(1 − αβ) = βc
Cross multiplying and simplifying we get the quadratic equation whose
solution is β = [1 − (1 − α2
)1/2
]/α
6.37
(i) t =
L
βc
(1)
N2 = N1 exp[−t/γ τ] = N1 exp

−
L
γβcτ

(2)
exp
*
L
γβcτ
+
= N1/N2
Take logarithm on both sides
L
γβcτ
= ln N1
N2

But γβ =

γ 2 − 1
Therefore τ = L
ln
N1
N2
√
γ 2−1c
(ii) γ = 1/(1 − β2
)1/2
= 1/(1 − 8/9)1/2
= 3
τ =
200
ln 10,000
8,983
√
32 − 1 × 3 × 108
= 2.2 × 10−6
s.

6.38 (a) β = v/c =
√
3
2
γ = (1 − β2
)−1/2
= (1 − 3/4)−1/2
= 2
Total energy of the particle
E = γ Mc2
= 2Mc2
(b) Distance traveled on an average
d = γβct0 = 2 ×
√
3
2
cτ =
√
3 cτ
(c) The sphere will shrink in the direction of motion but will not in the trans-
verse direction. Consequently, its shape would appear as that of a spheroid
as shown in Fig. 6.5
Fig. 6.5
6.39 ν′
= ν−u
1−uν/c2
put v = c
ν′
= c−u
1−uc/c2 = c−u
1−u/c
= c
6.40 (a) m = m0γ
γ = τ/τ0 = 6.6 × 10−6
/2.2 × 10−6
= 3
m = 3 × 207 = 621 me
(b) T = (γ − 1)m0c2
= (3 − 1)(207 × 0.51)
= 211 MeV
(c) Total energy E = mc2
= 621mec2
= 621 × 0.511 = 3.173 MeV
p = βE/c
β = (γ 2
− 1)1/2
/γ = (32
− 1)1/2
/3 = 0.9428
p = (0.9428)(317.3)/c
= 299 MeV/c
6.41 E = m0 c2
= (1 × 10−3
kg)(3 × 108
)2
= 9 × 1013
J
6.42 T = (γ − 1)m = m
γ = 2
β = (γ2
− 1)1/2
/γ = 0.866
The result is independent of the mass of the particle.

6.3 Solutions 343
6.43 (a) γ = 1 + T/m = 1 + 1, 000/940 = 2.064
β = (γ 2
− 1)1/2
/γ = [(2.064)2
− 1]1/2
/2.064 = 0.87
(b) γ = 1 + 1, 000/0.511 = 1, 958
β = [1, 9582
− 1]1/2
/1, 958 = 0.999973
6.44 (a) There are 6 protons and 6 neutrons in 12
C nucleus.
Δmc2
= [(1.007825 × 6 + 1.008665 × 6 − 12.000] × 931.5
= 92.16 MeV
(b) Δmc2
= [3 × 4.002603 − 12.000] × 931.5
= 7.27 MeV
(c) Δmc2
= [2 × (mn + mp) − mα] × 931.5
= [2 × (1.008665 + 1.007825) − 4.002603] × 931.5
= 28.296 MeV
Total energy required = 3 × 28.296 + 7.27 = 92.16 MeV
which is identical with that in (a)
6.45 c2
p2
= T 2
+ 2T mc2
m = (c2
p2
− T 2
)/2T = (3682
− 2502
)/2 × 250 = 145.85 MeV
Or m = 145.85/0.511 = 285.4 me
The particle is identified as the pion whose actual mass is 273 me
6.46 E0 = m0c2
= (9.1 × 10−31
kg)(3 × 108
ms−1
)2
= 8.19 × 10−14
J
= (8.19 × 10−14
J)(1 MeV/1.6 × 10−13
J)
= 0.51 MeV
6.47 γ = 1/(1 − β2
)1/2
= 1/(1 − 0.62
)1/2
= 1.25
Energy acquired by electron
T = (γ − 1)mec2
= (1.25 − 1) × 0.51 = 0.1275 MeV
1 eV energy is acquired when an electron (or any singly charged particle) is
accelerated from rest through a P.D of 1 V. Hence the required P.D is 0.1275
Mega volt or 127.5 kV.
6.48 (a) K(relativistic) = (γ − 1)m0c2
(1)
K(classical) = (1/2)m0v2
= (1/2)m0c2
β2
= (1/2)m0c2
(γ 2
− 1)/γ 2
(2)
ΔK
K
=
K(relativistic) − K(Classical)
K(relativistic)
=
γ − 1
2γ
(3)
where we have used (1) and (2)
Putting ΔK/K = 1/100, we find γ = 50/49. Using
β = (γ2
− 1)/γ (4)
we obtain β = 0.199
Or v = 0.199 c
(b) Putting ΔK/K = 10/100 = 1/10 in (3), we find γ = 5/4, Using (4) we
obtain β = 0.6
Or v = 0.6 c

6.49 T = (γ − 1)m0c2
= [1/(1 − β2
)1/2
− 1]m0c2
= [(1 − β2
)−1/2
− 1]m0c2
= [1 + 1
2
β2
+ 3
8
β4
+ . . . − 1]m0c2
= [1 + 3
4
β2
+ . . .](1
2
m0β2
c2
)
where we have expanded the radical binomially
For v/c ≪ 1 or β ≪ 1, obviously T ≪ m0c2
For small velocities T = m0β2
c2
/2 + small terms = m0v2
/2
6.50 (a) The photon energy Eγ = 1, 240/500 nm = 2.48 eV
= 2.48 × (1.6 × 10−19
) J
= 3.968 × 10−19
J
Effective mass, m = Eγ /c2
= 3.968 × 10−19
/(3 × 108
)2
= 4.4 × 10−36
kg
(b) E = 1,240/0.1 nm = 12,400 eV = 1.984 × 10−15
J
m = 1.984 × 10−15
/(3 × 108
)2
kg = 2.2 × 10−32
kg.
6.51 1amu = 1.66 × 10−27
kg
=
(1.66 × 10−27
kg)(c2
)
c2
= 1.66 × 10−27
× 2.998 × 108
)2
J/c2
= 1.492 × 10−10
J/c2
= 1.492 × 10−10
J/MeV.MeV/c2
= 1.492 × 10−10
/1.602 × 10−13
MeV/c2
= 931.3 MeV/c2
6.52 Number of Uranium atoms in 1.0 g is
N = N0/A = 6.02 × 1023
/235 = 2.56 × 1021
Number in 5 kg = 2.56 × 1021
× 5,000 = 1.28 × 1025
In each fission ∼ 200 MeV energy is released.
Therefore, total energy released
= 1.28 × 1025
× 200 = 2.56 × 1027
MeV
= (2.56 × 1027
MeV/J)(1.6 × 10−13
J)
= 4 × 1014
J
6.53 The analysis is similar to that for Compton scattering except for some approx-
imations.
Energy conservation gives
E = E′
+ T (1)
From momentum triangle (Fig 6.6)
p′2
= p2
+ pe
2
− 2ppe cos θ (2)
Using c2
p′2
= E′2
− M2
c4
(3)
c2
pe
2
= T 2
+ 2T mc2
(4)
p = γ Mβc (5)
γ = 1/(1 − β)1/2
(6)
Combining (1) – (6) and simplifying and using the fact that mc2
≪ E. we
easily obtain the desired result.

6.3 Solutions 345
Fig. 6.6 (a) Scattering of a
proton with a stationary
electron (b) Momentum
triangle
6.54 The energy released in the decay of pion is
Q = mπ c2
− (mμc2
+ mνc2
)
= (273 − 207 − 0)mec2
= 66 × 0.511 = 33.73 MeV
Tμ + Tν = 33.73 (1)
In order to conserve momentum, muon neutrino must move in opposite direc-
tion
pν = pμ (2)
Multiplying (2) by c and squaring
c2
p2
ν = T 2
ν = Kμ + 2Kμmμc2
(3)
Solving (1) and (3) and using
mμc2
= 207mec2
= 207 × 0.511 = 105.77 MeV
Kμ = 4.08 MeV, Kν = 29.65 MeV
Observe that the lighter particle carries greater energy.
6.55 Let the mass of the final single body be M which moves with a velocity βc.
m × 0.6c/(1 − 0.62
)1/2
= Mβc/(1 − β2
)1/2
Or 3m/4 = Mβ/(1 − β2
) (1)
Since the total energy is conserved
mc2
/(1 − 0.62
)1/2
+ mc2
= Mc2
/(1 − β2
)1/2
(2)
(a) Using (2) in (1), β = 1/3
(b) Using β = 1/3 in (2), M = 2.12 m

6.56 TK = [(MD − MK )2
− M2
π ]/2MD = [(1865 − 494)2
− 1402
]/2 × 1865
= 498.67 MeV
PK = (TK
2
+ 2MK TK )1/2
= 861 MeV/c
EK
2
= pK
2
+ mK
2
= (0.861)2
+ (0.494)2
= 0.9853 GeV2
EK = 0.9926 GeV/c
Eπ
2
= pπ
2
+ mπ
2
= (0.861)2
+ (0.140)2
= 0.7609 GeV2
Eπ = 0.8736 GeV
Mx = EK + Eπ = 1.8662 GeV/c2
It is a D0
meson.
6.57 The mass of neutrino is zero. Applying conservation laws of energy and
momentum
Eμ + Eν = mπ c2
(1)
pμ = pν (2)
Multiplying (2) by c and squaring
c2
pμ
2
= c2
pν
2
Or Eμ
2
− mμ
2
c4
= Eν
2
Or Eμ
2
− Eν
2
= mμ
2
c4
(3)
Solve (1) and (3)
γμ = (m2
π + m2
μ)/2mπ mμ
βμ = (1 − 1/γ 2
μ)1/2
= (mπ
2
− mμ
2
)/(mμ
2
+ mμ
2
)
6.58 EB + EC = mAc2
(energy conservation) (1)
PB = PC (momentum conservation) (2)
Or c2
PB
2
= c2
PC
2
(3)
Using the relativistic equations E2
= c2
p2
+ m2
c4
, (3) becomes
EB
2
− mB
2
c4
= EC
2
− mC
2
c4
(4)
Eliminating EC between (1) and (4), and simplifying
EB = (mA
2
+ mB
2
− mC
2
)c2
/2mA (5)
6.59 K+
→ e+
+ π◦
+ νe
The maximum energy of positron will correspond to a situation in which the
neutrino is at rest. In that case the total energy carried by electron will be
Ee(max) = (m2
K +m2
e −m2
π0 )/2mK = (4942
+0.52
+1352
)/2×494 = 228.5MeV
∴ Te(max) = 228 MeV
6.60 Let the incident particle carry momentum p0. As the scattering is symmetrical,
each particle carries kinetic energy T/2 and momentum P after scattering, and
makes an angle θ/2 with the incident direction.
Momentum conservation along the incident direction gives
p0 = p cos θ/2 + p cos θ/2 = 2p cos θ/2 (1)
Or (T 2
+ 2T mc2
)1/2
= 2(T2
/4 + 2T/2mc2
)1/2
cos θ/2 (2)
Squaring (2), and using the identity, cos2
θ/2 = (1 + cos θ)/2
We get the result cos θ = T/(T + 4mc2
)

6.3 Solutions 347
Fig. 6.7 Symmetrical elastic
collision between identical
particles
6.61 (γ − 1)mc2
= mc2
γ = 2
β = (1 − 1/γ 2
)1/2
=
√
3/2
pe = mγβc
cpe = mc2
γβ = mc2
× 2 ×
√
3/2 = mc2
×
√
3 MeV
pγ = pe =
√
3E0 MeV/c
6.62 For the three particles the energies (total) are equal.
E1 = E2 = E3 (3)
(masses are neglected)
The magnitude of momenta are also equal
p1 = p2 = p3
The momenta represented by the three vectors AC, CB and BA form the closed
Δ ABC.
180◦
− θ = 60◦
Therefore, θ = 120◦
Thus the paths of any two leptons are equally inclined to 120◦
Fig. 6.8 Decay of a muon at
rest into three leptons whose
masses are neglected
6.63 By Problem 6.53, T = 2mc2
β2
cos2
θ/[1 − β2
cos2
θ] (1)
T = 109
eV = 1, 000 MeV
γ = 1 + T/M = 1 + 1, 000/940 = 2.0638
β = [1 − (1/γ 2
)]1/2
= [1 − (1/2.0638)2
]1/2
= 0.875
Using mc2
= 0.511 MeV, β = 0.875 and θ = 3◦
in (1),
We find T = 3.3 MeV

6.64 (a) π+
→ μ+
+ νμ
T = Q(Q + 2mν)/2(mμ + mν + Q) = (mπ − mμ)2
/2mπ
(Q = mπ − mμ − mν = mπ − mμ and mν = 0)
mμ = 206.9me = 206.9 × 0.511 = 105.7 MeV
Therefore,
Tμ = 4.12 = (mπ − 105.7)2
/2mπ
Solving for mπ
mπ = 141.39 MeV/c2
= 141.39/0.511me
= 276.7me
(b) μ+
→ e+
+ νe + νμ
Tmax for electron is obtained when νe and νμ fly together in opposite direc-
tion. Thus, the three-body problem is reduced to a two-body one.
mμc2
= 206.9 × 0.511 = 105.72 MeV
Q = 105.72 − 0.51 = 105.21 MeV
Te(max) = Q2
/2(me + Q) = (105.72)2
/2(0.51 + 105.72) = 52.6 MeV
6.65 Q = 938.2 + 1,875.5 − (939.5 + 135.0) − 2.2 = 1,737 MeV
Tπ = Q(Q + 2mn)/2(Q + mπ + mn)
= 1,737(1,737 + 2 × 939.5)/2(1, 737 + 135 + 939.5) = 1, 117 MeV
Total energy Eπ = 1, 117 + 135 = 1, 252 MeV
6.66 M2
= m1
2
+ m2
2
+ 2(E1 E2 − P1 P2 cos θ) (1)
m1 = 966 me = 966 × 0.511 MeV = 493.6 MeV (2)
m2 = 273 me = 273 × 0.511 MeV = 139.5 MeV (3)
p1 = 394 MeV, p2 = 254 MeV (4)
E1 = (p1
2
+ m1
2
)1/2
= [(394)2
+ (93.6)2
]1/2
= 631.6 (5)
E2 = (p2
2
+ m2
2
)1/2
= [(254)2
+ (139.5)2
]1/2
= 289.5 (6)
cos θ = cos 154◦
= −0.898 (7)
Using (2) to (7) in (1), M = 899.4 MeV
6.67 Using the results of Problem 6.60, the angle between the outgoing particles
after the collision is given by
cos θ = T/(T + 4M)
Here T = 940 MeV = M
Therefore, cos θ = 0.2 → θ = 78.46◦
6.68 Using the invariance, E2
− |

p|2
= E∗2
E∗2
= (E + Ef)2
− (p2
+ p2
f + 2ppf cos θ)
= (E + Ef)2
− (E2
− M2
+ E2
f − M2
+ 2ppf cos θ
= 2M2
+ 2(E Ef − p.pf)
(a) For parallel momenta, θ = 0, p.pf = +ppf
(b) For anti-parallel momenta θ = π, p.pf = −ppf
(c) For orthogonal momenta θ = π/2, p.pf = 0
6.69 By problem 6.13 it is sufficient to show that tan θ tan ϕ = 2/(γ + 1)
p = γβm = (γ 2
− 1)1/2
m

6.3 Solutions 349
Therefore, γ = [1 + (p/m)2
]1/2
= [1 + (5/0.938)2
]1/2
= 5.41
tan θ tan ϕ = tan 82◦
tan 2◦
30′
= 7.115 × 0.04366 = 0.3106
2/(γ + 1) = 2/(5.41 + 1) = 0.3120
Hence the event is consistent with elastic scattering.
6.70 Let p, pe, pN be the momentum of the incident electron, scattered electron
and recoil nucleus, respectively. From the momentum triangle, Fig. 6.9.
pN
2
= pe
2
+ p2
− 2ppe cos θ = EN
2
+ 2EN M (1)
where we have put c = 1
From energy conservation
EN + Ee = E (2)
As
Ee ≈ pe (3)
E ≈ P (4)
(2) Can be written as
EN + pe = E (5)
Combiining (1), (3), (4) and (5), we get
EN = E2
(1 − cos θ)/M[1 + E/M(1 − cos θ)]
Restoring c2
, we get the desired result.
Fig. 6.9 Momentum triangle
6.71 Use the result of Problem 6.53,
T = 2mc2
β2
cos2
ϕ/(1 − β2
cos2
ϕ) (1)
Put c = 1, T = E − m (2)
P = Mβγ = Mβ/(1 − β2
)1/2
whence β2
= P2
/(P2
+ M2
) (3)
Use (2) and (3) in (1) and simplify to get the desired result.
6.72 The formula for the recoil energy of electron in Compton scattering is
T = (E2
/mc2
)(1 − cos θ)/[1 + α(1 − cos θ)]
Here neutrinos are assumed to be massless, so that the same formula which
is based on relativistic kinematics can be used.
The maximum recoil energy will occur when the neutrino is scattered
back, that is θ = 180◦
. Substituting E = 2 GeV for the incident neutrino
energy, mc2
= 0.511 MeV = 0.511 × 10−3
GeV, and α = E/mc2
=
2/0.511×10−3
= 3,914, we find the maximum energy transferred to electron
is 1.9997 GeV. The maximum momentum transfer
pmax = (T2
max + 2 mec2
.Tmax)1/2
= 2.0437 GeV/c

6.73 Expressing θ in terms of θ∗
tan θ = sin θ∗
/γc(cos θ∗
+ βc/β∗
) (1)
Differentiating with respect to θ∗
and setting ∂ tan θ/∂θ∗
= 0, gives
cos θ∗
= −β∗
/βc (2)
And sin θ∗
= (β2
c − β∗2
)1/2
/βc (3)
Using (2) and (3) in (1), and the equations
β∗2
γ ∗2
= γ ∗2
− 1 (4)
βc
2
γc
2
= γc
2
− 1 (5)
as well as
m1β∗
γ ∗
= m2βcγc (momentum conservation in the CMS) (6)
and simplifying, we get
tan θmax = [m2
2
/(m1
2
− m2
2
)]1/2
6.74 Let P1 be the electron four-momentum before the collision and P2 the final
four-momentum. If Q is the four-momrntum transfer then the conservation of
energy and momentum requires that
Q = P2 − P1 (1)
and Q2
= Q.Q = P2
2
+ P1
2
− 2P1P2 (2)
But P = (p, E/c) (3)
P2
c2
= E2
− p.pc2
= me
2
c4
(4)
For stationary electron, the initial four-momrntum is
P1 = (0, mec) (5)
and final four-momentum
P2 = (p2, mec + ν/c) (6)
Therefore
P1.P2 = mec(mec + ν/c) (7)
Substituting (7) in (2) and using (4)
Q2
= 2me
2
c4
− 2mec(mec + ν/c)
= −2 meν
6.75 (a) As the collision is elastic the total initial energy = total final energy.
E + Ee = E′
+ Ee
′
(1)
But p′
= p (2)
∴ E′
= E (3)
It follows that E′
e = Ee (4)
Consequently pe
′
= pe (5)
p − pe = −p′
+ pe
′
(6)

6.3 Solutions 351
Because of (2) and (5)
p′
e = p = E/c
(b) β = v/c = (1 − 1/γ 2
)1/2
= (1 − m2
/Ee
2
)1/2
= (1 − m2
/(pe
2
+ m2
))1/2
= (pe
2
/(pe
2
+ m2
))1/2
= (1 + (mc
2
/E)2
)−1/2
where Pe = E and c = 1.
6.76 Referring to Fig. 7.8, let Pγ and Pγ
′
be the four-momentum vectors of photon
before and after the scattering, respectively, Pe and Pe
′
the four vectors of
electron before and after scattering respectively. Form the scalar product of
the four-vector of the photon and the four-vector of photon + electron. Since
this scalar product is invariant.
Pγ .(Pγ + Pe) = Pγ
′
.(Pγ
′
+ Pe
′
) (1)
Further, total four-momentum is conserved.
Pγ + Pe = Pγ
′
+ Pe
′
(2)
Now, Pγ = [hν, 0, 0, ihν] (3)
Pe = [0, 0, 0, imc] (4)
Pγ
′
= [h ν′
cos θ, h ν′
sin θ, 0, ihν]
where m is the rest mass of the electron.
Using (1) to (4)
[hν, 0, 0, ihν].[hν, 0, 0, i(hν + mc2
)]
= [hν′
cos θ, hν′
sin θ, 0, ihν].[hν, 0, 0, i(hν + mc2
)]
Therefore, h2
ν2
− hν(hν + mc2
) = h2
νν′
cos θ − hν′
(hν + mc2
)
Simplifying
hν′
ν(1 − cos θ) = mc2
(ν − ν′
)
Or h/mc(1 − cos θ) = c(1/ν′
− 1/ν) = λ′
− λ
Or Δλ = λ′
− λ = (h/mc)(1 − cos θ)
This is the well known formula for Compton shift in wavelength (formula
7.37).
6.77 Let the initial and final four momenta of the electron be Pi = (Ei/c, pi) and
Pf = (Ef, pf), respectively. The squared four-momentum transfer is defined by
Q2
= (Pi − Pf)2
= −2m2
c2
+
2Ei Ef
c2
− 2Pi · Pf
However, Ei = Ef = E and |pi| = |pf| = E/c; so neglecting the electron
mass
Q2
= 2E2
(1 − cos θ)/c2

6.78
1
2
(M2
− m1
2
− m2
2
) = E1 E2 − p1 p2 cos θ
θ = 900
P± = 530 MeV/c
E± = (P2
± + mπ
2
)1/2
= 548
2 × 5482
+ 2 × 1402
= M2
M = 800 MeV/c
It is a ρ meson.
Fig. 6.10
6.79 Using the invariance of squared four-momentum before and after the decay
Ei
2
− Pi
2
= Ef
2
− |pf|2
M2
= (E1 + E2)2
− (p1
2
+ p2
2
+ 2p1 p2 cos θ)
= (E1
2
− p1
2
) + (E2
2
− p2
2
) + 2(E1 E2 − p1 p2 cos θ)
= m1
2
+ m2
2
+ 2(E1 E2 − p1 p2 cos θ)
Or E1 E2 − p1 p2 cos θ =
1
2
(M2
− m1
2
− m2
2
) = Invariant
6.80 s + t + u = (1/c2
)[(PA + PB)2
+ (PA − PC )2
+ (PA − PD)2
]
= (1/c2
)[3PA
2
+ PB
2
+ PC
2
+ PD
2
+ 2PA(PB − PC − PD)] (1)
From four-momentum conservation, PA + PB = PC + PD (2)
(1) becomes (s + t + u)c2
= (3mA
2
+ mB
2
+ mC
2
+ mD
2
) − 2PA
2
Using PA = mAc, PB = mBc, PC = mC c and PD = mDc
(s + t + u)c2
= (mA
2
+ mB
2
+ mC
2
+ mD
2
)c2
Or s + t + u =

i=A,B,C,D
m2
i
6.81 t = (PA − PC )2
/c2
= (1/c2
)(PA
2
+ PC
2
− 2PA PC )
= (1/c2
)[mA
2
c2
+ mC
2
c2
− 2(EA EC /c2
− PA.PC )]
For elastic scattering A ≡ C. Thus EA = EC and |PA| = |PC | = p
So that PA.PC = p2
cos θ.
c2
t = 2mA
2
c2
− 2(EA
2
/c2
− p2
cos θ)
But EA
2
= c2
p2
+ mA
2
c4
Therefore, t = −2p2
(1 − cos θ)/c2
6.82 sin ϕ/2 = mπ c2
/2(E1 E2)1/2
(see Prob. 6.103 and 6.104)
Minimum angle is ϕmin = 2/γ

6.3 Solutions 353
But γ = 10/0.14
Therefore, ϕmin = 0.028 rad or 1.6◦
6.83 Rest mass energy of ω0
= Total available energy – (total kinetic energy +
mass energy of π+
and π−
)
mω c2
= 2.29 − (1.22 + 0.14 + 0.14) = 0.79 GeV
m1 γ1 + m2 = M γ (1)
m1 γ1 β1 = M γ β (2)
Squaring (1)
m1
2
γ1
2
+ m2
2
+ 2γ1 m1 m2 = M2
γ 2
(3)
Squaring (2)
m1
2
γ1
2
β1
2
= M2
γ 2
β2
(4)
Using β1 = (1 − 1/γ1
2
)1/2
and β = (1 − 1/γ 2
)1/2
(4) becomes
m1
2
(γ1
2
− 1) = M2
(γ 2
− 1) (5)
Subtracting (5) from (3)
m1
2
+ m2
2
+ 2m1m2/(1 − v2
/c2
)1/2
= M2
6.85 E0 = Ep + Eπ (energy conservation) (1)
Q = m0 − (mp + mπ ) (2)
P0
2
= Pp
2
+ Pπ
2
+ 2Pp Pπ cos θ (3)
Or E0
2
− m0
2
= Ep
2
− mp
2
+ Eπ
2
− mπ
2
+ 2Pp Pπ cos θ (4)
Using (1) in (4) and simplifying
2Ep Eπ − 2Pp Pπ cos θ + mp
2
+ mπ
2
= m0
2
= (Q + mp + mπ )2
(5)
Or Q = (mp
2
+ mπ
2
+ 2Ep Eπ − 2Pp Pπ cos θ)1/2
− (mp + mπ ) (6)

Fig. 6.11 (a) Decay
∧ → P + π−
in flight. (b)
Momentum triangle
(a)
(b)
6.86 M0βγ = p1 sin θ + p2 cos θ (momentum conservation along x-axis) (1)
p1 cos θ = p2 sin θ (momentum conservation along y-axis) (2)
M0γ = E1 + E2 (Energy conservation) (3)
Solving (1), (2) and (3)
M0 = (m1
2
+ m2
2 + 2E1 E2)1/2
β = (p1
2
+ p2
2
)1/2
/(E1 + E2)
θ = tan−1
(p1/p2)
Fig. 6.12 Decay
M0 → m1 + m2
6.87 Under the assumption (a)
M2
= 2(Eπ+ + Eπ− − Pπ+ Pπ− cos θ) + mπ+
2
+ mπ−
2
(1)
Eπ+ = (pπ
2
+ mπ+
2
)1/2
(2)
Eπ− = (Pπ−
2
+ mπ−
2
)1/2
(3)
mπ+ = mπ− = 0.14 GeV/c2
, θ = 15◦
(4)
p+ = 1.67, p− = 0.408 GeV/c (5)
Using (2), (3), and (4) in (1) and solving for M, we find M = 0.239 GeV/c2
,
a value quite different from the standard value, mk0 = 0.498 GeV/c2
Under the assumption (b)

6.3 Solutions 355
M2
= 2(Ep Eπ− − Pp Pπ− cos θ) + mπ+
2
+ mπ−
2
(6)
Ep = (P+
2
+ mp
2
)1/2
(7)
mp = 0.938 GeV/c2
(8)
Using (3), (4), (5), (7) and (8) in (6) and solving for M, we find M =
1.109 GeV/c2
which is in good agreement with the mass mΛ = 1.115 GeV/c
Thus, the neutral particle is Λ.
6.88 Let the momenta of photons in the LS be p1 ad p2 ad energies E1 and E2. The
invariant mass W of the initial state is given by
W2
= E2
− p2
= m2
In the final state
E2
− p2
= (E1 + E2)2
− |(p1 + p2)|2
= 2E1 E2(1 − cos ϕ) = 4E1 E2 sin2
(ϕ/2) (because E1 = p1 and
E2 = p2 and p1.p2 = E1 E2 cos ϕ)
Invariance of E2
− p2
gives
Sin (ϕ/2) = mc2
/2(E1 E2)1/2
6.89 n → p + e−
+ ν
The proton will carry maximum energy when the neutrino with negligible
mass is at rest.
(qn − qp)2
= (En − Ep) − Pp
2
= (mn − Ep)2
− (Ep
2
− mp
2
); (because neutron is at rest)
= mn
2
+ mp
2
− 2mn Ep
But Pp = Pe → Pp
2
= Pe
2
Or Ep
2
− mp
2
= Ee
2
− me
2
= (mn − Ep)2
− me
2
= mn
2
− 2mn Ep + Ep
2
− me
2
∴ mn
2
+ mp
2
− 2mn Ep = me
2
Thus (qn − qp)2
= m2
e
Or qn − qp = mec2
= 0.511 MeV/c
6.90 λ = λ′√
(1 + β)/(1 − β)
β = v/c = 3 × 106
/3 × 108
= 0.01
λ′
= 6,563 Å
λ = 6,629 Å
Δλ = λ − λ′
= 6,629 − 6,563 = 66 Å

6.91 (a) λ/λ′
= [(1 + β)/(1 − β)]1/2
= 656/434 = 1.5115
v = βc = 1.17 × 108
ms−1
(b) the nebula is receding
6.92 λ/λ′
= [(1 + β)/(1 − β)]1/2
= (1 + β)1/2
(1 − β)−1/2
≈ (1 + β/2 + . . .)(1 + β/2 − . . .)
= 1 + β + . . . (neglecting higher order terms)
Δλ/λ′
= (λ/λ′
) − 1 = β = v/c
6.93 λ/λ′
= [(1 + β)/(1 − β)]1/2
= 670/525
β = 0.239
v = βc = 0.239 × 3 × 108
= 7.17 × 107
ms−1
= 7.17 × 104
kms−1
This speed exceeds the escape velocity. Hence the explanation is not valid.
6.94 λ′
= λ[(1 − β)/(1 + β)]1/2
= 589.3[(1 − 0.21)/(1 + 0.21)]1/2
= 476.2 nm
The color is blue
6.95 The source velocity is perpendicular to the line of sight. θ = 90◦
,
ν′
= νγ
λ = γ λ′
γ = 1/(1 − β2
)1/2
= 1/(1 − 0.052
)1/2
= 1.00125
Δλ = λ − λ′
= λ′
(γ − 1) = 589(1.00125 − 1)
= 0.736 nm = 7.36 Å
6.96 Let the electron recoil at angle ϕ with momentum p, and neutrino get scat-
tered with energy E′
and momentum p′
.
E0 + m = E + E′
(1)
From the momentum triangle
p′2
= p0
2
+ p2
− 2p0 p cos ϕ (2)
We can write p′
= E′
, p = E, p0 = E0, so that (2) becomes
E′2
= E0
2
+ E2
− 2E0 E cos ϕ (3)
Fig. 6.13 Collision of an energetic neutrino with a stationary electron

6.3 Solutions 357
From (1) we have
E′2
= (E0 − E + m)2
(4)
Comparing (3) and (4) and simplifying
cos ϕ = 1 − m(E0 − E − m)/E0 E ≈ 1 − m(E0 − E)/E0 E (5)
where we have neglected m in comparison with E0 − E.
For small angle, (5) becomes ϕ = [2m(E0 − E)/E0 E]1/2
(6)
For E0 = 2 GeV, E = 0.5 GeV, m = 0.51 × 10−3
GeV
ϕ = 0.039 radians = 2.24◦
6.97 Let the mass of the primary particle be M, and that of secondary particles m1
and m2. Let the total energy of the secondary particles in the LS be E1 and
E2, and momenta p1 and p2. Using the invariance of (total energy)2
− (total
momentum)2
M2
= (E1 + E2)2
− (p1
2
+ p2
2
+ 2p1 p2 cos θ)
= E1
2
− p1
2
+ E2
2
− p2
2
+ 2(E2
1 − p1
2
cos θ)
= m1
2
+ m2
2
+ 2(E1
2
− p1
2
cos θ) (Since E1 = E2, p1 = p2)
= 2m1
2
+ 2(m1
2
+ p1
2
− p1
2
cos θ)
= 4m1
2
+ 4p1
2
sin2
θ/2
= 4(140)2
+ 4(300)2
sin2
35◦
= 196,836
M = 444 MeV/c2
6.98 Consider one of the two γ -rays. From Lorentz transformation
cpx = γc(cpx
∗
+ βc E∗
)
where the energy and momentum refer to one of the two γ -rays and the
subscript C refers to π0
. Starred quantities refer to the rest system of π0
.
cp cos θ = γ (cp∗
cos θ∗
+ βE∗
)
where we have dropped off the subscript C. But for γ -rays cp∗
= E∗
and
cp = E, and because the two γ -rays share equal energy in the CMS, E∗
=
mc2
/2, where m is the rest mass of π◦
.
Therefore cp cos θ = (γ mc2
/2)(β + cos θ∗
) (1)
Also, E = γ (E∗
+ βcp∗
x ) = γ (E∗
+ βcp∗
cos θ∗
)
or
cp = E = (γ mc2
/2)(1 + β cos θ∗
) (2)
When θ∗
= 0
Emax =
1
2
Eπ◦ (1 + β) (3)
When θ∗
= π
Emin =
1
2
Eπ◦ (1 − β) (4)

From (3) and (4),(Emax − Emin)/(Emax + Emin) = β (5a)
From the measurement of Emax and Emin, the velocity of π0
can be deter-
mined.
6.99 In the solution of Problem 6.98 multiply (3) and (4) and write E = γ mπ c2
mπ c2
= 2(Emax Emin)1/2
(5b)
From the measurement of Emax and Emin, mass of π0
can be determined.
It Emax = 75 MeV and Emin = 60 MeV, then mc2
= 2 × (75 × 60)1/2
=
134.16 MeV
Hence the mass of π0
is 134.16/0.51 = 262.5me
6.100 dN/dE = (dN/dΩ∗
).dΩ∗
/dE = (1/4π).2 sin θ∗
dθ∗
/dE
= (1/2)d cos θ∗
/dE (6)
where we have put dΩ∗
= 2π sin θ∗
d θ∗
for the element of solid angle and
dN/dΩ∗
= 1/4π under the assumption of isotropy.
Differentiating (2) with respect to cos θ∗
dE/d cos θ∗
= γβmc2
/2
or d cos θ∗
/2dE = 1/γβmc2
(7)
Combining (6) and (7), the normalized distribution is
dN/dE = 1/γβmc2
= constant (8)
This implies that the energy spectrum is rectangular or uniform. It extends
from a minimum to maximum, Fig. 6.14.
From (3) and (4),
Emax − Emin = βEπ = γβmc2
(9)
Note that the area of the rectangle is height × length
(dN/dE) × (Emax − Emin) = (1/γβmc2
) × γβmc2
= 1
That is, the distribution is normalized as it should.
The higher the π0
energy the larger is the spread in the γ -ray energy spec-
trum. For mono-energetic source of π0
s, we will have a rectangular distribu-
tion of γ -ray energy as in Fig. 6.14. But if the γ -rays are observed from π0
s,
of varying energy, as in cosmic ray events the rectangular distributions may
be superimposed so that the resultant distributions may look like the solid
curve, shown in Fig. 6.15.
Fig. 6.14 γ -ray energy
spectrum from π0
decay at
fixed energy

6.3 Solutions 359
Note that if the π0
s were to decay at rest (γ = 1) then the rectangle would
have reduced to a spike at E = 67.5 MeV, half of rest energy of π0
.
6.101 The γ -rays of intensity I(θ∗
) which are emitted in the solid angle d Ω∗
in
the CMS will appear in the solid angle dΩ∗
in the LS with intensity I(θ).
Therefore
I(θ)d Ω = I(θ∗
)d Ω∗
or
I(θ) = I(θ∗
) sin θ∗
dθ∗
/ sin θdθ (10)
Fig. 6.15 γ -ray energy
spectrum from π0
decay in
cosmic ray events
From the Lorentz transformation
E∗
= γ E(1 − β cos θ)
= γ E∗
γ (1 + cos θ∗
)(1 − cos θ)
Where we have used (2)
1/γ 2
(1 − β cos θ) = 1 + β cos θ∗
Differentiating
− β sin θdθ/γ 2
(1 − β cos θ)2
= −β sin θ∗
dθ∗
Therefore sin θ∗
dθ∗
/ sin θdθ = 1/γ 2
(1 − β cos θ)2
(11)
Also I(θ∗
) = 1/4π (12)
because of assumption of isotropy of photons in the rest frame of π0
I(θ) = 1/4πγ 2
(1 − β cos θ)2
(13)

This shows that small emission angles of photons in the lab system are
favored.
6.102 In Fig. 6.16, AB and AD represent the momentum vectors of the two photons
in the LS. BC is drawn parallel to AD so that ABC forms the momentum
triangle, that is
AB + BC = AC
Fig. 6.16 Locus of the tip of
the momentum vector of
γ -rays from π0
decay is an
ellipse
From energy conservation
h ν1 + h ν2 = γ mc2
/2(1 + β cos θ∗
) + (γ mc2
/2)(1 + β cos(π − θ∗
))
= γ mc2
= const
where we have used the fact that the angles of emission of the two photons
in the rest frame of π0
, are supplementary.
Since momentum is given by p = h/c, it follows that AB + BC = constant,
which means that the locus of the tip of the momentum vector is an ellipse.
E = mc2
/2γ (1 − β cos θ) (14)
Compare this with the standard equation for ellipse
r = a(1 − ε2
)/(1 − ε cos θ)
We find ε = β
a(1 − β2
) = mc2
/2 γ
Or a = γ mc2
/2
The larger the velocity of π0
, the greater will be the eccentricity, ε.

6.3 Solutions 361
6.103 The angle between the two γ -rays in the LS can be found fom the formula
m2
c4
= m1
2
c4
+ m2
2
c4
+ 2(E1 E2 − c2
p1 p2 cos ϕ)
Putting m1 = m2 = 0, cp1 = E1, cp2 = E2
m2
c4
= 2E1 E2(1 − cos ϕ) = 4E1 E2 sin2
(ϕ/2)
sin (ϕ/2) = mc2
/2(E1 E2)1/2
(15)
6.104 For small angle ϕ,
ϕ = mc2
/(E1 E2)1/2
(16)
Set E2 = γ mc2
− E1
ϕ = mc2
/[E1(γ mc2
− E1)]1/2
(17)
For minimum angle d ϕ/dE1 = 0. This gives E1 = γ mc2
/2.
Using this value of E1 in (17), we obtain
ϕmin = 2/γ (18)
Measurement of ϕmin affords the determination of Eπ via γ .
ϕmin = 2mc2
/Eπ
6.105 E1 + E2 = E (energy conservation) (1)
p1 + p2 = p (momentum conservation) (2)
Taking the scalar product
(p1 + p2).(p1 + p2) = p.p
or
p1
2
+ p2
2
+ 2p1 p2 cos θ = p2
(3)
Using c = 1, Eq. (3) becomes
E1
2
+ E2
2
+ 2E1 E2 cos θ = E2
− m2
(4)
Let E2/E1 = D, (5)
the disparity factor. Then (1) becomes
E1(D + 1) = E (6)
2DE2
(1 − cos θ) = m2
Or sin θ/2 = [m/2E][
√
D + 1/
√
D]
For small θ,
θ = [m/E][
√
D + 1
√
D]

The minimum angle is found by setting dθ/dD = 0
This gives us D = 1, that is E1 = E2 = E/2.
θmin = 2mc2
/E
The Lorentz transformation of angles gives us the relation
E = [m γ/2][1 + β cos θ∗
)
We need to consider one of the photons in the forward hemisphere. The
fraction of photons emitted in the CMS (rest frame of π0
) within the angle θ∗
is (1 − cos θ∗
). This fraction is 1/2 for θ∗
= 600
, that is cos θ∗
= 1/2. When
one photon goes at θ∗
= 600
, the other photon will go at θ∗
= 1200
with the
direction of flight of π0
. Hence cos θ∗
= −1/2 for the second photon. The
disparity factor
D = E2/E1 = (1 + β/2)/(1 − β/2) = (2 + β)/(2 − β).
For relativistic pions β = 1 Hence D 1.
A quarter of pions will be emitted within an angle θ∗
= 41.40
, that is
cos θ∗
= 0.75. In this case
D = (1 + 3β/4)/(1 − 3β/4)
And the previous argument gives us D 7
6.106 First find E∗
the total energy available in the CMS
E∗2
= (Eπ + mp)2
− P2
π ≈ (Pπ + mp)2
− Pπ
2
(Because Eπ ≫ mπ )
E∗
= 4.436 GeV
Total energy carried by K∗
in the CMS
Ek
∗
= (E∗2
+ mk
∗2
− mγ 0 )/2E∗
= 1.942 GeV
γK
∗
= EK
∗
/mK
∗
= 1.942/0.89 = 2.18
βK
∗
= 0.8888
γc = (γ + ν)/(1 + 2γ ν + ν2
)1/2
γ = 10/0.14 = 71.4
ν = m2/m1 = 0.940/0.140 = 6.71
γc = 2.466, βc = 0.9141
tan θ = sin θ∗
/(cos θ∗
+ βc/β∗
) (1)
Differentiate with respect to θ∗
and set
∂ tan θ/∂θ∗
= 0.This gives cos θ∗
= −β∗
/βc
cos θ∗
= −0.8888/0.9141 = −0.9723
θ∗
= 166.50
Using the values of θ∗
, γc, and the ratio βc/β∗
in (1) we find θm = 59.3◦

6.3 Solutions 363
6.107 (a), (b) In the CMS, m2 will move with the velocity βc in a direction opposite
to that of m1. By definition, the total momentum in the CMS before and after
the collision is zero. In natural units c = 1.
m1 γ ∗
β∗
= m2 γc βc (1)
Squaring (1) and expressing the velocities in terms of Lorentz factors
m1
2
(γ ∗2
− 1) = m2
2
(γc
2
− 1) (2)
Using the invariance
(ΣE)2
− |ΣP|2
= (ΣE∗
)2
− |ΣP∗
|2
= (ΣE∗
)2
(3)
(because

P∗
= 0, in the CMS)
(m1 γ + m2)2
− m1
2
(γ 2
− 1) = (m1 γ ∗
+ m2 γc)2
(4)
Combining (2) and (4) and calling v = m2/m1
γc = (γ + ν)/(1 + 2γ ν + ν2
)1/2
(5)
γ ∗
= (γ + 1/ν)/(1 + 2γ/ν + 1/ν2
)1/2
(6)
For the special case, m1 = m2, as in the P–P collision
γc = γ ∗
= [(γ + 1)/2]1/2
(7)
In addition if γ ≫ 1
γc ≈ (γ/2)1/2
(8)
(c), (d)
The Lorentz transformations are
P cos θ = γc(p∗
cos θ∗
+ E∗
) (9)
P sin θ = p∗
sin θ∗
(10)
tan θ = p∗
sin θ∗
/γc(p∗
cos θ∗
+ βc E∗
) = sin θ∗
/γc(cos θ∗
+ βc/β∗
) (11)
(because p∗
/E∗
= β∗
)
From the inverse transformation
P∗
cos θ∗
= γc(P cos θ − βc E) (12)
and (10) we get
tan θ∗
= sin θ/γc(cos θ − βc/β) (13)
6.108 At the right angle to the direction of source velocity the Doppler shift in
wavelength is calculated from
ν′
= γ ν or λ′
= λ/γ
where γ is the Lorentz factor of the carbon atoms and T is the kinetic energy
of carbon and Mc2
is the approximate rest mass energy, the quantity

(λ′
− λ)/λ becomes (1 − γ )/γ
But γ = 1 + T/Mc2
= 1 + 120/12 × 10 = 1.01
Hence (λ′
− λ)/λ = −0.01/1.01 = −0.0099
6.109 The observed frequency v due to Doppler effect is given by
ν = γ ν′
(1 + β cos θ′
)
Where ν′
is the natural frequency
When the star is moving toward the observer θ′
= 0
β = v/c = (300 km/s)/3 × 105
km/s = 10−3
γ = 1/(1 − β2
)1/2
≈ 1 + (1/2)β2
= 1 + 5 × 10−7
Neglecting small terms, v = (1 + 10−3
)ν′
Fractional change in frequency
(ν − ν′
)/ν′
= 10−3
6.110 Use the formula for Lorentz transformation of angles from CMS to LS
tan θ = sin θ∗
/γc(cos θ∗
+ βc/β∗
). (1)
For one of the photons, in the rest system of π◦
.θ∗
= 90◦
.β∗
= 1. From
the given value βc = 0.8 we find γc = 1.6666. Inserting these values in (1)
tan θ = 0.75 or θ = 36.87◦
in the LS. From symmetry the second photon
will also be emitted at the same angle on the other side of the line of flight and
be coplanar. Hence the angle between the two photons will be 2θ = 73.75◦
6.111 Use the formula for the transformation of angles.
tan θ = sin θ∗
/γc(cos θ∗
+ βc/β∗
)
Use θ = 45◦
, θ∗
= 60◦
, β∗
= 1, γc = 1/(1 − βc
2
)1/2
in the above formula,
and simplify to obtain a quadratic equation in βc. On solving this equation
we find the velocity of the object
v = βcc = 0.535 c
6.112 (a) The y-component of the rod is unchanged that is Ly = Ly
′
or L sin θ = L′
sin θ′
(1)
Also Lx = L cos θ (2)
L′
x = L′
cos θ′
(3)
Eliminating L and L′
, (4)
Lx . tan θ = Lx
′
. tan θ′
(5)
(b) Lx
′
= Lx /γ
where γ is the Lorentz factor. Using (4) in (5)

6.3 Solutions 365
γ = Lx /Lx
′
= tan θ/ tan θ = tan 45◦
/ tan 30◦
=
√
3
β = (γ 2
− 1)1/2
/γ =
2
3
= 0.816
The speed at which the rod is moving is v = βc = 0.816 c
6.113 For the production reaction
m1 + m2 → m3 + m4
The threshold energy for m1 when m2 is at rest is
T1 = [(m3 + m4)2
− (m1 + m2)2
]/2m2
In the given reaction we can put
m3 + m4 = 4M, m1 = m2 = M
T1 = 6M or T1 = 6Mc2
6.114 Here m1 = 0, m2 = m, (m3 + m4) = 3m
T1 = 4mc2
6.115 If m1 is the projectile mass, m2 target mass, and m3 + m4 + m5, the mass of
product particles. The threshold is given by formula
T1 = [(m3 + m4 + m5)2
− (m1 + m2)2
]/2m2
= [(940 + 940 + 140)2
− (940 + 940)2
]/2 × 940
= 290.4 MeV
The threshold energy is thus slightly greater than twice the rest-mass
energy of pion (140 MeV). Non-relativistically, the result would be 280 MeV,
that is double the rest mass energy of Pion. The extra energy of 10 MeV is to
be regarded as relativistic correction
6.116 Use the invariance of E2
− P2
= E∗2
− P∗2
= E∗2
− 0 = E∗2
E∗2
= (4m)2
= (T + m + m + 0.025)2
− (P1 − 0.218)2
Putting m = 0.938, P1 = (T 2
+ 2T m)1/2
and
solving for T , we find that T (threshold) = 4.3 GeV
6.117 Tthreshold = [(m3 + m4)2
− (m1 + m2)2
]/2m2
= [(0.89 + 1.11)2
− (0 + 0.94)2
]/2 × 0.94
= 3.12 GeV

6.118 Tthreshold = [(mp + mp + mp)2
− (mp + mp)2
]/2mp by Eq. (6.53)
mp = 1,837 me = 1,837 × 0.00.00051 GeV = 0.937 GeV
M = 273 × 0.00051 GeV = 0.137 GeV
Using the above values we find Tthreshold = 0.167 GeV
6.119 Tthreshold = (mk + mΛ)2
− (mp + mp)2
/2mp
mk = 0.498 GeV, mΛ = 1.115 GeV, mπ = 0.140 GeV, mp = 0.938 GeV
Using these values, we find Tthreshold = 0.767 GeV = 767 MeV
Note that when pions are used as bombarding particles the threshold for
strange particle production is lowered then in N–N collisions. However, first
a beam of pions must be produced in N–N collisions.
6.120 Consider the reaction P + P → P + P + nπ
Tthreshold = [(mp + mp + nmπ )2
− (mp + mp)2
]/2mp
Simplifying we get the desired result
6.121 Tthreshold = [(mp + mπ0)2
− (mp + 0)2
]/2mp
Using mp = 940 MeV and mπ0 = 135 MeV, we
find Tthreshold = 145 MeV
Note that the threshold energy for pion production in collision with gamma
rays is only half of that for N–N collisions. But the cross-section is down by
two orders of magnitude as the interaction is electromagnetic.
6.122 Tthreshold = [(mΞ− + mk + mk0)2
− (mπ− + mp)2
]/2mp
= [(1,321 + 494 + 498)2
− (140 + 938)2
]/2 × 938
= 2,233 MeV
Note that for Ξ production, the threshold is much higher than that for
−
production as it has to be produced in association with two other strange
particles (see Chaps.9 and 10).
6.123 Using the invariance, E2
− |

p|2
= E∗2
− |

p∗
|2
At threshold: (E + Mp)2
− Eν
2
= (Mp + Mμ + Mw)2
− 0
(5 + 0.938)2
− 52
= (0.938 + 0.106 + Mw)2
Mw = 2.16 GeV
Since the reaction does not proceed, Mw 2.16 GeV
6.124 Tthr = [(mp + mΛ + mk)2
− (2mp)2
]/2mp
= [(0.938 + 1.115 + 0.494)2
− (2 × 0.938)2
]/2 × 0.938
= 1.58 GeV

6.3 Solutions 367
6.125 Tthr = [(mk0 + mk + mΩ−)2
− (mk− + mp)2
]/2mk−
= [(0.498 + 0.494 + 1.675)2
− (0.494 + 0.938)2
]/2 × 0.938 = 2.7 GeV
Minimum momentum pK− = (T 2
+ 2T m)1/2
= (2.72
+ 2 × 2.7 ×
0.494)1/2
= 3.15 GeV/c
Pthr = 3.15 GeV/c
EK = 2.7 + 0.494 = 3.194 GeV
γK = Ek/mk = 3.194/0.494 = 6.46
γc = (γ + m2/m1)/[1 + 2γ m2/m1 + (m2/m1)2
]1/2
m2/m1 = 938/494 = 1.9
γc = (6.46 + 1.9)/(1 + 2 × 6.46 × 1.9 + 1.92
)1/2
= 1.61
γΩ = γc = 1.61; βΩ = (γ 2
Ω − 1)1/2
/γΩ = 0.79
Proper time t0 = d/v = d/βc
Observed time t = γ t0 = γ d/βc = 1.61 × 0.03/0.79 × 3 × 108
=
2 × 10−10
s
Probability that Ω−
will travel 3 cm before decay.
= exp(−t/τ)
= exp(−2 × 10−10
/1.3 × 10−10
)
= 0.21
6.126 TF(max)p = (9/32π2
)2/3
4π2
(2
c2
/2mpc2
r2
0 )(Z/A)2/3
R = r0 A1/3
r0 = R/A1/3
= 5.17/(63)1/3
= 1.3 fm
TF(max)p = (9/32π2
)2/3
× 4π2
(197 MeV − fm)2
(29/63)2/3
/2
× 938 × (1.3)2
= 26.886 MeV
PF(max) = (T2
+ 2T m)1/2
= [(26.886)2
+ 2 × 26.886 × 938]1/2
= 226.2 MeV/c
E2
− (p1 + p2)2
= E∗2
(maximum energy will be available when p1 and
p2 are antiparallel)
E∗
= [(938 + 160 + 938 + 27)2
− (570.4 − 226.2)2
]1/2
= 2034 MeV
md + mπ = 938 + 939 − 2.2 + 139.5 = 2, 014 MeV
As the energy available in the CMS is in excess of the required energy, we
do expect the pions to be produced.

Chapter 7
Nuclear Physics – I
Solid angle
In two dimensions the angle in radians is defined as the ratio of the arc of the circle
and the radius, that is θ = s/r. In three dimensions, the element of solid angle dΩ
is defined as the elementary area A at a distance d from a point, perpendicular to
the line joining the point and the area, divided by the square of the distance, that is
d Ω = ΔA/d2
. For a ring of radii r and r + dr, located on the surface of a sphere of
radius R, the element of solid angle in polar coordinates is subtended at the centre
O is given by (Fig. 7.1)
d Ω = 2 π sin θ dθ (7.1)
We assume an azimuthal symmetry, that is scattering is independent of the
azimuthal angle β.
Fig. 7.1 Concept of solid
angle
369

370 7 Nuclear Physics – I
The solid angle for θ1 = 0 and θ2 = θ is
Ω =

d Ω =
θ

0
2 π sin θ d θ = 2 π(1 − cos θ)
The maximum solid angle Ω = 4π (for θ = π)
Kinematics of scattering
Relations between velocities, angles etc. in the Lab system (LS) and centre
of mass system (CMS)
In the centre of mass system the total momentum of particles is zero. Let a particle of
mass m1 moving with velocity u1 in the LS be scattered by the target particle of mass
initially at rest and be scattered at angle θ with velocity v1. The target particle recoils
with velocity v2 at angle ϕ the angles being measured with the incident direction
(Fig. 7.2). The corresponding angles in the CMS will be denoted by θ∗
and ϕ∗
.
Fig. 7.2 Scattering angle and recoil angle in the LS and CMS

tan θ = sin θ∗
/(cos θ∗
+ m1/m2) (7.2)
If m2 m1; 0 θ π
m2 = m1; 0 θ π/2
m2 m1; 0 θ θmax
where θmax = sin−1
(m2/m1) (7.3)
CM velocity combined with v1
∗
or v2
∗
gives v1 or v2, respectively
ϕ = ϕ∗
/2 (regardless of the ratio m1/m2) (7.4)
ϕmax = π/2 (7.5)
Total kinetic energy available in the CMS
T∗
= (1/2) μ v1
2
(7.6)
where μ is the reduced mass given by
μ = m1m2/(m1 + m2) (7.7)
Energy associated with the CMS is
(1/2)(m1 + m2)v2
c (7.8)
Scattering cross-section
Let I0 be the beam intensity of the projectiles, that is the number of incident particles
crossing unit area per second, and I be the intensity of the scattered particles going
into a solid angle dΩ per second, and n the number of target particles intercepting
the beam, then
I = I0 n σ(θ, ϕ) d Ω (7.9)
If we assume here an azimuthal symmetry, then we can omit the azimuth angle
and simply write σ(θ).
The constant of proportionality σ(θ), also written as dσ(θ)/dΩ, is known as the
differential cross-section. It is a measure of the probability of scattering in a given
direction (θ, β) per unit solid angle from the given target nucleus. The integral over
the solid angle is known as total scattering cross-section (Fig. 7.3).
σ =

σ (θ, ϕ) d Ω (7.10)

Fig. 7.3 Concept of
differential cross-section
The unit of σ is a Barn (B). 1 Barn = 10−24
cm2
, 1mB = 10−27
cm2
and 1 µB =
10−30
cm2
. The unit of σ(θ, ϕ) is Barn/Steradian, where steradian (sr) is the unit of
solid angle.
Relation between the differential cross-sections in the LS and CMS
σ(θ) =
(1 + γ 2
+ 2γ cos θ∗
)
3
2
|1 + γ cos θ∗|
σ(θ∗
) (7.11)
where γ = m1/m2
Note that the total cross-section is the same for both LS and CMS because the
occurrence of the total number of collisions is independent of the description of the
process.
Geometric cross-section
σg = π R2
(7.12)
This is the projected area of a sphere of radius R.
Rutherford Scattering
σ (θ) = [1.295(zZ/T )2
/ sin4
(θ/2)] mb/sr (7.13)
σ(θ′
, θ′′
= (π/4)R0
2
[cot2
(θ′
/2) − cot2
(θ′′
/2)] (7.14)
represents the cross-section for particles to be scattered between angles θ′
and θ′′
R0 (fm) = 1.44zZ/T 0(MeV) (7.15)

Fig. 7.4 Rutherford
scattering
R0 is also the minimum distance of approach in head-on collision for positively
charged particles of energy below Coulomb barrier (Fig. 7.4).
Impact parameter (b) and scattering angle (θ)
tan (θ/2) = R0/2b (7.16)
Minimum distance of approach
R0
2

1 +

1 +
4b2
R2
0
1/2
'
= (R0/2)[1 + cosec (θ/2)] (7.17)
Multiple scattering angle
The root mean square angle of multiple scattering
√
Θ2
= k
√
t ze/pv (7.18)
where k is the scattering constant
k = [8π N Z2
e2
ln(bmax/bmin)]1/2
t=thickness, N=number of atoms/cm3
, Ze and ze are the charge of nuclei of medium
and projectile, pv is momentum times the velocity of the incident particle. For pho-
tographic emulsions, ln(bmax/bmin) is of the order of 10.
Cross-section and mean free path
If n is the number of atoms/cm3
, then the macroscopic cross-section
Σ = n σ (cm−1
) (7.19)
And mean free path
λ = 1/Σ (7.20)
Also, n = N0 ρ/A (7.21)
where N0 =Avagadro’s number, ρ =density, A the atomic weight.

Ionization
Bethe’s quantum mechanical formula
− dE/dx = (4π z2
e4
n/mv2
)[ln (2mv2
/I) − ln(1 − β2
) − β2
] (7.22)
where n =number of electrons/cm3
, I =ionization potential, v = β c is the particle
velocity and ze is its charge, m is the mass of electron
Note that −dE/dx is independent of the mass of the incident particle (Fig. 7.5).
Fig. 7.5 Ionization
(−dE/dx) versus particle
energy
Range–Energy-relation
E = kz2n
M1−n
Rn
(7.23)
where k and n are empirical constants which depend on the nature of the absorber,
M is the mass of the particle in terms of proton mass.
If two particles of mass M1 and M2 and atomic number z1 and z2 enter the
absorber with the same velocity then the ratio of their ranges
R1/R2 = (M1/M2)(z2
2
/z1
2
) (7.24)
Range in air – Geiger’s rule
R = const. v3
R = 0.32 E3/2
(alphas in air) (7.25)
Valid for 4–10 MeV α particles. R is in cm and E in MeV
The Bragg–Kleeman rule
If R1, ρ1 and A1 are the range, density and atomic weight in medium 1, the corre-
sponding quantities R2, ρ2 and A2 in medium 2, then
R2/R1 = (ρ1 / ρ2) (A2/A1)1/2
(7.26)

Straggling
When a mono-energetic beam of charged particles traverses a fixed absorber thick-
ness, Δr, there will be fluctuations in the energy of the emerging beam about a
mean value due to finite number of collisions with the atoms of the medium along
the path. This phenomenon is known as Energy straggling, Fig. 7.6.
Fig. 7.6 Energy straggling
The variance of the energy distribution of the emerging particles is given by
σ2
= 4π n z2
e4
. Δr (7.27)
where n is the number of electrons/cm3
, ze is the charge of beam particles. When a
mono-energetic beam of particles is arrested in the absorber, there will be fluctuation
in the ranges of the paths of the particles about a mean value. If σR is the standard
deviation of the range distribution and R̄ the mean range, then the ratio σ/R̄ for the
particle of mass number A is related to that for α-particle by
(σ/R)A/(σ /R)α = (4/A)1/2
(7.28)
Delta rays
In the collision of a charged particle with the atoms one or more electrons are
ejected. The more energetic ones of these are called Delta rays are responsible for
the secondary ionization, that is the production of further ions due to collisions with
other atoms of the absorber.
The kinetic energy of the delta ray is given by
W = 2mv2
cos2
ϕ (7.29)

Motion of a charged particle in a magnetic field
Centripetal force = Magnetic force
mv2
/ρ = qvB (7.30)
Momentum P(GeV/c) = 0.3 Bρ (7.31)
ρ is in meters and B is in Tesla
1 T = 10 k G
Cerenkov radiation
Electromagnetic radiation is emitted when a charged particle on passing through
a medium with a velocity v = βc which exceeds the phase velocity of light c/n,
where n is the index of refraction, the radiation is instantaneous and possesses a
sharply pronounced spatial symmetry.
The radiation is soft and is mostly emitted in the blue part of the spectrum. The
radiation is emitted on a conical surface BDA, as in Fig. 7.7
cos θ = 1/ β n (7.32)
Fig. 7.7 Cerenkov radiation
The threshold velocity
β (thresh) = 1/n (7.33)
The number of photons radiated in the interval dE = h dν by a particle of charge
ze in track length dx is given by
d2
Nγ /dx dE = (α z2
/c) [1 − (1/β2
n2
)] (7.34)
Threshold Cerenkov counters can be used to discriminate between two relativis-
tic particles of the same momentum p and different masses m1 and m2, if the heavier
particle (m2) is just below the threshold. In that case
sin2
θ1 ≈ (m2
2
− m1
2
)/p2
(7.35)

Bremsstrahlung
When a relativistic particle of mass m, charge ze moves close to a target nucleus
of charge Ze, it undergoes acceleration which is proportional to zZ/m, and emits
radiation known as bremsstrahlung. Thus the radiation losses for electrons under
identical conditions, are 3 × 106
times greater than for protons. Since energy loss is
a one-shot process, the law of energy degradation is exponential. If E0 is the initial
energy then at distance x
E= E0 exp(−x/X0) (7.36)
Radiation length X0 is defined as that absorber thickness which reduces the particle
energy by a factor e.
Passage of radiation through matter
When electromagnetic radiation passes through matter the type of interaction depends
on (1) photon energy (2) Z of the material (3) particle or field with which the photon
interacts. The important processes in Nuclear physics are
1. Compton scattering
2. Photoelectric effect
3. Electron–positron pair production
4. Nuclear resonance fluorescence
The compton effect
The process of elastic scattering of photon by a free electron with reduced frequency
(or increased wavelength) is known as Compton scattering, Fig. 7.8
Fig. 7.8 Compton scattering
Shift in wavelength
Δ λ = λ − λ0 = (h/mc) (1 − cos θ) (7.37)
The Compton wavelength
λc = h/mc = 2.43 × 10−12
m (7.38a)

Maximum shift in wavelength
(Δ λ)max = 2 λc (7.38b)
Shift in frequency
ν = ν0/(1 + α(1 − cos θ)) (7.39a)
where α = h ν0 /mc2
(7.39b)
Δ ν = ν0 − ν = α ν0(1 − cos θ)/(1 + α(1 − cos θ))
Energy of recoil electron
T = h ν0 − h ν = α h ν0 (1 − cos θ)/(1 + α (1 − cos θ)) (7.40)
Angular relation
cot ϕ = (1 + α) tan (θ /2) (7.41)
Compton attenuation coefficients
I = I0 exp (−μc x) (7.42)
If x is in cm then μc is in cm−1
; if x is in g/cm2
which is obtained by multiplying
x cm by ρ (the density), then μ is in cm2
/g
Photoelectric effect
If the incident photon is absorbed by an electron in the atom or metal, and the
electron is ejected then the process is known as photoelectric effect. The kinetic
energy of the photoelectron
T = h ν − W ( Einstein’s equation) (7.43)
where W is the ionization energy of the ejected electron
eVo =
1
2
m ν2
max = hν − W (7.43a)
Threshold energy h ν0 = W (7.44)

At low Photon energy
σph ∝ Z5
/(hν)7/2
(7.45)
μph = N σph (7.46)
I = I0 exp (−μph x) (7.47)
Low photon energy can be measured from the observation of absorption edges.
E(eV) = 13.6 (Z − σ)2
/n2
(7.48)
where n = 1 for the K-series and n = 2 for the L-series etc, σ is the screening
constant and Z is the atomic number of the absorber.
The photoelectric absorption also follows the exponential law.
Pair production
At incident photon energies greater than 2mc2
(1.02 MeV), the electron–positron
pair production becomes important.
h ν = T− + T+ + 2mc2
(7.49)
μp ∝ Z2
(7.50)
μtotal = μc + μph + μβ (7.51)
Importance of the three processes is shown in Fig. 7.9
Fig. 7.9 Importance of the
three processes at increasing
photon energy

Nuclear resonance fluorescence
If E0 is the transition energy in an atom of mass M, the resonance energy Eγ = hν
is given by
Eγ = E0 − E2
γ /2Mc2
(7.52)
The recoil energy
ER = E2
γ /2Mc2
(7.53)
The width of the energy level Γ is the full width at half maximum, Fig. 7.10. Γ is
calculated from the mean life time by the uncertainty principle
Γ.τ = (7.54)
Fig. 7.10 Nuclear resonance
fluorescence
Compensation for the recoil energy loss
Source velocity v = E0/Mc (7.55)
Mosbauer Effect is the recoilless emission and absorption of nuclear radiation
Δ Er/Er = v/c. (7.56)
Radioactivity
Radioactivity is the spontaneous disintegration of an atomic nucleus. In natural
radioactivity, the decay may occur via alpha, beta or gamma emission. In artificial
radioactivity, neutron or proton may also be emitted.

Units
1 Curie (ci) = 3.7 × 1010
disintegrations/s
1 Rutherford = 106
disintegrations/s
1 Becquerel (Bq) = 1 disintegration/s
Radiation dose
1 Rad = 100 ergs/g absorbed
Radioactive law
dN/dt = −Nλ (7.57)
N = N0e−λ t
(7.58)
T1/2 = 0.693τ (7.59)
Activity = |dN/dt| = nλ (7.60)
Successive decays
A→B→C
dNB/dt = λA NA − λB NB (7.61)
NB =
λA N0
A
λB − λA
[exp(−λAt) − exp(−λBt)] (7.62)
Transient equilibrium (λA λB)
NB/NA = λA/(λB − λA) (7.63)
Secular equilibrium (λA λB)
NB/NA = λA/λB (7.64)
Alpha-decay (Gamow’s formula)
λ = (v/R) exp (−2π zZ/137β) (7.65)
Geiger–Nuttal law
log λ = K log x + C (7.66)
Beta-decay
λ =
G2
|Mif|2
E5
0
60π3(c)6
(E0 ≫ mec2
) (7.67)

Selection rules
Fermi rule : ΔI = 0, Ii = 0 → If = 0 allowed, Δπ = 0 (7.68)
GT rule : ΔI = 0, ±1, Ii = 0 → If = 0 forbidden, Δπ = 0 (7.69)
7.2 Problems
7.2.1 Kinematics of Scattering
7.1 A particle of mass M is elastically scattered from a stationary proton of
mass m. The proton is projected at an angle ϕ = 22.1◦
while the incident
particle is scattered through an angle θ = 5.6◦
with the incident direction.
Calculate M in atomic mass units. (This event was recorded in photographic
emulsions in the Wills Lab. Bristol).
7.2 A particle of mass M is elastically scattered through an angle θ from a target
particle of mass m initially at rest (M m). (a) Show that the largest possible
scattering angle θmax in the Lab. System is given by sin θmax = m/M, the
corresponding angle in the CMS being cos θ∗
max = −m/M. (b) Further show
that the maximum recoil angle for m is given by sin ϕmax = [(M −m)/2M]1/2
.
(c) Calculate the angle θmax + ϕmax for elastic collisions between the incident
deuterons and target protons.
7.3 A deuteron of velocity u collides with another deuteron initially at rest. The
collision results in the production of a proton and a triton (3
H), the former
moving at an angle 45◦
with the direction of incidence. Assuming that this
re-arrangement collision may be approximated to an elastic collision (quasi-
scattering), calculate the speed and direction of triton in the Lab and CM
system.
7.4 An α-particle from a radioactive source collides with a stationary proton and
continues with a deflection of 10◦
. Find the direction in which the proton
moves (α-mass = 4.004 amu; Proton mass = 1.008 amu).
7.5 When α-particles of kinetic energy 20 MeV pass through a gas, they are found
to be elastically scattered at angles up to 30◦
but not beyond. Explain this, and
identify the gas. In what way if any, does the limiting angle vary with energy?
[University of Bristol]
7.6 A perfectly smooth sphere of mass m1 moving with velocity v collides elas-
tically with a similar but initially stationary sphere of mass m2 (m1 m2)
and is deflected through an angle θL. Describe how this collision would
appear in the center of mass frame of reference and show that the relation

7.2 Problems 383
between θL and the angle of deflection θM, in the center of mass frame is
tan θL = sin θM/[m1/m2 + cos θM]
Show also that θL can not be greater than about 15◦
if m1/m2 = 4.
7.7 Show that the maximum velocity that can be imparted to a proton at rest by
non-relativistic alpha particle is 1.6 times the velocity of the incident alpha
particle.
7.8 Show that the differential cross section σ(θ) for scattering of protons by pro-
tons in the Lab system is related to σ(θ∗
) corresponding to the CMS by the
formula σ(θ) = 4 cos(θ∗
/2) σ(θ∗
).
7.9 If E0 is the neutron energy and σ the total cross-section for low energy n–p
scattering assumed to be isotropic in the CMS, then show that in the LS, the
proton energy distribution is given by dσp/dEp = σ/E0 = constant.
7.10 Particles of mass m are elastically scattered off target nuclei of mass M ini-
tially at rest. Assuming that the scattering in The CMS is isotropic show that
the angular distribution of M in the LS has cos ϕ dependence.
7.11 A beam of particles of negligible size is elastically scattered from an infinitely
heavy hard sphere of radius R. Assuming that the angle of reflection is equal
to the angle of incidence in any encounter, show that σ(θ) is constant, that is
scattering is isotropic and that the total cross-section is equal to the geometric
cross-section, πR2
. (Osmania University)
7.2.2 Rutherford Scattering
7.12 Show that for the Rutherford scattering the differential cross section for the
recoil nucleus in the Lab system is given by σ(ϕ) = (zZe2
/2T )2
/ cos3
ϕ
7.13 A beam of α-particles of kinetic energy 5 MeV passes through a thin foil of
4Be9
. The number of alphas scattered between 60◦
and 90◦
and between 90◦
and 120◦
is measured. What would be the ratio of these numbers?
7.14 If the probability of α-particles of energy 10 MeV to be scattered through an
angle greater than θ on passing through a thin foil is 10−3
, what is it for 5 MeV
protons passing through the same foil?
7.15 What α-particle energy would be necessary in order to explore the field of
force within a radius of 10−12
cm of the center of nucleus of atomic number
60, assuming classical mechanics to be adequate?
7.16 In an elastic collision with a heavy nucleus when the impact parameter b is
just equal to the collision radius R0/2, what is the value of the scattering angle
θ∗
in the CMS?

7.17 In the elastic scattering of deuterons of 11.8 MeV from 82Pb208
, the differen-
tial cross-section is observed to deviate from Rutherford’s classical prediction
at 52◦
. Use the simplest classical model to calculate the closest distance of
approach d to which this angle of scattering corresponds. You are given that
for an angle of scattering θ, d is given by (d0/2) [1 + cosec(θ/2)], where d0
is the value of d in a head-on collision.
[University of Manchester]
7.18 Given that the angle of scattering is 2 tan−1
(a/2b), where “a” is the least
possible distance of approach, and b is the impact parameter. Calculate what
fraction of a beam of 0.5 MeV deuterons will be scattered through more than
90◦
by a foil of thickness 10−5
cm of a metal of density 5 g cm−3
atomic
weight 100 and atomic number 50.
[University of Liverpool]
7.19 An electron of energy 1.0 keV approaches a bare nucleus (Z = 50) with an
impact parameter corresponding to an orbital momentum . Calculate the dis-
tance from the nucleus at which this has a minimum (take = 10−27
j-s, e =
1.6 × 10−19
C and m = 10−30
kg)
7.20 A beam of protons of 5 MeV kinetic energy traverses a gold foil, one particle
in 5 × 106
is scattered so as to hit a surface 0.5 cm2
in area at a distance 10 cm
from the foil and in a direction making an angle of 60◦
with the initial direction
of the beam. What is the thickness of the foil?
[Saha Institute]
7.21 A narrow beam of alpha particles falls normally on a silver foil behind which
a counter is set to register the scattered particles. On substitution of platinum
foil of the same mass thickness for the silver foil, the number of alpha parti-
cles registered per unit time increases 1.52 times. Find the atomic number of
platinum, assuming the atomic number of silver and the atomic masses of both
platinum and silver to be known.
7.22 Derive an expression for the differential cross-section for energy transfer in
elastic collision between a heavy charged particle and an electron.
7.23 If σg is the geometrical cross-section (πR2
) for neutrons interaction with a
nucleus of charge Z and radius R, then show that for positively charged
particles(+ze) the cross-section will decrease by a factor (1 − R0/R), where
R0 = zZe2
/4 π ε0 E0, and E0 is the neutron energy.
7.24 Alphas of 4.5 MeV bombarded an aluminum foil and undergo Rutherford scat-
tering. Calculate the minimum distance of approach if the scattered alphas are
observed at 60◦
with the beam direction.
7.25 If the radius of silver nucleus (Z = 47), is 7 × 10−15
m, what is the minimum
energy that the particle should have to just reach it? Give your answer in MeV.

7.2 Problems 385
7.26 The following counting rates (in arbitrary units) were obtained when α parti-
cles were scattered through 180◦
from a thin gold (Z = 79) target. Deduce a
value for the radius of a gold nucleus from these results.
Energy of α
particle
(MeV)
8 12 18 22 26 27 30 34
Counting rate 91,000 40,300 18,000 12,000 8,400 100 12 1.1
7.27 If a silver foil is bombarded by 5.0 MeV alpha particles, calculate the deflec-
tion of the alpha particles when the impact parameter is equal to the distance
of closest approach.
7.28 Calculate the minimum distance of approach of an alpha particle of energy
0.5 MeV from stationary 7
Li nucleus in a head-on collision. Take the nuclear
recoil into account.
7.29 A narrow beam of alpha particles with kinetic energy T = 500 keV falls nor-
mally on a golden foil incorporating 1.0 × 1019
nuclei cm−3
. Calculate the
fraction of alpha particles scattered through the angles θ θ0 = 30◦
7.30 A narrow beam of protons with kinetic energy T = 1.5 MeV falls normally
on a brass foil whose mass thickness ρt = 2.0 mg cm−2
. The weight ratio
of copper and zinc in the foil is equal to 7:3. Find the fraction of the protons
scattered through the angles exceeding θ = 45◦
. For copper, Z = 29 and
A = 63.55 and for zinc Z = 30 and A = 65.38
7.31 The effective cross-section of a gold nucleus corresponding to the scattering
of monoergic alpha particles at angles exceeding 90◦
is equal to Δσ = 0.6 kb.
Find (a) the energy of alpha particles (b) the differential cross-section σ(θ) at
θ = 90◦
7.32 Derive Darwin’s formula for scattering (modified Rutherford’s formula which
takes into account the recoil of the nucleus).
7.2.3 Ionization, Range and Straggling
7.33 Show that the order of magnitude of the ratio of the rate of loss of kinetic
energy by radiation for a 10-MeV deuteron and a 10-MeV electron passing
through lead is 10−7
.
7.34 Suppose at the sea level the central core of an extensive shower consists
of a narrow vertical beam of muons of energy 60 GeV which penetrate the
interior of the earth. Assuming that the ionization loss in rock is constant at
2 MeV g−1
cm2
, and the rock density is 3.0 g cm−3
, find the depth of the rock
through which the muons can penetrate.
7.35 Show that deuteron of energy E has twice the range of proton of energy E/2.

7.36 If the mean range of 10 MeV protons in lead is 0.316 mm, calculate the mean
range of 20 MeV deuterons and 40 MeVα-particles.
7.37 Show that the range of α-particles and protons of energy 1–10 MeV in alu-
minium is 1/1,600 of the range in air at 15 ◦
C, 760 mm of Hg.
7.38 Show that except for small ranges, the straggling of a beam of 3
He particles is
greater than that of a beam of 4
He particles of equal range.
[University of Cambridge]
7.39 The range of a 15 MeV proton is 1,100 µm in nuclear emulsions. A second
particle whose initial ionization is the same as the initial ionization of proton
has a range of 165 µm. What is the mass of the particle? (The rate at which a
singly ionized particle loses energy E, by ionization along its range is given
by dE/dR = K/(βc)2
MeV µm where βc is the velocity of the particle, and
K is a constant depending only on emulsion; the mass of proton is 1,837 mass
of electron)
7.40 α-particles and deuterons are accelerated in a cyclotron under identical con-
ditions. The extracted beam of particles is passed through an absorber. Show
that the range of deuteron will be approximately twice that of α-particles.
7.41 The α-particle from Th C′
have an initial energy of 8.8 MeV and a range in
standard air of 8.6 cm. Find their energy loss per cm in standard air at a point
4 cm distance from a thin source.
7.42 Compare the stopping power of a 4 MeV proton and a 8 MeV deuteron in the
same medium.
7.43 (a) Show that the specific ionization of 480 MeV α-particle is approximately
equal to that of 30 MeV proton
(b) Show that the rate of change of ionization with distance is different for the
two particles and indicate how this might be used to identify one particle,
assuming the identity of the other is known.
7.44 Calculate the thickness of aluminum in g cm−2
that is equivalent in stop-
ping power of 2 cm of air. Given the relative stopping power for aluminum
S = 1,700 and its density = 2.7 g cm−3
.
7.45 Calculate the minimum energy of an α-particle that can be counted with a
GM counter if the counter window is made of stainless steel (A ≈ 56) with
2.5 mg cm−2
thickness. Take the density of air as 1.226 g cm−3
, and atomic
weight as 14.6.
7.46 Calculate the range in aluminum of a 5 MeV α-particle if the relative stopping
power of aluminum is 1,700.

7.2 Problems 387
7.47 The range of 5 MeV α’s in air at NTP is 3.8 cm. Calculate the range of 10 MeV
α’s using the Geiger–Nuttal law.
7.48 Mean range of α-particles in air under standard conditions is given by the
formula R(cm) = 0.98 × 10−27
v0
3
, where v0 (cm s−1
) is the initial velocity
of an α-particle of 5.0 MeV, find (a) its mean range (b) the average number of
ion pairs formed by the α-particle over the whole path as well as over its first
half, given that the ion pair formation energy of an ion pair is 34 eV.
7.49 Protons and deuterons are accelerated to the same energy and passed through
a thin sheet of material. Compare their energy losses.
7.50 Protons and deuterons lose the same amount of energy in passing through a
thin sheet of material. How are their energies related?
7.51 Determine the average radiative energy loss of electrons of p = 2.7 GeV/c
crossing one radiation length of lead.
7.52 A beam of electrons of energy 500 MeV traverses normally a foil of lead
1/10th of a radiation length thick. Show that the angular distribution of
bremsstrahlung photons of energy 400 MeV is determined more by multiple
scattering of the electrons than by the angular distribution in the basic radiation
process.
7.2.4 Compton Scattering
7.53 In the Compton scattering, the photon of energy E0 = hν0 and momentum
P0 = hν0/c is scattered from a free electron of rest mass m. Show that
(a) the scattered photon will have energy E = E0/[1 + α(1 − cos θ)], where
θ is the angle through which the photon is scattered and α = h ν0/mc2
(b) the kinetic energy acquired by the electron is
T = α E0(1 − cos θ)/[1 + α(1 − cos θ)]
(c) tan (θ/2) = (1 + α) tan ϕ, where ϕ is the recoil angle of the electron.
7.54 Calculate the maximum fractional frequency shift for an incident photon of
wavelength λ = 1 Å scattering off a proton initially at rest (Compton scatter-
ing analogue with proton instead of electron)
[adopted from the University College, Dublin, Ireland 1967]
7.55 A 30 keV x-ray photon strikes the electron initially at rest and the photon is
scattered through an angle of 30◦
, what is the recoil velocity of electron?
[University of New Castle 1966]
7.56 A collimated beam of 1.5 MeV gamma rays strikes a thin tantalum foil. Elec-
trons of 0.7 MeV energy are observed to emerge from the foil. Are these due
to the photoelectric effect, Compton scattering or pair production? Assume
that any electrons produced in the initial interaction with the material of the
tantalum foil do not undergo a second interaction.

7.57 X-rays are Compton scattered at an angle of 60◦
. If the wavelength of the
scattered radiation is 0.312 Å, find the wavelength of the incident radiation.
7.58 Compare the energy loss of a photon in the following situations.
(a) One single Compton scattering through 180◦
(b) Two successive scatterings through 90◦
each
(c) Three successive scatterings through 60◦
each
7.59 For Aluminum, and a photon energy of 0.06 MeV, the atomic absorption cross-
section due to the Compton effect is 8.1 × 10−24
cm2
and due to the photo-
effect is 4.0 × 10−24
cm2
. Calculate how much the intensity of a given beam
is reduced by 3.7 g cm−2
of aluminum and state the ratio of the intensities
absorbed due to the Compton effect and due to the photo effect.
7.60 Calculate the maximum change in the wavelength of Compton scattered
radiation.
7.61 A photon is Compton scattered off a stationary electron through an angle of
45◦
and its final energy is half its initial energy. Calculate the value of the
initial energy in MeV
7.62 What is the range of energies of gamma rays from the annihilation radiation
which are Compton scattered?
7.2.5 Photoelectric Effect
7.63 The wavelength of the photoelectric threshold for silver is 3,250 ×10−10
m.
Determine the velocity of electron ejected from a silver surface by ultraviolet
light of wavelength 2,536 ×10−10
m
7.64 The gamma ray photon from 137
Cs when incident upon a piece of uranium
ejects photo-electrons from its K-shell. The momentum measured with a mag-
netic beta ray spectrometer, yields a value of Br = 3.08 × 10−3
Wb/m. The
binding energy of a K-electron in Uranium is 115.59 keV. Determine
(a) the kinetic energy of the photoelectrons
(b) the energy of the gamma ray photons
7.65 Ultraviolet light of wavelength 2,537 Å from a mercury arc falls upon a silver
photocathode. If the photoelectric threshold wavelength for silver is 3,250 Å,
calculate the least potential difference which must be applied between the
anode and the photo-cathode to prevent electrons from the photo-cathode.
7.66 Show that photoelectric effect can not take place with a free electron.

7.2 Problems 389
7.67 Estimate the thickness of lead (density 11.3 g cm−3
) required to absorb 90%
of gamma rays of energy 1 MeV. The absorption cross-section or gammas of
1 MeV in lead (A = 207) is 20 barns/atom.
7.68 An X-ray absorption survey of a specimen of silver shows a sharp absorption
edge at the expected λkα value for silver of 0.0485 nm and a smaller edge at
0.0424 nm due to an impurity. If the atomic number of silver is 47, identify
the impurity as being 44Ru, 45Rh, 46Pd, v48Cd, V49In or 50Sn.
7.69 A metal surface is illuminated with light of different wavelengths and the cor-
responding stopping potentials of the photoelectrons V , are found to be as
follows.
λ (Å) 3,660 4,050 4,360 4,920 5,460 5,790
V (V) 1.48 1.15 0.93 0.62 0.36 0.24
Determine Planck’s constant, the threshold wavelength and the work function.
7.70 A 4 cm diameter and 1 cm thick NaI is used to detect the 660 keV gammas
emitted by a 100 µCi point source of 137
Cs placed on its axis at a distance
of 1 m from its surface. Calculate separately the number of photoelectrons
and Compton electrons released in the crystal given that the linear absorp-
tion coefficients for photo and Compton processes are 0.03 and 0.24 per cm,
respectively. What is the number of 660 keV gammas that pass through the
crystal without interacting? (1 Curie = 3.7 × 1010
disintegration per second)
7.71 A photon incident upon a hydrogen atom ejects an electron with a kinetic
energy of 10.7 eV. If the ejected electron was in the first excited state, calculate
the energy of the photon. What kinetic energy would have been imparted to an
electron in the ground state?
7.72 Ultraviolet light of wavelengths, 800 Å and 700 Å, when allowed to fall on
hydrogen atoms in their ground state, are found to liberate electrons with
kinetic energy 1.8 and 4.0 eV, respectively. Find the value of Planck’s constant.
[Indian Institute of Technology 1983]
7.73 What is the maximum wavelength (in nm) of light required to produce any
current via the photoelectric effect if the anode is made of copper, which has
a work function of 4.7 eV?
7.74 Photons of energy 4.25 eV strike the surface of a metal A. The ejected photo-
electrons have maximum kinetic energy TA eV and de Broglie wavelength λA.
The maximum kinetic energy of photoelectrons liberated from another metal
by photons of energy 4.70 eV is TB = (TA − 1.50) eV.
If the de Broglie wavelength of the photoelectrons is λB = 2λA, then calcu-
late the kinetic energies TA and TB, and the work functions WA and WB.
[Indian Institute of Technology]

7.2.6 Pair Production
7.75 Calculate the maximum wavelength of γ-rays which in passing through matter,
can lead to the creation of electrons.
7.76 A positron and an electron with negligible kinetic energy meet and annihilate
one another, producing two γ-rays of equal energy. What is the wavelength of
these γ-rays?
7.77 Show that electron–positron pair cannot be created by an isolated photon.
7.2.7 Cerenkov Radiation
7.78 Pions and muons each of 150 MeV/c momentum pass through a transparent
material. Find the range of the index of refraction of this material over which
the muons alone give Cerenkov light. Assume mπ c2
= 140 MeV; mμ c2
=
106 MeV.
7.79 A beam of protons moves through a material whose refractive index is 1.8.
Cerenkov light is emitted at an angle of 11◦
to the beam. Find the kinetic
energy of the proton in MeV.
7.80 The rate of loss of energy by production of Cerenkov radiation is given by
the relation −dW/dl = (z2
e2
/c2
)

1 − 1
β2 μ2

ωdω erg cm−1
where βc is
the velocity, ze is the charge, μ is the refractive index of the medium and
ω/2π is the frequency of radiation. Make an order of magnitude estimate of
the number of photons emitted in the visible region, per cm of track, by a
particle having β = 0.9 passing through water. The fine structure constant
α = e2
/c = 1/137
7.2.8 Nuclear Resonance
7.81 The 129 keV gamma ray transition in 191
Ir was used in a Mösbauer experiment
in which a line shift equivalent to the full width at half maximum (Γ) was
observed for a source speed of 1 cm s−1
. Estimate the value of Γ and the mean
lifetime of the excited state in 191
Ir.
7.82 An excited atom of total mass M at rest with respect to a certain inertial system
emits a photon, thus going over into a lower state with an energy smaller by
Δw. Calculate the frequency of the photon emitted.

7.2 Problems 391
7.83 Calculate the spread in energy of the 661 keV internal conversion line of 137
Cs
due to the thermal motion of the source. Assume that all atoms move with the
root mean square velocity for a temperature of 15 ◦
C.
7.84 Pound and Rebeka at Harward performed an experiment to verify the Red shift
predicted by general theory of Relativity. The experiment consisted of the use
of 14 keV γ -ray 57
Fe source placed on the top of a tower 22.6 m high and the
absorber at the bottom. The red shift was detected by the Mosbauer technique.
What velocity of the absorber foil was required to compensate the red shift,
and in which direction?
7.2.9 Radioactivity (General)
7.85 The disintegration rate of a radioactive source was measured at intervals of
four minutes. The rate was found to be (in arbitrary units) 18.59, 13.27,
10.68, 9.34, 8.55, 8.03, 7.63, 7.30, 6.99, 6.71, and 6.44. Assuming that the
source contained only one or two types of radio nucleus, calculate the half
lives involved.
7.86 100 millicuries of radon which emits 5.5 MeV α-particles are contained in
a glass capillary tube 5 cm long with internal and external diameters 2 and
6 mm respectively. Neglecting end effects and assuming that the inside of the
tube is uniformly irradicated by the particles which are stopped at the surface,
calculate the temperature difference between the walls of a tube when steady
thermal conditions have been reached.
Thermal conductivity of glass = 0.025 Cal cm−2
s−1
C−1
Curie = 3.7 × 1010
disintegrations per second
J = 4.18 joule Cal−1
7.87 Radium being a member of the uranium series occurs in uranium ores. If the
half lives of uranium and radium are respectively 4.5 × 109
and 1,620 years,
calculate the relative proportions of these elements in a uranium ore, which
has attained equilibrium and from which none of the radioactive products
have escaped.
7.88 A sealed box was found which stated to have contained an alloy composed of
equal parts by weight of two metals A and B. These metals are radioactive,
with half lives of 12 years and 18 years, respectively and when the container
was opened it was found to contain 0.53 kg of A and 2.20 kg of B. Deduce
the age of the alloy.
[University of New Castle]

7.89 Determine the amount of 210
84 Po necessary to provide a source of α-particles
of 10 millicuries strength. Half life of Polonium = 138 D.
7.90 A radioactive substance of half life 100 days which emits β-particles of aver-
age energy 5×10−7
ergs is used to drive a thermoelectric cell. Assuming the
cell to have an efficiency 10%, calculate the amount (in gram-molecules) of
radioactive substance required to generate 5 W of electricity.
7.91 The radioactive isotope, 14
6 C does not occur naturally but it is found at con-
stant rate by the action of cosmic rays on the atmosphere. It is taken up by
plants and animals and deposited in the body structure along with natural
carbon, but this process stops at death. The charcoal from the fire pit of an
ancient camp has an activity due to 14
6 C of 12.9 disintegrations per minute,
per gram of carbon. If the percentage of 14
6 C compared with normal Carbon
in living trees is 1.35 × 10−10
%, the decay constant is 3.92 × 10−10
s−1
and
the atomic weight = 12.0, what is the age of the campsite?
7.92 Consider the decay scheme RaE
β
→ RaF
β
→ RaG (stable). A freshly purified
sample of RaE weighs 5×10−10
g at time t = 0. If the sample is undisturbed,
calculate the time at which the greatest number of atoms of RaF will be
present and find this number. Derive any necessary formula [Half life of RaE
210
83 Bi

= 5.0 days; Half life of RaF
210
84 Po

= 138 days]
7.93 It is found that a solution containing 1 g of the α– emitter radium (226
Ra)
never accumulates more than 6.4 × 10−6
g of its daughter element radon
which has a half life of 3.825 days. Explain how the half life of radium may
be deduced from this information and calculate its value.
7.94 Find the mean-life of 55
Co radionuclide if its activity is known to decrease
4.0% per hour. The decay product is non-radioactive.
7.95 What proportion of 235
U was present in a rock formed 3,000×106
years ago,
given that the present proportion of 235
U to 238
U is 140?
7.96 A source consisting of 1 µg of 242
Pu is spread thinly over one plate of an
ionization chamber. Alpha-particle pulses are observed at the rate of 80 per
second, and spontaneous fission pulses at the rate of 3 per hour. Calculate the
half life of 242
Pu and the partial decay constants for the two modes of decay.
7.97 90
Sr decays to 90
Y by β decay with a half-life of 28 years. 90
Y decays by β
decay to 90
Zr with a half-life of 64 h. A pure sample of 90
Sr is allowed to
decay. What is the composition after (a) 1 h (b) after 10 years?

7.2 Problems 393
7.98 Natural Uranium, as found on earth, consists of two isotopes in the ratio
of 235
92 U/238
92 U = 0.7%. Assuming that these two isotopes existed in equal
amounts at the time the earth was formed; calculate the age of the earth.
[Mean life times: 238
92 U = 6.52 × 109
years, 235
92 U = 1.02 × 109
years]
7.99 Calculate the activity (in Ci) of 2.0 µg of 224
ThX. ThX (T1/2 = 3.64 D)
7.100 Calculate the energy in calories absorbed by a 20 kg boy who has received a
whole body dose of 40 rad.
7.101 A small volume of solution, which contained a radioactive isotope of sodium
had an activity of 16,000 disintegrations per minute/cm3
when it was injected
into the blood stream of a patient. After 30 h, the activity of 1.0 cm3
of the
blood was found to be 0.8 disintegrations per minute. If the half-life of the
sodium isotope is taken as 15 h, estimate the volume of the blood in the
patient.
7.2.10 Alpha-Decay
7.102 If two α-emitting nuclei, with the same mass number, one with Z = 84 and
the other with Z = 82 had the same decay constant, and if the first emitted
α-particles of energy 5.3 MeV, estimate the energy of α-particles emitted by
the second.
7.103 Calculate the energy to be imparted to an α-particle to force it into the
nucleus of 238
92 U (r0 = 1.2 fm)
7.104 Radium, Polonium and RaC are all members of the same radioactive series.
Given that the range in air at S.T.P of the α-particles from Radium (half-life
time 1,622 Year) the range is 3.36 cm where as from polonium (half life time
138 D) the range is 3.85 cm. Calculate the half-life of RaC′
for which the
α-particle range at S.T.P is 6.97 cm assuming the Geiger Nuttal rule.
7.105 The α particles emitted in the decays of 88Ra226
and 90Th226
have energies
4.9 MeV and 6.5 MeV, respectively. Ignoring the difference in their nuclear
radii, find the ratio of their half life times.
7.2.11 Beta-Decay
7.106 Classify the following transitions (the spin parity, JP
, of the nuclear states are
given in brackets):-
14
O →14
N∗
+ e+
+ ν (0+
→ 0+
)
6
He → 6
Li + e−
+ ν (0+
→ 1+
)

Why is the transition
17
F → 17
O + e−
+ ν (5/2+
→ 5/2+
)
called a super-allowed transition?
7.107 Beta particles were counted from Mg nuclide. At time t1 = 2.0 s, the count-
ing rate was N1 and at t2 = 6.0 s, the counting rate was N2 = 2.66 N1.
Estimate the mean lifetime of the given nuclide.
7.108 Determine the half-life of β emitter 6
He whose end point energy is 3.5 MeV
and |Mif|2
= 6. Take G/(c)3
= 1.166 × 10−5
GeV−2
7.109 The maximum energy Emax of the electrons emitted in the decay of the iso-
tope 14
C is 0.156 MeV. If the number of electrons with energy between E
and E + dE is assumed to have the approximate form
N(E)dE ∝
√
E (Emax − E)2
dE
find the rate of evolution of heat by a source of 14
C emitting 3.7 × 107
elec-
trons per sec.
[University of Cambridge]
7.110 In the Kurie plot of the decay of the neutron, the end point energy of the elec-
tron is 0.79 MeV. What is the threshold energy required by an antineutrino
for the inverse reaction.
ν̄ + p → n + e+
7.111 36Kr88
decays to 37Rb88
with the emission of β-rays with a maximum energy
of 2.4 MeV. The track of a particular electron from the nuclear process has a
curvature in a field of 103
gauss of 6.1 cm. Determine
i. the energy of this electron in eV and that of the associated neutrino.
ii. the maximum possible kinetic energy of the recoiling nucleus
7.112 If the β-ray spectrum is represented by n(E)dE ∝
√
E (Emax −E)2
dE Show
that the most intense energy occurs at E = Emax/5
7.3 Solutions
7.1 We shall find an expression for the ratio of the masses M/m in terms of the
angles θ and ϕ. To this end we start with the equation for the transformation
of angle from CMS to LS.
tan θ = sin θ∗
/(cos θ∗
+ M/m) (1)
θ∗
= π − ϕ∗
= π − 2ϕ

7.3 Solutions 395
(because m and M are oppositely directed in the CMS, and the recoil angle of
the proton in the CMS is always twice the angle in the LS)
Therefore sin θ∗
= sin(π − 2ϕ) = sin 2ϕ
and cos θ∗
= cos(π − 2ϕ) = − cos 2ϕ
tan θ = sin θ/ cos θ = sin 2ϕ/(M/m − cos 2ϕ)
Cross multiplying the second equation and re-arranging
(M/m) sin θ = sin θ cos 2ϕ + cos θ sin 2ϕ = sin(θ + 2ϕ)
M/m = sin(θ + 2ϕ)/ sin θ
Using θ = 5.60
and ϕ = 22.10
, we find M = 7.8 m ≈ 7.8 amu.
7.2 (a) We can work out this problem in the LS, but we prefer the CMS for
convenience. Writing the equation for transformation of angles
tan θ = sin θ∗
/(cos θ∗
+ M/m) (1)
The condition for the maximum angle of scattering, θmax is d tan θ/dθ =0.
This gives us
cos θmax = −m/M (2)
sin θ∗
max = (M2
− m2
)1/2
/M (3)
When (2) and (3) are used in (1) we find cot θmax = (M2
− m2
)1/2
/m,
whence
sin θmax = m/M (4)
(b) sin ϕmax = sin(ϕ∗
max/2) = sin[(π − θ∗
max)/2] = cos(θ∗
max/2)
= [(1 + cos θ∗
max)/2]1/2
= [(1 − m/M)/2]1/2
= [(M − m)/2M]1/2
(5)
(c) Using m = 1 and M = 2 in (4) and (5)
we find θmax + ϕmax = 30◦
+ 30◦
= 60◦
7.3 We prefer to work in the CMS.
Let m be the proton mass.
Energy available in the CMS in the d-d collision is E∗
= 1/2 µ u2
= 1/2 mu2
The centre of mass velocity vc = u/2
The energy E∗
is partitioned between the product particles, proton and tri-
ton as follows.
Ep
∗
= 3E∗
/4 and Eθ
∗
= E∗
/4. The corresponding velocities in the CMS will
be, vp
∗
=
√
3u/2 and vt
∗
= u/2
√
3, respectively. Using the formula for the
transformation of angles from CM to LS (see formula 7.2)
tan θ =
sin θ∗
cos θ∗ + vc/v∗
(1)
And using θ = θp = 45◦
, vc = u/2 and v∗
= vp
∗
=
√
3u/2 and solving
for θ∗
, we find θp
∗
= θ∗
= 69◦
.

As triton will be emitted in the opposite direction in the CMS
θt
∗
= 111◦
Using formula (1) again, with the substitution.
θt
∗
= 111◦
, vc = u/2 and vt
∗
= u/2
√
3, we can solve for θ, and obtain
θt = 34◦
in the LS.
Finally, we can use the inverse transformation
tan θ∗
= sin θ
cos θ−vc/v
And substitute θ∗
= 111◦
, θ = 34◦
and vc = u/2 to find v = 0.48 u.
7.4 Use the result of Problem 7.1,
M/m = sin (θ + 2ϕ)/ sin θ
M = 4.004, m = 1.008, θ = 100
.
Substituting these values in the above equation and solving for ϕ, we find
ϕ = 16.8◦
which is the recoil angle of proton
7.5 Alphas of 20 MeV energy means that we are dealing with non-relativistic par-
ticles. From the results of Problem 7.2, the maximum scattering angle θmax for
m1 m2, is given by
sin θm = m2/m1
sin 300
= 0.5 = m2/4 or m2 = 2
The gas is deuterium. The limiting angle θmax is independent of the incident
energy.
7.6 The description of the scattering event in the CM system is shown in Fig. 7.11.
From the velocity triangle
vL sin θL = vM sin θM
vL cos θL = vM cos θM + vC
Dividing the two equations and simplifying
tan θL =
sin θM
cos θM + vC/vM
=
sin θM
cos θM + m1/m2

∵ vC =
vm1
(m1 + m2)
and vm =
vm2
(m1 + m2)

The maximum scattering angle θm is given by
sin θM = m2/m1 = 1/4 = 0.25
θM = 15◦
Fig. 7.11 Velocity triangle
for elastic scattering

7.3 Solutions 397
7.7 Let the alpha particle of mass m moving with velocity v0 and momentum
p0 collide with proton of mass m. After the collision the maximum velocity
v2 and momentum p2 will be acquired by proton when it is emitted in the
incident direction. Since the alpha particle is heavier than proton, it must also
proceed in the same direction as the proton with velocity v1 and momentum
p1. Assuming that the collision is elastic,
P0
2
/2m1 = p1
2
/2m1 + p2
2
/2m2 (energy conservation) (1)
P0 = p1 + p2 (momentum conservation) (2)
Noting m1 = 4m2, p1 can be eliminated between (1) and (2) to yield
p2 = 0.4p0
or m2 v2 = 0.4m1 v0 = 0.4 × 4m2v0
v2 = 1.6v0
7.8 Use the transformation equation for the differential cross-sections in the CMS
and LS.
σ (θ) = (1 + γ 2
+ 2γ cos θ∗
)3/2
σ(θ∗
)/|1 + γ cos θ∗
|
where γ = m1/m2 = m/m = 1. Then the equation simplifies to
σ (θ) = 4 cos(θ∗
/2) σ (θ∗
)
7.9 We can write by chain rule
d σp/dEp = (d σp/d Ω∗
).(dΩ∗
/dEp) (1)
Let the proton be scattered through an angle θ∗
in the CMS with the direc-
tion of incidence of neutron (left to right). The CMS velocity will be v0/2.
vc = v0 m1/(m1 + m2) = v0/2 as the masses of neutron and proton are
approximately equal. The proton has velocity vc in the CMS both before and
after scattering. The velocity of the scattered proton is combined with the CMS
velocity to give the velocity v0 in the LS as shown in Fig. 7.12. From the
velocity triangle we have
v2
= vc
2
+ vc
2
+ 2vc
2
cos θ∗
= 2 vc
2
(1 + cos θ∗
)
= v0
2
(1 + cos θ∗
)/2
Therefore, Ep = E0(1 + cos θ∗
)/2
Differentiating
dEp = −E0 sin θ∗
dθ∗
/2 = −E0 dΩ∗
/4π
The negative sign means that as θ∗
increases Ep decreases
Thus,
Fig. 7.12 n−p scattering
in CMS

dΩ∗
/dEp = 4π/E0 (2)
And dσp/dΩ∗
= σ/4π (3)
as the scattering is isotropic in the CMS.
Using (2) and (3) in (1) we get dσp/dEp = σ/E0
7.10 As the scattering is isotropic in the CMS the differential cross-section of the
recoiling nuclei is constant and is given by σ (ϕ∗
) = σ/4π = constant.
Now the differential cross-sections in the LS and CMS are related by
σ (ϕ) = (sin ϕ∗
dϕ∗
/ sin ϕ d ϕ). σ (ϕ∗
)
But ϕ∗
= 2 ϕ and dϕ∗
= 2 dϕ
σ (ϕ) = (sin 2 ϕ.2 dϕ / sin ϕ dϕ)(σ/4 π) = σ
π
cos ϕ
Thus, σ (ϕ) has cos ϕ dependence. It is of interest to note that

σ (ϕ) d Ω =
π/2
0 σ cos ϕ.2 π sin ϕ d ϕ/π = σ
as it should. The upper limit for the integration is confined to 90◦
as the target
nucleus can not recoil in the backward sphere in the LS.
7.11 In Fig. 7.13, b denotes the impact parameter. Consider particles I0 going
through a ring perpendicular to the central axis, its area being 2πb db. On
hitting the sphere, the same number of particles are scattered through a solid
angle dΩ = 2π sin θ dθ
Fig. 7.13 Scattering of
particles of negligible size
from an infinitely heavy hard
sphere of radius R
Therefore, I0 σ (θ) .2π sin θ dθ = −I0 2πb db
Or σ (θ) = −b db/ sin θ d θ (1)
The angles of incidence and reflection are measured with respect to the
normal at the point of scattering. From the geometry of the figure,
θ = π − (i + r) = π − 2i (2)
sin θ = sin (π − 2i) = sin 2i = 2 sin i cos i (3)
Since r = i and dθ = −2di
b = R sin i and db = R cos i di
We find σ(θ) = R2
/4 (4)
The right hand side of (4) is independent of the scattering angle θ; that is
the scattering is isotropic or equally in all directions.
The total cross-section

7.3 Solutions 399
σ =

σ(θ) d Ω =
π
0
(R2
/4). 2π sin θ d θ = π R2
This is called geometric cross-section.
7.12 Rutherford law for scattered particles in the CMS is
σ(θ∗
) = d σ(θ∗
)/d Ω∗
=
1
16

zZe2
T
2
1
sin4
θ∗
2

Now sin (θ∗
/2) = sin((π − ϕ∗
)/2) = cos(ϕ∗
/2) = cos ϕ
d Ω∗
= 2 π sin θ∗
d θ∗
= 2 π sin(π − ϕ∗
) d ϕ∗
= 2 π sin 2 ϕ.2 d ϕ = 8 π sin ϕ cos ϕ dϕ = 4 cos ϕ dΩ(ϕ)
Therefore d σ(ϕ)/dΩ(ϕ) = (zZe2
/T )2
.1/ cos3
ϕ
7.13 The Rutherford scattering cross-section for scattering between angle θ1 and
θ2 is given by σ(θ1, θ2) =

π R0
2
/4

[cot2
(θ1/2) − cot2
(θ2/2)]
where R0 = zZe2
/4π ε0 T = 1.44zZ/T
Therefore, σ(60◦
,90◦
)
σ(90◦,120◦)
= (cot2
30◦
− cot2
45◦
)/(cot2
45◦
− cot2
60◦
) = 3/1
7.14 Since the atomic number and foil thickness and angles are the same, the prob-
ability for scattering will be inversely proportional to the square of energy
and directly proportional to this square of the charge of the particle. Hence the
probability for scattering of 5 MeV α-particles will be 22
102 × 52
12 ×10−3
= 10−3
7.15 The minimum requirement is that the particle should be able to penetrate the
nucleus of radius R. Thus
T = 1.44zZ/R(fm) = 1.44 × 2 × 60/10 ≈ 17.3 MeV
7.16 tan(θ∗
/2) = r(collision)/2b
for b = 1
/2 r (collision), tan (θ∗
/2) = 1 = tan 45◦
Therefore, θ∗
= 90◦
7.17 Given d = d0
2
(1 + cosec (θ/2))
Here d = rmin, d0 = R0
Therefore, R0 = 1.44zZ/T = 1.44 × 1 × 82/11.8 = 10 fm
Rmin = (10/2)(1 + cosec 26◦
) = 16.4 fm
7.18 Given θ = 2 tan−1
(a/2b)
Therefore, tan (θ/2) = R0/2b (1)
R0 = 1.44zZ/T = 1.44 × 1 × 50/0.5 = 144 fm
From (1)
b = (R0/2) cot (θ/2) = (144 cot 450
)/2 = 72 fm
Where we have used θ = 90◦
σ(90◦
, 180◦
) = πb2
= π(72)2
= 16, 278 fm2
= 1.628 × 10−22
cm2
Macroscopic cross-section
Σ = σ N0 ρ/A = 1.628 × 10−22
× 6.02 × 1023
× 5/100 = 4.9 cm−1

Therefore, Mean free path λ = 1/Σ = 0.2 cm
Fraction of scattered particles = Probability of scattering through more than
900
is t/λ = 10−5
/0.2 = 5 × 10−5
.
7.19 rmin = R0
2

1 + +4b2
R2
0
1
2

R0 = 1.44zZ/T = 1.44 × 1 × 50/(1 × 10−3
) = 7.2 × 104
fm
bp = L =
b = /p = c/cp = 197.3 MeV fm/1 × 10−3
MeV = 1.973 × 105
fm
b/R0 = 1.973 × 105
/(7.2 × 104
) = 2.74
rmin = 7.2
2
× 104
*
1 +

1 + 4 × 2.742
1
2
+
= 2.365 × 105
fm
7.20 The mean free path λ = 1/Σ = 1/N.dσ (1)
where N is the number of atoms per cm3
N = N0ρ/A = 6.02 × 1023
× 19.3/197 = 5.9 × 1022
(2)
d σ = (1.44zZ/4T )2
d Ω/ sin4
(θ/2) (3)
Put z = 1, Z = 79, θ = 60◦
, dΩ = area/(distance)2
= 0.5/102
= 0.005
Using these values in (3), d σ = 2.588 fm2
= 2.588 × 10−26
cm2
(4)
Using (2) and (4) in (1) we find λ = 655 cm,
The probability of scattering at 60◦
,
P = t/λ (5)
where t is the foil thickness
t = pλ = (655 × 1/5 × 106
) cm = 1.31 μm.
7.21 The counting rate is dependent on the factor
(zZ/T )2
(N0/A)(ρt/ sin4
(θ/2))
In the problem z, T, (ρ t) and θ are unchanged. Hence, counting rate with
platinum/counting rate with silver
= (Zpt
2
/Apt)/(ZAg
2
/AAg) = 1.52
Substituting the known values: ZAg = 47, AAg = 108.87 and Apt = 195, the
above equation can be solved to yield Zpt = 77.55 or 78.
7.22 Let the particle of mass m1, charge z, velocity v and kinetic energy T collide
elastically with an electron of mass m2 = m. The electron velocity before
and after the collision is v2
∗
= vc in the CMS. The velocity v2
∗
is combined
vectorially with the CMS velocity vc to yield the LS velocity v2 at angle ϕ
with the incident direction

7.3 Solutions 401
Fig. 7.14 Collision of a
heavy charged particle with
an electron
From the velocity triangle which is isosceles as in Fig. 7.14,
v2 = 2vc cos ϕ = 2m1 v cos ϕ/(m1 + m2) ≈ 2v cos ϕ
(Because m2 = m m1)
The energy acquired by the electron
W = (m/2) (2v cos ϕ)2
= 2mv2
cos2
ϕ (1)
If the recoil angle of the electron is ϕ∗
in the CMS then ϕ = ϕ∗
/2 and
ϕ∗
= π–θ∗
, where θ∗
is the scattering angle of the incident particle in the
CMS. And so
cos2
ϕ = sin2
(θ∗
/2)/2 (2)
W = 2mv2
sin2
(θ∗
/2) (3)
dW = mv2
sin θ∗
d θ∗
(4)
But Rutherford’s formula for scattering in the CMS is
σ(θ∗
) = d σ/d Ω∗
= z2
e4
/4μ2
v2
sin4
θ∗
/2 (5)
where we have put Z = −1 for electron. Since the electron mass is negligible
compared to that of the incident particle, μ ≈ m. Further, the element of solid
angle dΩ∗
= 2π sin θ∗
d θ∗
.
Formula (5) becomes
d σ = (2π sin θ∗
dθ∗
z2
e4
)/(4m2
v4
sin4
(θ∗
/2)) (6)
d σ/dW = 2 z2
e4
/mv2
W2
(differential energy spectrum)
This gives us the cross-section for finding the delta rays(emitted electrons)
of energy W per unit energy interval.
7.23 When the charged particle just grazes the nucleus
rmin = R = 1/2R0[1 + (1 + 4b2
/R0
2
)1/2
] (1)
Solving for b, we obtain
b2
= R2
–RR0 (2)
Denoting the cross-section by σ = πb2
,
σ/σg = π b2
/π R2
= 1–R0/R
7.24 The closest distance of approach r is given by
rmin = (R0/2)(1 + cosec (θ/2)) = (R0/2)(1 + 2) = 3R0/2

Now R0 = 1.44zZ/T = 1.44 × 2 × 13/4.5 = 8.32
Hence rmin = 1.5 × 8.32 = 12.48 fm
7.25 The minimum distance of approach rmin is obtained in the head-on colli-
sion when the initial α particle energy is entirely converted into the potential
energy.
T = zZe2
/4 π ε0rmin
Putting rmin = R, the nuclear radius, numerically
T(MeV) = 1.44zZ/R(fm) = 1.44 × 2 × 47/7 = 19.34 MeV
7.26 For the same charges of interacting particles, target thickness and beam inten-
sity and fixed scattering angle, the Rutherford scattering depends inversely as
the square of particle energy. So long as the incident particle is outside the
target nucleus, the Rutherford scattering is expected to be valid. But when
the incident particle touches the nucleus, pure Rutherford scattering would be
invalidated. We can assume that the observed counting rate N8 is the expected
counting rate for the lowest energy (8 MeV). For higher energy T the counting
rate is expected to be NE = N8(8/T )2
. In the table below are displayed the
calculated counting rates as well as the observed ones.
T(MeV) 8 12 18 22 26 27 30 34
N(Cal) 91,000 40,400 17,970 12,030 8,600 8,000 6,471 5,038
N(Obs) 91,000 40,300 18,000 12,000 8,400 100 12 1.1
Comparison between the calculated and observed counting rates indicates
that Rutherford scattering begins to break down at 26 MeV. Since scattering
angle is θ = 180◦
, we are concerned with head-on collisions. Hence
R0 = R = 1.44 zZ/T = 1.44 × 2 × 79/26 = 8.75 fm.
Hence the radius of gold nucleus is 8.75 fm.
7.27 When distance of closest approach b = R0
tan(θ/2) = R0/2b = 0.5 = tan 26.56◦
Therefore, θ/2 = 26.56◦
or θ = 53.1◦
7.28 We work out in the CMS because 7
Li being a light nucleus will recoil in the
encounter. Equating the potential energy at the closest distance of approach
R0 to the initial kinetic energy,
1.44zZ/R0 = μ v2
/2 = m1 m2 v2
/(m1 + m2) = m2T/(m1 + m2)
where μ is the reduced mass. Solving for R0
R0 = (1.44zZ/T )(1 + m1/m2) = (1.44x2x3/0.5)(1 + 4/7) = 27.15 fm
7.29 Given nt = 1.0 × 1019
nuclei/cm2
The fraction ΔN/N = 1−

π R0
2
cot2
(θ/2).nt

/4 = 1−(π/4)(1.44zZ/T )2
cot2
(θ/2).nt = 1 − π(1.44 × 2 × 79/0.5)2
cot2
(30/2) × (1.0 × 1019
×
10−26
)/4 = 0.82
The factor 10−26
has been introduced to convert fm into cm2
.

7.3 Solutions 403
7.30 The fraction of particles scattered in brass at angles exceeding θ is given
by ΔN/N = (π/4)(1.442
/T 2
)

0.7Z1
2
/A1 + 0.3Z2
2
/A2

ρtN 0 cot2
θ/2 ×
10−26
where Z1 = 29 for copper and Z2 = 30 for Zinc, A1 = 63.55 and
A2 = 65.38 are the atomic masses, respectively, and N0 = 6.02 × 1023
is the
avagadro’s number, T = 1.5 MeV, ρt = 2 × 10−3
g cm−2
and θ = 450
. The
factor 10−26
is introduced to convert fm2
into cm2
. Using these values in the
above equation ΔN/N = 6.78 × 10−4
7.31 (a) Δσ = σ(90◦
, 180◦
) = (π/4)(1.44zZ/T )2
Using Δσ = 0.6 kb = 60 fm2
, z = 2, Z = 79 and solving for T , we get
T = 3.36 MeV
(b) σ(θ) = (1/16)(1.44zZ/T )2
1/ sin4
(θ/2)
Put θ = 90◦
and T = 3.36 MeV to find σ(90) = 1,146 fm2
/sr =
11.46 b/sr.
7.32 First we work out in the CMS and then transform σ(θ∗
) in the CMS to σ(θ)
in the LS.
Rutherford’s formula in CMS is
σ(θ∗
) = (1/4)(zZe2
/μv2
)2
.1/ sin4
(θ/2) (1)
where the reduced mass
μ = mM/(m + M) = m/(1 + γ ) (2)
where m and M are the masses of the incident and target masses, respectively,
and γ = m/M.
Further
sin4
(θ∗
/2) = (1/4)(sin4
θ∗
)/(1 + cos θ∗
)2
(3)
and
tan θ = sin θ∗
/(γ + cos θ∗
) (4)
Squaring (4) and expressing it as a quadratic equation and solving it
cos θ = γ sin2
θ ± cos θ(1 − γ 2
sin2
θ)1/2
(5)
Now σ(θ) = (1 + γ 2
+ 2 γ cos θ∗
)3/2
/|1 + γ cos θ∗
| (6)
Combining (1), (2), (3), (5) and (6) and after some algebraic manipulations
we get
σ(θ) =

zZe2
2T
2
1
sin4
θ
[cos θ ± (1 − γ 2
sin2
θ)1/2
]2
(1 − γ 2 sin2
θ)1/2
If γ 1, the positive sign only should be used before the square root. If
γ 1, the expression should be calculated for positive and negative signs and
the results added to obtain σ(θ). For γ = 1, σ(θ) = zZe2
T
2
cos θ
sin4
θ

7.33 The rate of loss of energy due to radiation is inversely proportional to the
square of particle mass and directly proportional to the square of charge of
the incident particle as well as the atomic number of the target nucleus and
directly proportional to the kinetic energy.
−(dE/dx)rad ∝ z2
Z2
T/m2
As the medium is identical and both e and d are singly charged with the
same kinetic energy the ratio of the radiation loss for deuteron and electron
will be(me/md)2
≈ (1/3, 670)2
= 7.4 × 10−8
or ≈ 10−7
7.34 Ionization loss of muons in the rock = 2 MeV g−1
cm2
= 2 MeV g−1
cm2
× ρ = 2 MeV g−1
cm2
× 3.0 g cm−3
= 6 MeV/cm.
The depth of the rock which will reduce 60 GeV to zero = 60×103
/6 cm =
104
cm = 100 m.
7.35 Ed = md vd
2
/2
Ep = mp vp
2
/2
Ed/Ep = E/(E/2) = 2 = md vd
2
/mp vp
2
= 2vd
2
/vp
2
Therefore, vd = vp
Deuteron and proton having the same initial speed will have their ranges in
the ratio of their masses. Therefore the deuteron has twice the range of proton.
7.36 R = (M/z2
) f(E/M)
The E/M ratio for the three given particles is identical because Ep/Mp =
10/1, Ed/Md = 20/2 and Eα/Mα = 40/4. Hence
Rd = (Rp Mdzp
2
)/(z2
d Mp) = (0.316 × 2 × 12
)/(1 × 12
) = 0.632 mm
Rα = (Rp Mαzp
2
)/(zα
2
Mp) = (0.316 × 4 × 12
)/(1 × 22
) = 0.316 mm
7.37 Apply the Bragg–Kleeman formula
RAl/Rair = (ρair/ρAl)
√
AAl/
√
AAir
Substitute the values: ρair = 1.226 × 10−3
g cm−3
, ρAl = 2.7 g cm−3
AAl = 27 and Aair = 14.5, we find RAl/RAir = 1/1, 614
7.38 The straggling of charged particles is given by the ratio σR/R, where R is the
mean range of a beam of particles in a given medium and σR is the standard
deviation of the ranges.
Now R = f (v0, I)/z2
where v0 is the initial velocity of beam of particles, I is
the ionization of the atoms of the absorber. Further, σR =
√
M F(v0, I)/z2
. So
σR/R = ϕ(v0, I)/
√
M

7.3 Solutions 405
As the straggling is inversely proportional to the square root of particle
mass, the straggling for 3
He will be greater than that for 4
He of equal range.
7.39 dE/dR = k/(βc)2
dR = [(β c)2
/k] d(Mv2
/2) = k′
M f (v) dv
where k′
is another constant
Integrating from zero to v0
R =

dR = k′
M

f (v) dv = k′
M F(v0)
If two single charge particles of masses M1 and M2, be selected so that their
initial velocities are identical then their residual ranges
R1/R2 = M1/M2
M2 = R2 M1/R1 = 165 × 1, 837 me/1, 100 = 275.5 me
It is a pion. Its accepted mass is 273 me
7.40 Balancing the centripetal force with the magnetic force at the point of
extraction
mv2
/r = zevB (1)
which gives us
v = zeBr/m
The ratio of velocities for α-particle and deuteron is
vd/vα = (zd.mα)/(md.zα) = (1 × 2)/(2 × 1) = 1
since mα ≈ 2md
As the initial velocities are identical the ratio
Rd/Rα = (mdzα
2
)/(mαzd
2
) = 22
/2 = 2
Therefore, Rd = 2Rα
7.41 Use Geiger’s rule
R = K E3/2
(1)
8.6 = K (8.8)3/2
K = 0.33
At a distance of 4 cm from the source, residual range is 4.6 cm. At this point
the energy can be found out by applying (1) again
4.6 = 0.33 × E
3/2
1
Or E1 = 5.79 MeV
Differentiating (1)
dR/dE = (3/2)K E1/2
dE1/dR = 2/3K E1
1/2
= 2/(3 × 0.33 ×
√
5.79)
= 0.84 MeV/cm

7.42 The proton velocity vp = (2Ep/mp)1/2
=

2 × 4/mp
The deuteron velocity vd = (2Ed/md)1/2
=
√
2 × 8/md = (2 × 4/mp)1/2
(because md ≈ 2mp)
Thus, the proton and deuteron have the same velocity, and both of them are
singly charged. Hence their stopping power is identical.
7.43 (a) dE/dR ∝ z2
/v2
or ∝ Mz2
/E
(dEα/dR)/(dEp/dR) = Mα zα
2
Ep/Mp zp
2
Eα = (Mα/Mp)(zα
2
/zp
2
)
(Ep/Eα) = 4 × 4 × 30/480 = 1
(b) The change in ionization over a given distance will be different for differ-
ent particles in a medium. Calibration curves can be drawn for particles
of different masses. The proton curve can be assumed to be the standard
curve. This method is particularly useful for those particles which are not
arrested within the emulsion stack or bubble chamber.
7.44 If S is the relative stopping power then
R(Al) = R(air)/S = 2/1, 700 cm
The thickness of aluminum is obtained by multiplying the range in air by
the density
Therefore, R(Al) = (2.7 × 2/1, 700) g cm−2
= 3.18 × 10−3
g cm−2
7.45 Apply the Bragg–Kleeman rule
Ra = Rs ρs
√
Aa/ρa
√
As (1)
Now Rs(cm) = Rs(g cm−2
)/ρs = 2.5 × 10−3
/ρs
Hence (1) becomes
Ra(cm) = 2.5 × 10−3√
14.5/1.226 × 10−3
×
√
56 = 1.04
Apply Geiger’s rule
E = (R/0.32)2/3
= (1.04/0.32)2/3
= 2.19 MeV
α’s of energy greater than 2.2 MeV will be registered.
7.46 R(Al) = R(air)/S = R(air)/1,700
We find R(air) by Geigers’s rule
R(air) = 0.32 E3/2
= 0.32 × 53/2
= 11.18 cm
Therefore R(Al) = 11.18/1, 700 = 0.00658 cm = 66 µm
7.47 Geiger’s rule is
R = 0.32(E)3/2
where R is in cm and E in MeV
3.8 = 0.32(5)3/2
R = 0.32(10)3/2
Therefore, R = 3.8 × (10/5)3/2
= 10.75 cm
7.48 (a) v0 = (2T/m)1/2
= c(2T/mc2
)1/2
= c(2 × 5/3728)1/2
= 0.0518 c
R = 0.98 × 10−27
× (3 × 1010
× 0.0518)3
= 3.678 cm

7.3 Solutions 407
(b) Over the whole path total number of ion pairs
= Total energy lost/Ionization energy of each pair = 5.0 × 106
/34
= 1.47 × 105
For R1 = 1.839 cm, we can find the velocity at the middle of the path by the
given formula
R1 = 1.839 = 0.98 × 10−27
v1
3
or v1 = 1.233 × 109
. The corresponding energy at the mid-path is
E1 = mv1
2
/2 = mc2
v1
2
/2c2
= (1/2) × 3728 × (1.233 × 109
/3 × 1010
)2
=
3.148 MeV
Energy lost in the first half of the path
E = 5.000 − 3.148 = 1.852 MeV.
Number of ion pairs produced over the first half of the path =
1.852 × 106
/34 = 5.45 × 104
7.49 v2
p = 2E/mp, v2
d = 2E/md = 2E/2mp = v2
p/2
= dE/dx ∝ z2
/v2
∴
(dE/dx)p
(dE/dx)d
=
v2
d
v2
p
= 2(∵ z = 1 for both p and d)
7.50 (−dE/dx)proton = (−dE/dx)deuteron
As their charges are identical, their velocity must be the same. Therefore,
the ratio of their kinetic energies must be equal to the ratio of their masses.
Ep/Ed = Mp/Md = 1/2
7.51 After crossing a radiation length of lead electrons emerge with an aver-
age energy of 2.7/e = 2.7/2.71 = 1.0 GeV. Thus, the average energy
loss = 1.7 GeV.
7.52 The root mean square multiple scattering angle is approximately given by
(θ2)
1
2 = 20
βp(MeV/c)
$
L
LR
For a traversal of a distance L. For P = 400 MeV/c and L = LR/10, this
angle is 15.8 mr. An electron emitting a photon will be emitted at an angle of
mec2
/Eγ with the direction of flight. This angle is 0.511/400 or 1.28 mr.
Thus, the angular distribution of Bremsstrahlung photons is determined
mainly by the multiple scattering of electrons.
7.53 (a) Let E = hν be the energy of the scattered photon and hν/c be its momen-
tum (Fig. 7.15).
h ν0 = h ν + T (1)
where T is the electron’s kinetic energy.
Balancing momentum along and perpendicular to the direction of
incidence

Fig. 7.15
h ν0/c = h ν cos θ/c + Pe cos ϕ (2)
0 = −h ν sin θ/c + Pe sin ϕ (3)
where Pe is the electron momentum
Re-arranging (2) and (3) and squaring
Pe
2
cos2
ϕ =

hν0
c
−
hν cos θ
c
2
(4)
Pe
2
sin2
ϕ =

h
c
ν sin θ
2
(5)
Add (4) and (5) and using the relativistic equation
c2
Pe
2
= T 2
+ 2T mc2
= h2
(ν0
2
+ ν2
− 2ν0ν cos θ) (6)
Eliminating T between (1) and (2) and simplifying we get
E = E0/[1 + α(1 − cos θ)] (7)
(b) T = E0 − E = E0 − E0/[1 + α(1 − cos θ)] = [αE0(1 − cos θ)]/
[1 + α(1 − cos θ)] (8)
(c) From (2) and (3), we get
Cot ϕ = (ν0 − ν cos θ)/ν sin θ
With the aid of (7), and re-arranging we find tan(θ/2) = (1 + α)
tan ϕ (9)
7.54 Fractional shift in frequency is
Δν/ν0 =
1
1 + Mc2
2hν0 sin2
(θ
2 )
hν0 = Eγ = 1, 241/λnm = 1, 241/0.1 = 12, 410 eV = 0.01241 MeV
Put θ = 180◦
and MC2
= 938.3 MeV to obtain
(Δν/ν0)max =
1
1 + 938.3
2×0.01241
= 2.645 × 10−5
7.55 The energy of scattered photon will be
h ν = h ν0/[1 + α(1 − cos θ)]

7.3 Solutions 409
For hν0 = 30 keV, α = hν0/mc2
= 30/511 = 0.0587 and θ = 30◦
We find hν = 29.766 keV
Therefore the kinetic energy of the electron
T = 30.0 − 29.766 = 0.234 keV
The velocity v = (2T/m)1/2
= c(2T/mc2
)1/2
= c(2×0.234/511)1/2
= 0.03 c
7.56 As the K-shell ionization energy for tantalum is under 80 keV, the ejected
electrons due to photoelectric effect are expected to have kinetic energy little
less than 1.5 MeV. Hence photoelectric effect is ruled out as the source of the
observed electrons.
For the e+
e−
pair production the threshold energy is 1.02 MeV. The com-
bined kinetic energy of the pair would be (1.5 − 1.02) or 0.48 MeV, a value
which falls short of the observed energy of 0.7 MeV for the electron. Thus the
observed electrons can not be due to this process.
In the Compton scattering the electrons can be imparted kinetic energy
ranging from zero to Tmax = h ν0/(1 + 1/2α), depending on the angle
of emission of the electron. Substituting the values, h ν0 = 1.5 MeV and
α = h ν0/mc2
= 1.5/0.51 = 2.94, we find Tmax = 1.28 MeV. Thus the
Compton scattering is the origin of the observed electrons.
7.57 For Compton scattering, if λ0 and λ are the wavelength of the incident photon
and scattered photon,
Δλ = λ − λ0 =
h
mc
(1 − cos θ) = 0.02425(1 − cos 60◦
) Å
= 0.01212 Å
Therefore, λ0 = λ − 0.01212 = 0.312 − 0.012 = 0.3Å
7.58 (a) The energy of the scattered photon through an angle θ is
E1 = E0/[(1 + α(1 − cos θ)] (1)
where α = E0/mc2
For θ = 1800
, E1 = E0/(1 + 2α)
Loss of energy Δ E = E0 − E1 = 2 α E0/(1 + 2 α) (2)
(b) The energy of photon after first scattering through 90◦
by the application
of (1) is E′
1 = E0/(1 + α)
The energy of this photon after the second scattering will be
E′
2 =
E′
1
(1 + α)(1 + α′)
=
E0
1 + 2 α

∵ α′
=
E′
1
mc2
=
E0
1 + α
mc2
=
α
1 + α

The total energy loss ΔE′
= E0 − E0/(1 + 2α) = 2αE0/(1 + 2α) (3)
(c) The energy of photon after the first scattering through 60◦
will be
E1
′′
= 2E0/(2 + α)
The photon energy after second scattering through 60◦
will be

E2
′′
=
2E′′
1
2 + α′′
=
4E0
(2 + α)(4)(1 + α)(2 + α)
=
E0
(1 + α)0
Photon energy after the third scattering through 60◦
will be
E3
′′
=
2E′′
2
2 + α′′
=
2 E0
(1 + α)(2 + 3 α)/(1 + α)
=
2E0
2 + 3 α
∴ Total energy loss
Δ E′′
= E0 − 2E0/(2 + 3 α) = (3αE0)/(2 + 3 α) (4)
Comparison of (2), (3) and (4) shows that the energy loss is equal for
Case (a) and (b), and each is greater than in case (c)
7.59 μc = σcρN0/A
μc x = σc (N0/A) (ρ x) = 8.1 × 10−24
× 6 × 1023
× 3.7/27 = 6.66
μph = 4 × 10−24
× 6 × 1023
× 3.7/27 = 3.288
μx = (μc + μph)χ = 6.66 + 3.288 = 9.948
The intensity will be reduced by
I/I0 = e−μx
= e−9.948
= 4.78 × 10−5
Ratio of intensities absorbed due to the Compton effect and due to the photo-
effect = (1 − exp(−μc x))/(1 − exp(−μph x)) = (1 − e−6.66
)/(1 − e−3.288
) =
1.037
7.60 The change in wavelength in Compton scattering is given by
Δλ = h
mc
(1 − cos θ)
where θ is the scattering angle. Δλ will be maximum for θ = 180◦
in
which case
Δλ(max) = 2h/mc = 4ℏc/mc2
= 4π × 197.3 MeV-fermi/0.511 MeV = 0.0485 Å
7.61 The energy E of the Compton scattered photon by the incident photon of
energy E0 is
E = E0/[(1 + α (1 − cos θ)]
where α = E0/mc2
and θ is the scattering angle.
E/E0 = 1/2 = 1/[(1 + α(1 − cos 45◦
)]
whence α = 3.415
Or E = (3.415((0.511) = 1.745 MeV
7.62 Energy of the scattered gamma rays
E = E0/[1 + α(1 − cos θ)] (1)
Where α = E0/mc2
In e+
− e−
annihilation, each gamma ray has energy E0 = 0.511 MeV
α = 0.511/0.511 = 1
E = 0.511/(2 − cos θ)
Emax is obtained by putting θ = 0 and Emin by putting θ = 180◦
. Thus
Emax = 0.511 MeV
Emin = 0.17 MeV

7.3 Solutions 411
7.63 According to Einsteins’s equation, kinetic energy of the photoelectron
T = h ν − h ν0 (1)
where ν is the frequency of the incident photon and ν0 is the threshold
frequency.
λ = 2,536 × 10−10
m = 253.6 nm
Corresponding energy E = h ν = 1,241/253.6 eV = 4.894 eV
λ0 = 3,250 × 10−10
m = 325 nm
E = h ν0 = 1,241/325 = 3.818
T = 4.894 − 3.818 = 1.076 eV
T = 1/2 mv2
= mc2
v2
/2c2
= 0.511 × 106
× v2
/2 c2
= 1.076
whence v = 2.05 × 10−3
c = 6.15 × 105
m s−1
7.64 The momentum p = 300 Br MeV/c
= 300 × 3.083 × 10−3
= 0.925 MeV/c
E2
= (T + m)2
= p2
+ m2
T = (p2
+ m2
)1/2
− m
Put p = 0.925 and m = 0.511
T = 0.546 MeV
(a) The kinetic energy of the photoelectrons is 0.546 MeV
(b) The energy of the gamma ray photons is 0.546 + 0.116 = 0.662 MeV
7.65 T = h ν − h ν0
h ν = 1,241/253.7 = 4.89 eV
h ν0 = 1,241/325 = 3.81 eV
T = eV = 4.89 − 3.81 = 1.08
The required potential is 1.08 V
7.66 Suppose the photoelectric effect does take place with a free electron due to the
absorption of a photon of energy T . The photoelectron must be ejected with
energy in the incident direction. Energy and momentum conservation give
T = h ν (1)
P = h ν /c (2)
Equation (1) can be written as the relativistic relation connecting momentum
and kinetic energy
T 2
= c2
p2
= T 2
+ 2T mc2
(3)
Using (1) and (2) in (3), we get
2h ν . mc2
= 0
Neither h nor mc2
is zero. We thus end up with an absurd situation. This
only means that both energy and momentum can not be conserved for photo-
electric effect with a free electron.

7.67 Number of gamma rays absorbed in the thickness x cm of lead
N = N0(1 − e−μ x
)
where N0 is the initial number and μ is the absorption coefficient expressed in
cm−1
.
Now μ = σ N0 ρ/A
where N0 is the Avagadro’s number, ρ is the density of lead and A is its atomic
weight.
μ = 20 × 10−24
× 6.02 × 1023
× 11.3/207 = 0.657 cm−1
n/n0 = 90/100 = 1 − e−0.657x
x = 3.5 cm
7.68 The K-shell absorption wavelength in Ag
λk = 0.0424 nm
The corresponding energy
EK(Ag) = 1,241/0.0485 = 25,567 eV
We use the formula
EK(Z) = 13.6(Z − σ)2
For silver 25.567 × 103
= 13.6(47 − σ)2
whence σ = 3.64
For the impurity X of atomic number Z = 50
Ek(Z) = 13.6(50 − 3.64)2
= 29,229 eV
= 29.23 keV
Now the wavelength of 0.0424 nm corresponds to E = 29.24 keV which is
in agreement with the calculated value. Thus the impurity is 50Sn
7.69 eV = hc/λ − W
A plot of V against 1/λ must be a straight line. The slope of the line gives
hc/e, hence h can be determined. The intercept multiplied by hc give W, the
work function. The threshold frequency is given by ν0 = W/h (Fig. 7.16).
Fig. 7.16
0.0
0.5
1.0
1.5
0.0 1.0 2.0 3.0
V

7.3 Solutions 413
V λ × 10−10
m (1/λ) × 106
1.48 3,660 2,732
1.15 4,050 2,469
0.93 4,360 2,293
0.62 4,920 2,032
0.36 5,460 1,831
0.24 5,790 1,727
Slope = 1.24/106
= hc/e
h = 1.24×10−6
×e/c = (1.24×10−6
×1.6×10−19
/(3×108
) = 6.6×10−34
J-s
Intercept = 1.5 × 10−6
m−1
W = hc × intercept = 6.6 × 10−34
× 3 × 108
× 1.5 × 106
= 3 × 10−19
J = 3 × 10−19
/1.6 × 10−19
eV = 1.9 eV
hν0 = W
ν0 = W/h = 3 × 10−19
/6.6 × 10−34
= 4.545 × 1014
λ0 = 6.6 × 10−7
m
7.70 The solid angle subtended at the source
Ω = area of the face of the crystal/square of the distamce from the source
= π R2
/d2
Assuming that the gamma rays are emitted isotropically, the fraction of
gamma rays entering the crystal
F = Ω/4π = R2
/4d2
= 22
/4 × 1002
= 10−4
The number of photons entering the crystal per second from the source of
strength S is
N = SF = 100 × 10−6
× 3.7 × 1010
× 10−4
= 370/s
Number of photons absorbed in the crystal of 1 cm thickness due to photo-
electric effect will be
Nph = N(1 − exp(−μph x) = 370(1 − e−0.03×1
) = 11
and the number absorbed due to Compton scattering will be
Nc = N(1 − exp(−μc x) = 370(1 − e−0.24×1
) = 79
Assuming that each photon that interacts in the crystal produces a photo
electron, Nph and Nc also denote the number of photoelectrons in the respec-
tive processes.
Number of photons that do not interact in 1sec in the crystal is 370 − (11 +
79) = 280
7.71 In the first excited state the ionization energy is 13.6/22
= 3.4 eV
T = h ν − W
h ν = T + W = 10.7 + 3.4 = 14.1 eV
In the ground state, W = 13.6 eV
T = 14.1 − 13.6 = 0.5 eV
7.72 T = h ν − W (Einstein’s equation)
T1 = hc/λ1 − W

T2 = hc/λ2 − W
T2 − T1 = hc(1/λ2 − 1/λ1)
(4.0 − 1.8) × 1.6 × 10−19
J = hc 1
700×10−10 − 1
800×10−10

Solving for h, we get h = 6.57 × 10−34
Js
The accepted value is h = 6.625 × 10−34
J-s
7.73 hν0 = W = 4.7 eV
λ0 = 1,241/4.7 = 264 nm.
7.74 TA = 4.25 − WA (1)
TB = 4.7 − WB (2)
TA
TB
=
P2
A
P2
B
=
λ2
B
λ2
A
= 4 (3)
TB = TA − 1.5 (4)
Solving the above equations,
TA = 2 eV and TB = 0.5 eV
WA = 2.25 eV and WB = 4.2 eV
7.75 The minimum photon energy required for the e−
e+
pair production is 2mc,
where m is the mass of electron. Therefore
h ν = 2mc2
= 2 × 0.511 = 1.022 MeV
The corresponding wavelength
λ = (1, 241/1.022 × 106
eV) nm = 1.214 × 10−12
m.
7.76 In the annihilation process energy released is equal to the sum of rest mass
energy of positron and electron which is 2mc2
. Because of momentum and
energy conservation, the two photons must carry equal energy. Therefore each
gamma ray carries energy
Eγ = mc2
= 0.511 MeV
The wavelength of each photon
λ = 1, 241/(0.511 × 106
)nm = 2.428 × 10−3
nm = 2.428 × 10−12
m
7.77 Let us suppose that the e+
e−
pair is produced by an isolated photon of energy
E = h ν and momentum h ν/c. Let the electron and positron be emitted with
momentum p− and p+, their total energy being E− and E+. Energy conserva-
tion gives
h ν0 = E+ + E− (1)
The momentum conservation implies that the three momenta vectors, p0, p+
and p− must form the sides of a closed triangle, as in Fig. 7.17. Now in any
triangle, any side is equal or smaller than the sum of the other sides. Thus

7.3 Solutions 415
Fig. 7.17
h ν/c ≤ p+ + p− (2)
Or (h ν)2
≤ (cp+ + cp−)2
By virtue of (1)
(E+ + E−)2
≤ c2
p+
2
+ c2
p−
2
+ 2c2
p+ p−
Or E+
2
+ E−
2
+ 2E+ E− ≤ E+
2
− m2
c4
+ E−
2
− m2
c4
+ 2[(E+
2
− m2
c4
)
(E−
2
− m2
c4
)]1/2
E+ E− ≤ [(E+
2
− m2
c4
)(E−
2
− m2
c4
)]1/2
− mc2
Now the left hand side of the inequality is greater than the value of the rad-
ical on the right hand side. It must be still greater than the right hand side. We
thus get into an absurdity, which has resulted from the assumption that both
momentum and energy are simultaneously conserved in this process. Thus an
electron–positron pair cannot be produced by an isolated photon.
7.78 Pμ = 150 MeV/c
Eμ = (p2
μ + m2
μ)1/2
= (1502
+ 1062
)1/2
= 183.67 MeV
βμ = Pµ/Eμ = 150/183.67 = 0.817
nμ = 1/βμ = 1.224
Pπ = 150 MeV/c
Eπ = (P2
π + m2
π )1/2
= (1502
+ 1402
)1/2
= 205.18
βπ = Pπ /Eπ = 150/205.18 = 0.731
nπ = 1/βπ = 1.368
Therefore the range of the index of refraction of the material over which the
muons above give Cerenkov light is 1.224 − 1.368.
7.79 cos θ = 1/β μ
β = 1/μ cos θ = 1/(1.8 × cos 11◦
) = 0.566
γ = (1 − β2
)−1/2
= 1.213
Kinetic energy of proton T = (γ − 1) mp c2
= (1.213 − 1) × 938.3
= 200 MeV

7.80 For electron z = −1. The given integral is easily evaluated assuming that this
refractive index μ is independent of frequency. Integrating between the limits
ν1 and ν2.
−dW/dl = (4 π2
e2
/c2
)(1 − 1/β2
μ2
)[(ν2
2
− ν1
2
)/2] (1)
Calling the average photon frequency as
ν = 1/2 (ν1 + ν2) (2)
the average number of photons emitted per second is given by
N = 1/h v̄ = (4 π2
e2
/hc)(1 − 1/β2
μ2
)(ν2 − ν1)
= (2 π /137)(1 − 1/ β2
μ2
)(1/ λ2 − 1/ λ1) (3)
where λ1 = c/ν1 and λ2 = c/ν2 are the vacuum wavelengths and μ is the
average index of refraction over the wavelength interval λ2 = 4,000 Å to
λ1 = 8,000 Å. Substituting β μ = 0.9 × 1.33 = 1.197,
λ1 = 8×10−5
cm and λ2 = 4×10−5
cm, in (3) we find the number of photons
emitted per cm, N = 173.
7.81 The condition for resonance fluorescence is
ΔEγ /Eγ = v/c
Put Δ Eγ = Γ
Γ = vEγ /c = (1 cm/s/3 × 1010
cm/s) × (129 keV) = 4.3 × 10−6
eV
The mean lifetime
τ = /Δ Eγ = /Γ = 1.05×10−34
/1.6×10−19
×4.3×10−6
= 1.5×10−10
s
h ν = ΔW − ER (1)
PR = h ν /c (2)
Therefore ER = PR
2
/2M = (h ν)2
/2Mc2
(3)
Eliminating ER between (1) and (3), we get a quadratic equation in hν.
(h ν)2
/2Mc2
+ h ν − ΔW = 0
Which has the solution
h ν = −Mc2
+ Mc2
(1 + 2ΔW/Mc2
)1/2
Expanding the radical binomially and retaining up to second power of ΔW,
and simplifying
h ν = ΔW(1 − ΔW/Mc2
)
ν = (ΔW/h)(1 − ΔW/Mc2
)
7.83 The root mean square velocity of 137
Cs atoms
√
v2
= v = (3kT/m)1/2
Substituting k = 1.38 × 10−23
J K−1
= 0.8625 × 10−10
MeV K−1
T = 288 K, mc2
= 137 × 931.5 = 1.276 × 105
, we find v/c = 7.64 × 10−7

7.3 Solutions 417
Optical Doppler effect is given by
h ν = h ν0(1 + β cos θ∗
)
The maximum and minimum energy of photons will be h ν0(1 + β) and
h ν0(1 − β) or 661 keV ± 0.5 eV
7.84 The gravitational red shift is due to the change in the energy of a photon as
it moves from one region of space to another differing gravitational poten-
tial. The photon carries an inertial as well as the gravitational mass given by
h ν/c2
. In its passage from a point where the gravitational potential is ϕ1
to another point where the potential is ϕ2 there will be expenditure of work
given by h ν/c2
times the potential difference (ϕ2 − ϕ1). This would result
in an equivalent decrease in the energy content of the photon and hence its
frequency.
ΔE = E(ϕ2 − ϕ1)/c2
A level difference of H near the earth’s surface would result in the fractional
shift of frequency
Δν/ν = gH /c
Now, Δ ν/ν = v/c = gH /c2
or v = gH /c = 9.8 × 22.6/3 × 108
= 7.38 × 10−7
m/s = 7.38 × 10−4
mm/s
Thus resonance fluorescence would occur for downward velocity of the
absorber of magnitude 7.38 mm/s.
7.85
T 0 4 8 12 16 20 24 28 32 36 40
dN/dt 18.59 13.27 10.68 9.34 8.55 8.03 7.63 7.30 6.99 6.71 6.44
ln(dN/dt) 2.923 2.585 2.368 2.234 2.146 2.083 2.032 1.988 1.944 1.904 1.863
The log-linear plot of dN/dt versus time (t) is not a straight line because
the source contains two types of radioactive material of different half-lives
described by the sum of two exponentials. If the two half-lives are widely dif-
ferent then it is possible to estimate the half-lives by the following
procedure.
Toward the end of the curve (Fig. 7.18), say for time 20–28 h, most of the
atoms of the shorter-lived substance with half-life T1 would have decayed
and the curve straightens up, corresponding to the single decay of longer
half-life of T2. If this straight line is extrapolated back up to the y-axis, then
the half-life T2 can be estimated in the usual way as from the slope of the
curve for a single source on the log-linear plot. In this example,
T2 =
0.693 Δt
ln
dN
dt

0
− ln
dN
dt

t
=
(0.693)(40)
(2.083 − 1.863)
= 63 min
For the shorter-lived substance, the contribution dN2/dt of the source 2
can be subtracted from the observed values in the initial portion of the curve
over suitable time interval and the procedure repeated. In this way we find
T1 = 10 min.

Fig. 7.18
7.86 The flow of heat in a material placed between the walls of a coaxial cylinder
is given by
dQ
dt
=
2π L
ln r2
r1
(T1 − T2) (1)
Number of decays of radon atoms per second
dN/dt = 100 × 10−3
× 3.7 × 1010
= 3.7 × 109
disintegration/second
Energy deposited by α′
s = 3.7 × 109
× 5.5 MeV/s
= 2.035 × 1010
MeV/s = 3.256 × 10−7
J = 0.779 × 10−3
Cal/s
Using the values, k = 0.025 Cal cm−2
s−1
C−1
, L = 5 cm, r1 = 2 mm and
r2 = 6 mm in (1), and solving for (T1 − T2) we find (T1 − T2) = 1.09 ◦
C
7.87 In the series decay A→B→C, if λA λB the transient equilibrium occurs
when
NB/NA = λA/(λB − λA)
Here A = Uranium and B = Radium
λA = 1/τA = 0.693/4.5 × 109
year−1
, λB = 1/τB = 0.693/1,620 year−1
N(Rad)/N(U) ≈ 1,620/4.5 × 109
= 1/2.78 × 106
7.88 Use the law of radioactivity
NA = NA
0
exp(−λA t) (1)
NB = NB
0
exp(−λB t) (2)
Dividing (2) by (1)
NB/NA = exp(λA − λB)t (Because NA
0
= NB
0
)
Take loge on both sides
t =
1
(λA − λB) ln

NB
NA

7.3 Solutions 419
Given NB/NA = 2.2/0.53 = 4.15
λA = 0.693/T1/2 (A) = 0.693/12 = 0.05775 year−1
λB = 0.693/T1/2 (B) = 0.693/18 = 0.0385 year−1
We find age of alloy t = 73.93 years.
7.89 Let W grams of 210
Po be required.
|dN/dt| = Nλ (1)
Required activity |dN/dt| = 10×10−3
×3.7×1010
= 3.7×108
disintegration/
second
N = 6.02 × 1023
W/210 = 2.867 × 1021
W (2)
λ = 0.693/T1/2 = 0.693/138 × 86,400 = 5.812 × 10−8
(3)
Use (2) and (3) in (1) and solve for W to obtain W = 2.22×10−6
= 2.22 µg
7.90 Decay constant, λ = 0.693/100 × 86,400 = 8 × 10−8
s−1
Let M be the gram molar weight of the substance.
Then number of atoms N = N0 M = 6.02 × 1023
M
|dN/dt| = N λ = N0 Mλ = 6.02 × 1023
× 8 × 10−8
M = 4.816 × 1016
M/s
Power P = 4.816 × 1016
M × 5 × 10−14
= 2408 M Watts
But 10% of this power is available, that is 240.8 M Watts. Equating this to
the required power
240.8 M = 5
Or M = 0.02 g-molecules
7.91 The present day activity
|dn/dt| = n0λ e−λt
(1)
n0 = 6.02 × 1023
× 1.35 × 10−10
/12 × 100 = 6.77 × 1010
λ = 3.92 × 10−10
s−1
dn/dt = 12.9/60 = 0.215 s−1
Using these values in (1) and solving for t, we get t = 1.228 × 1010
s or
390 years
λE λF
7.92 RaE → RaF → RaG
The rate of decay of RaE is given by
dNE/dt = −λE NE (1)
where NE is the number of atoms of RaE and λE its decay constant.
The net change of RaF is given by
dNF/dt = λE NE − λF NF (2)
The first term on the right side represents the rate of increase of RaF (Note
the positive sign) and the second term, the rate of decrease (Note the negative
sign). Here NE and NF are the number of atoms of E and F respectively at
time t.

Now at time t
NE = NE
0
exp(−λE t) (3)
Solution of (2) is
NF = A exp(−λE t) + B exp(−λF t) (4)
where A and B are constants which can be determined by the use of the
initial conditions. At t = 0, the initial number of F is zero, that is NF
0
= 0.
Using this condition in (4) gives B = −A and (4) becomes
NF = A[exp(−λE t) − exp(−λF t)] (5)
Further, dNF/dt = −λE A exp(−λE t) + λF A exp(−λF t)
At t = 0
dNF
0
/dt = λE NE
0
= −λE A + λF A
or A = λE NA
0
/(λB − λA) (6)
Using (6) in (5)
NF =
λE N0
E
λF − λE
[exp(−λEt) − exp(−λFt)] (7)
The time at which the greatest number of RaF atoms is obtained by differ-
entiating NF with respect to t in (7) and setting dNF/dt = 0
We find tmax =
1
λF − λE
ln

λF
λE

(8)
λE = 0.693/5 = 0.1386 day−1
, λF = 0.693/138 = 0.0052 day−1
Number of bismuth atoms = 6.02 × 1023
× 5 × 10−10
/210 = 1.43 × 1012
Using these values in (6) and (5) we find
T = 24.6 days
NF(max) = 1.37 × 1010
7.93 In the series decay, A → B → C, if λA λB, that is τA τB, secular
equilibrium is reached and NB/NA = λA/λB = T1/2(B)/T1/2(A)
Here A=Radium and B=Radon
T1/2(Radium) = N(radium).T1/2(Radon)/N(radon)
=

1
6.4 × 10−6

3.825
365

= 1,637 years
7.94 Activity |dN1/dt| = λN1
|dN2/dt| = λ N2 = λ N1 exp(−λt)
Fractional decrease of activity
= [λN1 − λN1 exp(−λt)]/λN1 = 1 − exp(−λt) = 4/100
Or exp(−λt) = 24/25
λ =
1
t
ln

25
24

Put t = 1 h, λ = 0.0408
τ = 1/λ = 24.5 h

7.3 Solutions 421
7.95 Let the proportion of 235
U and 238
U at t = 0 be x : (1 − x). Present day
radioactive atoms for the two components after time t will be
N235 = x N0 exp(−λ235t) (1)
N238 = (1 − x)N0 exp(−λ238t) (2)
Dividing (1) by (2)
N235/N238 = 1/140 =
xe−(λ235−λ238)t
1 − x
(3)
λ235 = 0.693/T1/2, T1/2 = 8.8 × 108
years
λ238 = 0.693/T1/2, T1/2 = 4.5 × 109
years
t = 3 × 109
years
Substituting these values in (3) and solving for x we get x = 1/22
Therefore, the ratio of 235
U and 238
U atoms 3 × 109
years ago was
1/22:21/22 or 1:21
7.96 |dN/dt| = Nλ
N = 1.0 × 10−6
× 6.02 × 1023
/242 = 2.487 × 1015
|dN/dt| = Nλ
80 + 3/3,600 = 2.487 × 1015
× 0.693/T1/2
Solving for T1/2, we find T1/2 = 2.15 × 1013
s or 6.8 × 105
years
For α-decay: 80 = Nλα = 2.487 × 105
λα
λα = 1.01 × 10−6
year−1
For fission 3/3,600 = 2.487 × 1015
λf
λf = 1.05 × 10−11
year−1
7.97 Nsr = Nsr
0
exp(−λsrt) (1)
Ny =
λsr N0
sr
λy − λsr
[exp(λsrt) − exp(−λyt)] (2)
Ny
Nsr
=
λsr
λy − λsr
[1 − exp(λsr − λy)t] (3)
λsr = 0.693/(28 × 365 × 24) = 2.825 × 10−6
h−1
λy = 0.693/64 = 0.0108 h−1
(a) For t = 1 h and using the values for the decay constants Nsr/Ny = 3.56×
105
(b) For t = 10 years, Nsr/Ny = 3,823
7.98 If N is the number of atoms of each component at t = 0, then at time t the
number of atoms of the two components will be
N235 = N0 exp(−λ235 t) (1)
N238 = N0 exp(−λ238 t) (2)
N235/N238 = (0.7/100) = exp −(λ235 − λ238) t (3)

Using λ238 = 1/τ238 = 1/6.52 × 109
, λ235 = 1/τ235 = 1/1.02 × 109
And solving (3) we get the age of the earth as 5.26 × 109
years
7.99 Number of Th atoms in 2 µg is
N = N0 W/A = 6.02 × 1023
× 2 × 10−6
/224 = 5.375 × 1015
λ = 0.693/(3.64 × 86, 400) = 2.2 × 10−6
Activity A = |dN/d λ| = N λ = (5.375 × 1015
)(2.2 × 10−6
) = 1.183 ×
1010
/s
= (1.188 × 1010
/3.7 × 1010
) ci = 0.319 ci
7.100 1 rad = 100 ergs g−1
Energy received = (40)(100)(20 × 103
)/107
J
= 8 J
= 8/4.18 = 1.91 Cal
7.101 Let V cm3
be the volume of blood, Initial activity is 16,000/V per minute
per cm3
. 0.8 = 16,000
V
exp(−0.693 × 30/15)
= 4,123
V
∴ V = 5000 cm3
7.3.10 Alpha-Decay
7.102 λ1 = 1021
exp(−2π zZ1e2
/ v1)
λ2 = 1021
exp(−2 πzZ2 e2
/v2)
Given λ1 = λ2
Z1/
√
E1 = Z2/
√
E2
E2 = E1 Z2
2
/Z1
2
= 5.3(80/82)2
= 5.04 MeV
Thus energy from the second nuclei is 5.04 MeV.
7.103 The energy required to force an α-particle(classically) into a nucleus of
charge Ze is equal to the potential energy at the barrier height and is given by
E = zZe2
/4 π ε0 R.
This is barely possible when α particle and the Uranium nucleus will just
touch each other and the kinetic energy of the bombarding particle is entirely
converted into potential energy.
R = R1 + R2 = r(A
1/3
1 + A2
1/3
)
= 1.2(41/3
+ 2381/3
) = 9.34 fm
E = 1.44zZ/R = 1.44 × 2 × 92/9.34 = 28.37 MeV
7.104 Geiger–Nuttal law is
log λ = k log x + c
where k and c are constants, λ is the decay constant and x is the range
log(0.693/T ) = k log x + c
k log x + log T = log 0.693 − c = c1 (1)

7.3 Solutions 423
where c1 is another constant.
k log 3.36 + log(1622 × 365) = c1 (2)
k log 3.85 + log 138 = c1 (3)
Solving (2) and (3), k = 61.46 and c1 = 36.5. Using the values of k and c1
in (1) 61.46 log 6.97 + log T = 36.5
Solving for T , we find T = 4.79 × 10−16
days = 4.14 × 10−11
s
7.105 For α-decay we use the equation
λ = 1/ τ = 1021
exp(−2 π zZ/137 β) (1)
T1/2(1)/T1/2(2) = τ1/τ2 = exp(2 π zZ1/137 β1)/ exp(2 π zZ2/137 β2) (2)
For 88Ra226
decay put Z1 = 86 for the daughter nucleus, z = 2 for α-particle
and β1 = (2E/Mc2
)1/2
= (2 × 4.9/3728)1/2
= 0.05127
For 90Th226
decay, put Z2 = 88 for daughter nucleus and z = 2 for α-particle
and β2 = (2 × 6.5/3728)1/2
= 0.05905
Using the values in (2), we find
τ1/τ2 = 5.19 × 107
7.3.11 Beta-Decay
7.106 The selection rules for allowed transitions in β-decay are:
Δ I = 0
Ii = 0 → If = 0 allowed
Δ π = 0
⎫
⎬
⎭
Fermi Rule
Δ I = 0, ±1
Ii = 0 → Ii = 0 forbidden
Δ π = 0
⎫
⎬
⎭
G.T. Rule
where I is the nuclear spin and π is the parity. In view of the above selection
rules the first transition is Fermi transition, the second one Gamow–Teller
transition. The third one occurs between two mirror nuclei 17
F and 17
O in
which the proton number and neutron number are interchanged. The con-
figuration of nucleus is very much similar in such nuclei, consequently the
wave functions are nearly identical. This leads to a large value for the overlap
integral. The log ft value for such transitions is small being in the range of
3–3.7. These are characterized by
ΔI = 0,±1 and Δπ = 0. The given transition 17
F →17
O is an example of
superallowed transition.
7.107 If the number of 23
Mg nuclides is N0 at t = 0, then at time t the number
decayed will be
N = N0[1 − exp(−λ t)] (1)
At t = t1, N1 = N0[1 − exp(−λ t)] (2)
At t = t2, N2 = N0[1 − exp(−λ t)] (3)

Dividing (3) by (2)
N2/N1 = [1 − exp(−λt2)]/[1 − exp(−λ t1)]
N2/N1 = 2.66 = [1 − exp(−6 λ)]/[1 − exp(−2λ)]
= (1 − x3
)/(1 − x) = 1 + x + x2
(4)
where x = exp(−2λ)
Solving the quadratic equation for x, we find x = 0.885 = exp(−2λ)
whence the mean life time τ = 1/λ = 15.93 s
7.108 According to Fermi’s theory of β-decay, for E0 mec2
, the decay constant
λ =
G2
|Mif|2
E5
0
60π3(c)6
(1)
So that with the value G/3
c3
= 1.166 × 10−5
GeV−2
, (1) becomes λ =
1.11E5
0 |Mif|2
104 S−1
= 1.11×(3.5)5
×6
104
= 0.3466 s−1
Therefore, T1/2 = 0.693/0.3466 = 2.0 s. The experimental value is 0.8 s.
7.109 The mean energy of electrons
E =
E max
0
E n(E)dE/
E max
0
n(E) dE
Given n(E)dE = c
√
E(Emax − E)2
dE
where c = constant
E =
c
E max
0 E
√
E(Emax − E)2
dE
c
E max
0
√
E(Emax − E)2 dE
=
Emax
3
If all the electrons emitted are absorbed then the kinetic energy of the
electrons is converted into heat.
Heat evolved/sec = (mean energy)(no. of electrons emitted/second)
= (0.156) × (3.7 × 107
)/3 MeV/s = 1.92 × 106
MeV/s
7.110 Let the threshold energy be Eν. In that case the particles in the CMS will be
at rest. Now, in the neutron decay
n → p + e−
+ ν̄ + 0.79 MeV
Mn − (Mp + Me) = 0.79
as ν̄ has zero rest mass
Mn + Me = Mp + 2me + 0.79 = Mp + 1.81
(the masses of electron and positron are identical, each equal to 0.511 MeV)
We use the invariance of
(Σ E)2
− |ΣP|2
= (Σ E∗
)2
− zero
where (∗
) refers to the CMS and total momentum of particles in CMS is zero
(Ev + Mp)2
= (Mn + Me)2
= (Mp + 1.81)2
Or Eν = 1.81 MeV.

7.3 Solutions 425
7.111 i. Momentum = 300 Br = (300)(0.1)(0.061) = 1.83 MeV/c
Use the relativistic equation to find total energy, E = (p2
+ m2
)1/2
=
(1.832
+ 0.5112
)1/2
= 1.90 MeV
Kinetic energy of electron
T = E − m = 1.90 − 0.51 = 1.39 MeV = 1.39 × 106
eV
Neglecting the energy of the recoiling nucleus the neutrino energy
Eν = 2.4 − 1.39 = 1.01 MeV
ii. Maximum kinetic energy will be carried by the nucleus when the nucleus
recoils opposite to β-particle and the neutrino is at rest. Momentum
conservation gives
PN
2
= 2MNTN = Pe
2
= Te
2
+ 2Te m (1)
TN + Te = 2.4 (2)
Eliminating Te, we find TN = 5 keV.
7.112 n(E) = C
√
E(Emax − E)2
where C = constant.
Differentiate n(E) with respect to E and set dn/dE = 0. We find that the
maximum occurs at E = Emax/5.

Chapter 8
Nuclear Physics – II
Nuclear models
Fermi gas model
Total kinetic energy of all protons
EZ =
3
5
Z EZ(max); EZ(max) =
K2
2M

Z
A
2
3
(8.1)
Total energy of all neutrons.
EN =
3
5
N EN (max); EN(max) =
K2
2M

N
A
2
3
(8.2)
where K =

r0

9π
4
1
3
(8.3)
Shell model Magic Numbers: 2, 8, 20, 28, 50, 82, 126. Table 8.1 shows the shell
number (Λ) occupation number (Ns) and the Number of neutrons and protons in
various shells (ΣNs) under the spin – orbit coupling scheme.
Table 8.1
Λ Ns ΣNs
5 N′
Λ = 44 126
4 N′
Λ = 32 82
3 N′
Λ = 22 50
3 2(Λ + 1) = 8 28
2 NΛ = 12 20
1 NΛ = 6 8
0 NΛ = 2 2
427

428 8 Nuclear Physics – II
The shell model energy levels are:
*
1s1
2
+ *
1p3
2
, 1p1
2
+ *
1d5
2
, 2s1
2
, 1d3
2
+ *
1 f 7
2
+ *
2p3
2
, 1 f 5
2
, 2p1
2
, 1g9
2
+
*
1g7
2
, 2d5
2
, 2d3
2
, 3s1
2
, 1h 11
2
+
. . . (8.4)
Liquid drop model (8.5)
M(atom) = Z MH + (A − Z)Mn − Δ (8.6)
Δ = mass defect
P =
M − A
A
= packing fraction. (8.7)
1 amu =
1
12
of atomic mass of 12
C atom (8.8)
f =
B.E
A
(8.9)
1 amu = 931.5 Mev (8.10)
1 amu = 1.66 × 10−27
kg (8.11)
The f-A curve is shown in Fig. 8.1. A more detailed diagram is shown in
Problem 8.40
Fig. 8.1 BE/A Versus A
Stability against decay
β−
− decay : M(Z, A) ≤ M(Z + 1, A) (8.12)
β+
− decay : M(Z + 1, A) ≤ M(Z, A) + 2Me (8.13)
e−
capture : M(Z + 1, A) ≤ M(Z, A) (8.14)
α − decay : M(Z, A) ≤ M(Z − 2, A − 4) + MHe4 (8.15)
Assuming that γ-ray precedes the decay, the energy released
Qβ− = [M(Z, A) − M(Z + 1, A)] c2
= Tmax + Tγ (8.16)
Qβ+ = [M(Z + 1, A) − M(Z, A)] c2
= 2Me c2
+ Tmax + Tγ (8.17)
QEC = [M(Z + 1, A) − M(Z, A)] c2
= Tv + Tγ (8.18)

Charge symmetry of nuclear forces: p − p = n − n force
Charge Independence of nuclear forces: p − p = p − n = n − n force
Isospin: A fictitious quantum number (T) which is used in the formalism of charge
independence The charge
Q
e
= T3 +
B
2
(8.19)
where Q
e
is the charge of the particle in terms of electron charge, T3 is the third
component of T , and B is the baryon number.
Thus, n and p of similar mass form an isospin doublet of nucleon. For proton,
T3 = +1
2
and for neutron T3 = −1
2
. For strong interaction of other particles refer to
Chap. 10. Although there is no connection between isospin and ordinary spin, the
algebra is the same. For a system of particles, the notation for isospin will be I.
Nuclear spin (J)
Odd A nuclei : J =
1
2
,
3
2
,
5
2
, . . . (8.20)
Even A nuclei : J = 0, 1, 2, . . . (8.21)
Nuclear parity: By convention n and p are assigned even(+) intrinsic parity. In
addition parity comes from the orbital angular momentum (l) and is given by (−1)l
.
Thus, for deuteron which is mainly in the s-state, this part of parity is +1. Parity
is multiplicative quantum number, so that for deuteron, P = +1.
Hyperfine structure of spectral lines
Fine structure of spectral lines is explained by the electron spin, while the hyperfine
structure is accounted for by the nuclear spin.
Nuclear magnetic moment (µ)
μ = gJ

e
2mpc

(8.22)
where J is the nuclear spin and g is the nuclear g factor.
In Rabi’s experiment the resonance technique is used. In a constant magnetic
field B, the magnetic moment precesses with Larmor’s frequency ν given by
v =
μB
Jh
(8.23)
If an alternating magnetic field of frequency f is superimposed there will be a
dip in the resonance curve when
v = f. (8.24)

Electric quadrupole moment
For nuclei with J = 0 or 1
2
, electric dipole moment is zero. For spherical nuclei
quadrupole moment is also zero. Only in non-spherical nuclei the quadrupole
moment (Q) exists.
The concept of quadrupole comes from the classical electrostatic potential theory
(Fig. 8.2).
Φ(r, θ) =
1
r
Σ∞
n=0
an
rn
Pn(cos θ) (8.25)
The quantity 2a2/e is known as the quadrupole.
Q
e
=

ψ∗
(3z2
− r2
)ψ dτ (8.26)
Fig. 8.2 Electric quadrupole
moment for nuclei of various
shapes
Nuclear reactions
x(a, b)y a + x → b + y
Q = (mx + ma−my − ma) c2
= Eb + Ey − Ea (8.27)
Q = Eb(1 + mb/my) − Ea(1 − ma/my) −
2
my
(mamb Ea Eb)
1
2 cos θ (8.28)
Exoergic reactions: Q = +ve
Endoergic reactions: Q = −ve;
Ea (threshold) = | − Q|(1 + ma/mx ) (8.29)
Elastic Scattering: Q = zero

Table 8.2 Comparison of compound nucleus reactions and direct reactions
Compound nuclear
Feature reactions Direct reactions
Times involved 10−14
–10−16
s ∼10−20
–10−21
s
Dominance of
reaction
Low energy High energy
Nature of reaction Surface
phenomenon
Nuclear interior
Cross - section ∼Barn ∼Millibarn
Angular distribution Isotropic in the
CMS
Peaked in the forward
hemisphere
Location of peaks Energy dependent Orbital angular
momentum
dependent
Fig. 8.3 Kinematics
of a collision
Inverse reactions and the reciprocity theorem
σ(b → a)
σ(a → b)
=
(2Ix + 1)(2Ia + 1)
(2Iy + 1)(2Ib + 1)
.
p2
a
p2
b
(8.30)
At the same CM energy.
σl
S =
π
k2
(2l + 1)|1 − ηl|2
(8.31)
σl
r =
π
k2
(2l + 1)(1 − |ηl |2
) (8.32)
The Breit – Wigner formulae – Reactions via compound nucleus formation.
σS =
π -
λ2
Γ2
s .g
(E − ER)2 + (Γ/2)2
(8.33)
σr =
π -
λ2
ΓRΓS.g
(E − ER)2 + (Γ/2)2
(8.34)
σt = σs + σr =
π -
λ2
ΓsΓ.g
(E − ER)2 + (Γ/2)2
(8.35)

Fig. 8.4 Resonance and
potential sacttering and
absorption cross-sections as a
function of neutron energy
where g =
(2Ic + 1)
(2Ia + 1)(2Ix + 1)
(8.36)
Γaτa = ; Γbτb = (8.37)
σel =
π
k2
%
%Ares + Apot
%
%2
(8.38)
with Ares =
iΓn
(En − ER) + 1
2
iΓ
(8.39)
and Apot = exp(2ikR) − 1 (8.40)
Optical model
V = −U − iW (8.41)
U = vk1 (8.42)
W =

2
vk (8.43)
where v =

2E/m (8.44)
is the velocity of the incident nucleon.
k = (2mE)1/2
/; (8.45)
k1 = [2m(E + U)]1/2
/ (8.46)
K =
1
λ
=
A σ
vol
=
3σ
4πr3
0
(8.47)
Direct reactions (Reactions without compound nucleus formation)
1. Inelastic scattering
2. Charge exchange reactions
3. Nucleon transfer reactions

4. Breakup reactions
5. Knock-out reactions
Pre-equilibrium reactions
Fig. 8.5 Energy spectrum of
particles emitted in various
types of reactions
The pre-equilibrium reactions take place before the formation of the compound
nucleus.
Heavy Ion reactions
Fig. 8.6 Collision between
heavy ions
Types of interactions
(a) The coulomb region with rmin RN , where RN is the distance for which the
nuclear interactions are ineffective.
(b) The deep inelastic and the incomplete fusion region with rmin = R1 + R2.
(c) The fusion region with 0 ≤ rmin ≤ R1 + R2.

If Ecm is the center of mass energy of the two interacting ions then the minimum
distance of approach is given by
rmin =
b
$
1 − V (rmin)
Ecm
(8.48)
8.2 Problems
8.2.1 Atomic Masses and Radii
8.1 Singly-charged lithium ions, liberated from a heated anode are accelerated by a
difference of 400 V between anode and cathode. They then pass through a hole
in the cathode into a uniform magnetic field perpendicular to their direction of
motion. The magnetic flux density is 8×10−2
Wb/m2
and the radii of the paths
of the ions are 8.83 and 9.54 cm, respectively. Calculate the mass numbers of
the lithium isotopes.
8.2 A narrow beam of singly charged 10
B and 11
B ions of energy 5 keV pass through
a slit of width 1 mm into a uniform magnetic field of 1,500 gauss and after a
deviation of 180◦
the ions are recorded on a photographic plate
(a) What is the spatial separation of the images.
(b) What is the mass resolution of the system?
8.3 Singly charged chlorine ions are accelerated through a fixed potential differ-
ence and then caused to travel in circular paths by means of a uniform field of
magnetic induction of 1,000 gauss. What increase in induction is necessary to
cause the mass 37 ion to follow the path previously taken by the mass 35 ion?
8.4 27
14Si and 27
13Al are mirror nuclei. The former is a positron emitter with Emax =
3.48 MeV. Determine r0.
8.5 Use the uncertainty relation to estimate the kinetic energy of the nucleon, the
nuclear radius is about 5 × 10−13
cm and the mass of a proton is about 2 ×
10−24
g.
8.6 14
O is a positron emitter decaying to an excited state of 14
N. The 14
N
γ - rays have an energy of 2.313 MeV and the maximum energy of the positron
is 1.835 MeV. The mass of 14
N is 14.003074 amu and that of electron is
0.000548 amu. Find the mass of 14
O in amu.

8.2 Problems 435
8.2.2 Electric Potential and Energy
8.7 Derive an expression for the electrostatic energy of a spherical nucleus of radius
R assuming that the charge q = Ze is uniformly distributed homogeneously in
the nuclear volume.
8.8 A charge q is uniformly distributed in a sphere of radius R. Obtain an expres-
sion for the potential V (r) at a point distant r from the centre (r R).
8.2.3 Nuclear Spin and Magnetic Moment
8.9 Suppose a proton is assumed to be a classical particle rotating with angular
velocity of 2.6 × 1023
rad/s about its axis. If it posseses a rotational energy of
537.5 MeV, then show that it has angular momentum equal to h.
8.10 (a) A D5/2 term in the optical spectrum of 39
19K has a hyperfine structure with
four components. Find the spin of the nucleus.
(b) In (a) what interval ratios in the hyperfine quadruplet are expected?
8.11 Given that the proton has a magnetic moment of 2.79 magnetons and a spin
quantum number of one half, what magnetic field strength would be required
to produce proton resonance at a frequency of 60 MHz in a nuclear magnetic
resonance spectrometer?
8.12 In a nuclear magnetic experiment for the nucleus 25Mn55
of dipole moment
3.46 μN , the magnetic field employed is 0.8 T. Find the resonance frequency.
You may assume J = 7/2, μN = 3.15 × 10−14
MeV T−1
8.2.4 Electric Quadrupole Moment
8.13 Show that for a homogeneous ellipsoid of semi axes a, b the quadrupole
moment is given by Q = 2
5
Ze(a2
− b2
)
8.14 Estimate the ratios of the major to minor axes of 181
73 Ta and 123
51 Sb. The
quadrupole moments are +6 × 10−24
cm2
for Ta and −1.2 × 10−24
cm2
for
Sb (Take R = 1.5 A1/3
fm)
[Saha Institute 1964]
8.15 Show that the electric quadrupole moment of a nucleus vanishes for
(a) Spherically symmetric charge distribution
(b) Nuclear spin I = 0 or I = 1/2
8.16 Show that for a rotational ellipsoid of small eccentricity and uniform charge
density the quadrupole moment is given by Q = 4
5
ZRΔR. Assuming that the
quadrupole moment of 181
71 Ta is 4.2 barns, estimate its size.

8.2.5 Nuclear Stability
8.17 In general only the heavier nuclei tend to show alpha decay. For large A it is
found that B/A = 9.402 − 7.7 × 10−3
A.
Given that the binding energy of alpha particles is 28.3 MeV, show that
alpha decay is energetically possible for A 151.
[University of Wales, Aberystwyth, 2003]
8.18 For the nucleus 16
O the neutron and proton separation energies are 15.7 and
12.2 MeV, respectively. Estimate the radius of this nucleus assuming that the
particles are removed from its surface and that the difference in separation
energies is due to the Coulomb potential energy of the proton.
8.19 By considering the general conditions of nucleus stability show that the
nucleus 229
90 Th will decay and decide whether the decay will take place by
α or β emission.
The atomic mass excesses of the relevant nuclei are:
Element 4
2He 225
88 Ra 229
89 Ac 229
90 Th 229
91 Pa
Mass excess amu ×10−6
2,603 23,528 32,800 31,652 32,022
8.20 The masses of 64
28Ni, 64
29Cu and 64
30Zn are as tabulated below. It follows that 64
29Cu
is radioactive. Detail the various possible decay modes to the ground state
of the daughter nucleus and give the maximum energy of each component,
ignoring the recoil energy. Compute the recoil energy in one of the cases and
verify that it was justifiable to ignore it.
Nuclide Atomic mass
64
28Ni 63.927959
64
29Cu 63.929759
64
30Zn 63.929145
8.21 28
13Al decays to 28
14Si via β−
emission with Tmax = 2.865 MeV. 28
14Si is in
the excited state which in turn decays to the ground state via γ -emission.
Find the γ -ray energy. Take the masses 28
Al = 27.981908 amu, 28
Si =
27.976929 amu.
8.22 22
11Na decays to 22
10Ne via β+
with Tmax = 0.542 MeV followed by γ -decay
with energy 1.277 MeV. If the mass of 22
Ne is 21.991385 amu, determine the
mass of 22
Na in amu.
8.23 7
4Be undergoes electron capture and decays to 7
3Li. Investigate if it can decay
by the competitive decay mode of β+
emission. Take masses 7
4Be = 7.016929
amu, 7
3Li = 7.016004 amu.

8.2 Problems 437
8.2.6 Fermi Gas Model
8.24 In the Fermi gas model the internal energy is given by U =
3
5
AEF, where
A is the mass number and EF is the Fermi energy. For a nucleus of volume
V with N = Z = A/2. A = KVEF
3/2
where K is a constant. Using the
thermodynamic relation, p = −

∂U
∂V

s
, show that the pressure is given by
p =
2
5
ρn EF, where ρn is the nucleon density.
8.25 Assuming that in a nucleus N = Z = A/2, calculate the Fermi momentum,
Fermi energy EF, and the well depth.
8.2.7 Shell Model
8.26 A certain odd-parity shell model state can accommodate up to a maximum of
12 nucleons. What are its j and l values?
8.27 The shell model energy levels are in the following way
[1s1/2][1p3/2, 1p1/2][1d5/2, 2s1/2, 1d3/2][1 f7/2][2p3/2, 1 f5/2, 2p1/2, 1g9/2]
[1g7/2, 2d5/2, 2d3/2, 3s1/2, 1h11/2] . . .
Assuming that the shells are filled in the order written, what spins and pari-
ties should be expected for the ground state of the following nuclei?
7
3Li, 16
8 O, 17
8 O, 39
19K, 45
21Sc.
8.28 Find the gap between the 1p1/2 and 1d5/2 neutron shells for nuclei with mass
number A ≈ 16 from the total binding energy of the 15
O (111.9556 MeV),
16
O (127.6193 MeV) and 17
O (131.7627 MeV) atoms.
8.29 Compute the expected shell-model quadruple moment of 209
Bi(9/2
−
)
8.30 From the shell model predictions find the ground state spin and parity of the
following nuclides:
3
2He; 20
10Ne; 27
13Al; 41
21Sc;
8.31 Making use of the shell model, write down the ground state configuration of
protons and neutrons for 12
6 C and the next three isotopes of increasing A. Give
the spin and parity assignment for the ground state of 13
6 C and compare this
with the equivalent assignments for the ground state of 15
6 C.
8.32 How does the shell model predict
7−
2
for the ground state spin parity of 41
20Ca.
What does the model predict about the spin and parities of the ground states
of 30
14Si and 14
7 N?

8.2.8 Liquid Drop Model
8.33 Deduce that with ac = 0.72 MeV and as = 23 MeV the ratio zmin/A is approx-
imately 0.5 for light nuclei and 0.4 for heavy nuclei.
[Royal Holloway, University of London 1998]
8.34 Determine the most stable isobar with mass number A = 64.
8.35 The masses (amu) of the mirror nuclei 27
13Al and 27
14Si are 26.981539 and
26.986704 respectively. Determine the Coulomb’s coefficient in the semi
emperical mass formula.
8.36 If the binding energies of the mirror nuclei 41
21Sc and 41
20Ca are 343.143 V and
350.420 MeV respectively, estimate the radii of the two nuclei by using the
semi empirical mass formula [e2
/4πε0 = 1.44 MeV fm]
8.37 The empirical mass formula (neglecting a term representing the odd – even
effect) is M(A, Z) = Z(mp + me) + (A − Z)mn − αA + β A2/3
+ γ (A −
2Z)2
/A + εZ2
A−1/3
where α, β, γ and ε are constants. By finding the min-
imum in M(A, Z) for constant A obtain the expression Zmin = 0.5A(1 +
0.25A2/3
ε/γ )−1
for the value of Z which corresponds to the most stable
nucleus for a set of isobars of mass number A.
[Royal Holloway, University of London 1998]
8.38 The binding energy of a nucleus with atomic number Z and mass number A
can be expressed by Weisacker’s semi – empirical formula
B = av A − as A2/3
−
ac Z2
A1/3
− aa
(N − Z)2
A
−
ap
A1/2
where av = 15.56 MeV, as = 17.23 MeV,
ac = 0.697 MeV, aa = 23.285 MeV, ap = −12, 0, 12 MeV
for even-even, odd-even (or even-odd) or odd-odd nucleus respectively.
Estimate the energy needed to remove one neutron from nucleus 40
20Ca.
8.39 (a) Consider the alpha particle decay 230
90 Th →226
88 Ra + α and use the follow-
ing expression to calculate the values of the binding energy B for the two
heavy nuclei involved in this process.
B = av A − as A2/3
− ac
Z(Z − 1)
A1/3
− aa
(N − Z)2
A
− ap A−3/4
where values for the constants av, as, ac, aa and ap are respectively 15.5,
16.8, 0.72, 23.0 and −34.5 MeV. Given that the total binding energy of the
alpha particle is 28.3 MeV, find the energy Q released in the decay.
(b) This energy appears as the kinetic energy of the products of the decay.
If the original thorium nucleus was at rest, use conservation of momen-
tum and conservation of energy to find the kinetic energy of the daughter
nucleus 226
Ra.

8.2 Problems 439
8.40 The following plot (Fig. 8.7) shows the average binding energy per nucleon
for stable nuclei as a function of mass number.
Explain how the mass of a nucleus can be calculated from this plot and esti-
mate the mass of 235
92 U.
Briefly describe the main features of the plot in the context of nuclear
models such as the liquid Drop Model, the Fermi Gas Model and the Nuclear
Shell Model.
In terms of the Liquid Drop Model, explain why nuclear fission and fusion
are possible and estimate the energy released when a nucleus of 235
92 U under-
goes fission into the fragments 87
35Br and 145
57 La with the release of three prompt
neutrons.
Fig. 8.7 B/A versus A plot
8.41 Investigate using liquid drop model, the β stability of the isobars 127
53 I and
127
54 Xe given that
127
51 Sb → β−
+127
52 Te + 1.60 MeV
127
55 Cr → β+
+127
54 Xe + 1.06 MeV
8.42 The empirical mass formula is
A
Z M = 0.99198 A − 0.000841 Z + 0.01968 A2/3
+ 0.0007668 Z2
A−1/3
+0.09966(Z − A/2)2
A−1
− δ
in atomic mass units, where δ = ±0.01204 A−1/2
or 0.
Determine whether or not the nuclide 27
12Mg is stable to β decay.
[University of Newcastle 1966]
8.2.9 Optical Model
8.43 Show that the imaginary part of the complex potential V = −(U + iW) in the
optical model has the effect of removing particle flux from the elastic channel.

8.44 For neutrons with kinetic energy 100 MeV incident on nuclei with mass num-
ber A = 120, the real and imaginary parts of the complex potential are approx-
imately −25 and −10 MeV, respectively. On the basis of these data, estimate
(i) the deBroglie wavelength of the neutron inside the nucleus
(ii) the probability that the neutron is absorbed in passing through the nucleus
8.2.10 Nuclear Reactions (General)
8.45 13
N is a positron emitter with an end point energy of 1.2 MeV. Determine the
threshold of the reaction p +13
C →13
N + n, if the neutron – hydrogen atom
mass difference is 0.78 MeV.
8.46 The reaction, p +7
3 Li →7
4 Be + n, is known to be endothermic by 1.62 MeV.
Find the total energy released when 7
Be decays by K capture and calculate the
energy carried off by the neutrino and recoil nucleus, respectively.
(Mpc2
= 938.23 MeV, Mnc2
= 939.52 MeV, Mec2
= 0.51 MeV)
8.47 If a target nucleus has mass number 24 and a level at 1.37 MeV excita-
tion, what is the minimum proton energy required to observe scattering from
this level.
8.48 The nuclear reaction which results from the incidence of sufficiently energetic
α-particles on nitrogen nuclei is 4
2He +14
7 N → X +1
1 H. What is the decay
product X? What is the minimum α-particle kinetic energy (in the laboratory
frame) required to initiate the above reaction?
(Atomic masses in amu: 1
H = 1.0081; 4
He = 4.0039; 14
N = 14.0075; X =
17.0045)
8.49 The Q values for the reactions 2
H(d, n) 3
He and 2
H(d, p) 3
H are 3.27 MeV and
4.03 MeV, respectively. Show that the difference between the binding energy
of the 3
H nucleus and that of the 3
He nucleus is 0.76 MeV and verify that this
is approximately the magnitude of Coulomb energy due to the two protons of
the 3
He nucleus. (Distance between the protons in the nucleus 31/3
× 1.3 fm).
8.50 Thermal neutrons are captured by 10
5 B to form 11
5 B which decays by α-particle
emission to Li. Write down the reaction equation and calculate
(a) The Q-Value of the decay in MeV
(b) The Kinetic energy of the α-particles in MeV.
(Atomic masses: 10
5 B = 10.01611 amu; 1
0n = 1.008987 amu; 7
3Li
= 7.01822 amu; 4
2He = 4.003879 amu; 1 amu = 931 MeV)

8.2 Problems 441
8.51 Consider the reaction 27
Al(p, n)27
Si. The positrons from the decay of 27
Si are
observed and their spectrum is found to have an end – point energy of 3.5 MeV.
Derive the Q value of the (p, n) reaction, and find the threshold proton energy
for the reaction given that the neutron – proton mass difference is 0.8 MeV.
8.52 The Q value of the 3
H(p, n)3
He reaction is −0.764 MeV. Calculate (a) the
threshold energy for appearance of neutrons in the forward direction (b) the
threshold for the appearance of neutron in the 90◦
direction.
[University of Liverpool 1960]
8.53 When 30
Si is bombarded with a deuteron, 31
Si is formed in its ground state
with the emission of a proton. Determine the energy released in this reaction
from the following information:-
31
Si →31
p + β−
+ 1.51 MeV
30
Si + d → 31
p + n + 5.10 MeV
n → p + β−
+ ν + 0.78 MeV
8.54 The nucleus 12
C has an excited state at 4.43 MeV. You wish to investigate
whether this state can be produced in inelastic scattering of protons through
90◦
by a carbon target. If you have access to a beam of protons of kinetic
energy 15 MeV, what is the kinetic energy of the scattered protons for which
you must look?
8.55 A thin hydrogenous target is bombarded with 5 MeV neutrons, and a detec-
tor is arranged to collect those protons emitted in the same direction as the
neutron beam. The neutron beam is replaced by a beam of γ -rays; calculate
the photon energy needed to produce protons of the same energy as with the
neutron beam.
8.56 Protons of energy 5 MeV scattering from 10
5 B at an angle of 45◦
show a peak
in the energy spectrum of the scattered protons at an energy of 3.0 MeV
(a) To what excitation energy of 10
5 B does this correspond?
(b) What is the expected energy of the scattered protons if the scattering is
elastic?
8.57 Calculate the energy of protons detected at 90◦
when 2.1 MeV deutrons are
incident on 27
Al to produce 28
Al with an energy difference Q = 5.5 MeV.
[Royal Holloway University of London 1998]
8.58 The reaction 2
H +1
H →3
He + γ + 5.3 MeV occurs with the deuterons and
proton at rest. Estimate the energy of the helium nucleus.
If the reaction occurs in a region of the sun where the temperature is about
1.7 × 107
K, estimate how close the deuteron and proton must approach for
fusion to occur.

8.59 The Q- value of the reaction 16
O(d, n)17
F is −1.631 MeV, while that of the
reaction 16
O(d, p)17
O is +1.918 MeV. Which is the unstable member of the
pair 17
O−17
F, and what is the maximum energy of β-particles it emits? (n−1
H
mass difference is 0.782 MeV)
8.60 An aluminum target is bombarded by α-particles of energy 7.68 MeV, and the
resultant proton groups at 90◦
were found to possess energies 8.63, 6.41, 5.15
and 3.98 meV. Draw an energy level diagram of the residual nucleus, using the
above information.
8.61 If the Q-value for the 3
H(p, n)3
He reaction is −0.7637 MeV and tritium (3
H)
emits negative β-particles of end point energy 18.5 KeV, calculate the differ-
ence in mass between the neutron and the hydrogen atom.
[Andhra University]
8.62 The reaction 3
H(d, n)4
He has Q value of 17.6 MeV. What is the range of neu-
tron energies that may be obtained from this reaction for an incident deuteron
beam of 300 KeV?
8.63 A target of 181
Ta is bombarded with 5 MeV protons to form 182
W in an excited
state. Calculate the energy of the excited state (ignore the coulomb barrier and
assume the target nuclei at rest). If 182
W in the same excited state were pro-
duced by bombarding a hydrogen target with energetic Ta nuclei, what energy
would be needed? The atomic masses in amu are: 1
H1 = 1.007825; 181
73 Ta =
180.948007; 182
74 W = 181.948301
8.64 In the reaction 48
Ca +16
O →49
Sc +15
N, the Q-value is −7.83 MeV. What is
the minimum kinetic energy of bombarding 16
O ions to initiate the reaction?
At this energy, estimate the orbital angular momentum in units of of the ions
for grazing collision. Take R = 1.1 A1/3
fm.
8.2.11 Cross-sections
8.65 Calculate the thickness of Indium foil which will absorb 1% of neutrons inci-
dent at the resonance energy for Indium (1.44 eV) where σ = 28, 000 barns.
At. Wt of Indium = 114.7 amu, density of Indium = 7.3 g/cm3
.
[Andhra University]
8.66 60
Co is produced from natural cobalt in a reactor with a thermal neutron flux
density of 5 × 1012
n cm−2
s−1
. Determine the maximum specific activity.
Given σact = 20 b.

8.2 Problems 443
8.67 Natural Cobalt is irradiated in a reactor with a thermal neutron flux density of
3 × 1012
n cm−2
s−1
. How long an irradiation will be required to reach 20%
of the maximum activity? Given T1/2 = 5.3 years
8.68 In a scattering experiment an aluminum foil of thickness 10 µm is placed in
a beam of intensity 8 × 1012
particles per second. The differential scattering
cross-section is known to be of the form
dσ
dΩ
= A + B cos2
θ
where A, B are constants, θ is scattering angle and Ω is the solid angle.
With a detector of area 0.01 m2
placed at a distance of 6 m from the foil, it is
found that the mean counting rate is 50 s−1
when θ is 30◦
and 40 s−1
when θ
is 60◦
. Find the values of A and B. The mass number of aluminum is 27 and
its density is 2.7 g/cm2
.
8.69 A thin target of 48
Ca with 1.3×1019
nuclei per cm2
is bombarded with a 10 nA
beam of α particles. A detector, subtending a solid angle of 2×10−3
steradians,
records 15 protons per second. If the angular distribution is measured to be
isotropic, determine the total cross section for the 48
Ca(α, p) reaction.
8.2.12 Nuclear Reactions via Compound Nucleus
8.70 Cadmium has a resonance for neutrons of energy 0.178 eV and the peak value
of the total cross-section is about 7,000 b. Estimate the contribution of scatter-
ing to this resonance.
8.71 A nucleus has a neutron resonance at 65 eV and no other resonances nearby.
For this resonance, Γn = 4.2 eV, Γγ = 1.3 eV and Γα = 2.7 eV, and all
other partial widths are negligible. Find the cross-section for (n, γ ) and (n, α)
reactions at 70 eV.
8.72 Neutrons incident on a heavy nucleus with spin JN = 0 show a resonance at an
incident energy ER = 250 eV in the total cross-section with a peak magnitude
of 1,300 barns, the observed width of the peak being Γ = 20 eV. Find the
elastic partial width of the resonance.
8.2.13 Direct Reactions
8.73 The reaction d +14
N → α +12
C has been used to test the principle of detailed
balance which relates the cross-section σab for a reaction a + x → b + y, to
the cross-section for the inverse reaction and has the form
(2Sa + 1) (2Sx + 1) Pa
2
σab = (2Sb + 1) (2Sy + 1) Pb
2
σba

The reaction above will take place for low energy incident deuterons in a
s-wave state leaving the 12
C nucleus in the ground state.
Given that the deuteron has spin 1 and positive parity while the alpha par-
ticle has zero spin and positive parity, estimate the spin of 14
N in the ground
state.
Can the alpha particle come off with orbital angular momentum l = 1?
If the incident kinetic energy of the deuteron is 20 MeV in the laboratory
frame, calculate the laboratory kinetic energy at which α’s should scatter from
12
C to test the principle of detailed balance. What is the expected ratio between
the cross-sections for the direct and inverse reactions? Atomic masses of
14
N, 2
H and 4
He are 14.003074, 2.014102, 4.002603 amu respectively. 1 amu
= 931.44 MeV.
8.74 A beam of 460 MeV deuterons impinges on a target of bismuth. Given the
binding energy of the deuteron is 2.2 MeV, compute the mean energy, spread
in energy and the angle of the cone in which the neutrons are emitted.
8.2.14 Fission and Nuclear Reactors
8.75 1.0 g of 23
Na of density 0.97 is placed in a reactor at a region where the ther-
mal flux is 1011
/cm2
/s. Set up the equation for the production of 24
Na and
determine the saturation activity that can be produced. The half-life of 24
Na is
15 h, and the activation cross-section of 23
Na is 536 millibarns.
8.76 Suppose 100 mg of gold (197
79 Au) foil are exposed to a thermal neutron flux
of 1012
neutrons/cm2
/s in a reactor. Calculate the activity and the number
of atoms of 198
Au in the sample at equilibrium [Thermal neutron activation
cross-section for 197
Au is 98 barns and half-life for 197
Au is 2.7 h]
8.77 Estimate the energy released in fission of 238
92 U nucleus, given ac = 0.59 MeV
and as = 14.0 MeV.
8.78 A small container of Ra–Be is embedded in the middle of a sphere of paraffin
wax of a few cm radius so as to form a source of (predominantly) thermal
neutrons. This source is placed at the centre of a very large block of graphite.
Derive an expression for the density of thermal neutrons at a large distance r
from the source in terms of the source strength Q, the diffusion coefficient D
and diffusion length L.
A small BF3 counter is placed in the graphite at a distance of 3 m from
the above source contains 1020
atoms of 10
B. The cross-section of 10
B for
the thermal neutron capture follows a 1/v law and has a magnitude of 3,000

8.2 Problems 445
barns for a neutron velocity v = 2, 200 m/s. If the counting rate is 250/min,
calculate the value of Q. Given L = 50 cm; D = 5 × 105
cm2
/s.
8.79 Calculate the thermal utilization factor for a heterogeneous lattice made up of
cylindrical uranium rods of diameter 3 cm and pitch 18 cm in graphite
Take the flux ratio φm/φU as 1.6
Densities : Uranium = 18.7 × 103
kg m−3
, Graphite = 1.62 × 103
kg m−3
Absorption cross-sections σav = 7.68 b; σam = 4.5 × 10−3
b.
8.80 Calculate approximately, using one-group theory results, the critical size of a
bare spherical reactor, given k∞ = 1.54 and Migration area M2
= 250 cm2
.
8.81 Assuming the energy released per fission of 235
U is 200 MeV, calculate the
amount of 235
U consumed per day in Canada India reactor “Cirus” operating
at 40 MW of power.
8.82 Assuming that the energy released per fission of 235
92 U is 200 MeV, calculate the
number of fission processes that should occur per second in a nuclear reactor
to operate at a power level of 20,000 kW. What is the corresponding rate of
consumption of 235
92 U.
8.83 (a) Assume that in each fission of 235
U, 200 MeV is released. Assuming that
5% of the energy is wasted in neutrinos, calculate the amount of 235
U
burned which would be necessary to supply at 30% efficiency, the whole
annual electricity consumption in Britain 50 × 109
kWh
(b) A thermal reactor contains 100 tons of natural uranium (density 19) and
operates at a power of 100 MW (heat). Assuming that the thermal cross-
section of 235
U is 550 barns and that the uranium contains 0.7 % of 235
U.
Calculate the neutron flux near the centre of the reactor by neglecting neu-
tron losses from the outside, and assuming flux constant through out the
lattice.
[University of Liverpool 1959]
8.84 If the elastic scattering of neutrons by hydrogen nuclei is isotropic in the centre
of mass system, show that
ln(E1/E2) = 1
where E1 and E2 are respectively the kinetic energies of a neutron before and
after the collision.
8.85 (a) Estimate the average number of collisions required to reduce fast fission
neutrons of initial energy 2 MeV to thermal energy (0.025 eV) in graphite
moderator.
(b) Calculate the corresponding slowing-down time given that Σs=0.385 cm−1

8.86 Show that a homogeneous, natural uranium-graphite moderated assembly can
not become critical. Use the following data:
400 moles of graphite per mole of uranium
Natural uranium Graphite
σa(U) = 7.68 b σa(M) = 0.0032 b
σs (U) = 8.3 b σs(M) = 4.8 b
ε = 1.0; η = 1.34 ξ = 0.158
8.87 A point source of thermal neutrons is placed at the centre of a large sphere of
beryllium.
Deduce the spatial distribution of neutron density in the sphere. Estimate
what its radius must be if less than 1% of the neutrons are to escape through
the surface. Find also the neutron density near the surface in this case in terms
of the source strength.
At. Wt of beryllium = 9
Density of beryllium = 1.85 g/cc
Avagadro number = 6 × 1023
atoms/g atom
Thermal neutron scattering cross-section on beryllium = 5.6 barns
Thermal neutron capture cross-section on beryllium = 10 mb (at velocity
v = 2,200 m/s)
8.88 Calculate the steady state neutron flux distribution about a plane source emit-
ting Q neutrons/s/cm2
in an infinite homogeneous diffusion medium. Assume
that neutrons are not produced in any region of interest.
8.89 Calculate the thermal diffusion time for graphite. Use the data:
σa(C) = 0.003 b, ρc = 1.62 g cm−3
. Average thermal neutron speed
= 2,200 m/s.
8.90 Estimate the generation time for neutrons in a critical reactor employing 235
U
and graphite. Use the following data:
Σ
a
= 0.0006 cm−1
; B2
= 0.0003; L2
= 870 cm2
; v = 2200 ms.
8.91 The spatial distribution of thermal neutrons from a plane neutron source kept
at a face of a semi-infinite medium of graphite was determined and found to
fit e−0.03x
law where x is the distance along the normal to the plane of the
source. If the only impurity in the graphite is boron, calculate the number of
atoms of boron per cm3
in the graphite if the mean free path for scattering
and absorption in graphite are 2.7 and 2,700 cm, respectively. The absorption
cross-section of boron is 755 barns.

8.3 Solutions 447
8.92 Assuming that the elastic scattering of low energy neutrons is isotropic,
show that the mean energy of the neutron after each collision will be
Ef = (A2
+ 1)Ei/(A + 1)2
where A is the mass number of the target nucleus. Determine the number of
collisions needed to thermalize fission neutrons (2 MeV) in graphite (A = 12).
8.2.15 Fusion
8.93 Determine the range of neutrino energies in the solar fusion reaction, p+ p →
d + e+
+ ν. Assume the initial protons have negligible kinetic energy and that
the binding energy of the deuteron is 2.22 MeV, mp = 938.3 MeV/c2
and
md = 1875.7 MeV/c2
and me = 0.51 MeV/c2
.
8.94 If the kinetic energy of the deuterons in the fusion reaction D + D →3
2He +
n + 3.2 MeV can be neglected, what is the kinetic energy of the neutron?
8.95 (a) It is estimated that the deutrons have to come within 100 fm of each other
for fusion to proceed. Calculate the energy that the deuterons must possess
to overcome the electrostatic repulsion.
(b) If the energy is supplied by the thermal energy of the deutrons, what is the
temperature of the deuteron?
[e2
/4πε0 = 1.44 MeV fm, Boltzmann constant k = 1.38 × 10−23
J K−1
]
(c) In (b) the actual required temperature is lower than the estimated value.
Explain the mechanism by which the fusion reaction may proceed.
8.96 In a fusion reactor, the D-T reaction with Q value of 17.62 MeV is employed.
Assuming that the deuteron density is 7×1018
m−3
and the experimental value
σDT · v = 10−22
m3
s
−1
and that equal number of deuterons and tritons
exist in the plasma at energy 10 keV, calculate the confinement time if the
Lawson criterion is just satisfied.
8.3 Solutions
8.1 An ion of charge q will pick up kinetic energy, T = qV in dropping through a
P.D of V volts. In a magnetic induction B perpendicular to its path, the ion of
momentum p will describe a circular path of radius r given by
p = qBr =
√
2MT =

2MqV
M =
q B2
r2
2V
(1)
For the first ion, q = 1.6 × 10−19
C, B = 0.08, r = 0.0883 m and V =
400 V. Substituting these values in (1) we get
M1 = 9.98 × 10−27
kg

The mass of this ion is then
9.98 × 10−27
kg
1.66 × 10−27 kg/amu
= 6.012 amu
Therefore the mass number is 6.
For the second ion, the only change is the radius of the orbit which is
0.0954 m. The mass of the second ion is
m2 = m1 ×
r2
r
2
= 6.012 ×

0.0954
0.0883
2
= 7.0176 amu
Therefore the mass number is 7.
8.2 The radius of curvature of an ion in the magnetic induction B, perpendicular to
the orbit, with kinetic energy qV will be
r =

2MV
q B2
1/2
(1)
After a deviation of 180◦
the two ions will be separated by d, the difference
between the diameters of the circular path.
(a) d = 2(r11 − r10) =

8V
q B2
1/2
M1 −

M2

(2)
Substitute V = 5, 000 V, q = 1.6 × 10−19
C, B = 0.15 T, M1 = 11 ×
1.66 × 10−27
Kg and M2 = 10 × 1.66 × 10−27
Kg to find the separation
d = 0.021 m or 2.1 cm.
(b) From (1), we get
ΔM
M
= 2
Δr
r
The mass resolution is
δ =
M
ΔM
=
r
2Δr
=
r
d
=
22 cm
2.1 cm
= 10.5
where r is the mean radius of the ions, r11 and r10, determined from (1) as
22.52 cm and 21.47 cm.
8.3 p =
√
2Tm = qBr
Here T, q and r are fixed.
∴
B2
B1
=

m2
m1

=
37
35
Increase in induction
ΔB = B2 − B1 = B1

37
35
− 1

=
2B1
35
=
2 × 0.1
35
= 0.57 × 10−3
T or 5.7 G
8.4 27
Si →27
Al + β+
+ ν + Tmax
MSi − MAl = 2me + Tmax = 2 × 0.511 + 3.48 = 4.5 MeV
The transition is between two mirror nuclei of charge Z + 1 and Z. The
difference in Coulomb energy is

8.3 Solutions 449
ΔEc =
3
5
.
1
4πε0
.
e2
R
[Z(Z + 1) − Z(Z − 1)]
=
6Ze2
5R
.
e2
4πε0
= 1.2 × 1.44
Z
R
MeV-fm
Equating ΔEc to MSi − MA = 3.48 + 2 × 0.51 = 4.5 MeV, and Z = 13,
we find R from which ro = R/A1/3
can be determined, where A = 27. Thus
ro = 1.66 fm.
8.5 ΔPx .Δx = (uncertainty principle)
or cΔPx =
c
Δx
=
197.3 MeV − fm
5 fm
= 39.6 MeV
The kinetic energy T =
c2
p2
2Mc2
=
(39.6)2
2 × 940
= 0.83 MeV
8.6 The mass-energy equation for positron decay is
M(14
O) − M(14
N) = 2me +
Eβ (max) + Eγ
931.5
= 2 × 0.000548 +
1.835 + 2.313
931.5
= 0.005549
or M(14
O) = 14.003074 + 0.005549 = 14.008623 amu
8.7 The charge density ρ = 3ze/4π R3
. Consider a spherical shell of radii r and
r + dr.
The volume of the shell is 4π r2
dr. The charge in the shell q′
= (4πr2
dr)ρ.
The electrostatic energy due to the charge q′
and the charge (q′′
) of the sphere
of radius r, which is 4
3
πr3
ρ, is calculated by imagining q′′
to be deposited at
the centre. The charge outside the sphere of radius r does not contribute to this
energy (Fig. 8.8). Thus
dU =
q′
q′′
4πε0r
=
(4πr2
drρ)(4πr3
ρ/3)
4πε0r
=
4π
ε0
ρ2
r4
dr
The total electrostatic energy is obtained by integrating the above expression
in the limits 0 to R,
Fig. 8.8 Spherical shell of
radius R

U =

dU =
4πρ2
3ε0
R

0
r4
dr =
4πρ2
R5
1.5ε0
=
3z2
e2
20πε0 R
where we have substituted the value of ρ.
8.8 Choose a point at distance r (R) from the centre of the nucleus. Let q′
be the
charge within the sphere of radius r.
Then q′
= q
r
R
3
The electric field will be E =
q′
4πε0r2
=
qr
4πε0 R3
The potential V (r) = −

Edr = −
qr
4πε0 R3
+ C
= −
qr2
8πε0 R3
+ C (1)
where C is a constant.
At r = R, the point is just on the surface and the potential will be given by
Coulomb’s law.
V (R) =
q
4πε0 R
(2)
Using (2) in (1), the value of C is determined as C =
3
2
q
4πε0 R
and (1) becomes
V (r) =
q
8πε0 R

3 −
r2
R2

8.9 The rotational kinetic energy is given by
ER =
1
2
Iω2
(1)
where ER is the rotational energy, I the rotational inertia and ω is the angular
velocity
The angular momentum is given by
J = Iω (2)
ER =
1
2
Jω

8.3 Solutions 451
Therefore J =
2ER
ω
=
2 × 537.5 × 1.6 × 10−13
2.6 × 1023
= 6.61 × 10−34
Js = h
8.10 (a) If J be the electronic angular momentum and I the nuclear spin, the mul-
tiplicity is given by (2J + 1) or (2I + 1), whichever is smaller.
Now, 2J + 1 = 2 ×
5
2
+ 1 = 6. But only four terms are found. Therefore,
the multiplicity is given by 2I + 1 = 4, whence I = 3/2.
(b) The magnetic field produced by the electron interacts with the nuclear
magnetic moment resulting in the energy shift in hyperfine structure.
ΔE ≈ 2I. J = F(F + 1) − I(I + 1) − J(J + 1)
where, F = I + J takes on integral values from 4 to 1.
F = 4,ΔE = 20 − 25/2
F = 3,ΔE = 12 − 25/2
F = 2,ΔE = 6 − 25/2
F = 1,ΔE = 2 − 25/2
The intervals are 8, 6 and 4 in the ratio 4:3:2
8.11 The resonance frequency ν is given by
ν =
μβ
Ih
where μ is the magnetic moment, B the magnetic induction, I the nuclear spin
in units of . A nuclear magneton μN = 5.05 × 10−27
JT−1
.
B =
νIh
μ
=
60 × 106
× (1/2) × 6.625 × 10−34
2.79 × 5.05 × 10−27
= 1.4 T
8.12 f =
μB
Jh
=
3.46 × 3.15 × 10−14
× 1.6 × 10−13
× 0.8

7
2

× 6.63 × 10−34
= 6.012 × 106
Hz
= 6.012 MHz
8.3.4 Electric Quadrupole Moment
8.13 Quadrupole Moment
Assume that the charge is uniformly distributed over an ellipsoid of revolution,
with the axis of symmetry along the z/
-axis, the semi-major axis of length a,
and the other two semi-axes of length b. The charge density ρ is
ρ =
Ze
volume
=
Ze
4π
3
ab2

Fig. 8.9 Ellipsoid of
revolution
The equation of the ellipsoid is

x′2
b2

+

y′2
b2

+

z′2
a2

= 1
In cylindrical coordinates (z′
, s′
, ϕ′
) the equation is

s′2
b2

+

z′2
a2

= 1, where s′2
= x′2
+ y′2
Q =

ρ(3z′2
− r′2
)dτ′
ρ =
Ze
4π
3

ab2
Q =
3Ze
4ab2
a
0
dz/
b
√
1−(z/2/a2)
0
(2z/2
− s/2
)s/
ds/
2π
0
dϕ/
=
2
5
Ze(a2
− b2
)
8.14 The quadrupole Q = 4
5
ηZR2
η =
a − b
R
Tantalum:
R = 1.5 × (181)1/3
= 8.48fm
η =
5Q
4ZR2
=
5 × 6 × 100 fm2
4 × 73 × (8.48)2 fm2
= 0.143
The ratio of major to minor axes is
a
b
∼
= 1 + η = 1.143
Antimony: R = 1.5 × (123)1/3
= 7.458 fm
η =
5
4
(−1.2 × 100 fm2
)
51 × (7.458)2 fm2
= −0.053
a
b
= 1 − 0.053 = 0.947

8.3 Solutions 453
8.15 (a) The electrical quadrupole moment is the expectation value of the operator
Qij =
z

k=1
ek(3xi xj − δijr2
)k (1)
where δij is the kronecker delta.
The Qzz – component,
Qzz =
z

k=1
ek(3z2
k − r2
k ) (2)
is zero for a spherically charge distribution. This feature also becomes obvi-
ous from the formula
Q =
2
5
ze(a2
− b2
) for a homogenous ellipsoid of semi axes a, b
(Problem 8.54).
For a sphere a = b and therefore Q = 0.
(b) The quadrupole moment which is a tensor has the property that it is sym-
metric, that is Qij = Qji and that its trace (the sum of the diagonal ele-
ments), Qxx + Qyy + Qzz = 0. Using these two proiperties, Qij can be
expressed in terms of the spin vector I which specifies the quantized state
of the nucleus.
Qij = C

Ii Ij + Ij Ii −
2
3
I2
δij

(3)
where C is a constant. Substituting I2
= I(I + 1) in (3)
Q =
2
3
CI(2I − 1) (4)
which is zero for I = 0 or I = 1/2
8.16 From the results of Problem 8.15
Q = 2
5

a2
− b2

Z
We can write
Q =
4
5
Z

a + b
2

(a − b)
Calling R =

a + b
2

and ΔR = a − b, we get
Q =
4
5
ZRΔR
Q = 4.2 barn = 4.2 × 10−24
cm2
= 420 fm2
=
2
5
× 71(a2
− b2
)

Therefore (a2
− b2
) = 14.79 fm2
(1)
Now, A =
4
3
πab2
ρ (2)
The nuclear charge density, ρ = 0.17 fm−3
(3)
Using the value of ρ and A = 181 in (2), we get
ab2
= 254.3 fm3
(4)
Solving (2) and (4) we find a = 7.1 fm, b = 6.0 fm
8.17 Given B/A = 9.402 − 7.7 × 10−3
A
For the parent nucleus
B(A, Z) = 9.402A − 7.7 × 10−3
A2
For the product nucleus
B(A − 4, Z − 2) = 9.402(A − 4) − 7.7 × 10−3
(A − 4)2
B(∝) = 28.3
Condition that alpha decay is just energetically possible is
B(A, Z) = B(A − 4, Z − 2) + B(∝)
Or 9.402A − 7.7 × 10−3
A2
= 9.402(A − 4) − 7.7 × 10−3
(A − 4)2
+ 28.3
Simplifying and solving for A, we find that A = 153. Thus, alpha decay is
energetically possible for A 153.
8.18 Sn − Sp = 15.7 − 12.2 = 3.5 =
3
5
×
e2
4πε0 R
[Z2
− (Z − 1)2
]
Substitute e2
/4πεo = 1.44 MeV fm and Z = 8 to find R = 3.7 fm.
8.19 An atom of mass M1 will decay into the product of mass M2 and α particle of
mass mα if M1 M2 + mα. Now the mass excess Δ = M − A.
For 229
Th, M1 = 229 + 0.031652 = 229.031652 amu. For α decay, the prod-
uct atom would be 225
Ra, and M2 = 225 + 0.023528 = 225.023528.
For α - particle, mα = 4 + 0.002603 = 4.002603
M2 + mα = 229.026131
Since, M1 M2 + mα, 229
Th will decay via α emission.
For β−
decay, M2 = 229 + 0.032022 = 229.032022
As M1 M2, decay via β−
emission is not possible. By the same argument
the decay of 229
Th to 229
Ac is not possible via β+
emission as M1 M2+2me.

8.3 Solutions 455
8.20 Three types of decays are possible
64
29Cu →64
30 Zn + β−
+ ν̄e (1)
64
29Cu → 64
28Ni + β+
+ νe (2)
64
29Cu + e−
→ 64
28Ni + νe (3)
The energy released for the three processes are as follows:
Q1 = mcu − mzn = (63.929759 − 63.929145) × 931.5 = 0.572 MeV
Q2 = mcu − mNi − 2me = (63.929759 − 63.927959) × 931.5 − 2 × 0.511
= 0.655 MeV
Q3 = mcu − mNi = (63.929759 − 63.927959) × 931.5 = 1.677 MeV
We calculate the recoil energy in the process (3) and show that it is
negligible.
TNi + Tν = Q = 1.677 (energy conservation) (4)
PNi = Pν
or P2
Ni = 2M NiTNi = P2
ν = T 2
ν (5)
Solve (4) and (5) to obtain TNi = 6 eV
8.21 Q = (MAl − MSi) × 931.5
= Emax + Eγ
= (27.981908 − 27.976929) × 931.5
= 4.638 MeV
= Emax + Eγ
Therefore, Eγ = Q − Emax = 4.638 − 2.865
= 1.773 MeV
Fig. 8.10 β−
decay of 28
13Al
followed by γ decay

8.22 The mass-energy equation for the positron decay gives
MNa = MNe + 2me +
Tmax + Tγ
931.5
= 21.991385 + 2 × 0.511 +
0.542 + 1.277
931.5
= 23.015338 amu
8.23 For electron capture 7
Be + e−
→7
Li + νe,
Q = (mBe − mLi) × 931.5 = (7.016929 − 7.016004) × 931.5 = 0.8616 MeV
which is positive
For positron emission the Q-value must be atleast 1.02 meV which is not
available. Therefore, positron emission is not possible.
8.24 U =
3
5
AEF (1)
p = −
∂U
∂V
=
3
5
A
∂EF
∂V
(2)
From Fermi gas model
A = KVE
3/2
F (3)
Differentiating (3) with respect to V
3
2
V

EF
∂EF
∂V
+ E
3/2
F = 0
whence
∂EF
∂V
= −
2
3
EF
V
(4)
Using (4) in (2)
p =
2
5
A
V
EF =
2
5
ρN EF
where ρN = A/V , is the nucleon density.
8.25 The Fermi momentum for N = Z = A/2, is
pF(n) = pF(p) = (/r0)(9π/8)1/3
cpF = (c/r0)(9π/8)1/3
= (197.3 MeV.fm/1.3 fm)(9π/8)1/3
= 231 MeV
pF = 231 MeV/c
EF = pF
2
/2M = (231)2
/(2 × 940) = 28 MeV
If B is the binding energy of a nucleon
V = EF + B = 28 + 8 = 36 MeV

8.3 Solutions 457
8.3.7 Shell Model
8.26 A state with quantum number j can accommodate a maximum number of
Nj = 2(2 j + 1) nucleons. Now j = l ± 1
2
and for Nj = 12, j = 5/2 and
l = 3 or 2. However, because the parity is odd and P = (−1)l
, it follows
that l = 3.
8.27 In the shell model the nuclear spin is predicted as due to excess or deficit of
a particle (proton or neutron) when the shell is filled. Its parity is determined
by the l value of the angular momentum, and is given by (−1)l
. For s- state,
l = 0, p – state, l = 1, d-state l = 2, f-state l = 3 etc.
For the ground state of the nuclei:
7
3Li: Spin is due to the third proton in P3/2 state.
Therefore Jπ
= (3/2)−
(∵ l = 1)
16
8 O: This is a doubly magic nucleus, and Jπ
= 0+
17
8 O: Spin is due to the 9th neutron in d5/2 state.
Therefore Jπ
= (5/2)+
(∵ l = 2)
39
19K: Spin is due to the proton hole in the d3/2 state.
Therefore Jπ
= (3/2)+
(∵ l = 2)
45
21Sc: spin is due to the 21st proton in the f7/2 state.
Therefore Jπ
= (7/2)−
(∵ l = 3)
8.28 The 15
O nucleus in the 1p1/2 shell is an 16
O nucleus deficit in one neutron,
its energy being B(15)–B(16), while 17
O in the 1 f5/2 shell is an 16
O nucleus
with a surplus neutron, its energy being B(16)–B(17). Thus the gap between
the shells is
E(1 f5/2) − E(1 p1/2) = B(16) − B(17) − [B(15) − B(16)]
= 2 B(16) − B(17) − B(15)
= 2 × 127.6193 − 131.7627 − 111.9556
= 11.52 MeV
8.29 Q = −

2 j − 1
2 j + 2

r2

For 209
Bi, j = 9/2, r2
= 3
5
R2
= 3
5
r2
0 A2/3
= 3
5
(1.2)2
(209)2/3
= 30.42 fm2
= 0.3 b
Q = −0.22 b
8.30 The spin and parity are determined as in Problem 8.27. The ground state spin
and parity for the following nuclides are:

3
2He : Jπ
= (1/2)+
The state due to neutron hole is 1s1/2
20
10Ne : Jπ
= (0)+
The protons and neutrons complete the sub-shell.
27
13Al : Jπ
= (5/2)+
The state of an extra neutron is 1d5/2
41
21Sc : Jπ
= (7/2)−
The state of an extra proton is 1 f7/2
8.31 For the nuclei 12
6 C, 13
6 C, 14
6 C and 15
6 C the ground state configuration of protons
is

1s1/2, 1p3/2

For neutrons it is
(1s1/2, 1p3/2), (1s1/21p3/2, 1p1/2), (1s1/2, 1p3/2, 1p1/2) and
(1s1/2, 1p3/2, 1p1/2, 1D5/2) respectively
The spin and parity assignment for 13
6 C is (1/2)−
and that for 15
6 C, it is
(1/2)+
.
8.32 Twenty neutrons and twenty protons fill up the third shell. The extra neutron
goes into the 1 f7/2 state. The spin and parity are determined by this extra
neutron. Therefore the spin is 7/2. The parity is determined by the l - value
which is 3 for the f-state. Hence the parity is (−1)l
= (−1)3
= −1.
The model predicts Jπ
= (0)+
for 30
14Si and Jπ
= (1)+
for 14
7 N nuclides.
8.33 Using the mass formula one can deduce the atomic number of the most stable
isobar. It is given by
Zmin =
A
2 + (ac/2as)A2/3
Zmin
A
=
1
2 + 0.0156A2/3
For light nuclei, say A = 10, the second term in the denominators is small,
and zmin/A = 0.48. For heavy nuclei, say A = 200, Zmin/A = 0.4
8.34 The most stable isobar is given by
Z0 =
A
2 + 0.015 A2/3
=
64
2 + 0.015 × 642/3
= 28.57 or 29
8.35 Δm = (26.986704 − 26.981539) × 931.5 = 4.76 MeV.
The difference in binding energy is due to mass difference of neutron and
proton. (1.29 MeV) plus Δm, that is 6.05. The difference in the masses of mir-
ror nuclei is assumed to be due to difference in proton number. Then equating
the Coulomb energy difference to the mass difference
ΔB = ac
[Z(Z + 1) − Z(Z − 1)]
A1/3
=
2ac Z
A1/3
ac =
A1/3
ΔB
2Z
=
3 × 6.05
2 × 13
= 0.7

8.3 Solutions 459
8.36 The difference in the binding energies of the two mirror nuclei is assumed to
be the difference in the electrostatic energy, the mass number being the same.
B(Ca) − B(Se) =
3
5
[Z(Z + 1) − Z(Z − 1)]e2
4πε0 R
e2
/4πε0 = 1.44 MeV fm, and Z = 20
The left hand side is 350.420 − 343.143 = 7.227 MeV, R is calculated as
4.75 fm.
8.37 The empirical mass formula is
M(A, Z) = Z(mp+me)+(A−Z)mn−αA+β A2/3
+γ (A−2Z)2
/A+εZ2
A−1/3
where α, β, γ and ε are constants. Holding A as constant, differentiate M(A, Z)
with respect to Z and set ∂M
∂ Z
= 0.
∂M
∂ Z
= mp + me − mn − 4γ (A − 2Z)/A + 2εZA−1/3
= 0
The terms mp+me−mn
∼
= mH, the mass of hydrogen atom which is neglected.
Rearranging the remaining terms, we obtain
Zmin =
A
2 + (ε/2γ )A2/3
which is identical with the given expression.
8.38 B
40
20 Ca

= 15.56 × 40 − 17.23 × 402/3
−
0.697 × 202
401/3
− 0 +
12
401/2
(1)
B
39
20 Ca

= 15.56 × 39 − 17.23 × 392/3
−
0.697 × 202
391/3
−
23.285
39
− 0 (2)
Subtracting (2) from (1) gives us the binding energy of neutron B(n) =
15.38 MeV.
This is the energy needed to separate one neutron from the nucleus.
8.39 (a) B(Th) = 15.5×230−16.8×2302/3
−0.72×
90 × 89
2301/3
−23.0×
502
230
+
34.5
2303/4
= 1743.70 MeV
B(Ra) = 15.5×226−16.8×2262/3
−0.72×
88 × 87
2261/3
−23.0×
482
226
+
34.5
2263/4
= 1740.88 MeV
B(α) = 28.3
Q = B(Ra) + B(∝) − B(Th)
= 25.48 MeV
(b) E(Ra) = Q×4
4+226
= 0.44 MeV
8.40 For A = 235 the diagram (Fig. 8.7) indicates the binding energy per nucleon,
B
A
= 7.6 MeV. Therefore, the total binding energy of 235
U nucleus B =
7.6 A = 7.6 × 235 = 1,786 MeV.

Hence the rest mass energy of the nucleus
M(A, Z)c2
= Zmpc2
+ (A − Z)mnc2
− B
= 92 × 938.3 + 143 × 939.5 − 1, 786
= 218, 929 MeV
Therefore the mass is
218,929
931.5
= 235.028 amu
The plot is based on the semi-empirical mass formula obtained in the Liquid
Drop Model. In this formula the lowering of binding energy at low mass num-
bers due to surface tension effects as well as at high Z (and hence large A) due
to coulomb energy, are predicted by this model. The jumps in the curves at low
mass numbers (A = 2–20) are attributed to the shell effects explained by the
shell Model. The asymmetry term occurring in the mass formula is explained
by the Fermi Gas Model. From the plot the binding energy per nucleon for 87
Br
is found to be 8.7 MeV and that for 145
La it is 8.2 MeV. The energy released
in the fission is
Q = [B(Br) + B(La)] − B(U)
= 8.7 × 87 + 8.2 × 145 − 1786
= 160 MeV
8.41 The liquid drop model gives the value of Z for the most stable isobar of mass
number A by
Z0 =
A
2 + 0.015A2/3
For A = 127, Z0 = 53.38, the nearest being Z = 53. Hence 127
53 I is stable.
But 127
54 Xe is unstable against β+
decay or e−
capture.
8.42 27
12Mg −27
13 Al = −0.000841 (12–13) + 0.0007668 × 27−1/3
(122
–132
)
+0.09966

12 −
27
2
2
−

13 −
27
2
2
'
= +0.225331 amu
As the right hand side is positive 27
12Mg is heavier than 27
13Al, and therefore it is
unstable against β−
decay.
8.3.9 Optical Model
8.43 Introduce the complex potential V = −(U +iW) in the Schrodinger’s equation
∇2
ψ +
2m
2
(E + U + iW)ψ = 0 (1)

8.3 Solutions 461
Multiply (1) by ψ∗
ψ∗
∇2
ψ +
2m
2
(E + U + iW)ψ∗
ψ = 0 (2)
Form the complex conjugate equation of (1) and multiply by ψ
ψ ∇2
ψ∗
+
2m
2
(E + U − iW)ψ∗
ψ = 0 (3)
Subtract (2) from (3)
ψ∗
∇2
ψ − ψ ∇2
ψ∗
=
−4imW
2
ψψ∗
(4)
Now the quantum mechanical expression for the current density is
j =

2im
(ψ∗
∇2
ψ − ψ ∇2
ψ∗
) (5)
so that (4) becomes
div j = −
2

Wψ∗
ψ (6)
Since ψ∗
ψ is the probability density and W = 1/2vK where K is the
absorption coefficient, Eq. (6) is equivalent to the classical continuity equation
∂ρ
∂t
+ div j = −
v
λ
ρ (7)
where v is the particle velocity and the mean free path λ = 1/K. When steady
state has reached the first term on the LHS of (7) vanishes. Provided W 0,
the imaginary part of the complex potential has the effect of absorbing flux
from the incident channel.
8.44 (i) cp =
√
2m(E − U)
λ =
h
p
=
2πc

2mc2(E − U)
=
2π × 197.3
√
2 × 939.6 × (100 + 25)
= 2.56 fm
(ii) W =
1
2
vK =
1
2
cβK
β =
2E
mc2
=
2 × 100
939.6
= 0.46
K =
2W
cβ
=
2 × 10
197.3 × 0.46
= 0.22 fm−1
2R = 2ro A
1
/3
= 2 × 1.3(120)
1
/3
= 12.82
2KR = 2.82
Probability that the neutron will be absorbed in passing diametrically
through the nucleus = (1 − e−2KR
) = (1 − e−2.82
) = 0.94

8.45 The given decay is
13
N → 13
C + β+
+ ν + 1.2 MeV (1)
MN − MC − 2me = 1.2 MeV (2)
where masses are atomic and c = 1
For the reaction
p + 13
C → 13
N + n
Q = Mp + MC − MN − Mn
where the masses are nuclear.
Add and subtract 7me to get Q in atomic masses
Q = MH + MC − MN − Mn = −[(Mn − MH ) + (MN − MC )]
Q = −[0.78 + 1.2 + 2me] = −3 MeV
where we have used (2) and the mass difference Mn − MH = 0.78 MeV and
me = 0.51 MeV
Ethreshold = |Q|

1 +
mp
mc

= 3 ×

1 +
1
13

= 3.23 MeV
8.46 Given reaction is
p +7
Li →7
Be + n − 1.62 MeV (1)
MLi + MP − MBe − Mn = −1.62 MeV (2)
where the masses are nuclear.
It follows that
MBe − MLi = 1.62 + 938.23 − 939.52 = 0.33 MeV
Add and subtract 4me to get the Q mass difference of Be and Li
MBe − MLi − me = 0.33
or MBe − MLi = 0.33 + me = 0.84 MeV
where me = 0.51 MeV. Thus the total energy released in the electron capture
e−
+7
Be →7
Li + ν is 0.84 MeV
This energy is shared between the neutrino and the recoil nucleus. Energy
and momentum conservation give
EN + Eυ = 0.84 MeV (1)
P2
υ = E2
υ = P2
N = 2 MN EN (2)
Using MN
∼
= 7 amu = 6520.5 MeV, (1) and (2) can be solved to obtain EN =
7.5 keV and Eυ
∼
= 0.84 MeV.
8.47 The Q-value is −1.37 MeV. Minimum energy required is
Ea = |Q|

1 +
ma
mx

= 1.37

1 +
1
24

= 1.427 MeV

8.3 Solutions 463
8.48 Given reaction is
4
2He +14
7 N →A
Z X +1
1 H
As the atomic number (Z) and mass number (A) are conserved
2 + 7 = Z + 1, or Z = 8
4 + 14 = A + 1 or A = 17
Therefore the product X is 17
8 O.
Q = [(4.0039 + 14.0075)–(17.0045 + 1.0081)] × 931.5
Q = −1.118 MeV
Thus it is an endoergic reaction for which the minimum α - particle kinetic
energy required to initiate the above reaction is
Ethreshold = |Q|

1 +
MHe
MN

= 1.118 ×

1 +
4
14

= 1.437 MeV
8.49 Given reactions are
2
H +2
H → n +3
He + 3.27 MeV
2
H +2
H → p +3
H + 4.03 MeV
It follows that
Mn + MHe3 + 3.27 = Mp + MH3 + 4.03
MH3 − MHe3 =

Mn − Mp

+ 3.27 − 4.03
= 1.29 + 3.27 − 4.03 = 0.53 MeV
Binding energies are given by
B(H3
) = mp + 2mn − M(H3
)
B(He3
) = 2mp + mn − M(He3
)
∴ B(H3
) − B(He3
) = mn − mp −
,
M(H3
) − M(He3
)
-
= 1.29 − 0.53 = 0.76 MeV
The Coulomb energy of two protons
Ec =
1.44 × 1 × 1
31

3
× 1.3
= 0.769 MeV
8.50 Given reaction can be written down as
n +10
5 B →11
5 B → α +7
3 Li + Q
(a) Q = (Mn + MB10 − Mα − MLi) × 931 MeV
= (1.008987 + 10.01611 − 4.003879–7.01822) × 931.5 MeV
= 2.79 MeV
(b) The energy released is partitioned as follows

Eα =
QMLi
Mα + MLi
=
2.79 × 7.018
4.004 + 7.018
= 1.78 MeV
ELi = Q − Eα = 2.79 − 1.78 = 1.01 MeV
8.51 p +27
Al → n +27
Si + Q (1)
27
Si →27
Al + β+
+ ν (2)
Msi − MAl = Emax + 2me
where masses are atomic. In terms of nuclear masses
MSi − MAl = Emax + 2me = 3.5 + 0.51 = 4.01 MeV
In (1), Q = mp + mAl − mn − msi
where masses are nuclear
Q = −

mn − mp

− (mSi − mAl) = −0.8 − 4.01 = −4.81 MeV
EThreshold = |Q|

1 +
mp
mAl

= 4.81 ×

1 +
1
27

= 5.0 MeV
8.52 (a) The threshold energy for appearance of neutron in the forward direction is
Ep(threshold) = |Q|

1 +
mp
mH3

= 0.764 ×

1 +
1
3

= 1.019 MeV
(b) The threshold for the appearance of neutrons in the 90◦
direction is
Ep(threshold) =
|Q|mHe
mHe − mp
= 0.764 ×

3
3 − 1

= 1.146 MeV
8.53 d +30
Si →31
Si + p (1)
Q = Md + MSi30 − MSi31 − Mp (2)
Given
MSi30 + Md = Mp31 + Mn + 5.1 (3)
MSi31 = Mp31 + Me + 1.51 (4)
Subtract (4) from (3)
MSi30 + Md − MSi31 = Mn − Me + 3.59 (5)
Further
Mn = Mp + Me + 0 + 0.78 (6)
Add (5) and (6)
MSi30 + Md − MSi31 − Mp = Q = 3.59 + 0.78 = 4.37 MeV

8.3 Solutions 465
8.54 Inelastic scattering is like an endoergic reaction except that the identity of
particles is unchanged.
Q = Eb

1 +
mb
my

− Ea

1 −
ma
my

∵ θ = 90◦
Substitute Q = −4.4, ma = 1, mb = 1, my = 12, Ea = 15 MeV to obtain
Eb = 8.63 MeV
8.55 In the head-on collision the proton will receive full energy of the incident
neutron, that is 5 MeV and neutron will stop. If the neutron was replaced by
the gamma ray then the proton will be emitted in the forward direction as
before but the gamma ray will be scattered backward at 180◦
. The kinetic
energy imparted to the proton can be found out by the use of the formula
employed for Compton scattering except now α would mean
α =
hνo
Mpc2
=
Eo
Mc2
.
T =
αEo(1 − cos θ)
1 + α(1 − cos θ)
Tmax =
2αEo
1 + 2α
=
2Eo
2 + Mc2

Eo
= 5 MeV
where we have put θ = 180◦
. Putting M = 938 MeV, the above equation is
easily solved to yield Eo = hν = 51 MeV.
8.56 (a) Q = Eb

1 +
mb
my

− Ea

1 −
ma
my

−
2
my
√
mamb Ea Eb cos θ
Here ma = mb = 1, my = 10, Ea = 5, Eb = 3 and θ = 450
. Substituting
these values, we find Q = −1.75 MeV. Thus, the excitation energy of 10
B
is 1.75 MeV.
(b) For elastic scattering Q = 0. Substituting the necessary values in the above
equation which is quadratic in
√
Eb, we find
√
Eb = 2.171 so that Eb =
4.715 MeV.
Thus the expected energy of elastically scattered protons will be 4.715 MeV.
8.57 For the reaction X(a, b)Y,
Q = Eb

1 +
mb
mY

− Ea

1 −
ma
mY

−
2
mY

mamb Ea EbCosθ
Here b = p, a = d, Y = 28
Al, θ = 90o
5.5 = Ep

1 +
1
28

− 2.1

1 −
2
28

Ep = 7.19 MeV

8.58 E(He) + Eγ = 5.3 MeV (energy conservation) (1)
PHe =

2M(He) E(He) = Pγ = Eγ/c (momentum conservation) (2)
Or Eγ =
$
2M(He)c2 E(He) (3)
Use M(He)c2
= 3, 728 MeV and solve (1) and (3) to find
E(He) = 3.775 × 10−3
MeV
= 3.78 keV
Heat energy =
3
2
kT =
3
2
× 1.38 × 10−23
× 1.7 × 107
= 3.5 × 10−16
J = 2.19 × 10−3
MeV
Equating this to the electrostatic energy
(1.44 × 1 × 1)/r = 2.19 × 10−3
we get r = 640 fm
8.59 d +16
O → n +17
F − 1.631
d +16
O → p +17
O + 1.918
It follows that
mn + mF − 1.631 = mp + mO + 1.918
or mF − mO = 3.549 − (mn − mp)
= 3.549 − (mn − mH + me)
= 3.549 − (0.782 + 0.511)
= 2.256 MeV
Therefore 17
F is heavier than 17
O. Actually 17
F decays to 17
O by β+
emission.
17
F →17
O + β+
+ ν
Q = Eβ(max) + 2me = Emax + 1.022
Emax = Q − 1.022 = 2.256 − 1.022 = 1.234 MeV
8.60 In the reaction X(a, b)Y at 90◦
,
Q = Eb 1 + mb
my

− Ea 1 − ma
my

= Ep

1 + 1
30

− 7.68

1 − 4
30

For Ep = 8.63 MeV, Q0 = 2.262
Ep = 6.41 MeV, Q1 = −0.032
Ep = 5.15 MeV, Q2 = −1.334
Ep = 3.98 MeV, Q3 = −2.543
The energy levels are determined by
E0 = Q0 − Q0 = 0 (ground state)
E1 = Q0 − Q1 = 2.294
E2 = Q0 − Q2 = 3.596
E3 = Q0 − Q3 = 4.805
The energy levels of 30
Si are shown in Fig. 8.11.

8.3 Solutions 467
Fig. 8.11 Energy levels
8.61 For the nuclear reaction
3
H + p →3
He + n − 0.7637 MeV (1)
3
H −3
He = n − p − 0.7637 (2)
∴ Mass difference
mn − mH = mH3 − mHe3 − me + 0.7637 (3)
where the masses are atomic.
Add and subtract me in the right hand side so that
mn − mH = mH3 − mHe3 − me + 0.7637 MeV (4)
The masses are now atomic.
Now consider the decay
3
H →3
He + β−
+ ν + 18.5 keV (5)
On the atomic scale
mH3 − mHe3 = 18.5 keV = 0.0185 MeV (6)
use (6) in (4) to find
mn − mH = (0.0185 + 0.7637) MeV
= 0.7822 MeV.
8.62 For the reaction X(a, b)Y,
Q = Eb

1 +
mb
mY

− Ea

1 −
ma
mY

−
2
mY

mamb Ea Eb cos θ
For the reaction 3
H(d, n)4
He, we identify a = d, x =3
H, b = n, Y =4
He.
Substitute Ea = 0.3 MeV. Maximum neutron energy is obtained by putting
θ = 0o
and minimum energy for θ = 180◦
in the above equation.
En(max) = 15.41 MeV and En(min) = 13.08 MeV.
Thus, the range of neutron energy will be 13.08 – 15.41 MeV, the energy at
other angle of emission will be in between.
8.63 Energy available in the CMS
−W∗
=
EpmTa
mTa + mp
(1)
where Ep is the Lab proton kinetic energy. Using the values of the masses of
Ea and mp and Ep, we find the excited level W∗
= 4.972 MeV. The energy of
the excited state will be mw + W∗
= 169, 490 MeV.

When the target and the projectile are interchanged, the same excitation
energy W∗
produced with Ta is given by
−W∗
= ETa
mp
mp + mTa
(2)
Comparing (1) and (2)
ETa + Ep
mTa
mp
= 897.7 MeV
8.64 Eth = |Q|

1 +
mo
mca

= 7.83

1 + 16
48

= 10.44 MeV
The velocity, β =
υ
c
=
2T
mc2
=
2 × 10.44
16 × 931
= 3.74 × 10−2
The impact parameter, b = R1 + R2 = 1.1

16
1/3 + 48
1/3

= 6.153 fm
J = Mo υb = n
n =
Mo υb

=
Moc2
βb
c
=
16 × 931 × 3.74 × 10−2
× 6.153
197.3
= 17.37 or 17
8.3.11 Cross-sections
8.65 If 1% of neutrons are absorbed then 99% are transmitted. The transmitted
number I are related to the incident number by
I = I0 exp(−μx) (1)
where μ is the absorption coefficient and x is the thickness of the foil.
μ = Σ = σ N = σ N0ρ/A
where σ is the microscopic cross-section, N0 is the Avagardro’s number, ρ the
density and A the atomic weight.
μ =
28000 × 10−24
× 6.02 × 1023
× 7.3
114.7
cm−1
= 1,073 cm−1
I/I0 = 99/100 = exp(−1,073x)
x = 9.37 × 10−6
cm or 9.37 µm
8.66 Specific activity, that is activity per gram
|dQ/dt| = Qmaxλ = φ Σact = φσa N0/A
= 5 × 1012
× 20 × 10−24
× 6.02 × 1023
/60
= 1012
disintegrations per second per gram.
8.67 |dQ/dt| = Qλ = φΣact(1 − e−λt
)
Given |dQ/dt|/|dQs/dt| = 20/100 = 1 − e−0.693t/5.3
or exp(0.693t/5.3) = 1.25
Take loge to find t = 1.7 years.

8.3 Solutions 469
8.68
dσ
dΩ
=
I
I0 NdΩ
= A + B cos2
θ
I0 = 8 × 1012
/m2
− s
N = number of target atoms intercepting the beam
=
N0ρt
A
=
6.02 × 1023
× 2.7 × 10−3
27
= 6.02 × 1019
dΩ =
0.01
62
= 2.78 × 10−4
A + Bcos2
30◦
=
50
8 × 108 × 6.02 × 1019 × 2.78 × 10−4
= 3.73 × 10−24
A + Bcos2
45◦
=
40
8 × 108 × 6.02 × 1019 × 2.78 × 10−4
= 2.99 × 10−24
Solving the above equations we find
A = 1.57 b/Sr
B = 2.88 b/Sr
8.69 σ(total) =
Total number of particles scattered/sec
(beam intensity) (number of target particles within the beam)
As the scattering is assumed to be isotropic total number of particles scattered
= (Observed number) (4π/dΩ) = 15 × 4π/2 × 10−3
= 9.42 × 104
/s
Beam intensity, that is number of beam particles passing through unit area
per second
=
beam current
charge on each proton
=
10 × 10−9
A
1.6 × 10−19C
= 6.25 × 1010
/cm2
s
Therefore σ(total) =
9.42 × 104
6.25 × 1010 × 1.3 × 1019
= 1.159 × 10−25
cm2
= 116 mb
8.70 Breit–Wigner formulae are
σt = π
-
λ2
ΓsΓ.g
(E − ER)2 + Γ2
4
(1)
σs = π
-
λ2
Γs
2
.g
(E − ER)2 + Γ2
4
(2)
Dividing (2) by (1)
σs
σt
=
Γs
Γ
(3)

Ignoring the statistical factor g, at resonance
σt = 7,000 × 10−24
cm2
=
λ2
π
.
Γs
Γ
(4)
But λ =
0.286
√
E
=
0.286
√
0.178
= 0.678Å = 0.678 × 10−8
cm
From (3) and (4) we get
Γs/Γ = 47.815 × 10−5
(5)
Inserting (5) and the value of σt in (3), we find
σs = 3.35b
8.71 Γ = Γn + Γγ + Γα = 4.2 + 1.3 + 2.7 = 8.2 eV
σ(n, γ ) =
λ2
4π
.
Γγ Γn
(E − ER)2 + Γ2
4
λ =
0.286
√
70
= 3.418 × 10−10
cm
ER = 60 eV, E = 70 eV, Γγ = 1.3 eV and Γn = 4.2 eV
Ignoring the g - factor, we find σ(n, γ ) = 1215 b.
σ(n, α) = σ(n, γ ).
Γα
Γγ
= 1215 ×
2.7
1.3
= 2523 b
8.72 Breit–Wigner’s formula is
σtotal =
π-
λ2
ΓsΓg
(E − ER)2 + Γ2
4
(1)
For spin zero target nucleus, the statistical factor g = 1. At resonance
energy (1) reduces to
Γs =
πΓσtotal
λ2
(2)
λ =
0.286
√
E
Å =
0.286
√
250
= 0.018Å = 1.8 × 10−10
cm
Substituting, σtotal = 1300×10−24
cm2
, Γ = 20 eV and λ = 1.8×10−10
cm
in (2), we find Γs = 2.5 eV.
8.73 Q = [(md + mN) − (mα + mC)] × 931.44 MeV
= [(2.014102 + 14.003074) − (14.002603 + 12.0)] × 931.44
= 13.57 MeV

8.3 Solutions 471
For the forward reaction, energy available in the CMS is
E∗
= Q + T ∗
d + T∗
N = Q +
TdmN
mN + md
= 13.57 +
20 × 14
14 + 2
= 31.07
The energy of 31.07 MeV is shared between α and 12
C. Using the energy
and momentum conservation, we find P∗
α = 416.8 MeV/c.
The inverse reaction will be endoergic, and so an energy of 31.07+13.57 =
44.64 MeV must be provided. This corresponds to the Lab kinetic energy for
α given by
Tα = 44.64 ×
12 + 4
12
= 59.52 MeV
In the CMS the energy of 31.07 MeV is shared between 2
H and 14
N. The
momentum of deuteron would be 441.9 MeV/c.
σdN
σαC
=
(2Sα + 1)(2SC + 1)
(2Sd + 1)(2SN + 1)
P∗2
α
P∗2
d
=
1 × 1
3 × (2SN + 1)
×

441.9
319.4
2
=
1.91
2SN + 1
since Sα = Sc = 0 and Sd = 1. The spin of 14
N can be determined from
the experimental value of the cross-sections. For SN = 1, the ratio of cross-
sections is expected to be 0.64.
The intrinsic parities of all the four particles is positive. If the α′
s are cap-
tured in the s-state for which the parity will be positive as it is given by (−1)l
,
the α′
s can not be produced in the l = 1 state for which the parity would be
negative, resulting in the violation of parity.
8.74 According to Butler’s theory, the neutron energy
En =
1
2
Ed = 0.5 × 460 = 230 MeV
The spread in energy
ΔEn = 1.5

Bd Ed = 1.5
√
2.2 × 460 = 47.7 MeV
And the angular spread is
Δθ = 1.6
0
Bd
Ed
= 1.6
2.2
460
= 0.11 radians
8.75 If Q is the number of atoms of 23
Na per gram at any time t, the net rate of
production of 24
Na is
dQ
dt
= φΣa − λQ

where the first term on the right hand side denotes the absorption rate of neu-
trons in 23
Na, and if it is assumed that each neutron thus absorbed produces a
24
Na atom, then this also represents the production rate of 24
Na. The second
term represents the decay rate, so that dQ/dt denotes the rate of change of
atoms of 24
Na.
The saturation activity is obtained by setting dQ/dt = 0. Then
λQs = φΣa = φσa
Noρ
A
=
1011
× 536 × 10−27
× 6.02 × 1023
× 0.97
23
= 1.36 × 109
s−1
8.76 At equilibrium number of 198
Au atoms is
Qs =
φσa NoWT1/2
0.0693A
=
1012
× 98 × 10−24
× 6.02 × 1023
× 0.1 × 2.7 × 3, 600
0.693 × 197
= 4.2 × 1014
Activity = Qsλ =
Qs × 0.693
T1/2
=
4.2 × 1014
× 0.693
2.7 × 3, 600
= 3 × 1012
s−1
8.77 Consider a binary fission, that is a heavy nucleus of mass number A and atomic
number Z breaking into two equal fragments each characterized by A
2
and Z
2
.
In this problem the only terms in the mass formula which are relevant are the
Coulomb term and the surface tension term.
M(A, Z) =
ac Z2
A1/3
+ as A2/3
Energy released is equal to the difference in energy of the parent nucleus
and that of the two fragments
Q = M(Z, A) − 2M

Z
2
,
A
2

=

as A2/3
+ ac
Z2
A1/3

− 2

as

A
2
2/3
+ ac
(Z/2)2
(A/2)1/3
'
= as A2/3
(1 − 21/3
) + ac
Z2
A1/3

1 −
1
22/3

Inserting A = 238, Z = 92, ac = 0.59 and as = 14 we find Q ∼
= 160 MeV.
8.78 The diffusion equation for the steady state in the absence of sources at the
point of interest is
∇2
φ − K2
φ = 0 (1)
where K2
= 3Σa/λtr

8.3 Solutions 473
Writing the Laplacian for spherical geometry (1) becomes
d2
φ
dr2
+
2
r
dφ
dr
− K2
φ = 0 (2)
Equation (2) is easily solved, the solution being
φ =
C1eKr
r
+
C2e−Kr
r
(3)
As K is positive, the first term on the right hand side tends to ∞ as r → ∞.
Therefore, C1 = 0 if the flux is required to be finite everywhere including at
∞.
φ =
C2e−Kr
r
(4)
We can calculate the constant C2 by considering the current J through a
small sphere of radius r with its centre at the source.
The net current
J = −
λtr
3
∂φ
∂r
=
λtr
3r2
C2(Kr + 1)e−Kr
(5)
where we have used (4)
The net number of neutrons leaving the sphere per second is
4πr2
J =
4
3
πλtrC2(Kr + 1)e−Kr
(6)
But as r → 0, the total number of neutrons leaving the sphere per second
must be equal to the source strength Q. Thus from (6)
Q =
4
3
πλtrC2
or C2 =
3Q
4πλtr
(7)
φ =
3Qe−Kr
4πrλtr
(8)
Therefore the neutron density
n(r) =
φ
ν
=
3Qe−Kr
4πλtrνr
=
Qe−r/L
4πDr
(9)
where
λtr v
3
= D, is the diffusion coefficient and K = 1/L, L being the
diffusion length.

The counting rate,
R = Nσνn(r) (10)
As the absorption obeys the 1/v law, the product σ.v = constant. We then have
250
60
= 1020
× 3,000 × 10−24
× 2.2 × 105
n(r)
Or n(r) = 6.313 × 10−5
/cm3
Substituting the values: D = 5 × 105
, r = 300, L = 50 and R = 250/60 in
(9), we find
Q = 4.8 × 107
/s.
8.79 The absorption rate in the fuel is
ΣauφuVu
where V is the volume, and the absorption rate in the moderator is
ΣamφmVm
The fraction of thermal neutrons absorbed by the Uranium fuel as compared
to the total number of thermal neutron absorptions in the assembly is known
as the thermal utilization factor f and is given by
f =
ΣauφuVu
ΣauφuVu + ΣamφmVm
=
1
1 +
ΣamφmVm
ΣauφuVu
Fig. 8.12 shows a unit cell of a heterogeneous assembly in which the ura-
nium rods of radius r, are placed at regular intervals (pitch). The equivalent
cell radius r1 is also indicated.
Fig. 8.12 Unit cell of a
heterogeneous assembly

8.3 Solutions 475
Area of the unit cell A = πr2
1
Now,
Σam
Σau
=
Nm
Nu
×
σam
σau
=
ρm/Am
ρu/Au
×
σam
σau
=
1.62/12
18.7/238
×
4.5 × 10−3
7.68
= 1.01 × 10−3
Vm
Vu
=
182
− π(1.5)2
π(1.5)2
= 44.8
φm
φu
= 1.6
Hence, f =
1
1 + (1.01 × 10−3) × 44.8 × 1.6
= 0.933
8.80 The equation for a critical reactor is
∇2
φ + B2
φ = 0 (1)
Where φ is the neutron flux and B2
is the buckling.
For spherical geometry, Eq. (1) becomes
d2
φ
dr2
+
2
r
dφ
dr
+ B2
φ = 0 (2)
φ =
A
r
sin(πr/R) (3)
where A is the constant of integration and R is the radius of the bare reactor
dφ
dr
= −
A
r2
sin
πr
R

+
π A
Rr
cos π
r
R

(4)
d2
φ
dr2
=
2A
r3
sin π
r
R

−
π A
Rr2
cos
πr
R

−
Aπ2
R2r
sin π
r
R

(5)
−
Aπ2
dr2
sin
πr
R

+
B2
A
r
sin π
r
R

= 0
Therefore, B2
= π2
R2
Or the critical radius,
R =
π
B
(6)
B2
=
k∞ − 1
M2
=
1.54 − 1
250
= 2.16 × 10−3
cm−2
B = 0.04647
R =
π
0.04647
= 67 cm

The actual radius will be shorter by d = 0.71 λtr where d = extrapolated
distance.
8.81 Let n fissions take place per second.
Energy released per second = 200 n MeV = (200 n) (1.6 × 10−13
) J
200 × 1.6 × 10−13
n = 40 × 106
Or n = 1.25 × 1018
/s
Number of atoms in 1.0 g of 235
U =
6.02 × 1023
235
= 2.562 × 1021
Mass of uranium consumed per second =
1.25 × 1018
2.562 × 1021
= 4.88 × 10−4
g
Mass consumed in 1 day = (4.88 × 10−4
)(86, 400) = 42.16 g
8.82 Let n be the number of fissions occurring per second in the nuclear reactor.
Energy released = 200 n MeV s−1
= (200 n)(1.6 × 10−13
)J-s−1
= power = 2 × 107
W
Therefore, n = 6.25 × 1017
s−1
In 1 g there are N0/A = 6.02 × 1023
/235 = 2.56 × 1021
atoms of 235
U.
Therefore consumption rate of 235
U will be
6.25 × 1017
2.56 × 1021
= 2.44 × 10−4
g s−1
8.83 (a) Let n fissions occur per second. Then energy available will be 200 n MeV
or 200 n × 1.6 × 10−13
J. Allowing for 5% wastage and 30% efficiency, net
power used is P = 200 n × 1.6 × 10−13
× 0.95 × 0.3 = 9.12 × 10−12
n W.
Required energy in 1 s that is power =
50 × 109
× 103
× 3, 600
3.15 × 107
=
5.714 × 109
Equating the two powers
9.12 × 10−12
n = 5.714 × 109
n = 6.26 × 1020
/s
Number of atoms in 1 g of 235
U =
6.02 × 1023
235
= 2.56 × 1021
Mass consumed per second =
6.26 × 1020
2.56 × 1021
= 0.244 g
Mass consumed in 1 year = 0.244×3.15×107
g = 7.7×106
g = 7.7 tons
(b) Volume, V =
M
ρ
=
108
19000
= 526.3 m3
Power density (Power per unit volume)
P/V =
100 × 106
526.3
= 1.9 × 105
W/m3

8.3 Solutions 477
Number of 235
U atoms/cm3
,
N =
N0ρ
A
×
0.7
100
=
6 × 1023
× 19
238
×
0.7
100
= 3.353 × 1020
Σa = σa N = 550 × 10−24
× 3.353 × 1020
= 0.1844 cm−1
If φ is the neutron flux and 200 MeV is released per fission, then energy
released in 1 cm3
/s will be 200φΣa MeV or 200 × 0.1844 × 1.6 × 10−13
φ =
5.9 × 10−12
φ = 7.648 Therefore φ = 1.3 × 1012
/cm2
− s
8.84 Let the Lab kinetic energy of neutron be E1 and E2 before and after the scat-
tering. The neutron and proton mass is approximately identical.
The neutron velocity in the CMS, is v∗
1 = v1/2 as the masses of projectile
and target are nearly identical. As the scattering is elastic, the velocity of the
neutron v∗
2 = v∗
1 in magnitude. Let the neutron be scattered at angle θ∗
in
the CMS. The velocity v∗
2 is combined with vc to yield v2 in the Lab. From
Fig. 8.13,
Fig. 8.13 Kinematics of
scattering
v2
2 = v∗2
2 + v2
c + 2v∗
2vc cos θ∗
(1)
v2
2 = v2
1(1 + cos θ∗
)/2
(∵ vc = v∗
2 = v1/2)
or
E2 =
E1
2
(1 + cos θ∗
) (2)
Let the neutrons scattered between the angles θ∗
and θ∗
+ dθ∗
in the CMS
appear with energy between E2 and E2 + dE in the LS.
Differentiating (2), holding E1 as constant.
dE2 =
E1
2
d cos θ∗
(3)
The mean value
5
ln
E1
E2
6
=

ln

E1
E2

2π sin θ∗
dθ∗

2π sin θ∗ dθ∗
(4)

Writing sin θ∗
dθ∗
= −d(cos θ∗
), and using (3), (4) becomes
5
ln
E1
E2
6
=
E1

0
ln

E1
E2

dE2
E1
= −

ln

E2
E1

dE2
E1
= −
E2
E1
ln
E2
E1
+
E2
E1
%
%
%
%
E1
0
At the upper limit the value is 1. At the lower limit, the second term gives
0. The first term also contributes zero because x ln x in the limit x → 0 gives
zero. Thus,
5
ln
E1
E2
6
= 1.
8.85 (a) The number of collisions required is
n =
1
ξ
ln
E0
En
The average logarithmic energy decrement
ξ = 1 +
(A − 1)2
2A
ln
A − 1
A + 1
For graphite (A = 12), ξ = 0.158
∴ n =
1
0.158
ln

2 × 106
0.025

= 115
(b) Slowing down time
t =
√
2m
ξΣs
[E
−1/2
f − E
−1/2
i ]
∼
=

2mc2/E f
cξΣs
(∵ Ei Ef)
Inserting the values, mc2
= 940 × 106
eV, Ef = 0.025 eV, ξ = 0.158
and Σs = 0.385 cm−1
for graphite, we find the slowing down time t =
1.5 × 10−4
s.
8.86 If Nu is the number of Uranium atom per cm3
and N0 is the number of 238
U
atoms per cm3
, then N0 = 139
140
Nu. Further, Nm/Nu = 400.
Therefore, Nm/N0 = 400 × 140/139 = 402.9
Thermal utilization factor ( f );
f =
Σa(U)
Σa(U) + Σa(m)
=
Nuσa(U)
Nuσa(U) + Nmσa(m)
=
1
1 + Nmσa(m)
Nuσa(U)
=
1
1 + 402.9×0.0032
7.68
= 0.857

8.3 Solutions 479
We can now calculate p, the resonance escape probability.
Σs/N0 =
Σs(U) + Σs(m)
N0
= σs(U) +
Nm
Nu
σa(M)
= 8.3 + 402.9 × 4.8 = 1942 b
We can use the empirical relation for the effective resonance integral (ERI)
E0

E
(σa)eff
dE
E
= 3.85 (Σs/N0)0.415
= 3.85(1942)0.145
= 89 b
We have ignored the contribution of Uranium to the scattering as its inclu-
sion hardly changes the result. Thus Σ0/N0 ≃ Σs/N0 =
Nm
N0
σs(M) =
402.9 × 4.8 = 1934
p = exp

−(ERI)
3
Σ0ξ
N0

= exp

−
89
1934 × 0.158

= 0.747
The reproduction factor
k∞ = ξnfp = (1.0)(1.34)(0.857)(0.747) = 0.858
Thus, k∞ 1, and so the reactor cannot go critical.
8.87 The spatial distribution was derived for Problem 8.78.
φ(r) =
3Q
4πλtr
e−r/L
r
(1)
If 1% of the neutrons are to escape then
φ(r)
Q
=
1
100
=
3
4πλtr
e−r/L
r
(2)
L =

λtrλa
3
1/2
λs =
1
Σs
=
A
σs N0ρ
=
9
5.6 × 10−24 × 6.02 × 1023 × 1.85
= 1.443 cm
λs =
1
Σa
=
A
σa N0ρ
=
9
10 × 10−27 × 6.02 × 1023 × 1.85
= 808 cm
λtr =
λs
1 − 2
3A
=
1.443
1 −
2
3 × 9
= 1.564 cm
L =

1.564 × 808
3
1/2
= 20.52 cm
Inserting the values of λtr and L in (2) and solving for r, we get r = 9.6 cm.
Thus the radius ought to be greater than 9.6 cm.

At the surface the neutron density corresponding to r = 9.6 cm and mean
neutron velocity 2.2 × 105
cm s−1
n(r) =
φ
v
=
3Q
4πλtrv
e−9.6/L
9.6
= 2.93 × 10−9
Q cm−3
8.88 Consider the diffusion equation
∂n
∂t
= S +
λtr
3
∇2
φ − φΣa (1)
where n is the neutron density, S is the rate of production of neutrons/
cm3
/s, φΣa is the absorption rate/ cm3
/s and
λtr∇2
φ
3
represents the leakage
of neutrons. Σa is the macroscopic cross-section, φ is the neutron flux and λtr
is the transport mean free path.
Since it is a steady state,
∂n
∂t
= 0. Further S = 0 because neutrons are not
produced in the region of interest. As we are interested only in the x-direction
the Laplacian reduces to d2
dx2 . Thus (1) becomes
λtr
3
d2
φ
dx2
− φΣa = 0 (2)
or
d2
φ
dx2
− K2
φ = 0 (3)
where
K2
=
3Σa
λtr
=
3
λaλtr
(4)
λa being the absorption mean free path.The solution of (3) is
φ = C1eKx
+ C2e−Kx
(5)
where C1 and C2 are constants of integration. The condition that the flux
should be finite at any point including at infinity means that C1 = 0. Therefore,
(5) becomes
φ = C2e−Kx
(6)
We can now determine C2. Consider a unit area located in the YZ plane
at a distance x from the plane source as in Fig. 8.14. On an average half
of the neutrons will be travelling along the positive x-direction. As x → 0,
the net current flowing in the positive x-direction would be equal to 1
2
Q; the
diffusion of neutrons through unit area at x = 0 would have a cancelling
effect because from symmetry equal number of neutrons would diffuse in the
opposite direction.

8.3 Solutions 481
Fig. 8.14
The net current,
J = −
λtr
3
∂φ
∂x
= A2 K
λtr
3
e−kx
As x → 0, J =
Q
2
= A2 K
λtr
3
whence A2 =
3Q
2Kλtr
φ =
3Q
2Kλtr
e−K x
(7)
8.89 Thermal diffusion time is given by
t =
λa
v
=
1
vΣa
Now Σa = σa
N0ρ
A
=
0.003 × 10−24
× 6.02 × 1023
× 1.62
12
= 2.44 × 10−3
cm−1
t =
1
2.2 × 103 × 2.44 × 10−3
= 0.19 s
8.90 The generation time for neutrons in a critical reactor is calculated from the
formula
t =
1
Σa v (1 + L2 B2)
=
1
0.0006 × 2.2 × 105(1 + 870 × 0.0003)
= 6 × 10−3
s
8.91 Let N be the number of boron atoms/cm3
. Ignore the scattering of neutrons
in boron.

Σa(graphite) =
1
λa(C)
=
1
2, 700
= 3.7 × 10−4
cm−1
Σa = Σa(C) + Σa(B) = Σa(C) + σa(B)N
= 3.7 × 10−4
+ 755 × 10−24
N (1)
λtr =
λs
1 − 2
3A
=
2.7
1 − 2
3×12
= 2.859 cm (2)
The attenuation dependence of e−0.03x
implies that the diffusion length
L =
1
0.03
= 33.33 cm (3)
But L2
=
λtrλa
3
=
λtr
3Σa
(4)
Combining (1), (2), (3), and (4) and solving for N, we find
N = 5.83 × 1017
boron atoms/cm3
8.92 Refer to solution of Problem 8.84 with the change of notation.
v1
2
= v1
∗2
+ vc
2
+ 2 v1
∗
vc cos θ∗
=
v2
0
(A + 1)2
(A2
+ 1 + 2A cos θ∗
)
where we have used the relations
v1
∗
= A V0/(A + 1) and vc = v0/(A + 1)
We can then write
E1
E0
=
A2
+ 2A cos θ∗
+ 1
(A + 1)2
E1/E0 =
1
(A + 1)2
+1
−1
(A2
+ 2A cos θ∗
+ 1)
1
2
d(cos∗
)
=
(A2
+ 1)
(A + 1)2
Therefore, Ef = Ei
(A2
+1)
(A+1)2
Let the neutron energy be En after n collisions
En
E0
=
E1
E0
.
E2
E1
.
E3
E2
. . .
En
En−1
=

E1
E0
n
Therefore 0.025 eV
2×106 eV
=
*
A2
+1
(A+1)2
+n
=
145
169
n
where we have put A = 12 for graphite. Taking logarithm on both sides and
solving for n, we obtain n = 119. Compare this value with n = 115 obtained
from the average logarithmic decrement (Problem 8.85).

8.3 Solutions 483
8.3.15 Fusion
8.93 The minimum energy of neutrino is zero when d and e+
are emitted in oppo-
site direction and the neutrino carries zero energy. The maximum energy of
neutrino corresponds to a situation in which e+
is at rest and d moves in a
direction opposite to neutrino.
Q = 2 × 938.3 − 1875.7 − 0.51 = 0.39 MeV
Td + Tν = Q = 0.39 (energy conservation) (1)
Ped
2
= Pν
2
(momentum conservation) (2)
2 mdTd = Tν
2
(3)
Equations (1) and (3) can be solved to give Tν(max) = 0.38996 MeV. Thus
neutrino energy will range from zero to 0.39 MeV.
8.94 En = QmHe
mHe+mn
= 3.2×3
3+1
= 2.4 MeV
8.95 (a) E = 1.44z1z2
r
= 1.44×1×1
80
= 0.018MeV = 18 keV
(b) Equating the kinetic energy to heat energy
E = 3
2
kT
18 × 1.6 × 10−16
J = 3
2
× 1.38 × 10−23
T
T = 1.39 × 108
K
(c) The temperature can be lowered because with smaller energy Coulomb bar-
rier penetration becomes possible.
8.96 The Lawson criterion is just satisfied if
L =
energy output
energy input
=
n2
d σdt v tc Q
6ndkT
=
nd σdt v tc Q
6kT
= 1
Substitute nd = 7 × 1018
m−3
, kT = 10 keV = 104
eV
σdT .v ≥ 10−22
m3
s
−1
, Q = 17.62 × 106
eV to find tc = 4.86 s

Chapter 9
Particle Physics – I
Interactions and decay
Probability for scattering
P = t/λ (9.1)
where t is the thickness and λ is the mean free path.
Attenuation of beam intensity due to interaction
I = Ioe−μx
(9.2)
The absorption coefficient µ is in metre−1
if x is in metre.
μ is in cm2
/g if x is in g/cm2
.
x in g/cm2
= (x in cm)(density in g/cm3
)
µ = Σ(macroscopic cross - section)
Σ = σ N0ρ/A
(9.3)
where σ = microscopic cross-section, No = Avogadro’s number, ρ = density and
A = Atomic or molecular weight.
Attenuation of beam intensity due to decay
I = Ioe−t/τ
= Ioe−s/νγ τ0
(9.4)
where ν = βc is the particle velocity, γ is the Lorentz factor, s is the distance
travelled and τo is the proper life time, that is the life time in the rest frame of the
particle.
485

486 9 Particle Physics – I
Energy in the center of mass system (CMS)
In the collision of particle of mass m1, of total energy E1 in the lab system with m2
initially at rest, total energy E∗
available in the CMS is given by
E∗
=

m2
1 + m2
2 + 2m2 E1
1
/2
(9.5)
Division of energy in the decay A → B + C, at rest.
Total energy carried by B,
EB =
m2
A + m2
B − m2
C
2mA
(9.6)
Circular accelerators
In static magnetic field, a charged particle is not accelerated but is only bent into
a circular path if the field is perpendicular to the plane of the path. Otherwise the
particle goes into a helical path.
The radius of curvature r is related to the momentum p by
P = 0.3 Br (9.7)
where p is in Gev/c, r in meters, and the field B in Tesla. (1 Tesla = 104
gauss).
If n = γ − 1 then R =
mc
qB
(n2
+ 2n)
1
/2
(9.8)
Betatron
Energy gained
ΔT = e
∆Φ
∆t
(9.9)
where Δϕ/Δt is the rate of change of flux.
Cyclotron
ω0 = qB
m
(Resonance condition)
where ω0 = 2π f0
Energy at the extraction point
T =
(qBR)2
2m
(9.10)
Synchrocyclotron
ω0 =
qB
m0
(9.11)
ω =
qB
m
(9.12)

ω
ω0
=
m0
m
=
1
γ
(9.13)
Synchrotron
The synchrotron radiation loss per turn
ΔE =
4π
3R
e2
4πǫ0

E
mc2
4
(9.14)
Linear Accelerator
Total length
L =
1
2 f

2 eV
m
1
/2
Σ
√
n (9.15)
where n is the number of drift tubes.
Colliders
Luminosity
L =
nN1 N2 f
A
(9.16)
The number of interactions per second
N = Lσ (9.17)
where σ is the interaction cross-section of a given type
N1 or N2 = the number of particles/bunch in each beam
A = area of cross-section of intersecting beams
n = number of bunches/beam
f = frequency of revolution
Detectors
Proportional counter
Gas multiplication factor
M =
CV
Ne
(9.18)
where N is number of ion pairs released.

G.M counter
E =
V
r ln(b/a)
(9.19)
where E is the electric field, V is the applied voltage, r is the distance from the
anode, b and a are the diameters of the cathode and anode.
Dead time of a counter(τ)
n =
n0
1 − n0τ
(9.20)
where n = true counts, n0 = observed counts.
Double source method
τ =
N1 + N2 − N12 + B
2(N1 − B)(N2 − B)
(9.21)
where N1 and N2 are the counting rates from individual sources, N12 is the counting
rate from the combined sources and B is the background counting rate.
Cerenkov counter
cos θ =
1
βn
(9.22)
where θ is the opening angle, n = index of refraction
Threshold velocity β =
1
n
(9.23)
9.2 Problems
9.2.1 System of Units
9.1 Show that 1 kg = 5.6 × 1026
GeV/c2
9.2 In the natural system of units ( = c = 1) show that
(a) 1 m = 5.068 × 1015
GeV−1
(b) 1 GeV−2
= 0.389 mb
(c) 1 s = 1.5 × 1024
GeV−1
9.3 In the natural system of units show that
(a) The Compton wavelength for an electron is λc = 1/me
(b) The Bohr radius of a hydrogen atom is 1/α me
(c) The velocity of an electron in the ground state of hydrogen atom is αc

9.2 Problems 489
Also, calculate the numerical values for the above expressions where α =
1/137 is the fine structure constant. The electron mass is me = 0.511 MeV.
9.4 One of the bound states of positronium has a lifetime given in natural units by
τ = 2/mα5
where m is the mass of the electron and α is the fine structure con-
stant. Using dimensional arguments introduce the factors and c and determine
τ in seconds.
9.5 The V-A theory gives the formula for the width (Γμ) of the muon decay in
natural units.
Γμ = /τ = GF
2
mμ
5
/192π3
Convert the above formula in practical units and calculate the mean life time
of muon
[(GF/(c)3
= 1.116 × 10−5
GeV−2
, mμ c2
= 105.659 MeV]
9.2.2 Production
9.6 An ultra high energy electron (β ≈ 1) emits a photon. (a) Derive an expression
to express the emission angle θ in the lab system in terms of θ∗
, the angle
of emission in the rest frame of the electron. Also, (b) Show that half of the
photons are emitted within a cone of half angle
θ ≈ 1/γ.
9.7 Show that in a fixed target experiments, the energy available in the CMS goes
as square root of the particle energy (relativistic) in the Lab system.
9.8 A positron with laboratory energy 50 GeV interacts with the atomic electrons
in a lead target to produce μ+
μ−
pairs. If the cross-section for this process is
given by σ = 4πα2
2
c2
/3(ECM)2
, calculate the positron’s interaction length.
The density of lead is ρ = 1.14 × 104
kg m−3
9.9 It is desired to investigate the interaction of e+
and e−
in flight, yielding a
nucleon-antinucleon pair according to the equation of e+
+ e−
→ p + p−
.
(a) To what energy must the positrons be accelerated for the reaction to be
energetically possible in collisions with stationary electrons. (b) How do the
energy requirements change if the electrons are moving, for example in the
form of a high energy beam? (c) What is the minimum energy requirement?
(me c2
= 0.51 MeV, Mp c2
= 938 MeV)
9.2.3 Interaction
9.10 A proton with kinetic energy 200 MeV is incident on a liquid hydrogen target.
Calculate the centre-of-mass energy of its collision with a nucleus of hydro-
gen. What kinds of particles could be produced in this collision?

9.11 Estimate the thickness of iron through which a beam of neutrinos with energy
200 GeV must travel if 1 in 109
of them is to interact. Assume that at high
energies the neutrino-nucleon total cross-section is 10−42
Eν m2
, where Eν is
the neutrino energy is in GeV. The density of iron is 7,900 kg m−3
. [Mass of
nucleon = 1.67 × 10−27
kg.]
9.12 (a) The cross-section for scattering of muons with air at atmospheric pressure
is 0.1 barns, and the natural lifetime of muons 2.2 × 10−6
s. Explain what
is meant by the terms elastic scattering and lifetime. Which of these fac-
tors limits the distance over which a beam of muons can travel in air in a
laboratory, if the muon velocity is 106
ms−1
?
(Assume number density of air at STP = 2.69 × 1025
m−3
)
(b) Given your answer to (a), why is it possible to detect showers of muons at
ground level caused by the impact of primary cosmic ray particles with air
at around 12 km altitude? Given that the mean energy of muons detected
at ground level is ≈ 4 GeV, calculate the distance (in air) over which the
number of such muons in a beam would reduce by a factor of e. What kind
of interactions contribute to the scattering cross-section for these particles?
[1 barn = 10−28
m2
; mass of muon = 105 MeV/c2
; number density of air
at STP = 2.69 × 1025
m−3
]
9.13 Obtain an approximate value for the interaction length (in cm) of a fast proton
in lead from the following data: r0 = 1.3×10−13
cm, Atomic weight of lead =
207, Density of lead = 11.3 g cm−3
, Avogadro’s number = 6×1023
molecules
mole−1
9.14 A liquid hydrogen target of volume 125 cm3
and density 0.071 g cm−3
is
bombarded with a mono-energetic beam of negative pions with a flux 2 ×
107
m−2
s−1
and the reaction π−
+ p → π0
+ n observed by detecting the
photons from the decay of the π0
. Calculate the number of photons emitted
from the target per second if the cross-section is 40 mb.
9.15 A beam of π+
mesons contaminated with μ+
mesons is passed through an
iron absorber. Given that the interaction cross-section of π+
mesons with
iron is 600 mb/nucleus, calculate the thickness of iron necessary to attenuate
the π+
beam by a factor of 500. Explain why muons will be reduced to a
much less extent. (The density of iron is 7,900 kg m−2
, the atomic mass being
55.85 amu)
9.16 A neutrino of high energy (E0 m) undergoes an elastic scattering with
stationary electron of mass m. The electron recoils at an angle φ with energy
T . Show that for small angle, φ = (2m/T )1/2
9.17 Why the following reactions can not proceed as strong interactions
(a) π−
+ p → K−
+ Σ+
(b) d + d →4
He + π0
?

9.2 Problems 491
9.18 From the data given calculate the mass of the hyper-fragment ΛHe5
Mass of Λ hyperon is 1115.58 MeV/c2
, Mass of He4
is 3727.32 MeV/c2
Binding energy BΛ for He5
is 3.08 MeV
9.19 Show that for a high energy electron scattering at an angle θ, the value of Q2
is given approximately by
Q2
= 2Ei Ef(1 − cos θ)
where Ei and Ef are the initial and final values of the electron energy and
Q2
is the four-momentum transfer squared. State when this approximation is
justified.
9.2.4 Decay
9.20 Consider the decay process K+
→ π+
π0
with the K+
at rest. Find
(a) the total energy of the π0
meson
(b) its relativistic kinetic energy
The rest mass energy is 494 MeV for K+
, 140 MeV for π+
, 135 MeV for π0
.
9.21 A collimated beam of π-mesons of 100 MeV energy passes through a liquid
hydrogen bubble chamber. The intensity of the beam is found to decrease with
distance s along its path as exp(-as) with a = 9.1×10−2
m−1
. Hence calculate
the life time of the π-meson (Rest energy of the π-meson 139 MeV)
9.22 One of the decay modes of K+
mesons is K+
→ π+
+ π+
+ π−
. What is the
maximum kinetic energy that any of the pions can have, if the K+
decays at
rest? Given mK = 966.7 me and mπ = 273.2 me
9.23 Show that the energy of the neutrino in the pion rest frame, E′
ν, can be written
in natural units as
E′
ν =

m2
π − m2
μ

/2mπ
where mπ and mμ are the masses of the charged pion and the muon, respec-
tively. (You may assume here that the neutrino mass is negligible)
9.24 Find the maximum energy of neutron in the decay of Σ+
hyperon at rest via the
scheme, Σ+
→ n +μ+
+νμ. The masses of Σ+
, n, μ+
and νμ are 1,189 MeV,
939 MeV, 106 MeV and zero, respectively.
9.25 A charged kaon, with an energy of 500 MeV decays into a muon and a neu-
trino. Sketch the decay configuration which leads to the neutrino having the
maximum possible momentum, and calculate the magnitude of this value
(mass of charged kaon is 494 MeV/c2
, mass of muon is 106 MeV/c2
)

9.26 Heavy mesons, M = 950 me produced in nuclear interactions initiated by a
more energetic beam of π-meson, have an energy of 50 MeV. Their tracks up
to the point of decay have a mean length of 1.7 m. Calculate their mean life
time.
9.27 A pion beam from an accelerator target has momentum 10 GeV/c. What frac-
tion of the particles will not have decayed into muons in a pathlength of
100 m?
Out of the pions decays one muon of a 8 GeV/c and a neutrino are produced
at the beginning of the flight path. Assuming that the decay particles follow
the same path, calculate the difference in arrival times at the end of the path.
(Pion mass = 139.6 MeV, muon mass = 105.7 MeV. Mean life time of pion
= 2.6 × 10−8
s, c = 3 × 108
m s−1
)
9.28 Pions in a beam of energy 5 GeV decay in flight. What are the maximum and
minimum energies of the muons from these decays? (mπ = 139.5 MeV/c2
;
mμ = 105.7 MeV/c2
)
9.29 Assuming that a drop in intensity by a factor less than 10 is tolerable, show
that a 1 GeV/c K+
meson beam can be transported over 10 m without a serious
loss of intensity due to decay, while a Λ -hyperon beam of the same momen-
tum after the same distance will not have useful intensity (Take masses of K+
meson and Λ-hyperon to be 0.5 and 1 GeV/c2
, respectively and their lifetimes
10−8
and 2.5 × 10−10
s respectively.
9.30 If a particle has rest mass m0 and momentum p, show that the distance traveled
in one lifetime is d = pT0/m0 where T0 is the life time in the frame of
reference in which the particle is at rest.
9.31 A beam of muon neutrinos is produced from the decay of charged pions of
Eπ = 20 GeV. Show that the relationship between the neutrino energy in the
laboratory frame, Eν, and its angle relative to the pion beam θ, for small θ, is
Eν =
Eπ
(1 + γ 2θ2)
=
1 − mμ
2
/mπ
2

where Eπ is the energy of the pion and γ = Eπ /mπ 1
9.32 It is intended to use a charged mono-energetic hyperon beam to perform scat-
tering experiments off liquid hydrogen. Assuming that the beam transport sys-
tem must have a minimum length of 20 m, calculate the minimum momentum
of a Σ beam such that 1% of the hyperons accepted by the transport system
arrive at the hydrogen target (τ = 0.8 × 10−10
s, mΣ = 1.19 GeV/c2
). What

9.2 Problems 493
measurements would you make to identify the elastic scattering in a (Σ−
p)
collision?
9.33 The energy spectrum for the electrons emitted in muon decay is given by
dω
dEe
=
2G2
F(mμc2
)2
E2
e
(2π)3(c)6

1 −
4Ee
3mμc2

where the electron mass is neglected. Calculate the most probable energy for
the electron. Show on a diagram the orientation of momenta of the decay
product particles and their helicitis when Ee ≈ mμc2
/2. Furthermore, show
the helicity of the muon. Integrate the energy spectrum to find the total
decay width of the muon. Hence compute the muon mean lifetime in seconds
[GF/(c)3
= 1.166 × 10−5
GeV−2
]
9.34 (a) Explain the following statements. The mean lifetime of the π+
meson is
2.6 × 10−8
s while that of π0
is 0.8 × 10−16
s.
(b) The π+
and π−
mesons are of equal mass, but the Σ+
and Σ−
baryon
masses differ by 8 MeV/c2
(c) The mean life of the Σ0
baryon is many orders of magnitude smaller than
those of the Λ and Ξ0
baryons.
9.35 In a bubble chamber two tracks originate from a common point, one caused
by a proton of 440 MeV/c and the other one by π−
meson of momentum
126 MeV/c. The angle between the tracks is 64◦
. Determine the mass of the
unknown particle and identify it.
9.2.5 Ionization Chamber, GM Counter and Proportional Counters
9.36 The dead time of a counter system is to be determined by taking measurements
on two radioactive sources individually and collectively. If the pulse counts
over a time interval t are, respectively, N1, N2 and N12, what is the value of
the dead time?
9.37 An ionization chamber is connected to an electrometer of capacitance 0.5 µµ F
and voltage sensitivity of 4 divisions per volt. A beam of α-particles causes a
deflection of 0.8 divisions. Calculate the number of ion pairs required and
the energy of the source of the α-particles (1 ion pair requires energy of
35 eV, e = 1.6 × 10−19
Coulomb)
9.38 An ionization chamber is used with an electrometer capable of measuring 7 ×
10−11
A to assay a source of 0.49 MeV beta particles. Assuming saturation
conditions and that all the particles are stopped within the chamber, calculate
the rate at which the beta particles must enter the chamber to just produce a
measurable response. Given the ionization potential for the gas atoms is 35 eV.

9.39 Estimate the gas multiplication required to count a 2 MeV proton which gives
up all its energy to the chamber gas in a proportional counter. Assume that the
amplifier input capacitance in parallel with the counter is 1.5×10−9
F and that
its input sensitivity is 1 mV. Energy required to produce one ion pair is 35 eV.
9.40 Calculate the pulse height obtained from a proportional counter when a 14 keV
electron gives up all its energy to the gas. The gas multiplication factor of the
proportional counter is 600, capacitance of the circuit is 1.0 × 10−12
F and
energy required to produce an ion pair is 35 eV
9.41 The plateau of a G.M. Counter working at 800 V has a slope of 2.0% count
rate per 100 V. By how much can the working voltage be allowed to vary if the
count rate is to be limited to 0.1%?
9.42 An organic-quenched G.M. tube has the following characteristics.
Working voltage 1,000 V, Diameter of anode 0.2 mm, Diameter of cathode
2.0 cm. Maximum life time 109
counts.
What is the maximum radial field and how long will it last if used for 30 h
per week of 3,000 counts per minute?
[University of New Castle]
9.43 A G.M. tube with a cathode and anode of 2 cm and 0.10 mm radii respectively
is filled with Argon gas to 10 cm Hg pressure. If the tube has 1.0 kV applied
across it, estimate the distance from the anode, at which electron gains just
enough energy in one mean free path to ionize Argon. Ionization potential of
Argon is 15.7 eV and mean free path in Argon is 2 × 10−4
cm at 76 cm Hg
pressure.
9.44 A S35
containing solution had a specific activity of 1m Curie per ml. A 25 ml
sample of this solution mass was assayed.
(a) in a Geiger-Muller counter, when it registered 2,000 cpm with a back-
ground count of 750 in 5 min; and
(b) in a liquid scintillation counter, where it registered 9,300 cpm with a back-
ground count of 300 in 5 min.
Calculate the efficiency as applied to the measurement of radioactivity and
discuss the factors responsible for the difference in efficiency of these two
types of counters.
[University of Dublin]
9.45 The dead time of a G.M. counter is 100 µs. Find the true counting rate if the
measured rate is 10,000 counts per min.
9.46 A pocket dosimeter has a capacitance of 5.0 PF and is fully charged by a
potential of 100 V. What value of leakage resistance can be tolerated if the
meter is not to lose more than 1% of full charge in 1 day?
9.47 A G.M tube with a cathode 4.0 cm in diameter and a wire diameter of 0.01 cm
is filled with argon in which the mean-free-path is 8×10−4
cm. Given that the

9.2 Problems 495
ionizing potential for Argon is 15.7 V, calculate the maximum voltage which
must be applied to produce the avalanche.
9.48 An electrometer with a capacitance 0.4 PF and 5.0 PF chamber are both
charged to a potential of 600 V. Calculate the potential of the combined system
after 8 × 109
ions have been collected in the chamber.
9.2.6 Scintillation Counter
9.49 Pions and protons, both with momentum 4 GeV/c, travel between two scintil-
lation counters distance L m apart. What is the minimum value of L necessary
to differentiate between the particles if the time of flight can be measured with
an accuracy of 100 ps?
9.50 In the historical discovery of antiproton, negatively charged particles of pro-
tonic mass had to be isolated from a heavy background of K−
, π−
and μ−
.
The negatively charged particles which originated from the bombardment of a
target with 5.3 GeV protons, were subject to momentum analysis in a magnetic
field which permitted only those particles with momentum p = 1.19 GeV/c
to pass through a telescope system comprising two scintillation counters in
coincidence with a separation of d =12 metres between the detectors. Identify
the particles whose time of flight in the telescope arrangement was determined
as t = 51 ± 1 ns. What would have been the time of flight of π mesons?
9.51 A scintillation spectrometer consists of an anthracene crystal and a 10-stage
photomultiplier tube. The crystal yields about 15 photons for each 1 keV of
energy dissipated. The photo-cathode of the photomultiplier tube generates
one photo-electron for every 10 photons striking it, and each dynode produces
3 secondary electrons. Estimate the pulse height observed at the output of
the spectrometer if a 1 MeV electron deposits its energy in the crystal. The
capacitance of the output circuit is 1.2 × 10−10
F.
9.52 A sodium iodide crystal is used with a ten-stage photomultiplier to observe
protons of energy 5 MeV. The phosphor gives one photon per 100 eV of energy
loss. If the optical collection efficiency is 60% and the conversion efficiency of
the photo-cathode is 5%, calculate the average size and the standard deviation
of the output voltage pulses when the mean gain per stage of the multiplier is
3 and the collector capacity is 12 PF.
9.53 A 400-channel pulse-height analyzer has a dead time τ = (17 + 0.5 K) µs
when it registers counts in channel K. How large may the pulse frequencies
become if in channel 100 the dead time correction is not to exceed 10%?
Repeat the calculations for the channel 400.
9.54 An anthracene crystal and a 12-stage photomultiplier tube are to be used as
a scintillation spectrometer for β-rays. The phototube output circuit has a

combined capacitance of 45 PF. If an 8-mV output pulse is desired whenever a
55-keV beta particle is incident on the crystal, calculate the electron multipli-
cation required per stage. Assume perfect light collection and a photo-cathode
efficiency of 5% (assume 550 photons per beta particle)
9.55 Figure 9.1 shows the gamma-ray spectrum of 22
Na in NaI scintillator. Indicate
with explanation the origin of the parts labeled as A, B, C, D and E, Given that
22
Na is a positron emitter and emits a γ -ray of 1.275 MeV.
Fig. 9.1
9.56 Assuming that the shape of the photpeak in the scintillation counter is described
by the normal distribution, show that the half width at half-maximum is
HWHM = 1.177 σ
9.57 The peak response to the 661 keV gamma rays from 137
Cs occurs in energy
channel 298 and 316. Calculate the standard deviation of the energy, and the
coefficient of variation of the energy determination, assuming the pulse ana-
lyzer to be linear.
9.2.7 Cerenkov Counter
9.58 Explain what is meant by Cerenkov radiation. How may a Cerenkov detector
distinguish between a kaon and a pion with the same energy? A pion of energy
20 GeV passes through a chamber containing CO2 at STP. Calculate the angle
to the electron’s path with which the Cerenkov radiation is emitted. (Use the
result that the velocity of a relativistic particle v = c(1 − γ −2
)1/2
. [Mass of
pion = 140 MeV, Mp = 938 MeV; refractive index of CO2 at STP = 1.0004]
9.59 An electron incident on a glass block of refractive index 1.5 emits Cerenkov
radiation at an angle 45◦
to its direction of motion. At what speed is the elec-
tron travelling?
9.60 Consider Cerenkov radiation emitted at angle θ relative to the direction of a
charged particle in a medium of refractive index n. Show that its rest mass
energy mc2
is related to its momentum by mc2
= pc(n2
cos2
θ − 1)1/2

9.2 Problems 497
9.61 In an experiment using a Cerenkov counter, one measures the kinetic energy of
a given particle species as E(kin) = 420 MeV and observes that the Cerenkov
angle in flint glass of refractive index (μ = 1.88) is θ = arc cos(0.55). What
particles are being detected (calculate their mass in me units)
9.62 Calculate the number of Cerenkov photons produced by a particle travelling
at β = 0.95 in water (n = 1.33) in the response range (3500–5500 A) per unit
path length.
9.63 Estimate the minimum length of a gas Cerenkov counter that could be used
in threshold mode to distinguish between charged pions and charged kaons
with momentum 15 GeV/c. Assume that a minimum of 175 photons need to
be radiated to ensure a high probability of detection. Assume also that the radi-
ation covers the whole visible spectrum between 400 and 700 nm and neglect
the variation with wavelength of the refractive index of the gas (n = 1.0004)
9.64 What type of material would you choose for a threshold Cerenkov counter
which is to be sensitive to 900 MeV/c pions but not to 900 MeV/c protons.
9.2.8 Solid State Detector
9.65 A depletion-layer detector has an electrical capacitance determined by the
thickness of the insulating dielectric. Estimate the capacitance of a silicon
detector with the following characteristics: area 1.5 cm2
, dielectric constant
10, depletion layer 40 µm. What potential will be developed across the capac-
itance by the absorption of a 5.0 MeV alpha particle which produces one ion
pair for each 3.5 eV dissipated?
9.2.9 Emulsions
9.66 The range of protons in C2 emulsion is given in the following table (Range in
microns, energy in MeV). Draw a graph of the Range – Energy-Relation for
Deuterons and 3
He particles.
R 0 50 100 150 200 250 300 350 400 450 500
E 0 2.32 3.59 4.61 5.48 6.27 7.01 7.69 8.32 8.91 9.47
9.2.10 Motion of Charged Particles in Magnetic Field
9.67 For a relativistic particle of charge e moving in a circular orbit of radius r in
a magnetic field B perpendicular to the orbital plane, show that p = 0.3 Br,
where the momentum p is in GeV/c and B in Tesla and r in metres.

9.68 A 10 GeV proton is directed toward the centre of the earth when it is at dis-
tance of 103
earth radius away. Through what mean transverse magnetic field
(assumed uniform) must it move if it is to just miss the earth?
9.69 A proton of kinetic energy 20 keV enters a region where there is uniform elec-
tric field of 500 V/cm, acting perpendicularly to the velocity of the proton.
What is the magnitude of a superimposed magnetic field that will result in no
deflection of the proton?
9.70 A particle of known charge but unknown rest mass is accelerated from rest by
an electric field E into a cloud chamber. After it has traveled a distance d, it
leaves the electric field and enters a magnetic field B directed at right angles
to E. From the radius of curvature of its path, determine the rest mass of the
particle and the time it had taken to traverse the distance d.
9.71 A proton accelerated to 500 keV enters a uniform transverse magnetic field of
0.51 T. The field extends over a region of space d = 10 cm thickness. Find
the angle α through which the proton deviates from the initial direction of its
motion.
9.72 An electron is projected up, at an angle of 30◦
with respect to the x-axis, with a
speed of 8 × 105
m/s. Where does the electron recross the x-axis in a constant
electric field of 50 N/C, directed vertically upwards?
9.73 A proton moving with a velocity 3×105
ms−1
enters a magnetic field of 0.3 T,
at an angle of 30◦
with the field. What will the radius of curvature of its path
be (e/m for proton ≈ 108
C kg−1
)?
9.74 What energy must a proton have to circle the earth at the magnetic equator?
Assume a 1 G magnetic field.
9.75 The average flux of primary cosmic rays over earth’s surface is approxi-
mately 1 cm−2
s−1
and their average kinetic energy is 3 GeV. Calculate the
power delivered to the earth from cosmic rays in giga watts. (Earth’s radius
= 6,400 km)
9.76 Kaons with momentum 25 GeV/c and deflected through a collimator slit at a
distance of 10 m from a bending magnet 1.5 m long which produces a field of
1.2 T. What should be minimum width of the slit so that it accepts particles of
momenta within 1% of the central value?
9.2.11 Betatron
9.77 The orbit of electrons in a betatron has a radius of 1m. If the magnetic field
in which the electrons move is changing at the rate of 50 W m−2
s−1
, calculate
the energy acquired by an electron in one rotation. Express your answer in
electron volts.

9.2 Problems 499
9.78 In a betatron of diameter 72 in operating at 50 c/s the maximum magnetic field
is 1.2 Wb/m2
. Calculate
(a) the number of revolutions made each quarter cycle
(b) the maximum energy of the electrons
(c) the average energy gained per revolution
9.2.12 Cyclotron
9.79 A cyclotron is designed to accelerate protons to an energy of 6 MeV. Deduce
the expression for the kinetic energy in terms of radius, magnetic field etc and
calculate the maximum radius attained.
9.80 If the acceleration potential (peak voltage, 50,000) has a frequency of 1.2 ×
107
c/s, find the field strength for cyclotron resonance when deuterons and
alpha particles, respectively are accelerated. For how long are the parti-
cles accelerated if the radius of orbit at ejection is 30.0 cm? Deuteron rest
mass = 2.01 amu. α-particle rest mass = 4.00 amu
9.81 The original (uniform magnetic field) type of fixed frequency cyclotron was
limited in energy because of the relativistic increase of mass of the accelerated
particle with energy. What percentage increment to the magnetic flux density
at extreme radius would be necessary to preserve the resonance condition for
protons of energy 20 MeV?
9.82 A cyclotron is powered by a 50,000 V, 5 Mc/s radio frequency source. If its
diameter is 1.524 m. what magnetic field satisfies the resonance conditions for
(a) protons (b) deuterons (c) alpha-particles? Also what energies will these
particles attain?
9.83 In a synchrocyclotron, the magnetic flux density decreases from 15,000 G at
the centre of the magnet to 14,300 G at the limiting radius of 206 cm. Calculate
the range of frequency modulation required for deuteron acceleration and the
maximum kinetic energy of the deuteron. (The rest mass of the deuteron is
3.34 × 10−24
g)
9.84 Calculate the magnetic field, B and the Dee radius of a cyclotron which will
accelerate protons to a maximum energy of 5 MeV if a radio frequency of
8 MHz is available.
9.85 When a cyclotron shifts from deuterons to alpha particles it is necessary to
drop the magnetic field slightly. If the atomic masses of 2
H and 4
He are
2.014102 amu and 4.002603 amu, what is the percentage decrease in field
strength that is required?

9.86 A synchro-cyclotron has a pole diameter of 4 m and a magnetic field of
1.5 W m−2
(15,000 G). What is the maximum energy that can be transmitted
to electrons which are struck by protons extracted from this accelerator?
9.87 If the frequency of the dee voltage at the beginning of an accelerating sequence
is 20 Mc/s, what must be the final frequency if the protons in the pulse have
an energy of 469 MeV?
9.2.13 Synchrotron
9.88 At what radius do 30 GeV protons circulate in a synchrotron if the guide field
is 1 W m−2
9.89 Calculate the orbit radius for a synchrotron designed to accelerate protons to
3 GeV assuming a guide field of 14 kG
9.90 What percentage depth of modulation must be applied to the dee voltage of
a synchrotron in order to accelerate protons to 313 MeV assuming that the
magnetic field has a 5% radial decrease in magnitude.
9.91 An electron synchrotron with a radius of 1 m accelerates electrons to 300 MeV.
Calculate the energy lost by a single electron per revolution when it has
reached maximum energy.
[Andhra University 1966]
9.92 Show that the radius R of the final orbit of a particle of charge q and rest mass
m0 moving perpendicular to a uniform field of magnetic induction B with a
kinetic energy n times its own rest mass energy is given by
R = m0 c (n2
+ 2n)1/2
/qB
9.93 Protons of kinetic energy 50 MeV are injected into a synchrotron when the
magnetic field is 147 G. They are accelerated by an alternating electric field
as the magnetic field rises. Calculate the energy at the moment when the mag-
netic field reaches 12,000 G (rest energy of proton = 938 MeV)
9.94 A synchrotron (an accelerator with an annular magnetic field) accelerates pro-
tons (mass number A = 1) to a kinetic energy of 1,000 MeV. What kinetic
energy could be reached by deuteron (A = 2) or 3
He (A = 3, Z = 2)
when accelerated in this machine? Take the proton mass to be equivalent to
1,000 MeV.

9.2 Problems 501
9.95 A synchrocyclotron accelerates protons to 500 MeV. B = 18 kG, V = 10 kV
and ϕs = 300
. To find
(a) the radius of the orbit at extraction
(b) Energy of ions for acceptance
(c) the initial electric frequency limits
(d) the range of frequency modulation
9.96 Explain how a synchrotron accelerates particles. What is the main energy loss
mechanism in these devices? How much more power is needed to maintain a
beam of 500 GeV electrons in a synchrotron of radius 1 km than to maintain a
beam of protons of the same energy? Is this feasible?
[University of Aberyswyth 2003]
9.97 Protons are accelerated in a synchrotron in the orbit of 10 m. At one moment in
the cycle of acceleration, protons are making one revolution per microsecond.
Calculate the value at this moment of the kinetic energy of each proton in MeV.
9.98 Electrons are accelerated to an energy of 10 MeV in a linear accelerator, and
then injected into a synchrotron of radius 15 m, from which they are acceler-
ated with an energy of 5 GeV. The energy gain per revolution is 1keV
(a) Calculate the initial frequency of the RF source. Will it be necessary to
change this frequency?
(b) How many turns will the electron make?
(c) Calculate the time between injection and extraction of the electrons
(d) What distance do the electrons travel within the synchrotron?
9.2.14 Linear Accelerator
9.99 Protons of 2 MeV energy enter a linear accelerator which has 97 drift tubes
connected alternately to a 200 MHz oscillator. The final energy of the pro-
tons is 50 MeV (a) What are the lengths of the second cylinder and the last
cylinder (b) How many additional tubes would be needed to produce 80 MeV
protons in this accelerator?
9.100 The Stanford linear accelerator produces 50 pulses per second of about 5 ×
1011
electrons with a final energy of 2 GeV. Calculate (a) the average beam
current (b) the power output.
9.101 A section of linear accelerator has five drift tubes and is driven by a 50 Mc
oscillator. Assuming that the protons are injected into the first drift tube at
100 kV and gain 100 kV in every gap crossing
(a) What is the output energy after the fifth drift tube?
(b) What is the total length of the whole section? (Ignore the gap length)
[AEC 1966 Trombay]

9.102 What is the length L of the longest drift tube in a linac which operating at a
frequency of f = 25 MHz is capable of accelerating 12
C ions to a maximum
energy of E = 80 MeV?
9.2.15 Colliders
9.103 Two beams of particles consisting of n bunches with N1 and N2 particles
in each circulate in a collider and make head-on-collisions. A, the cross-
sectional area of the beam and f is the frequency with which the particles
circulate, obtain an expression for the luminosity L.
9.104 In an electron-positron collider the particles circulate in short cylindrical
bunches of radius 1.2 mm. The number of particles per bunch is 6 × 1011
and
the bunches collide at a frequency of 2 MHz. The cross-section for μ+
μ−
creation at 8 GeV total energy is, 1.4 × 10−33
cm2
; how many μ+
μ−
pairs
are created per second?
9.105 (a) Show that in a head-on-collision of a beam of relativistic particles of
energy E1 with one of energy E2, the square of the energy in the CMS is 4
E1 E2 and that for a crossing angle θ between the beam this is reduced by
a factor (1+cos θ)/2, (neglect the masses of beam particles in comparison
with the energy)
(b) Show that the available kinetic energy in the head-on-collision with two
25 GeV protons is equal to that in the collision of a 1,300 GeV proton with
a fixed hydrogen target. (Courtesy D.H. Perkins, Cambridge University
Press)
9.106 Head-on collisions are observed between protons each moving with velocity
(relative to the fixed observer) corresponding to 1010
eV. If one of the protons
were to be at rest relative to the observer, what would the energy of the other
need to be so as to produce the same collision energy as before. The rest
energy of the proton is 109
eV.
9.107 It is required to cause protons to collide with an energy measured in their
centre of mass frame, of 4 M0c2
in excess of their rest energy 2 M0c2
. This
can be achieved by firing protons at one another with two accelerators each of
which imparts a kinetic energy of 2 M0c2
. Alternatively, protons can be fired
from an accelerator at protons at rest. How much energy must this single
machine be capable of imparting to a proton? What is the significance of this
result for experiments in high energy nuclear physics?
9.108 The HERA accelerator in Hamburg provided head-on collisions between
30 GeV electrons and 820 GeV protons. Calculate the centre of mass energy
that was produced in each collision.
[Manchester 2008]

9.3 Solutions 503
9.109 At a collider, a 20 GeV electron beam collides with a 300 GeV proton beam at
a crossing angle of 10◦
. Evaluate the total centre of mass energy and calculate
what beam energy would be required in a fixed-target electron machine to
achieve the same total centre-of-mass energy.
9.3 Solutions
9.1 E = Mc2
= 1 × (3 × 108
)2
= 9 × 1016
J
= 9 × 1016
J/(1.6 × 10−10
J/GeV) = 5.63 × 1026
GeV
9.2 (a) c = 197.3 MeV-fm
In natural units, = c = 1
Therefore 1 = 197.3 MeV-fm = 0.1973 GeV-10−15
m
Therefore 10−15
m = 1/0.1973 GeV−1
or 1 m = 1015
/0.1973 = 5.068 × 1015
GeV−1
(b) From (a) we have
1 m2
= (5.068)2
× 1030
GeV−2
= 25.6846 × 1030
GeV−2
Therefore 1 GeV−2
= 10−30
/25.6846 m2
= 0.389 mb
(c) = 1.055 × 10−34
J-s
1 = 1.055 × 10−34
J-s/1.6 × 10−10
J/GeV
Therefore 1 s = (1.6 × 10−10
/1.055 × 10−34
) GeV−1
= 1.5 × 1024
GeV−1
9.3 (a) In practical units λc = /me c
Put = c = 1
In natural units λc = 1/me
(b) In practical units Bohr’s radius of hydrogen atom is
a0 = ε0 2
/π me2
= ε0 c/πm ce2
In natural units a0 = 1/αme
(c) In the ground state velocity
v = /ma0 = α
where we have used the results of (b)
Put = 1 to find v = α in natural units.
Numerical values:
λc = /mec = (c/mec2
) MeV.fm/MeV = 197.3/0.511 fm = 385 fm
= 0.00385 × 10−10
m
= 0.00385 Å
a0 = 4πε0 2
/e2
m = (4 πε0/e2
) ( c)2
/mc2
= [1/(1.44 MeV-fm)] × (197.3 MeV-fm)2
/0.511 MeV
= 53,000 fm = 0.53 × 10−10
m
v = αc = (3 × 108
/137) ms−1
= 2.19 × 106
ms−1

9.4 In natural units τ = 2/mα5
. Introduce the factors and c
Let τ = 2x
cy
/m α5
Find the dimensions: [] = [Energy × Time], [c] = [velocity]
[τ] = [1/m][ML2
T−1
]x
[LT−1
]y
[T ] = [M]x−1
[L]2x+y
[T ]−x−y
Equating the coefficients on both sides of the equation,
−x − y = 1, 2x + y = 0, x − 1 = 0
we get x = 1, y = −2
Therefore τ = 2c−2
/mα5
= 2 /mc2
α5
τ = 2 × (0.659 × 10−21
MeV-s) × (137)5
/0.511 MeV
= 1.245 × 10−10
s
9.5 Γμ = GF
2
mμ
5
/192π3
(1)
Now, the dimensional formula for GF is
[Energy]−2
[ c]3
and [Γμ] = [Energy]
Introduce and c in (1) and take dimensions on both sides.
Γμ = GF
2
mμ5
x
cy
/192π3
(2)
[M L2
T−2
] = [M L2
T −2
]−4
[M]5
[M L2
T −1
]x+6
[L T−1
]y+6
or [M L2
T −2
] = [M]7+x
[L]10+2x+y
[T ]−4−x−y
Equating powers of M, L and T , we find
x = −6 and y = 4
and (2) becomes
Γμ = GF
2
mμ
5
−6
c4
/192π3
Γμ = /τ = GF
2
(mμc2
)5
/192 (c)6
π3
= (1.116 × 10−5
GeV−2
)2
(105.659 × 10−3
)5
/192π3
τ = 2.39 × 10−6
s
9.3.2 Production
9.6 (a) The relation for lab angle θ and the C.M.S. angle θ∗
is given by
tan θ = sin θ∗
/γc(cos θ∗
+ βc/β∗
) (1)
where βc = vc/c is the CMS velocity and γc is the corresponding Lorentz
factor (see summary of Chap. 6). For photon β∗
= 1 and βc = 1 as the
electron is ultra relativistic. Dropping off the subscript c, (1) becomes
tan θ = sin θ∗
/γ (cos θ∗
+ 1) = tan (θ∗
/2)/γ (2)
(b) Assuming that the photons are emitted isotropically, half of the photons will
be contained in the forward hemisphere in the CMS, that is within θ∗
= 90◦
.
Substituting θ∗
= 90◦
in (2)
tan θ = 1/γ
As the incident electron is ultrarelativistic the photons in the lab would come
off at small angles so that tan θ ≈ θ = 1/γ . Thus half of the photons will be
emitted within a cone of half angle θ ≈ 1/γ .

9.3 Solutions 505
9.7 If a particle m1 moving with total energy E1 collides with the target of mass
m2, then the total energy in the CMS will be
E∗
=

m1
2
+ m2
2
+ 2 m2 E1
1/2
If E1 m1 or m2, then E∗
∝
√
E1
9.8 ECM = E∗
≈ (2mELab)1/2
= (2 × 0.511 × 10−3
× 50)1/2
= 0.226 GeV = 226 MeV
σ = 4πα2
2
c2
/3(ECM)2
= (4π/1372
) × (197.3 MeV-fm)2
/(3 × 2262
) = 1.7 × 10−4
fm2
= 1.7 × 10−30
cm2
Macroscopic cross-section per atom
Σ = σ N0ρ/A = 1.7×10−30
×6.02×1023
×11.4/207 = 0.5636×10−7
cm−1
Σe = Z.Σatom = 82 × 0.5636 × 10−7
= 4.62 × 10−6
cm−1
Interaction mean free path λ = 1/Σe = 1/4.62 × 10−6
= 2.16 × 105
cm
= 2.16 km
9.9 (a) Minimum energy required in the C.M.S. is
E∗
= M + M = 2 × 0.938 = 1.876 GeV
If E is the positron energy required in the LS, then
E = E∗2
/2me = (1.876)2
/2 × 0.511 × 10−3
= 3,444 GeV
(b) Energy requirements are drastically reduced
(c) Each beam of energy mp − me ≈ 938 MeV, need to be oppositely directed.
9.3.3 Interaction
9.10 In the collision of a particle of mass m1 of total energy E1 with m at rest, the
centre-of-mass energy is
E∗
=

m1
2
+ m2
2
+ 2 m2 E1
1/2
(1)
Here, m1 = 938 MeV, m2 = 938 MeV, E1 = 938 + 200 = 1,138 MeV
Using these values in (1), we find E∗
= 1973 MeV
Useful energy for particle production
= E∗
− (m1 + m2) = 1, 973 − (938 + 938) = 97 MeV
Pion can not be produced as the threshold energy is 290 MeV. Muons and
electron – positron pairs are not produced in strong interactions. The reso-
nance Δ(1, 236 MeV) also can not be produced.
9.11 σ = 10−42
E = 10−42
× 200 = 2 × 10−40
m2
Number of nucleons/m3
, n = ρ/mp = 7, 900/1.67 × 10−27
= 4.7 × 1030
Σ = nσ = 4.7 × 1030
× 2 × 10−40
= 9.4 × 10−10
m−1
The mean free path λ = 1/Σ = 1/9.4 × 10−10
= 1.06 × 109
m.
If t is the thickness of iron then the probability for interaction
P = t/λ = t/1.06 × 109
= 1/109
Required thickness of iron, t ≈ 1.0 m.

9.12 (a) In elastic scattering, total kinetic energy is conserved. Life-time means
mean-life-time i.e., the time in which the population of unstable particles
is reduced by a factor e. The important interactions that the muons undergo
are elastic scattering and the reaction, μ−
+ p → n + νµ.
Let the muons travel a distance d metres. Then their intensity will be
reduced due to decay by a factor
I/I0 = exp(−d/vγ τ0) (1)
γ = 1/(1 − v2
/c2
)1/2
= 1/
,
1 − (106
/3 × 108
)2
-1/2
= 1.0033
I/I0 = exp −(d/(1.0033 × 106
× 2.2 × 10−6
))
= exp −(0.453 d) (2)
The intensity reduction due to interaction will be
I/I0 = exp(−dΣ) (3)
Σ = σ n = 0.1 × 10−28
× 2.69 × 1025
= 2.69 × 10−4
I/I0 = exp(−2.69 × 10−4
) d (4)
Comparing (2) and (4), reduction due to decay will be by far greater than
by interaction for any value of d.
(b) Repetition of calculation for T = 4 GeV, γ = 38.7, β ≈ 1, and d =
12,000 m, gives I/I0 = e−0.47
= 0.625, so that a large number of muons
survive at the ground level. The distance at which muons of energy 4 GeV
will be reduced by e is given by d = vγ τ0 = 3×108
×38.7×2.2×10−6
=
2.55 × 104
m = 25.5 km.
9.13 The geometrical cross-section is
σg = π R2
= π(r0 A1/3
)2
= π r0
2
A2/3
= π(1.3 × 10−13
)2
(208)2/3
= 1.86 × 10−24
cm2
Number of lead atoms/cm3
, n = N0 ρ/A = 6 × 1023
× 11.3/207 =
3.275 × 1022
Macroscopic cross-section
Σ = n σ = (3.275 × 1022
) (1.86 × 10−24
) = 0.06 cm−1
The interaction length, λ = 1/Σ = 1/0.06 = 16.7 cm
9.14 Number of interactions per second in volume V
I = ΣVI0
= σ N0 ρV I0/A
= (40 × 10−27
cm2
)(6.02 × 1023
× 0.071/2) × 125 × 2 × 103
= 213.7 neutral pions/s are produced.
Each pion decays into two photons. Therefore number of photons produced
per second = 427
9.15 I/I0 = exp(−Σx)
Σ = nσ = Nρσ/A

9.3 Solutions 507
= 6.02 × 1023
× 7.9 × 600 × 10−27
/55.95 = 0.051 cm−1
I/I0 = 1/500 = exp(−0.051 x)
Take loge on both sides of the equation and solve for x.
We find x = 122 cm = 1.22 m.
Thus a thickness of 1.22 m iron will reduce the pion beam by a factor of
500. This does not include the reduction due to the decay of pions. Muons have
much less interaction cross-section so that much greater thickness is required.
9.16 From the momentum triangle (Fig. 9.2)
Pν
2
= p0
2
+ pe
2
− 2 p0 pe cos φ (momentum conservation)
Eν = E0 − T (energy conservation)
Since mν = 0, pν = Eν
Therefore
Pν
2
= Eν
2
= (E0 − T )2
= E0
2
+ T2
+ 2mT − 2E0(T 2
+ 2mT)1/2
cos φ
Simplifying,
E0(T 2
+ 2mT)1/2
cos φ = T (E0 − m) ≈ T E0 (because E0 m)
Squaring and simplifying
tan φ ≈ φ = (2m/T )1/2
Fig. 9.2
9.17 (a) In this reaction initially the total strangeness quantum number
Si = Sπ + Sp = 0 + 0 = 0, while for the final state
Sf = SK + SΣ = −1 + (−1) = −2
ΔS = −2. Therefore, the rule ΔS = 0 for strong interactions is violated.
(b) The total isospin for the initial state I = 0+0 = 0, while for the final state
I = 0 + 1 = 1.
ΔI = 1. Therefore, the rule ΔI = 0 for strong interactions is violated.
9.18 M(5
HeΛ) = M(Λ) + M (He4
) − B(Λ)
= 1,115.58 + 3,727.32 − 3.08 = 4,839.82 MeV/c2
9.19 Let Pi and Pf be the initial and final four-momentum transfer.
Pi = (pi, iEi)
Pf = (pf, iEf)
Q = Pi − Pf = (pi − pf) + i(Ei − Ef)
Q2
= pi
2
+ pf
2
− 2pi.pf − Ei
2
− Ef
2
+ 2 Ei Ef
= −m2
− m2
+ 2 Ei Ef − 2pi pf cos θ
If Ei m and Ef m, then pi ≈ Ei, and pf ≈ Ef.

Q2
= 2 Ei Ef (1 − cos θ)
Thus, the approximation is valid when the particle mass is much smaller
than the energies.
9.3.4 Decay
9.20 (a) If a particle A at rest decays into B and C then the total energy of B will be
EB =

mA
2
+ mB
2
− mC
2

/2mA
Inserting the values mA = 494 MeV, mB = 135 MeV and mC =
140 MeV, we find E(π0
) = 245.6 MeV.
(b) Kinetic energy of π0
is Eπ0 − mπ0 = 245.6 − 135 = 110.6 MeV
9.21 If I0 is the original intensity and I the observed intensity, then
I = I0 e−s/vτ
= I0 e−as
γ = 1 + T/m = 1 + 100/139 = 1.719
β = (γ 2
− 1)1/2
/γ = (1.7192
− 1)1/2
/1.719 = 0.813
a = 1/vτ → τ = 1/va = 1/β ca = 1/(0.813 × 3 × 108
× 9.1 × 10−2
)
τ = 4.5 × 10−8
s
Proper life time τ0 = τ/γ = 4.5 × 10−8
/1.719 = 2.62 × 10−8
s
9.22 mK c2
= 966.7 × 0.511 = 494 MeV
mπ c2
= 273.2 × 0.511 = 139.5 MeV
Energy released in the decay
Q = mK − 3mπ = 494.0 − 3 × 139.5 = 75 MeV
Maximum kinetic energy of one pion will occur when the other two pions
go together in the opposite direction, that is energy is shared between one π
and 2 π’s
Non-relativistically
Tπ (max) = Q × 2mπ /(2mπ + mπ ) = 75 × 2/3 = 50 MeV
Relativistically
Tπ (max) = Eπ − mπ =
,
mK
2
+ mπ
2
− (2mπ )2

/2mK
-
− mπ
Inserting the values of mK and mπ , we find Tπ (max) = 48.4 MeV. The dif-
ference in the two results is 3.3%.
9.23 In the decay A → B + C,
EB =

mA
2
+ mB
2
− mC
2

/2mA
Here B = ν, A = π, C = μ. It follows that
E′
ν =

m2
π − m2
μ

/2mπ (because mν = 0)
9.24 The maximum energy of neutron is obtained when the neutrino has zero
energy, neutron and muon being emitted in the opposite direction.
The energy released in the decay is
Q = mΣ − (mn + mμ + mν) = 1,189 − (939 + 106 + 0) = 144 MeV
Non-relativistically,
Tn = Q mμ/(mμ + mn) = 144 × 106/(106 + 939) = 14.6 MeV
Relativistically,

9.3 Solutions 509
En =

mΣ
2
+ mn
2
− mμ
2

/2mΣ = (1,1892
+ 9392
− 1062
)/2 × 1,189 =
960.6 MeV
Therefore Tn = En − mn = 960.6 − 939 = 21.6 MeV
9.25 Maximum neutrino momentum in the L-system is obtained when the neutrino
is emitted in the forward direction, that is at θ∗
= 0 in the rest system of
K-meson (Fig. 9.3).
The energy of the neutrino in the rest frame of K-meson is
Eν
∗
=

mK
2
− mμ
2

/2mK = 235.6 MeV
Pν
∗
= 235.6 MeV/c (because mν = 0)
Pν = γK pν
∗
(1 + βKβν
∗
cos θ∗
)
βν
∗
= 1; θ∗
= 0; pν
∗
= 235.6 MeV/c
γK = 1 + TK/mK = 1 + 500/494 = 2.012
βK = 0.868
pν = (2.012)(235.6)(1 + 0.868) = 885.5 MeV/c
Fig. 9.3 Decay configuration
9.26 Rest mass energy of the heavy meson, M = 965 × 0.511 = 493 MeV
γ = 1 + T/M = 1 + 50/493 = 1.101
β = (γ 2
− 1)1/2
/γ = (1.1012
− 1)1/2
/1.101 = 0.418
Observed mean lifetime τ is related to proper lifetime by
τ = τ0γ
τ = d/βc = 1.7/0.418 × 3 × 108
= 1.355 × 10−8
s = 1.101 τ0
Therefore τ0 = 1.23 × 10−8
s
9.27 Fraction of pions that will not decay
I/I0 = exp(−t/τ0) = exp(−d/β c γ τ0)
γ = E/m = (p2
+ m2
)1/2
/m = [(p2
/m2
) + 1]1/2
= [(10/0.1396)2
+ 1]1/2
=
71.64
β ≈ 1
Therefore I/I0 =exp(−100/1×3×108
×71.64×2.6×10−8
)= exp(−0.179) =
0.836
γμ = ((p2
/m2
) + 1)1/2
= ((82
/0.10572
) + 1)1/2
= 75.69
βμ ≈ 1
tμ = dγ/βc = 100 × 75.69/1 × 3 × 108
= 25.23 × 10−6
s
tν = d/c = 100/3 × 108
= 0.333 × 10−6
s
Therefore Δt = tμ − tν = (25.23 − 0.33) × 10−6
s
= 24.9 µs
9.28 In the rest system of pion the muon is emitted with kinetic energy
Tμ
∗
= 4 MeV (see Problem 6.54)
Its energy in the LS will be maximum when θ∗
= 0 and minimum when
θ∗
= 180 in the CMS. Applying the relativistic transformation equation

Eμ = Eμ
∗
γπ (1 + βμ
∗
βπ cos θ∗
)
Eμ
∗
= 105.7 + 4.0 = 109.7 MeV; γμ
∗
= 1.0378; βμ
∗
= 0.2676
γπ = 5/0.1395 = 35.84; βπ ≈ 1
Eμ (max) = 4.98 GeV; Eμ (min) = 2.88 GeV
9.29 The decay equation is
I = I0 exp (−t/γ τ) = I0 exp(−d/β c γ τ) (1)
For K+
meson beam
βK = 1/
,
1 + mK
2
/pK
2
-1/2
Substituting mK = 0.5 GeV/c2
and pK = 1 GeV/c
βK = 0.894; γK = 2.23; τK = 10−8
s; d = 10 m
Using these values in Eq. (1)
I/I0 = exp(−1.672) = 0.188
which is tolerable.
For Λ-hyperon beam
βΛ = 1/
,
1 + mΛ
2
/pΛ
2
-1/2
= 0.7
γΛ = 1.96, τΛ = 2.5 × 10−10
s; d = 10 m
Using the above values in (1) we find
I/I0 = exp(−97) ≈ zero
which is not at all useful.
9.30 d = vt (1)
t = γ T0 (2)
p = m0 v γ (3)
d = pT0/m0
9.31 Inverse Lorentz transformations gives
Eν
∗
= γ Eν(1 − β cos θ) = Eν[γ − (γ 2
− 1)1/2
cos θ]
= Eν[γ − γ (1 − 1/2γ 2
)(1 − θ2
/2)]
Eν
∗
≈ (Eν/2γ )(1 + γ 2
θ2
) (1)
where we have neglected terms θ2
/2γ for γ 1 and considered small θ.
But Eν
∗
=

mπ
2
− mμ
2

/2mπ and Eπ = γ mπ (2)
Using (2) in (1) we get the desired result.
9.32 I/I0 = 1/100 = exp(−d/βc γ τ)
Put d = 20 m, c = 3 × 108
m/s and τ = 0.8 × 10−10
s and take loge on
both sides to solve for βγ . We find βγ = 181. Therefore, the momentum of
hyperon,
cp = mβγ = 1.19 × 181 = 215 GeV
Thus, the minimum momentum required is P = 215 GeV/c
The elastic scattering of Σ−
hyperons with protons can be recorded in the
hydrogen bubble chamber from the kinematical fits of events.

9.3 Solutions 511
It may be noted that the beams of sigma hyperons are available only in
recent times. This has been possible because of time dilation for high momenta
particles. Even beams of omega minus have been used.
9.33
dω
dEe
=
2G2
F

mμc2
2
E2
e
(2π)3(c)6

1 −
4Ee
3mμc2

This is the differential energy distribution of electrons from the decay of muon.
The distribution has a maximum, which is easily found out by differentiating
the right hand side of the above expression with respect to Ee and setting it
equal to zero. We easily find the maximum of distribution to occur at Ee =
mμ c2
/2.
Now, if we regard the electron mass to be negligible then the maximum
value of the electron energy will occur when it is emitted in a direction oppo-
site to the two neutrinos, its maximum energy being also mμ c2
/2. A rough
plot of the electron energy spectrum is shown in Fig. 9.4.
Fig. 9.4
The orientation of momenta and helicities of the particles in the decay are
shown in Fig. 9.5.
Fig. 9.5 Pion rest-frame
sketch indicating sense of
spin polarization in pion

The integration of the expression
mμc2
2
0
E2
e

1 − 4Ee/3mμ c2

dEe
gives (mμ c2
)3
/48 so that the full expression for full width is
Γ = GF
2
(mc2
)5
/(c)6
192π3
which is identical with the one stated in Problem 9.5. The mean life time is
obtained from τ = /Γ as in the solution of 9.5.
9.34 (a) The charged pions π±
decay by weak interactions π → μ + ν, and so
their mean lifetime is relatively longer, while the neutral pion π0
decays
via π0
→ 2γ , electromagnetically and therefore their mean life time is
shorter.
(b) The π+
and π−
are particle and antiparticle pair and they are expected to
have the same mass by the CPT theorem. On the other hand Σ+
and Σ−
hyperons are not particle-antiparticle. Actually Σ
−−
is the antiparticle of
Σ+
. Σ+
and Σ−
are the members of the isospin triplet (Σ+
, Σ0
, Σ−
) and
because of difference in their charges can slightly differ in their mass sim-
ilar to the masses of neutron and proton for the isospin doublet of nucleon.
(c) Λ-hyperon decays via Λ → p + π−
or n + π0
, the interaction is weak.
Similarly the hyperon Ξ0
decays via Ξ0
→ Λ + π0
, which is also a weak
decay. In both the cases the lifetimes are relatively long on the nuclear
scale. On the other hand the decay of Σ0
-hyperon via Σ0
→ Λ + γ is
electromagnetic. The explanation is the same as in (a)
9.35 Apply the formula
M2
= mp
2
+ mπ
2
+ 2(Ep Eπ − Pp Pπ cos θ) (1)
Use the values:
mp = 0.939 GeV, mπ = 0.139 GeV, Pp = 0.44 GeV
Pπ = 0.126 GeV, Ep = 1.036 GeV, Eπ = 0.188 GeV, θ = 640
and find
M = 1.114 GeV/c2
. The particle is Λ hyperon
9.36 The problem is based on the double source method for the determination of the
dead time of a G.M. counter. Two radioactive sources of comparable strength
are chosen. In all four counts are taken. First, the background rate B per second
is found out when neither source is present. One of the sources is placed in
a suitable position so that a high counting rate N1 is registered. While this
counter is in the same position, the second source is placed by its side to get
the counting rate N12. Finally, the first source is removed so that the second
counter above gives a count of N2. If n1, n2 and n12 are the true counting rates
then we expect
(n1 + B) + (n2 + B) = n12 + B

9.3 Solutions 513
or
n1 + n2 = n12 − B (1)
Now, for each particle that is counted, on an average there will be a dead
time τ during which particles are not counted. Then if N is the counting rate
(number of counts/second) then the time lost will be Nτ, so that the true
counting rate
n = N/(1 − Nτ)
Thus n1 = N1/(1 − Nτ) etc.
We can then write (1) as
N1/(1 − N1 τ) + N2/(1 − N2 τ) = N12/(1 − N12 τ) − B/(1 − Bτ) (2)
Now, in practice N1 and N2 will be of the order of 100 per second, N12 is of
200 per second, B ≈ 1 per second and τ ≈ 10−4
second, so that N1 τ 1,
etc. We can then expand the denominators binomially and write to a good
approximation
N1(1 + N1τ) + N2(1 + N2τ) = N12(1 + N12τ) − B(1 + Bτ)
N1 + N2 − N12 + B = τ
,
N12
2
− N1
2
− N2
2
-
≈ τ
,
(N1 + N2)2
− N1
2
− N2
2
-
Or τ = (N1 + N2 − N12 + B)/2N1 N2
9.37 The voltage sensitivity is 4 divisions per volt.
So 0.8 divisions correspond to 0.2 V.
The charge deposited, Q = CV = 0.5 × 10−12
× 0.2 = 10−13
Coulomb
If n ion pairs are released then ne = Q
Therefore n = Q/e = 10−13
/1.6 × 10−19
= 6.25 × 105
ion pairs.
Energy of alpha particles = (number of ion pairs) × (Ionization energy)
= (6.25 × 105
)(35) = 21.87 × 106
eV = 21.87 MeV.
9.38 Let n beta particles enter the ionization chamber per second and stopped.
Number of ion pairs released when ionization energy is I is given by
n = N/I = 0.49 × 106
/35 = 1.4 × 104
The ionization current, i = Nne
Therefore, N = i/ne = 7 × 10−11
/1.4 × 104
× 1.6 × 10−19
= 3.125 × 104
beta particles/second
9.39 If M is the gas multiplication factor and N is the number of ion pairs released,
V the voltage developed, C the capacitance and e the electronic charge then
the charge Q deposited will be
Q = CV = MNe
Or M = CV /Ne = 1.5 × 10−9
× 10−3
/(2 × 106
/35) × 1.6 × 10−19
= 164
9.40 Number of primary ions produced by a 14 keV electron, N = 14,000/35 =
400
Charge collected Q = MNe
where M is the gas multiplication and e = 1.6 × 10−19
Coulomb the elemen-
tary charge. If C is the capacitance of the circuit the pulse height will be
V = Q/C = MNe/C = 600 × 400 × 1.6 × 10−19
/10 × 10−12
= 3.84 ×
10−3
V = 3.84 mV.

9.41 Slope =
2%
100 V
=
0.1%
5 V
The voltage should not vary more than ±5 V from the operating voltage of
800 V.
9.42 For the cylindrical geometry of the G.M. tube the electric field is given by
E = V/r ln(b/a)
where V is the applied voltage, r is the distance of a point from the anode, b
and a are the diameters of the cathode and the anode wire.
The field will be maximum close to the anode.
r = 0.1 mm = 10−4
m
Emax = 1, 000/10−4
ln(20/0.2) = 2.17 × 106
V/m
The lifetime of G.M. tube in years is given by dividing total number of
possible counts by counts per year.
t = 109
/(52 × 30 × 60 × 3, 000) = 3.56 years.
In practice, the G.M. tube will not work properly long before the above esti-
mate.
9.43 As the mean free path is inversely proportional to the pressure, the mean free
path at 10 cm pressure will be 2 × 10−4
× (76/10) or 1.52 × 10−3
cm.
At a distance r = mean free path = 1.52 × 10−3
cm. from the anode,
the electric field should be such that the electron acquires sufficient energy to
ionize Argon for which the ionization energy is 15.7 eV. The required value of
E is
E = 15.7/1.52 × 10−3
= 1.03 × 104
V/cm
r = V/E ln(b/a) = 1000/1.03 × 104
ln(20/0.1) = 0.0183 cm = 0.183 mm
9.44 The source gives 3.7 × 107
× 25 = 9.25 × 108
disintegrations/second
G.M. Counting rate of beta rays plus gamma rays
= (2,000/60) − (750/300) = 30.83/s
Therefore, efficiency = 30.83/(9.25 × 108
) = 3.33 × 10−8
Scintillation counter counting rate of beta particles = (9,300/60) − (300/300)
= 154/s
Therefore, efficiency = 154/(9.25 × 108
) = 16.65 × 10−8
For both the counters the efficiency is low because of small solid angle of
acceptance. Both the counters would register beta particles as well as gamma
rays. But the efficiency for counting gamma rays in G.M. counter will be quite
low, being of the order of 1%. This is because gamma rays cause ionization
only indirectly by hitting the walls of the GM counter and ejecting electrons.
The efficiency is relatively higher in the scintillation counter. For this reason,
in the given situation the efficiency for scintillation counter is approximately
five times greater.
9.45 The true counting rate n is related to the observed counting rate n0 by
n = n0/(1 − n0 τ)

9.3 Solutions 515
where τ is the dead time. τ = 100 µs = 10−4
s.
n0 = 104
/ min = 166.7/s
n = (104
/ min)/(1 − 166.7 × 10−4
) = 10,169/ min
9.46 V = V0 e−t/RC
et/RC
= V0/V = 100/99
R = t/C ln(100/99) = 86,400/(5 × 10−12
)(0.01) = 1.73 × 1018
ohm.
9.47 V = Er ln(d1/d2) (1)
Put r = d2/2 = 0.01/2 = 0.005 cm
E = 15.7/λ = 15.7/8 × 10−4
V/cm = 19,625 V/cm
d1/d2 = 4/0.01 = 400
V = (19,625)(0.005) ln 400 = 588 V.
9.48 Q = CV
Initial chamber charge:
Qch = 5 × 10−12
× 600 = 3.0 × 10−9
Coulomb
Initial electrometer charge:
Qel = 0.4 × 10−12
× 600 = 0.24 × 10−9
Coulomb
Chamber charge lost:
8 × 109
× 1.6 × 10−19
= 1.28 × 10−9
Coulomb
Final chamber charge:
Qch
′
(3.0 − 1.28) × 10−9
= 1.72 × 10−9
Final electrometer charge:
Q′
el = 0.24 × 10−19
Coulomb
Final system charge
Q = (1.72 + 0.24) × 10−9
= 1.96 × 10−9
Coulomb
Total capacitance C = (5.0 + 0.4) × 10−12
F = (5.4 × 10−12
)
Final potential= 1.96 × 10−9
/5.4 × 10−12
= 363 V .
9.49 The velocity of a particle
β = p/E = p/(p2
+ m2
)1/2
The time taken to cross a distance L is
t = L/βc = (L/c)[(p2
+ m2
)/p2
]1/2
= (L/c)[(1 + (m2
/p2
)]1/2
The difference in time taken for proton and pion is
Δt = tp − tπ = (L/c)
.,
1 +

mp
2
/p2
-1/2
−
,
1 +

mπ
2
/p2
-1/2
/
Substituting, Δt = 100 × 10−12
s, c = 3 × 108
ms−1
mp = 0.94 GeV/c2
, mπ = 0.14 GeV/c2
and p = 4 GeV/c, and solving for
L, we find the minimum value of L = 1.14 m
9.50 t =
d
c

1 +
m2
p2
1/2
Put t = 51 × 10−9
s, c = 3 × 108
m/s, d = 12 m and p = 1.19 GeV/c and
solve for m to find m p− = 0.94 GeV/c2
for the anti proton mass.

Eπ =

P2
+ mπ
2
1/2
= (1.192
+ 0.142
)1/2
= 1.198 GeV
βπ = P/Eπ = 1.19/1.198 = 0.9933
tπ = d/βπ c = 12/(0.9933 × 3 × 108
) = 40.26 × 10−9
s
= 40.26 ns
9.51 Number of photons emitted due to absorption of 1 MeV electron is
15 × 106
/103
= 15,000.
Number of photo – electrons emitted = 15,000/10 = 1,500
The electron multiplication factor M = 310
because the photomultiplier
tube has 10 dynodes and each dynode produces 3 secondary electrons. The
charge collected at the output is
q = 1,500 × 310
× 1.6 × 10−19
= 1.417 × 10−11
Coulomb
The pulse height will be
V = q/C = 1.417 × 10−11
/1.2 × 10−10
= 0.118 V
9.52 The number of electrons liberated by the phosphor when 5 MeV proton is
stopped
n = 5 × 106
/100 = 5 × 104
Allowing for light collection efficiency (η) and conversion efficiency (ε) of
the photocathode, number of electrons released from the cathode
N = nηε = 5 × 104
× (60/100)(5/100) = 1,500
After going through 10 stages the number of electrons reaching the anode
becomes with a gain (G) of 3/stage
N G = 1, 500 × 310
= 8.85 × 107
The charge collected at the anode
q = NGe = 8.85 × 107
× 1.6 × 10−19
= 14.16 × 10−12
Coulomb
The voltage developed
V = q/C = 14.16 × 10−12
/12 × 10−12
= 1.18 V.
9.53 τ = (17 + 0.5 k)µs
For channel 100, τ = 17 + 0.5 × 100 = 67 µs
True counting rate N = N0/(1 − 67 × 10−6
f)
N0/N = 90/100 = 1 − 67 × 10−6
f
Or f = 1,490 s−1
Similarly for channel 400, f = 460 s−1
9.54 Let the gain/stage be G, so that the net gain due to electron multiplication will
be G12
(because there are 12 stages).
Number of photons producing electrons from the cathode for each beta particle
absorbed, with 5% photo-cathode efficiency = 550 × 5/100 = 27.5
Number of electrons reaching the anode = 27.5 × G12
Charge collected, q = 27.5 × G12
× 1.6 × 10−19
The voltage developed, V = q/C = 27 × G12
× 1.6 × 10−19
/45 × 10−12
=
8 × 10−3
Solving for G, we get the gain/stage, G = 2.567

9.3 Solutions 517
9.55 Label A is due to photopeak of 1.275 MeV γ -rays. Some of the primary pho-
tons will be absorbed within the crystal after undergoing Compton scattering.
Such events merely enhance the photopeak. In other cases, the Compton-
scattered photon will escape from the crystal, the light output will now be
proportional to the energy of the recoil electron which will be absorbed in a
large crystal. There will be a energy continuum of the recoil electrons with
energy ranging from zero to maximum. The label B represents the Compton
shoulder. The strong peak labeled C marks the photopeak at 0.511 MeV due
to electron-positron annihilation leading to absorption of one of the photons.
The annihilation may take place from the positron emitted by the source or
by the pair production caused by the primary photon. Now the total kinetic
energy available by the electron-positron pair will be (hν − 1.02) MeV. When
all of the energy is dissipated, the positron will be annihilated. If both the
annihilation photons escape from the crystal, a peak (called Escape peak) will
occur at an energy (hν−1.02) MeV = 1.275−1.02 = 0.255 MeV, represented
by label D. If one of the photons escapes, another peak will occur at an energy
(hν −0.511) MeV = 1.275−0.511 = 0.764 MeV (not shown in the Fig. 9.6).
If both the photons are absorbed, full output will be realized and this will be
added up to the photo peak.
Fig. 9.6
If the annihilation peak at 0.511 MeV and the photo peak due to primary
photon occur simultaneously within the resolving time of the instrument then
their energies are added and the events are recorded as a single event, known
as sum peak at energy 1.275 + 0.511 = 1.786 MeV (label E), usually with
small amplitude.
9.56 The normal distribution is
p(x) = σ
√
2π
−1
exp[−(x̄ − x)/2σ2
] (1)

At x̄,
p(x̄) = σ
√
2π
−1
(2)
Let the HWHM points be located at
x̄ − x = kσ (3)
Then p(kσ) = σ
√
2π
−1
exp(−k2
/2) (4)
By definition,
p(kσ)
p(x̄)
=
1
2
= exp

−
k2
2

(5)
where we have used (2) and (4). Taking loge on both sides of (5)
k2
/2 = ln 2 = 0.69315
∴ k = 1.1774
Therefore HWHM = 1.177σ
9.57 Channel 298 corresponds to 661 keV energy (Fig. 9.7)
FWHM (full width at half maximum)= 316 − 281 = 35 channels
HWHM (half width at half maximum)= 35/2 = 17.5 channels
HWHM corresponds to (661 keV)(17.5)/298 = 38.8 keV
Now HWHM = 1.177 σ
Therefore σ = HWHM/1.177
= 38.8/1.177 = 33 keV
The coefficient of the energy determination = (33)(100)/661 = 5%
Fig. 9.7 Photo peak 661 keV
γ -rays
9.58 When a charged particle moves through a medium with velocity greater than
the light velocity in the medium then light is emitted known as Cerenkov radi-
ation, after the discoverer. The radiation is emitted around the surface of a
cone with its axis along the particle’s path and with a half-angle given by

9.3 Solutions 519
cos θ = 1/β n (1)
where n is the refractive index and βc is the particle velocity. The threshold
corresponds to θ = 00
and is given by
β = 1/n (2)
Kaon and pion of the same energy would have different velocities, the pion
velocity being higher than the kaon velocity. If the medium be chosen such
that βk 1/n but βπ 1/n, then the pion will be counted by the Cerenkov
counter but not the kaon: In practice there can be several modifications of
this principle as by providing coincidence/anticoincidence with a scintillation
counter, and by restricting the emitted light to a prescribed angular interval.
γπ = 1 + (T/mπ ) = 1 + (20/0.14) = 143.857
βπ =
,
1 − (1/γπ
2
-1/2
= 0.9999758
cos θ = 1/βπ n = 0.999624
θ = 1.57◦
9.59 cos θ = 1/β n
β = 1/n cos θ = 1/(1.5 × cos 45◦
) = 0.943
v = βc = 0.943 × 3 × 108
= 2.828 × 108
ms−1
9.60 β = 1/n cos θ (1)
Also β = cp/E (2)
where E is the total energy.
E = cpn cos θ (3)
The relativistic equation is
E2
= c2
p2
+ m2
c4
(4)
Eliminating E between (3) and (4), we get mc2
= pc (n2
cos2
θ − 1)1/2
9.61 θ = 0.55 radians = 0.55 × 57.3◦
= 31.515◦
β = 1/n cos θ = 1/1.88 × 0.55 = 0.967
γ = 1/(1 − β2
)1/2
= 3.925
γ = 1 + (T/m) → m = T/(γ − 1)
m = 420/(3.925 − 1) = 107 MeV/c2
= (107/0.511) me = 209 me
It is a muon
9.62 The number of photons N(λ) dλ radiated per unit path in a wavelength interval
dλ can be shown to be
N(λ) dλ = 2πα (1 − 1/β2
n2
) (1/λ1 − 1/λ2)
where α = 1/137, is the fine structure constant.
Inserting β = 0.95, n = 1.33, λ1 = 3500×10−8
cm and λ2 = 5500×10−8
cm
in the above expression, we find N(λ) dλ = 178 per cm.

9.63 Number of photons emitted per unit length
N = 2πα(1 − 1/β2
n2
)(1/λ1 − 1/λ2) (1)
Pions: βπ = p/(p2
+ m2
)1/2
= 15/(152
+ 0.142
)1/2
= 0.999956
Kaons: βk = 15/(152
+ 0.4942
)1/2
= 0.999458
If the signal is to be given by pions but not kaons, the condition on the refrac-
tive index is
βπ 1/n βk
0.999956 1/n 0.999458
For a value of n = 1.0004, 1/n = 0.9996, the above condition is satisfied.
Inserting
α = 1/137, βπ = 0.999956, n = 1.0004,
λ1 = 4 × 10−7
m, and λ2 = 7 × 10−7
m in (1), we find N = 35 photons/m.
Therefore to obtain 175 photons a length of 5.0 m is required.
9.64 Pions: E = (p2
+ m2
)1/2
= (9002
+ 1402
)1/2
= 911 MeV
βπ = p/Eπ = 900/911 = 0.9879
For threshold, nπ = 1/β = 1/0.9879 = 1.012
Protons: E = (9002
+ 9382
)1/2
= 1,300 MeV
βp = 900/1300 = 0.6923
For threshold np = 1.444
The material chosen must have the refraction index 1.012 n 1.444
For n = 1.012, Cerenkov light will come off at 0◦
with the path. If a higher
index of refraction is chosen, light will come off at wider angle. The n must
be less than 1.444, otherwise protons will be counted.
9.65 If A is the area, d the thickness of depletion layer and K the dielectric constant
then the capacitance is
C = εAK/d = 8.8 × 10−12
× 1.5 × 10−4
× 10/40 × 10−6
= 3.3 × 10−10
F
The charge liberated
q = 5 × 106
× 1.6 × 10−19
/3.5 = 2.286 × 10−13
Coulomb
Potential developed
V = q/C = 2.286 × 10−13
/3.3 × 10−10
= 0.69 × 10−3
V
= 0.69 mV
9.3.9 Emulsions
9.66 The Range-Energy-Relation must be such that it ensures that the ionization
−dE/dR is a function of z2
f(R/M). One such relation is E = K z2n
M1−n
Rn
where E is in MeV and R in microns, K and n are empirical constants which

9.3 Solutions 521
depend on the composition of emulsions, z is the charge of the ion and M its
mass in terms of proton mass. The data on R and E for protons have been
used to determine n and R. We find n = 0.6 and K = 0.2276. Using these
values of n and K, the energy for various values of R have been determined
for deuterons and 3
He as tabulated below and the corresponding graphs are
drawn, Fig. 9.8.
Fig. 9.8
R(µm) 0 50 100 150 200 250 300 350 400 450 500
D E(MeV) 0 3.06 6.31 34.2 45.6 57.0 68.6 80.1 91.4 102.6 113.7
3
He 0 46.2 70.0 89.3 106.2 121.4 135.4 148.5 160.9 172.7 184.0
9.67 For a circular orbit of a charged particle of charge q and momentum p moving
in a magnetic field of B Tesla perpendicular to the orbit, the radius r in metres
is related by the formula
P = qBr → cp = qBrc J
cp = (1.6 × 10−19
)(3 × 108
) Br J
= 4.8 × 10−11
Br J
= (4.8 × 10−11
/1.6 × 10−10
) Br GeV
P = 0.3 Br GeV/c
9.68 Let the proton be at a distance d from S, the earth’s centre. Under the influence
of magnetic field it will describe an arc of a circle of radius r.
From Fig. 9.9, it is clear that
(R + r)2
= d2
+ r2
Or 2 R r = d2
− R2
≈ d2
(Because R d)
Therefore r = d2
/2R = (1, 000 R)2
/2R
= 5 × 105
R
The momentum, p = 0.3 Br
P = (E2
− m2
)1/2
= (10.942
− 0.942
)1/2
= 10.9 GeV/c
B = p/0.3r = 10.9/0.3 × 5 × 105
× 6.4 × 106

= 1.13 × 10−11
T
= 1.13 × 10−7
G
Fig. 9.9
9.69 Electric force = qE
Magnetic force = qvB
If these two forces cross, that is
qE = qvB
then the condition for null deflection is
v = E/B
Now v = (2T/m)1/2
= c(2T/mc2
)1/2
= c(2 × 20/940 × 1,000)1/2
=
0.00652 c
E = 500 V/cm = 5 × 104
V/m
Therefore B = E/v = 5 × 104
/0.00652 × 3 × 108
= 0.0256 T = 256 G
9.70 p2
= 2m T = 2m qV = 2mqEd (1)
Also p = qBr (2)
m = qB2
r2
/2Ed (3)
Further d = at2
/2 = qE t2
/2m (4)
Using (3) in (4) and solving for t,
t = Br/E
9.71 Drop a perpendicular DA at D and extend the path DP to meet the extension
of the initial path HC in E. From the geometry of the figure (Fig. 9.10) angle
CAD = θ. Drop a perpendicular DG on CA. In the triangle AGD, sin θ =
GD/AD = d/R, where R is the radius of curvature.
R =
√
2mK
qB
=
√
2 × 1.67 × 10−27 × 5 × 105 × 1.6 × 10−19
1.6 × 10−19 × 0.51
≈ 0.2 m.
sin θ = d/R = 0.1/0.2 = 0.5
θ = 30◦

9.3 Solutions 523
Fig. 9.10
9.72 t =
2u sin α
α
=
2u sin α
Ee
.m
Where the acceleration α = Ee/m
t =
2 × 8 × 105
sin 30◦
× 9.1 × 10−31
50 × 1.6 × 10−19
= 91 × 10−9
s = 91 ns
9.73 The component of the velocity, ⊥ to the field is v sin θ. Equating the cen-
tripetal force to the magnetic force
mv2
⊥ = qv⊥ BR
R = mv⊥/qB = v sin θ/(q/m)B
=
3 × 105
sin 300
108 × 0.3
= 0.5 × 10−2
m
= 0.5 cm
9.74 p = 0.3BR = (0.3)(10−4
)(6.4 × 106
) = 192 GeV/c
T ≈ 192 GeV
9.75 Cosmic ray flux = 1 cm−2
s−1
= 104
m−2
s−1
Earth’s surface area, A = 4 π R2
= 4π(64 × 105
)2
= 5.144 × 1014
m2
Cosmic rays incident on earth’s surface
= 5.144 × 1014
× 104
= 5.144 × 1018
m−2
s−1
Cosmic rays energy delivered to earth
= 5.144 × 1018
× 3 GeV s−1
= 1.5432 × 1019
GeV s−1
= 1.5432 × 1019
× 1.6 × 10−10
J s−1
= 2.47 × 109
W
= 2.47 GW.

9.76 p = 0.3BR (1)
where p is in GeV/c, B in Tesla and R in metres.
Now θ ≈ L/R = 0.3 BL/p (2)
where L is the length of the magnet, θ is the angle of deflection, R is the radius
of curvature of the circular arc of the path in the magnetic field, and l is the
length of the straight path from the slits (Fig. 9.11).
Δθ = s/l = 0.3 BLΔ p/p2
Δp/p = 1/100
S = 0.3 × 1.2 × 1.5 × 10 × (1/100)(1/25)
= 2.16 × 10−3
m = 2.16 mm
Fig. 9.11
9.3.11 Betatron
9.77 ΔT = eΔϕ/Δt = 1.6 × 10−19
× 50 J
= 50 eV
9.78 (a) If N is the number of revolutions then
(2πRN )(4 f ) = c
The factor 4 arises due to the fact that the duty cycle is over a quarter of a
period.
N = c/8π Rf = 3 × 108
/8π × 0.9 × 50 = 2.65 × 105
(b) Radius, R = 0.9 m
Tmax = BRec = 1.2 × 0.9 × 1.6 × 10−19
× 3 × 108
= 5.184 × 10−11
J
= 5.184 × 10−11
/1.6 × 10−13
= 324 MeV
(c) The average energy gained per revolution
ΔT = Tmax/N = 324 MeV/2.65 × 105
= 122.2 × 10−5
MeV
= 1.222 keV
9.3.12 Cyclotron
9.79 The resonance condition is
ω = qB/m

9.3 Solutions 525
Further p = qBR
T = p2
/2m = q2
B2
R2
/2m
Maximum radius
R(max) = (2mTmax)1/2
/qB
= (2 × 1.67 × 10−27
× 6 × 1.6 × 10−13
)1/2
/(1.6 × 10−19
B)
= 0.354/B m
9.80 The resonance condition is
ω = 2π f = qB/m
B = 2π fm/q
For deuterons, Bd = 2 π × 1.2 × 107
× 2.01 × 1.66 × 10−27
/1.6 × 10−19
= 1.5715 T
For alpha particles, Bα = 2 π×1.2×107
×4.00×1.66×10−27
/2×1.6×10−19
= 1.5637 T
Deuterons: At ejection kinetic energy
Ef = (qBR)2
/2m = (1.6×10−19
×1.5715×0.3)2
/(2×2.01×1.66×10−27
)J
= 0.085 × 10−11
J
= 5.3 MeV
If N is the number of orbits, t total time, and T0 the time period then the
mean energy increment per orbit is E/N and the average time for each orbit
(supposed to be constant)
T0 = 2π/ω0 = 1/f0
N = t/T0 = t f 0
Therefore Ef/N = Ef/tf 0 = 2 eV (the factor 2 is introduced because there
are two gaps)
t = Ef/(2eV) f0 = 5.3 × 106
/(2 × 5 × 104
) × 1.2 × 107
= 4.42 × 10−6
s
= 4.42 µs
α-particles: At ejection Ef = (qBR)2
/2m
= (2 × 1.6 × 10−19
× 1.5637 × 0.3)2
/(2 × 4.0 × 1.66 × 10−27
)
= 0.1697 × 10 J
= 10.6 MeV
Total time t = Ef/(2 × 2 eV) f0
= 10.6 × 106
/(4 × 5 × 104
× 1.2 × 107
)
= 4.42 × 10−6
s = 4.42 µs
9.81 Cyclotron resonance condition is
ω = qB/m
Because of relativistic increase of mass the resonance condition would be
ω′
= qB′
/mγ
If ω′
= ω, then B′
= Bγ
Fractional increase of magnetic flux density required is
(B′
− B)/B = ΔB/B = γ − 1
For protons of 20 MeV. γ = (T/m) + 1 = (20/940) + 1 = 1.0213
Therefore percentage increase of B is (γ − 1) × 100
= (1.0213 − 1) × 100 = 2.13

9.82 (a) Resonance condition for protons is
B = 2π fm/q = (2π × 5 × 106
)(1.6726 × 10−27
)/(1.6 × 10−19
)
= 0.3284 T
T = (1/2)(Bqr)2
/m = (0.3284)2
(1.6 × 10−19
× 0.762)2
/(2 × 1.6726 ×
10−27
)
= 4.79 × 10−13
J = 3 MeV
(b) For deuteron, the charge is the same as that of proton but mass is approx-
imately double, the required magnetic field will be that for (a). So B =
0.655 T.
The kinetic energy ∝ B2
/m, so that it will be (22
/2) × 3 or 6 MeV
(c) For alpha particle the mass is approximately four times and charge is dou-
ble compared to proton, so that the required magnetic field is twice that for
proton, that is B = 0.655 T, and kinetic energy will be (2 × 2)2
/4, that is,
four times the proton energy or 12 MeV.
9.83 The resonance frequency at the beginning
f0 = B0 q/2 π m = 1.5×1.6×10−19
/(2π ×3.34×10−27
) = 11.44×106
c/s
= 11.44 Mc/s
The resonance frequency at the limiting radius is
f = qB/2π m = 1.43 × 1.6 × 10−19
/2 π × 3.34 × 10−27
= 10.91 × 106
c/s = 10.91 Mc/s
Range of frequency modulation is 11.44–10.91 Mc/s.
Tmax = q2
B2
r2
/2 m = (1.6 × 10−19
)2
(1.43)2
(2.06)2
/2 × 3.34 × 10−27
= 3.3256 × 10−11
J
= 3.3256 × 10−11
/1.6 × 10−13
MeV = 207.8 MeV
9.84 B = ω m/q = (2π × 8 × 106
)(1.66 × 10−27
)/(1.6 × 10−19
)
= 0.52 T
r = (2T/mω2
)1/2
= c(2T/mc2
ω2
)1/2
= 3 × 108
(2 × 5/938 × 4π2
× 82
× 1012
)1/2
= 0.616 m
9.85 The cyclotron resonance condition is ω = qB/m
For deuterons, ωd = 1 × Bd/md
For alpha particles, ωα = 2 × Bα/mα
If the resonance frequency is to remain unaltered ωd = ωα
Bd/Bα = 2 × (2.014102/4.002603) = 1.006396
Fractional decrease of magnetic field
(Bα − Bd)/Bd = −0.006355
Percentage decrease = 0.6355%
9.86 The energy of protons extracted from the accelerator is calculated from the
equations
E2
= p2
+ m2
p = 0.3 BR = 0.3 × 1.5 × 2 = 0.9 GeV/c
= 900 MeV/c.

9.3 Solutions 527
E = (p2
+ m2
)1/2
= (9002
+ 9382
)1/2
= 1, 300 MeV
γ = 1, 300/938 = 1.386
β = 0.692
Maximum energy transferred to electron is
E(max) = 2mβ2
γ 2
= 2 × 0.511 × 0.6922
× 1.3862
= 0.94 MeV
9.87 Initially ω0 = qB/m
ω = qB/mγ = ω0/γ
γ = 1 + T/m = 1 + 469/938 = 1.5
f = f0/γ = 20/1.5 = 13.33 Mc/s
9.3.13 Synchrotron
9.88 P = 0.3 BR
P = (T 2
+ 2 Tm)1/2
= (302
+ 2 × 30 × 0.938)1/2
= 30.924 GeV/c
R = p/0.3 B = 30.924/0.3 × 1 = 103.08 m
9.89 p = 0.3 BR
P = (T2
+ 2Tm)1/2
= [32
+ (2 × 3 × 0.938]1/2
= 3.825 GeV/c
R = p/0.3 B = 3.825/(0.3 × 1.4) = 9.1 m
9.90 Initially ω0 = B0 e/m
Finally ω = 0.95 B0 e/(m + T )
ω/ω0 = 0.95m/(m + T )
ω0 − ω
ω0
=
T + 0.05 m
T + m
=
313 + 0.05 × 938
313 + 938
= 0.288
Therefore Depth of modulation is 28.8 %
9.91 Radiation loss per revolution
ΔE = (4π e2
/3R)(E/mc2
)4
.1/4 πε0
= (1.44 × 4π/3R)(E/mc2
)4
MeV-fm
Substituting E = 300 MeV, mc2
= 0.511 MeV
R = 1.0 × 1015
fm
ΔE = 716 × 10−6
MeV = 716 eV
9.92 p = qBR
R = p/qB = m0γβ c/qB = m0 c (γ 2
− 1)1/2
/qB
But γ = (T/m0c2
) + 1 = n + 1
∴ γ 2
− 1 = n2
+ 2n
∴ R = (m0 c/qB)(n2
+ 2n)1/2

9.93 Using the result of Problem 9.92
R =
m0c
qB
(n2
+ 2n)
1
2
=
m0c
0.0147q

50
940
2
+

2 × 50
940
'1/2
= 22.48
m0c
q
(1)
As the radius of synchrotron does not change, we can use the same relation at
higher energy
R =

m0c
1.2q

(N2
+ 2N)1/2
(2)
Combining (1) and (2) and solving for n, we find N = 26
∴ T = Nm0 c2
= 26 × 0.938 = 24.39 GeV
9.94 Using the results of Problem 9.92
q/m0 = (c/BR)(n2
+ 2n)1/2
(1)
Proton: q/m0 = 1/1 = 1
n = T/mp = 1,000/1,000 = 1
Using the above values in (1)
c/BR = 1/
√
3 (2)
Deuteron: q/m0 = 1/2
Therefore 1/2 = (n2
+ 2n)1/2
/
√
3
Solving for n, we find n = 0.3229
Kinetic energy of deuteron = nmd
= 0.3229 × 2,000 MeV
= 646 MeV
3
He : q/m0 = 2/3
Therefore 2/3 = (1/
√
3)(n2
+ 2n)1/2
Solving for n, we find n = 0.527
Therefore Kinetic energy of 3
He = nmHe3
= 0.527 × 3, 000 MeV
= 1, 583 MeV.
9.95 (a) p = 0.3 BR (GeV/c if B is in Tesla and R in metres)
P = (T 2
+ 2T mc2
)1/2
= (0.52
+ 2 × 0.5 × 0.938)1/2
= 1.09 GeV/c
B = 18 kG = 1.8 T
R = p/0.3 B = 1.09/(0.3 × 1.8) = 2.02 m.
(b) Energy of ions for acceptance
Ts = ±(2eV.mc2
/π)1/2
[(ϕs − π/2) sin ϕs + cos ϕs]1/2
(1)
Substitute in (1), 10 keV = 0.01 MeV
mc2
= 938 MeV; ϕs = 300
= 0.5236 radians
Ts = ±2.5 MeV
(c) The initial frequency
f = Bqc2
/2π (mc2
+ Ts) (2)

9.3 Solutions 529
B = 1.8 T; q = 1.6 × 10−19
C; c = 3 × 108
m/s; mc2
= 938 × 1.6 ×
10−13
J
Ts = ±2.5 MeV = ±2.5 × 1.6 × 10−13
J
Substituting the above values in (2) we find
f1 = 27.43; f2 = 27.51 Mc
(d) The required electrical frequency for 500 MeV protons is
f = Bqc2
/2π(mc2
+ T ) (3)
Put (mc2
+T ) = (938+500)×1.6×10−13
J, B = 1.8 T, c = 3×108
m/s
and q = 1.6 × 10−19
C, in (3) to obtain f = 17.94 Mc.
Thus the range of frequency modulation is 27.51 − 17.94 Mc
9.96 The particles are constrained to move in a vacuum pipe bent into a torus that
threads a series of electromagnets, providing a field normal to the orbit. The
particles are accelerated once or more per revolution by radio frequency cav-
ities. Both the magnetic field and the R.F. frequency must increase and the
synchronized with the particle velocity as it increases.
The major energy loss is caused by the emission of synchrotron radiation.
The synchrotron radiation loss per turn is
ΔE = (4π/3R)(e2
/4πε0)(E/mc2
)4
For electron
∆E =

4π
3
(1.44 MeV.fm)
(103 × 1015 fm)

500 × 103
0.511
4
= 5,526 × 103
MeV
= 5,526 GeV
an energy loss which is an order of magnitude greater than the electron energy
to which the electrons are to be accelerated, which is impossible. On the other
hand for protons the energy loss per turn will be smaller by a factor (1, 836)4
or 1.1 × 1013
Thus for protons ΔE = 5, 526/1.1 × 1013
= 5 × 10−10
GeV
= 0.5 eV which is quite small.
As the synchrotron radiation losses for electrons in circular machines are
much beyond tolerable limits, linear accelerators are employed which are
capable of accelerating electrons up to 30–40 GeV.
9.97 Orbital frequency f = 1/10−6
= 106
c/s
P = 0.3 Br
f = qB/2π m
∴ p = 0.3×2π m f r/q = 0.3×2π×1.67×10−27
×106
×10/(1.6×10−19
)
= 0.1966 GeV/c
T 2
+ 2mT = p2
Using m = 0.938 GeV/c2
and solving for T , we find T = 0.02 GeV or
20 MeV

9.98 (a) At injection γ = 1 + 10
0.511
= 20.57
β = 0.9988
At extraction γ = 1 + 5,000
0.511
= 9, 786
β ≈ 1
Initial frequency f1 =
βc
2πr
=
(0.9988) × 3 × 108
2π × 15
c = 3.1809 Mc
Final frequency f2 =
1 × 3 × 108
2π × 15
= 3.1847 Mc
As the initial and final frequencies are nearly the same there is hardly
any need to change the R.F. frequency.
(b) Total energy gain= Ef − Ei = 5,000 − 10 = 4,990 MeV
Energy gain per turn = 1 keV
∴ Number of turns, n = 4,990
10−3 = 4.99 × 106
(c) The period of revolution
T0 = 1/f = 1/3.18 × 106
= 3.14 × 10−7
s
Time between injection and extraction is
T = nT0 = 4.99 × 106
× 3.14 × 10−7
= 1.567 s
(d) The total distance traveled by the electron is
d = 2πrn = 2π × 15 × 4.99 × 106
= 4.7 × 108
m = 4.7 × 105
km.
9.99 (a) As there are 97 drift tubes, there will be 96 gaps. The energy gain per
gap is ΔE = (50–2)/96 = 0.5 MeV/gap
After crossing the first gap, the protons are still non-relativistic and
their velocity will be in the second drift tube will be
v2 = (2T/m)1/2
= c [2 × (2 + 0.5)/938]1/2
= 0.073 c
The length of the second tube
l2 = v2/2 f = (0.073 × 3 × 1010
/2 × 2 × 108
) cm
= 5.48 cm
In the last tube v is calculated relativistically.
γ = 1 + T/m = 1 + 50/938 = 1.053
β = (γ 2
− 1)1/2
/γ = 0.313
lf = βc/2 f = 0.313 × 3 × 1010
/2 × 2 × 108
= 23.48 cm
(b) To produce 80 MeV protons, number of additional tubes required is (80−
50)/0.5 = 60
9.100 (a) Beam current i = q/t = 50 × 5 × 1011
× 1.6 × 10−19
= 4 × 10−6
amp
= 4µA

9.3 Solutions 531
(b) Power output W = iV
The current is obtained from (a) and the voltage V corresponding to final
voltage of 2 GeV would be 2 × 109
.
Therefore W = 4 × 10−6
× 2 × 109
= 8,000 W
9.101 There are five drift tubes in the section but only four gaps. Initial energy is
100 keV. Therefore, the final energy will be
(a) E = 100 keV + n eV
= 100 + 4 × 100 = 500 keV
(b) L = (1/2 f )(2eV/m)1/2
4
1
√
n
= 1
2×50×106
2×1.6×10−19
×105
1.67×10−27
1/2
(1 +
√
2 +
√
3 +
√
4)
= 0.269 m = 26.9 cm
9.102 The longest drift tube will be in the end of the linear accelerator where max-
imum ion energy has been achieved.
L = v/2 f
We can do non-relativistic calculations for v as the ion energy is not large
v = (2T/M)1/2
= c (2T/Mc2
)1/2
= c (2 × 80/12 × 938)1/2
= 0.119 c
L = 0.119 × 3 × 108
/2 × 25 × 106
= 0.714 m.
9.3.15 Colliders
9.103 In the colliding beam experiments, for two colliding beams the reaction rate
is written in terms of the “luminosity” L. The number of interactions per
second is
N = Lσ
where σ is the interaction cross-section in question.
If N1 or N2 = number of particles/bunch in each beam
A = area of cross-section of intersecting beams
N = number of bunches/beam
and f = frequency of revolution
then luminosity is simply given by
L = n N 1 N2 f/A
9.104 Number of beam electrons/cm2
/s, N1 = nf /πr2
= 6 × 1011
× 2 × 106
/π(0.12)2
= 2.65 × 1019
Number of beam positrons/s, N2 = 6 × 1011
Expected production rate of μ+
μ−
pairs/second
= N1 N2σ(e+
e−
→ μ+
μ−
)
= (2.65 × 1019
)(6 × 1011
)(1.4 × 10−33
)
= 0.022

9.105 (a) E∗2
= (E1 + E2)2
− (p1 − p2)2
= E1
2
+ 2 E1 E2 + E2
2
− p1
2
+ 2 p1 p2 − p2
2
=

E1
2
− p1
2

+

E2
2
− p2
2

+ 2(E1 E2 + p1 p2)
= m1
2
+ m2
2
+ 2(E1 E2 + p1 p2)
If E1 m1 and E2 m2, m1
2
and m2
2
can be neglected and p1 ≈
E1, p2 ≈ E2
Therefore E∗2
≈ 4E1 E2
If the beams cross at an angle θ then
E∗2
= (E1 + E2)2
−

p1
2
+ p2
2
− 2 p1 p2 cos θ

≈ 2E1 E2 (1 + cos θ) = 4 E1 E2 (1 + cos θ)/2
Thus the available energy in the CMS is reduced by a factor of (1 +
cos θ)/2 compared to head-on-collision.
(b) In the CMS E∗
= 25 + 25 = 50 GeV
With the fixed proton target
E∗
= (m2
+ m2
+ 2 T1 m)1/2
= 50 GeV
Substituting m = 0.94 GeV, we find T1 = 1329 GeV
9.106 When the protons travel toward each other with equal energy, and therefore
with the same speed, their net momentum is zero. In that case the Lab system
is reduced to the C-M system, and the observer is watching the events sitting
in the C.M. system. The total energy is then,
E∗
= 1010
eV + 1010
eV + 109
eV + 109
eV = 22 × 109
or 22 GeV
Let E1 be the energy of a proton in the lab system moving toward the target
proton originally at rest, then if the total energy available in the CMS has to
be the same as E∗
= 22 GeV,

m1
2
+ m2
2
+ 2 E1 M1
1/2
= E∗
= 22 GeV
(12
+ 12
+ 2 E1 × 1)1/2
= 22
Or E1 = 241 GeV
Therefore required kinetic energy = 241 − 1 = 240 GeV
9.107 Total energy available in the CMS is
E∗
=

m1
2
+ m2
2
+ 2 E1 m2
1/2
where E1 is the total energy of projectile of mass m1 and m2 is the mass of
the target.
E∗
= 4M0 + 2 M0 = 6M0
m1 = m2 = M0
6M0 =

M0
2
+ M0
2
+ 2 E1 M0
1/2
whence E1 = 17 M0
Or T1 = E1 − M0 = 16 M0
The significance of this result is that in the colliders a lot more energy is
available than in fixed target experiments.

9.3 Solutions 533
9.108 The CMS energy is calculated from the invariance of E2
− p2
E∗2
= E2
− p2
=

Ep + Ee
2
−

pp − pe
2
= E2
p − p2
p + E2
e − p2
e + 2

Ep Ee + pp pe

≈ mp
2
+ me
2
+ 4Ep Ee
Since mp Ep and me Ee
E∗
≈

4Ep Ee =
√
4 × 820 × 30 = 314 GeV
Note that the HERA accelerator is different from other colliders in that the
energy of the colliding particles (protons and electrons) is quite asymmet-
rical. It has been possible to achieve high momentum transfer in the CMS
(20,000 GeV2
), necessary for the studies of proton structure.
9.109 Total CMS energy, E∗
≈ [2 E1 E2 (1 + cos θ)/2]1/2
Substituting E1 = 20 GeV, E2 = 300 GeV, and θ = 100
we find E∗
= 109 GeV
If the same energy (E∗
= 109 GeV) is to be achieved in a fixed target
experiment, then the electron energy in the lab-system would be
(2 E M + M2
+ m2
)1/2
= E∗
Neglecting M2
and m2
E = (E∗
)2
/2M = (109)2
/(2 × 0.94) ≈ 6, 300 GeV

Chapter 10
Particle Physics – II
Classification of particles
Table 10.1 gives the mass, mean lifetimes (τ) and common decay modes of ele-
mentary particles excluding resonances. Their classification into hadrons, photon
and leptons is also indicated. Further subdivision of hadrons into mesons (pions and
kaons) and baryons (nucleons and hyperons) is also shown. Electron (e−
), muon
(μ−
), Tauon (τ−
), and the three neutrinos νe, νμ, and ντ constitute the class of
leptons. A hadron stands for a strongly interacting particle distinguished from lepton
which has only weak or electromagnetic interactions. Photon is the massless carrier
of the electromagnetic field.
In the fourth family, graviton a massless particle of spin 2, the quantum of gravi-
tation is not yet discovered.
Mesons and photon are Bosons (a particle of integral spin, 0, 1, 2, . . .).
Bosons obey Bose-Einstein statistics, the wave function describing two identical
bosons is symmetric under particle exchange. The baryons and leptons are Fermions
(a particle with half integral spin, 1
2
, 3
2
, . . .). Fermions obey Fermi-Dirac statis-
tics, for which the wave function of two identical particle is anti symmetric (changes
sign under particle exchange).
Antiparticle: Every particle has in association an antiparticle, with exactly the same
mass and lifetime but opposite values of electric charge, magnetic moment, baryon
number, lepton number, and flavor. Thus positron (e+
) is the anti particle of electron
(e−
), antiproton (p−
) that of proton (p), ¯
νe that of νe etc. Photon is the antiparticle
of itself, so also π0
.
Fundamental interactions
1. Strong (nuclear) interaction
2. Electromagnetic interaction
3. Weak (nuclear) interaction
4. Gravitational interaction.
535

536 10 Particle Physics – II
Table 10.1
Mass Common
Particle (MeV/c2
) τ(s) decay mode
Hadrons
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
Mesons
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
Pions

π−
, π+
139 2.5 × 10−8
μν
π0
135 1.8 × 10−16
γ γ
Kaons
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
K−
, K+
494 1.2 × 10−8
μν
K0
498 π±
π0
Mixture of K1, K2
K1 0.89 × 10−10
π+
π−
π0
π0
K2 5.18 × 10−8
π0
π0
π0
π+
π−
π0
πμν
πμeν̄
η 550 10−18
γ + γ
π+
+ π−
+ π0
Baryons
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
Nucleons

p 938.2 1038
stable
n 939.5 103
pe−
ν
Hyperons
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
Λ 1,115 2.6 × 10−10
pπ−
, nπ0
Σ+
1,189 0.8 × 10−10
pπ0
, nπ+
Σ0
1,192 10−20
Λγ
Σ−
1,197 1.6 × 10−10
n π−
Ξ0
1,314 3 × 10−10
Λπ0
Ξ−
1,321 1.8 × 10−10
Λπ−
Ω−
1,675 1.3 × 10−10
Ξπ
ΛK−
Photon γ 0 ∞ Stable
Leptons
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
τ−
1,784 3.4 × 10−13
Electrons
and neutrinos
μ−
105 2 × 10−6
eνν̄
e−
0.51 ∞ Stable
νe 0 ∞ Stable
νμ 0.5 Stable
ντ 164 ∞ Stable
Graviton ? 0 – Stable
Here we will be concerned with only the first three types. Table 10.2 summarizes
the characteristics of the interactions.
Coupling constant: Particles interact through strong electromagnetic or weak charges.
The square of the charge is known as the coupling constant. It enters the interaction
matrix which determines the cross-sections and decay rates. Strictly speaking, the
coupling constants are not constant but vary very gradually with the particle energy.
They are called running constants.
QED (quantum electrodynamics) is the quantum field theory of the electro mag-
netic interaction whose predictions have been verified to a precision of one part in a
billion.
QCD (quantum chromo dynamics) is the field theory of the strong color interaction
between quarks.

Table 10.2 Characteristics of the fundamental interactions
Strong Electromagnetic Weak Gravitational
Carrier of field
spin-parity(Jp
) of
quantum
Gluon 1−
Photon 1−
W±
, Z0
1−
, 1+
Gravitation ? 2+
Coupling constant αs ≤ 1
α =
e2
4πc
=
1
137
G

Mc2
2
(c)3
GN M2
4πc
= 5 × 10−40
Mass 0 0 80, 90 GeV 0
Relative strength 1 10−2
≤10−5
10−38
Time scale 10−23
s 10−18
− 10−20
s 10−13
s
Range ≤10−15
m ∞ 10−18
m ∞
Source Colour charge Electric charge Weak charge Mass
Standard model: As of today, the physics embodied in electrodynamics, chromo
dynamics and electro-weak interaction is termed as the standard model of elemen-
tary particles.
Resonances or resonant states are the analogs of excited states of atoms. They are the
excited states of familiar hadrons. Some of them are so short lived (∼10−23
−10−24
s)
that their direct detection is not possible. They ultimately decay into more familiar
particles, like nucleons, mesons, leptons and photons. Because of their short lives
their energy (mass) spread is enormous, due to uncertainty principle.
Baryon number (B) is the generalization of mass number. For nucleons and hyperons
B = +1, for anti baryons, B = −1, for pions, kaons and other particles B = 0.B is
an additive quantum and is conserved in all the three types of interactions.
Isospin (T or I) is a quantum number applicable to hadrons and is conserved in
strong interactions. It results from the near equality of u and d-quarks. This is
reflected in the near equality of masses of charged multiplets such as (n, p), (π+
, π0
,
π−
), (K+
, K0
) etc, as well as for the atomic nuclei once the coulomb interaction
is removed. It is thus named because its mathematical description is entirely analo-
gous to ordinary spin or angular momentum in quantum mechanics. The tables for
Clebsch – Gordon coefficients for the addition of angular momenta (displayed in the
summary of Chap. 3) can be directly used for the isospins. T is the additive quantum
number. The charge multiplicity is given by 2T +1. The antiparticle has the same T
as the particle but opposite T3. T3 is analogous to Lz for angular momentum. Total
isospin (I) is conserved in strong interactions but breaks down in em and weak
interactions. The third component (I3) of a system of hadrons is conserved in strong
and em interactions but is violated in weak interactions ΔI3 = ±1
2

.
The generalized pauli principle
(−1)l+s+I
= −1 (10.1)
where l is the orbital angular momentum, s the spin and I the isospin.

Strangeness and strange particles: Heavy unstable particles such as kaons and
hyperons which are produced copiously but decay slowly are named as strange
particles. A new quantum number S, strangeness is introduced to distinguish them
from other particles. Gellmann’s formula
Q
e
= T3 +
S + B
2
(10.2)
where B is the baryon number. K+
and K0
are assigned S = +1, while K−
and K0
have S = −1. The Σ hyperons and Λ hyperons have S = −1, Ξ−
and Ξ0
have
S = −2, Ω−
has S = −3. The ordinary particles, n, p, π+
, π0
, π−
have S = 0.
The anti particles have opposite strangeness. Strangeness S is an additive quantum
number. Table 10.3 summarises the strangeness S for various hadron multiplets.
Table 10.3 T and S assignments
T/S −3 −2 −1 0 1 2 3
0 Ω−
Λ Λ Ω
+
p
Ξ0
K−
n K+
Ξ
0
1/2 Ξ−
K0
p K0
Ξ
+
n
Σ+
π+
Σ
−
1 Σ0
π0
Σ
0
Σ−
π−
Σ
+
Strangeness S is conserved in strong and electromagnetic interactions, that is
ΔS = 0, but breaks down in weak interactions, such as decays, the rule being
ΔS = ±1.
Leptons: The electron, the muon and the tauon and their respective neutrinos as well
as their antiparticles constitute the family of leptons. Leptons are assigned lepton
number, L = +1 and antileptons, L = −1. The numbers Le, Lμ and Lτ are
separately conserved in all the three types of interactions. The lepton numbers are
shown in Table 10.4.
Table 10.4 Lepton numbers
Q/e Le = 1 Lμ = 1 Lτ = 1
0
−1

νe
e−

νμ
μ−

ντ
τ−

Q/e Le = −1 Lμ = −1 Lτ = −1
0
+1

νe
e+

νμ
μ+

ντ
τ+

Helicity or handedness: The helicity H is defined as the ratio Jz/J where Jz is
the component of spin along the momentum vector of the particle and J is the
total spin. Massless particles have spin components Jz = +J only. Thus H = +1
or −1. Neutrinos have H = −1 (left-handed) and anti-neutrinos have H = +1
(right-handed). Massive particles are not in pure helicity eigen states and contain
both LH and RH components.

Parity (p or π): The concept of parity was mentioned in Chap. 3. The absolute
intrinsic parity cannot be determined. Parity of a particle can be stated only rela-
tive to another particle. By convention baryons are assigned positive parity. All the
antifermions have parity opposite to the fermions. On the other hand, bosons have
the same parity for particle and antiparticle. Pions and Kaons have odd parity.
Parity is a multiplicative number, so that the parity of a composite system is equal
to the parities of the parts. Thus, for a system comprising of particles A and B,
P(AB) = p(A) . p(B) . p(orbital motion) (10.3)
where p(orbital motion) = (−1)l
is the parity associated with the relative motion of
the particles, l being the orbital angular momentum quantum number (0, 1, 2, . . .).
Overall parity is conserved in strong and em interactions but is violated in weak
interactions.
Charge – conjugation: (C-parity) is the process of replacing a particle by an antipar-
ticle or a system of particles by the anti particle (s).
In general, a system whose charge is not zero cannot be an eigen function of C.
However if Q = B = S = 0, the effect of C is to produce eigen value ±1.
C is conserved in strong and em interactions but not in weak interactions. For
π0
, C = +1. For photon C = −1 and for n-photons
C = (−1)n
(10.4)
G-parity: The operation G consists of rotation of 1800
about the y-axis or z-axis in
isospin space followed by charge conjugation.
G-parity for the pion is −1 and for baryon it is zero. It is a multiplicative quantum
number. For a system of n pions
G = (−1)n
(10.5)
G - parity is conserved in strong interaction and is a good quantum number for
non-strange mesons. For a N − N system,
G = (−1)l+S+I
(10.6)
Time reversal means changing the sign of time. Strong interactions are invariant
under time reversal as evidenced by the absence of electric dipole moment of neu-
tron and verification of the predicted ratio of forward and backward reactions at the
same energy in the CMS.
If A + B → C + D,
then
σAB→C D
σC D→AB
=
(2sA + 1)(2sB + 1)P∗2
C
(2sC + 1)(2sD + 1)P∗2
A
(10.7)
where s is the spin of the particles, and p∗
is the momentum, the particle beams
being unpolarised.

The TCP theorem: A lagrangian which is invariant under proper Lorentz transfor-
mation is invariant with respect to the combined operation CPT, taken in any order.
The predictions of the TCP theorem which have been verified are
(i) the existence of an anti particle for every particle.
(ii) The equality of masses, lifetimes, and magnetic moments of particles and anti
particles.
Table 10.5 Conservation laws for the three types of interactions
Quantity Strong Electromagnetic Weak
1. Q (Charge) Yes Yes Yes
2. B (Baryon no.) Yes Yes Yes
3. J (Angular momentum) Yes Yes Yes
4. Mass + Energy Yes Yes Yes
5. Linear momentum Yes Yes Yes
6. I (Isospin) Yes No No
7. I3 (Third component of I) Yes Yes No ΔI3 = ±1/2
8. S (Strangeness) Yes Yes No ΔS = ±1
9. p (parity) Yes Yes No
10. C (Charge conjugation) Yes Yes No
11. G (G-parity) Yes No No
12. L (Lepton number) Yes Yes Yes
Quarks are the structureless fermions from which all the strongly interacting parti-
cles (hadrons) are built. The quarks occur with fractional baryon number, B = 1
3
and charges +2
3
e or −1
3
e. Baryons are built with three quarks (u, d, s) and mesons
with a quark-antiquark pair. Their characteristics are shown in Table 10.6.
Table 10.6 Characteristics of quarks
Quark Symbol Mass(GeV/c2
) Q/e I S C B∗
T
Down d ∼ 0.3 −
1
3
1/2 0 0 0 0
Up u ∼ 0.3 +
2
3
1/2 0 0 0 0
Strange s 0.5 −
1
3
0 −1 0 0 0
Charmed c 1.6 +
2
3
0 0 +1 0 0
Bottom b 4.5 −
1
3
0 0 0 −1 0
Top t 175 +
2
3
0 0 0 0 +1
For all the quarks spin-parity Jp
= 1/2
+
. The quark structure of some of the
hadrons is as follows.
p = uud; n = udd; π+
= ud̄; π−
= ūd; Σ+
= uus; k̄◦
= d̄s;
Ξ−
= dss; Ω−
= sss; K+
= us̄; D+
= cd̄; π0
=
uū − dd
√
2
; Δ++
= uuu

Gellman’s equation is generalized as:
Q/e = I3 + 1/2(B + S + C + B∗
+ T ) (10.8)
where B denotes Baryon number, C the charm, B∗
the beauty or bottom and T
the top.
Quarks are not observed as free particles as they are confined in hadrons.
In order to save pauli’s principle, a new quantum number called “colour” is
assigned to quarks. This has nothing to do with ordinary colour. The quarks appear
in three colours, red, blue, and green. The antiquarks have anticolour. The observed
hadrons are colourless. Color plays a role in strong interactions similar to charge in
electromagnetic interaction.
The strong color field between quarks is mediated by massless gluons analogous
to electro-magnetic field mediated by photons. While a photon does not carry elec-
tric charge, gluon itself carries color charge. There are eight types of gluons.
Charmonium (cc) is the state formed from the charmed anti-charmed quark pair.
D mesons (D0
, D±
) contain a charmed quark or antiquark. They are pseudoscalar
like pions (Jp
= 0−
) and decay weakly predominantly into non-charmed strange
mesons.
Flavor is a generic name to describe different types of quark and lepton.
Generation: The six flavors of quarks and of leptons are grouped into three gener-
ations or families. The quarks (d, u), (s, c) and (b, t) are of first, second and third
generations; the corresponding leptons being (e−
, νe), (μ−
, νμ) and (τ−
, ντ )
Cabibbo – Kobayashi – Maskowa (CKM) matrix
Vi j =
⎛
⎝
Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
⎞
⎠
The probability for a transition from a quark q to a quark q′
is proportional to
%
%Vqq′
%
%2
, the square of the magnitude of the matrix element. The diagonal elements
of this matrix, Vud , Vcs, Vtb which correspond to transitions within a family are
short of unity by only a few percent. Hence, transitions u → d, c → s, and t → b
are Cabibbo favoured.
The elements Vus , Vcd , Vcb, and Vts are small but not zero. Hence transitions,
s → u, c → d, b → u and t → s are Cabibbo suppressed.
The elements Vub and Vtd are nearly zero. Hence transitions b → u and t → d
are Cabibbo forbidden.
The boson propagator: The rate of a particular reaction mediated by boson exchange
is proportional to the square of the amplitude f (q2
) multiplied by a phase factor
and determines the cross-section or the decay of an unstable particle. Here q2
is the
square of the four-momentum transfer.
f (q2
) ∝
1
q2 + m2
(10.9)

where m is the mass of the exchanged boson. The quantity (q2
+ m2
)−1
is known
as the propagator term. For low momentum transfers the propagator is insensitive to
q, but in high energy collisions for large momentum transfer, f (q2
) decreases with
increasing q.
Spurion is a hypothetical particle which is introduced into the initial state to convert
the weak decay into a strong interaction.
Weak interaction- characteristics
Weak decays have long lifetimes (10−13
s) and small interaction cross-sections,
typically ∼10−39
cm2
. Charged leptons experience both weak and em interaction
while neutrinos only weak interaction. Strangeness and parity are not conserved.
Depending on the extent to which leptons are involved, the weak decays are
divided into three classes.
(1) Leptonic decays in which the decay products are leptons only, as in the decay
μ−
→ e−
+ νe + νμ
(2) Semi leptonic decays which involve both hadrons and leptons. Examples are
(a) n → p + e−
+ νe for ΔS = 0
(b) K+
→ π0
+ e+
+ νe for |ΔS| = 1
(3) Non-leptonic decays which do not involve leptons. Parity or strangeness are not
conserved, the selection rules being ΔS = ±1 and ΔI = ±1
2
as in the decay,
Λ → p + π−
In weak decays the flavors of the quark changes in contrast with strong or electro-
magnetic decays where the flavour is conserved. For example the decay of neutron,
n → p + e−
+ νe is represented by udd → udu + e−
+ νe in which a d-quark is
converted into a u-quark, d → u + e−
+ νe.
Charge current weak interaction is mediated by massive bosons W±
. The W±
exchange results in the change of the lepton charge as in the anti neutrino absorption
νe + p → n + e+
.
Neutral current weak interaction is mediated by the massive boson Z0
. The exchange
of Z0
does not cause the change of lepton as in νμ + e−
→ νμ + e−
.
Electro – Weak interaction
The electro magnetic interaction and weak interaction are two aspects of a single
interaction called electro-weak interaction. The corresponding charges are related
by θw the Weinberg angle

e = gsinθw (10.10)
sin2
θw = 0.2319 (10.11)
Mw/Mz = 0.88 (10.12)
Feynman diagrams are short-hand for writing down individual terms in the calcula-
tion of transition matrix elements in various processes pictorially. As an example,
consider the diagram in Fig. 10.1. The solid lines are the fermion lines. The con-
vention used here is that time runs from left to right. The top solid lines represent
electron and bottom lines µ−
. The diagram represents the scattering of a muon with
electron, μ−
+e−
→ μ−
+e−
by electromagnetic interaction. The dots correspond-
ing to vertices are points where interactions occur. The wriggly line is the photon
line. At the vertex, the electron emits a photon which is absorbed by the muon or
viceversa. This corresponds to the lowest order of perturbation theory and is known
as the first order Feynman diagram or leading Feynman diagram.
Fig. 10.1 Muon-electron
scattering
The arrows on the solid lines towards the vertices indicate the direction of
Fermions in time, the antifermions are indicated by reversed arrows, moving back-
ward in time. The photon being an antiparticle of itself does not need any arrow on
the photon line. The lines which begin and end within the diagrams are the internal
lines correspond to virtual particles that is those which are not observed. The lines
which enter or leave the diagram are the external lines which represent the observed
or real particles. The external lines show the physical process of an event, while the
internal lines indicate its mechanism. The direction of fermions is such that charge
is conserved on each vertex. Also, the four-momentum is conserved at each vertex.
The virtual particles which are exchanged (photon in Fig. 10.1) are not present in
the initial or final state. They exist briefly during the interaction. In this short time
τ energy can be violated compatible with the uncertainty principle, τ . ΔE ≈ .
Consequently, a virtual particle is not required to satisfy the relativistic relation,
E2
= p2
+ m2
. A virtual particle can be endowed with any mass which is different
from that of a free particle. In Fig. 10.1, the exchanged photon couples to the charge
of one electron at the top vertex and the second one at the bottom vertex. For each
vertex the transition amplitude carries a factor which is proportional to e that is
√
α
(square-root of fine structure constant). The transition matrix will be proportional
to
√
α
√
α or α. The exchanged particle also introduces a propagation term in the

matrix element, the general form being (q2
+ m2
)−1
, where m is the mass of the
exchanged particle and q is the four-momentum transfer.
In general, the same end result is obtained from a number of Feynman diagrams.
The transition matrix element includes the superposition of amplitudes of all such
diagrams.
For the weak interaction the exchanged particle is a massive boson, W±
or Z0
,
indicated by a wavy or broken line. For the strong interaction, the exchanged particle
is a gluon indicated by a spring.
10.2 Problems
10.2.1 Conservation Laws
10.1 Are the following particle interactions allowed by the conservation
rules? If so, state which force is involved and draw a Feynman diagram for
the process.
[University of Aberystwyth 2003]
(i) μ−
→ e−
+ νμ + νe
(ii) Λ → π+
+ π−
(iii) νe + n → p + e−
(iv) π0
→ τ+
+ τ−
(v) e+
+ e−
→ μ+
+ μ−
10.2 Indicate, with an explanation, whether the following interactions proceed
through the strong, electromagnetic or weak interactions, or whether they
do not occur.
(i) π−
→ μ−
+ νμ
(ii) τ−
→ μ−
+ ντ
(iii) Σ0
→ Λ + γ
(iv) p → n + e+
+ νe
(v) π−
+ p → π0
+ Σ0
(vi) π−
+ p → K0
+ Σ0
(vii) e+
+ e−
→ μ+
+ μ−
10.3 Consider the decay of K0
meson of momentum P0 into π+
and π−
of
momenta p+ and p− in the opposite direction such that p+ = 2p−. Deter-
mine p0.
[Mass of K0
is 498 MeV/c2
; mass of π±
is 140 MeV/c2
]
10.4 Indicate, with an explanation, whether the following interactions proceed
through the strong, electromagnetic or weak interactions, or whether they
do not occur.
(i) Ξ−
→ Σ−
+ π0
(ii) τ−
→ e−
+ νe + ντ
(iii) τ+
→ μ+
+ γ

10.2 Problems 545
(iv) μ+
+ μ−
→ τ+
+ τ−
(v) p → e+
+ π0
(vi) π0
→ γ + γ
(vii) π−
+ p → K+
+ Σ−
(viii) π−
+ p → K−
+ Σ+
10.5 The ρ0
meson is known to have an intrinsic spin of , and pions zero spin.
Show that the requirements of symmetry on the total wave function of the
final state permit the decay
ρ0
→ π+
π−
but not ρ0
→ π0
π0
10.6 The baryon Ω−
has a mass 1,672 MeV/c2
and strangeness s = −3. Which
of the following decay modes are possible?
(a) Ω−
→ Ξ−
+ π0
(mΞ− = 1,321 MeV/c2
, S = −2, mπ0 = 135 MeV/c2
)
(b) Ω−
→ Σ0
+ π−
(mΣ0 = 1,192 MeV/c2
, S = −1, mπ− = 139 MeV/c2
)
(c) Ω−
→ Λ + K−

mΛ = 1,115 MeV/c2
, S = −1
mK− = 494 MeV/c2
, S = −1

(d) Ω−
→ n + K−
+ K0(mK0 = 498 MeV/c2
, S = −1)
10.7 The following transitions have Q-values and mean lifetimes as indicated
Transition
Q-value
(MeV) Lifetimes(s)
(a) Δ++
→ p + π+
120 10−23
(b) π0
→ 2γ 135 10−17
(c) μ+
→ e+
+ νe + νμ 105 2.2 × 10−6
(d) μ+
+12
C →12
B + νμ 93 2 × 10−6
State which interactions are responsible in each case and estimate the relative
coupling strengths.
10.8 Indicate how the following quantities will transform under the P (space
inversion) and T (time reversal) operation:
(a) Position coordinate r
(b) Momentum vector P
(c) Spin or angular momentum vector σ = r × P
(d) Electric field E = −∇V
(e) Magnetic field B = i × r
(f) Electric dipole moment σ.E
(g) Magnetic dipole moment σ.B
10.9 The deuteron is a bound state of neutron and proton and has spin 1 and posi-
tive parity. Prove that it can exist only in the 3
S1 and 3
D1 states.

10.10 Show that (a) ρ → η + π is forbidden as a decay through strong interaction.
(b) ω → η + π is forbidden as an electro-magnetic or strong decay.
10.11 (a) The ρ0
and K0
mesons both decay predominantly to π+
+ π−
. Explain
why the mean lifetime of the ρ0
is 10−23
s, while that of K0
is 0.89 ×
10−10
s.
(b) The Δ0
and the Λ both decay to proton and π−
meson. Explain why the
Δ0
meson lifetime is ∼10−23
s while that of Λ is 2.6 × 10−10
s.
10.12 State and give reasons for the following decay modes of the ρ-meson
(Jp
= 1−
, I = 1) which are allowed by the strong or electromagnetic inter-
action
(a) ρ0
→ π+
π−
(b) ρ0
→ π0
π0
(c) ρ0
→ η0
π0
(d) ρ0
→ π0
γ
10.13 Conventionally nucleon is given positive parity. What does one say about
deuteron’s parity and the intrinsic parities of u and d-quarks?
10.14 Which of the following hyperon decays are allowed in lowest order weak
interactions?
(a) Ξ0
→ Σ−
+ e+
+ νe
(b) Ξ0
→ p + π−
+ π0
(c) Ω−
→ Ξ0
+ e−
+ νe
(d) Ω−
→ Ξ−
+ π+
+ π−
10.15 Explain how the parity of K−
meson has been determined.
10.16 (a) Briefly define the following terms giving two examples of each:
(i) Hadron
(ii) Lepton
(iii) Baryon
(iv) Meson
(b) A π0
meson at rest decays into two photons of equal energy. What is the
wavelength (in m) of the photons? [The mass of the π0
is 135 MeV/c2
]
10.2.2 Strong Interactions
10.17 (a) The Δ++
Resonance has a full width of Γ = 120 MeV. How far on aver-
age would such a particle of energy 200 GeV travel before decaying?
(b) Given that the width for W-boson decay is less than 6.5 MeV, estimate the
limit for the corresponding lifetime.
10.18 Analyze the pion – proton scattering data in terms of isospin amplitudes a1/2
and a3/2 for the reactions:

10.2 Problems 547
π+
+ p → π+
+ p (1)
π−
+ p = π−
+ p (2)
π−
+ p = π0
+ n (3)
Show that if a1/2 ≪ a3/2 then σ1 : σ2 : σ3 = 9 : 1 : 2 and if a1/2 ≫ a3/2,
then σ1 : σ2 : σ3 = 0 : 2 : 1
10.19 Use the results of π-N scattering at the same energy,
σ+
(π+
p → π+
p) =
%
%a3/2
%
%2
σ−
(π−
p → π−
p) =
1
9
%
%a3/2 + 2a1/2
%
%2
σ0
(π−
p → π0
n) =
2
9
%
%a3/2 − a1/2
%
%2
to deduce the inequality
√
σ+ +
√
σ− −
√
2σ0 ≥ 0
10.20 Calculate the branching ratio for the decay of the resonance Δ+
(1232) which
has two decay modes
Δ+
→ pπ0
→ nπ+
10.21 A resonance X+
(1520) decays by the strong interaction to the final states nπ+
and pπ0
with branching ratios of approximately 36 and 18% respectively.
What is its isospin?
10.22 Given that the ρ-meson has a width of 158 MeV/C2
in its mass, how would
you classify the interaction for its decay?
10.23 In which isospin states can (a) π+
π−
π0
(b) π0
π0
π0
exist?
10.24 The particles X and Y can be produced by strong interaction
K−
+ p → K+
+ X
K−
+ p → π0
+ Y
Identify the particles X (1,321 MeV) and Y(1,192 MeV) and deduce their
quark content. If their decay schemes are X → Λ + π−
and Y → Λ + γ ,
give a rough estimate of their lifetime.
10.25 The scattering of pions by proton shows evidence of a resonance at a centre
of mass system momentum of 230 MeV/c. At this momentum, the cross-
section for scattering of positive pions reaches a peak cross-section of 190 mb
while that of negative pions is only 70 mb. What can you deduce about the
properties of the resonance (a) from the ratio of the two cross-sections (b)
from the magnitude of the larger?

10.26 Consider the formation of the resonance Δ (1,236) due to the incidence of
π+
and π−
on p. Assuming that at the resonance energy the I = 1/2 contri-
bution to the π−
+ p interaction is negligible show that at the resonance peak
σ(π+
+ p → Δ)
σ(π− + p → Δ)
= 3
10.27 Consider the reactions at the same energy
π+
+ p → Σ+
+ K+
π−
+ p → Σ−
+ K+
π−
+ p → Σ0
+ K0
Assuming that the isospin amplitude a1/2 ≪ a3/2, show that the cross
sections for the reactions will be in the ratio 9:1:2
10.28 Calculate the ratio of the cross sections for the reaction π−
p → π−
p and
π−
p → π0
n on the assumption that the two I spin amplitudes are equal in
magnitude but differ in phase by 30◦
.
10.29 Negative pions almost at rest are absorbed by deuterium atoms and undergo
the following reaction
π−
+ d → n + n
which is established by the direct observation of the neutrons which have
a unique energy for this process. Assuming that the parity of neutron and
deuteron is positive, show how the existence of the above reaction affords
the determination of parity of negative pion.
10.30 The cross-section for K−
+ p shows a resonance at PK ≈ 400MeV/c. This
resonance appears in the reactions
K−
+ p → Σ + π
→ Λ + π + π
But not in the reaction
K−
+ p → Λ + π0
What conclusion can you draw on the isospin value of the resonance?
10.31 K−
mesons are incident with equal frequency on protons and neutrons and
the following reactions are observed:
K−
+ p → Σ+
+ π−
→ Σ0
+ π0
→ Σ−
+ π+
K−
+ n → Σ−
+ π0
→ Σ0
+ π−

10.2 Problems 549
Show that the number of charged Σ’s will be equal to twice the number of
neutral Σ’s.
10.32 In the rest system of the B+
-meson, the products of the strong interaction
decay, B+
→ ω0
+π+
are found to be formed in an s-state. Deduce the spin,
parity and isospin of the B-meson. What difference would it make to your
conclusion if the decay took place by the week interaction (spin and parity
are respectively 0−
for the charge triplet pion and 1−
for the ω0
).
10.33 The Δ (1,232) is a resonance with I = 3/2. What is the predicted branch-
ing ratio for (Δ0
→ pπ−
)/(Δ0
→ nπ0
)? What would be the ratio for a
resonance with I = 1/2?
10.34 Show the position of pseudo scalar mesons π+
, π−
, π0
, K0
, K0, K+
, K−
and η on the S − I3 diagram.
10.35 Show the position of vector mesons ρ−
, ρ0
, ρ+
, ϕ, ω, k∗0
, k∗+
, k∗−
and k∗0
on S . I3 diagram.
10.36 Show the position of Baryons p, n, Ξ−
, Ξ0
, Λ, Σ+
, Σ−
, and Σ0
particles
with spin-parity
1
2
+
on the Y − I3 diagram.
10.37 Describe the (3

2)+
baryon decuplet on Y − I3 diagram.
10.38 (a) Explain why at the same energy the total cross-sections
σ(π−
+ p) ∼
= σ(π+
+ n), while σ(K−
+ p) = σ(K+
+ n)
(b) How can the neutral K-mesons, K0
and K0 be distinguished?
10.39 A hyper nucleus is formed when a neutron is replaced by a Λ-hyperon. In
the reactions of K−
in a helium bubble chamber, the mirror hyper nuclei
4
ΛHe and 4
ΛH are produced
K−
+ 4
He → 4
ΛHe + π−
→ 4
ΛH + π0
Determine the ratio of the cross sections.
10.40 In the reaction K−
+4
He → 4
ΛH + π0
, the isotropy of the decay products
has established J(4
ΛH) = 0. Show that this implies a negative parity for the
K−
- meson, regardless of the angular momentum state from which the K−
-
meson is captured.
10.41 Show that the reaction π−
+ d → n + n + π0
cannot occur for pions at rest.
10.42 At 600 MeV the cross sections for the reactions p + p → d + π+
and
p + n → d + π0
are σ+
= 3.15 mb and σ0
= 1.5 mb. Show that the ratio of
the cross sections is in agreement with the iso-spin predictions.
10.43 Explain which of the following combination of particles can or cannot exist
in I = 1 state

(a) π0
π0
(b) π+
π+
(c) π+
π−
(d) Λπ0
(e) Σ0
π0
(f) π−
π−
?
10.2.3 Quarks
10.44 Describe the phenomena when a quark is struck by a high energy electron
with a high enough momentum transfer.
10.45 The B−
meson is the lightest particle consisting of a b quark and u antiquark.
Which type of interaction causes its decay. Describe with explanation its
decay chain.
10.46 The 3γ decay of positronium (the bound state of e+
e−
) has a width that in
QED is predicted to be Γ(3γ ) = 2(π2
− 9)α6
mec2
/9π, where α is the fine
structure constant.
(a) If the hadronic decay of the c c bound state J/ψ (3,100) proceeds via an
analogous mechanism, but involving three gluons, use the experimental
hadronic width (fragmentation into hadrons occurring with probability
unity) Γ(3g) = 80 keV to estimate the effective strong interaction cou-
pling constant αs.
(b) Determine αs from the radiative width Γ(ggγ) = 0.16 keV of the b b
bound state γ (9,460)
10.47 Calculate the ratio R of the cross section for e+
e−
→ QQ → hadrons to that
for the reaction e+
e−
→ μ+
μ−
as a function of increasing CMS energy up
to 400 GeV. Assume the quark masses in GeV/c2
up or down 0.31, strange
0.5, charm 1.6, bottom 4.6 and top 175.
10.48 Show that the quark model predicts the following cross-section relation
σ(Σ−
n) = σ(pp) + σ(K−
p) − σ(π−
p)
10.49 Using the additive quark model the total interaction cross-section is assumed
to result from the sum of the cross-sections of various pairs. Assuming that
σ(qq) = σ(qq) prove the relation σ(Λp) = σ(pp) + σ(K−
n) − σ(π+
p)
10.50 The coulomb self- energy of a hadron with charge +e or −e is about 1 MeV.
The quark content and rest energies (in MeV)of some hadrons are
n(udd)940, p(uud)938, Σ−
(dds)1197, Σ0
(uds)1192, Σ+
(uus)1189,
K0
(ds)498, K+
(us)494
The u and d quarks make different contribution to the rest energy. Estimate
this difference.

10.2 Problems 551
10.51 (a) What are the quark constituents of the states Δ−
, Δ0
, Δ+
, Δ++
?
(b) Assuming the quarks are in states of zero angular momentum, what fun-
damental difficulty appears to be associated with the Δ states, which have
I = 3/2 and how is it resolved?
(c) How do you explain the occurrence of excited states of the nucleons with
the higher values of J. What parities would the higher states have?
10.52 Use the quark model to determine the quark composition of (a) Σ+
, Σ−
, n
and p (b) K+
, K−
, π+
, π−
mesons.
10.53 (a) What are the three particles described by taking the three identical quarks?
(b) What are the quantum numbers of the b quark?
10.54 Draw the quark flow diagrams for the decays (a) ϕ → K+
K−
(b) ω →
π+
π−
π0
(c) Show that the decay ϕ → π+
π−
π0
is suppressed.
10.55 At a beam energy of 60 GeV, σ(π+
p) ∼
= σ(π−
p) = 25 mb while σ(pp) ∼
=
σ(pn) = 38 mb. Show that the ratio of cross sections σ(π N)/σ(NN ) can be
explained by simple quark model.
10.56 The production of a leptonic pair in a pion–nucleon collision is explained by
the Drell–Yan mechanism which consists of the annihilation of the anti-quark
from the pion with a quark from the nucleon, producing a virtual photon that
transforms to a muon pair. Show that the cross section in π-12
C collisions
away from heavy meson resonances is predicted as
σ(π−
C)/σ(π+
C) = 4 : 1
[Courtesy D.H. Perkins, Introduction to High Energy Physics, University of
Cambridge Press]
10.57 Below the production threshold of charm particles, the cross-section for the
reaction e+
e−
→ μ+
μ−
is 20 nb. Estimate the cross-section for hadron pro-
duction
10.58 In the quark model, a meson is described as a bound quark-antiquark state.
It is usual to represent the potential energy between q and q by V (r) =
− A
r
+ Br where A and B are positive constants. At r few fermis, A is
negligible. Use the method of variation, with the trial function ψ(r) = e−r/a
to show that the ground state energy is given by E0 = 2.96 B2
2
mq
1
/3
, where
mq is the quark mass.
10.2.4 Electromagnetic Interactions
10.59 The reaction e+
e−
→ μ+
μ−
is studied using colliding beams each of energy
10 GeV and at these energies the reaction is predominantly electromagnetic.
Draw the lowest order Feynman diagram. The differential cross-section is
given by

dσ
dΩ
=
α2
2
c2
4E2
CM
(1 + cos2
θ)
where ECM is the total centre of mass energy and θ is the scattering angle
with respect to the beam direction. Calculate the total cross-section at this
energy.
10.60 (a) The Σ0
hyperon decays to Λ + γ with a mean lifetime of 7.4 × 10−20
s.
Estimate its width.
(b) Explain why the absence of the decay K+
→ π+
+ γ can be considered
an argument in favor of spin zero for K+
meson.
10.61 Positronium is the bound state of positron and electron. It is found either in
the singlet s-state (para-positronium) or in a triplet s-state (ortho-positronium).
Show how the C-invariance restricts the number of photons into which the
positronium can annihilate for these two types of systems
10.62 Which of the following processes are allowed in electro-magnetic interac-
tions, and which are allowed in weak interactions via the exchange of a single
W±
or Z0
?
(a) K+
→ π0
+ e+
+ νe
(b) Σ0
→ Λ + νe + νe
10.2.5 Weak Interactions
10.63 Estimate the rate of decay for D+
(1,869) → e+
+ anything, D0
(1,864) →
e+
+anything. Given the branching fractions B = 19%, and 8% respectively,
τD+ = 10.6 × 10−13
s, τD0 = 4.2 × 10−13
s
10.64 It is observed that the cross section for neutrino-electron scattering falls by
20% as the momentum transfer increases from very small values to 30 GeV/c.
Deduce the mass of the exchanged boson.
10.65 Estimate the number of W+
→ e+
νe events produced in 109
pp−
interactions
[The cross-sections σ(pp−
→ W+
) = 1.8 nb and σ(pp−
→ anything) =
70 mb]
(University of Cambridge, Tripos 2004)
10.66 Use Cabibbo theory to explain the difference in the decays D+
→ K0μ+
νμ
and D+
→ π0
μ+
νμ. Given that the D+
consists of a c quark and d antiquark.
10.67 Show that the ratio of decay rates
R ≡
Γ(Σ−
→ n + e−
+ νe)
Γ(Σ− → Λ + e− + νe)
∼
= 17

10.2 Problems 553
10.68 Use lepton universality and lepton-quark symmetry to estimate the branching
fraction for the decay τ−
→ e−
+νe +ντ . Ignore final states that are cabibbo-
suppressed relative to the lepton modes.
10.69 The following decays are all ascribed to the weak interaction, resulting in
three final state particles. For each process, the available energy Q in the
decay is given as well as the mean lifetime.
Q(MeV) τ(s)
(a) μ+
→ e+
+ νe + νμ 106 2.19 × 10−6
(b) n → p + e−
+ νe 0.78 900
(c) τ+
→ e+
+ νe + ντ 1776 3.4 × 10−13
(d) π+
→ π0
+ e+
+ νe 4.1 2.56
(e) 14
O →14
N∗
+ e+
+ νe 1.81 198
Apply Sergent’s law of radioactivity to show that the Q-values and mean
lifetimes are compatible with the same weak coupling.
10.70 Which of the following reactions are allowed and which forbidden as under
weak interactions?
(a) νμ + p → μ+
+ n
(b) νe + p → e−
+ π+
+ p
(c) K+
→ π0
+ μ+
+ νμ
(d) Λ → π+
+ e−
+ νe
10.71 Introduce a fictitious particle called Spurion to calculate the ratio of decay
rates Ξ−
→ Λ + π−
/Ξ0
→ Λ + π0
10.72 Consider the following decays of Σ hyperons
Σ+
→ n + π+
Σ−
→ n + π−
Σ+
→ p + π0
Using the ΔI = 1/2 rule, show that the triangle relation for the amplitudes is
a+ +
√
2a0 = a−
10.73 Λ hyperon can decay predominantly through the non-leptonic modes Λ →
p + π−
and Λ → n + π0
. Introduce a fictitious particle called spurion of
isospin 1/2 to convert the decay into a reaction and determine the branching
ratios for these modes.
10.74 Assuming that the entire energy resulting from the p–p chain reaction escapes
from the sun’s surface, calculate the flux of neutrinos received on earth. Take
the earth-sun distance as 1.5×108
km. Assume that the total energy output of
the sun is LΘ = 3.83 × 1026
Js−1
, and that each α-particle produced implies
the generation of 26.72 MeV
10.75 The observation of neutrinos emitted by the supernova SN 1987A 170,000
years ago provided a rough estimate of neutrino’s mass. Assume that the

neutrino energy was spread between 5 and 15 MeV over a period of 4 s.
Estimate the upper limit for the mass of neutrino.
10.76 A particle X decays at rest weakly as follows
X → π0
+ μ+
+ νμ
Determine the following properties of X
(a) Charge
(b) baryon number
(c) Lepton number
(d) Isospin
(e) Strangeness
(f) spin
(g) boson or fermion
(h) lower limit on its mass in MeV/c2
(i) Identity of X
10.77 The α-decay of an excited 2−
state in 16
O to the ground 0+
state of 12
C is
found to have a width Tα
∼
= 1.0×10−10
eV . Explain why this decay indicates
a parity-violating potential.
10.78 Given the mean life time of μ+
meson is 2.197 µs and its branching fraction
for μ+
→ e+
+ νe + νμ is 100%, estimate the mean lifetime of τ+
if the
branching fraction B for the decay τ+
→ e+
+νe +ντ is 17.7%. The masses
of muon and τ-lepton are 105.658 and 1,784 MeV.
10.79 State with reasoning which of the following particles may undergo two-pion
decay?
(a) ω0
(J pI
= 1−0
)
(b) η0
(J pI
= 0−0
)
(c) f 0
(J pI
= 2+0
)
10.80 Why is the decay η → 4π not observed?
10.81 Van Royen-Weisskopf proposed a formula for the partial width of the lep-
tonic decays of the vector mesons. For the vector mesons ρ0
(765), ω0
(785)
and Φ0
(1,020) which have similar masses, the partial width Γ ∝ Q2
where
Q2
=
%
%
ai Qi
%
%2
is the squared sum of the charges of the quarks in the
meson. Show that
Γ(ρ0
) : Γ(ω0
) : Γ(Φ0
) = 9 : 1 : 2
Cambridge Press]
10.82 Classify the following semi-leptonic decays of the D+
(1,869) = cd meson
as Cabibbo-allowed, Cabibbo-suppressed or forbidden in lowest order weak
interactions.
(a) D+
→ K+
+ π−
+ e+
+ νe
(b) D+
→ π+
+ π−
+ e+
+ νe

10.2 Problems 555
10.83 Classify the following semileptonic decays of the D+
(1,869) = cd meson
as Cabibbo-allowed, Cabibbo-suppressed or forbidden in lowest order weak
interactions.
(a) D+
→ K−
+ π+
+ e+
+ υe
(b) D+
→ π+
+ π+
+ e−
+ νe
10.84 Which of the following decays are allowed in lowest order weak interac-
tions?
(a) K−
→ π+
+ π−
+ e−
+ νe
(b) Ξ0
→ Σ−
+ e+
+ νe
(c) Ω−
→ Ξ−
+ π+
+ π−
10.85 Which of the following decays are allowed and which are forbidden ?
(a) K0
→ π−
e+
νe
(b) K0
→ π+
e−
νe
(c) K0 → π+
e−
νe
(d) K0 → π−
e+
νe
10.86 A muon neutrino is generated at time t = 0 at a particle accelerator. Show
that at a later time t the probability that it is still a muon neutrino is, in natural
units and in the neutrino rest frame
Pμ(t) = 1 − sin2
2θ sin2

(E2 − E1)t
2

Cambridge Press]
10.87 (a) In Problem 10.86, write down an expression for the probability Pe(t) that
at the same time t, the neutrino has oscillated into an electron neutrino.
(b) Derive the expression for the time at which the probabilities Pμ(t) and
Pe(t) are first equal. Assuming that mνe
= 2 eV and mνμ
= 3 eV and
θ = 340
; find time t when beam energy is 1 GeV
10.88 Show how the following data prove the universality of the weak coupling
constant. τμ = 2.197 × 10−6
s, ττ = 2.91 × 10−13
s, the branching fraction
of the tauon, B(τ+
→ e+
νeντ ) = 0.178, mμ = 105.658 MeV/c2
, mτ =
1,777 MeV. Note that Γ(μ → eνeνμ) =
G2
m5
μ
192π3
in natural units and the
Fermi constant G ∝ g2
, where g is the weak charge also known as the
coupling amplitude.
10.89 Consider the semi leptonic weak decays (a) Σ−
→ n + e−
+ νe (b) Σ+
→
n + e+
+ νe Explain why the reaction (a) is observed but (b) is not.
10.90 The D+
meson (cd) decays via the weak interaction to K0 μ+
νμ
Alternatively the D+
can decay to π0
μ+
νμ. What are the predictions of
Cabibbo’s theory for the relative rates of the two decays?

10.91 The neutral kaons K0
and K0 are the charge conjugate of each other and are
distinguished by their strangeness. However, they decay similarly and mixing
can occur by a virtual process like K0
⇔ π+
+ π−
⇔ K0. Starting with a
pure beam of K0
’s at t = 0 obtain the intensity of K0
’s and K0’s at time t,
in terms of the mean lifetimes τL(0.9 × 10−10
s) and τs(0.5 × 10−7
s) for the
components KL and Ks, long lived and short lived respectively.
10.92 Suppose one starts with a pure beam of K0
’s which traverses in vacuum
for a time of the order of 100 Ks mean lives so that all the Ks- component
has decayed and one is left with KL only. If now the KL traverses a carbon
screen some of the KS states are regenerated. Explain this phenomenon of
regeneration.
10.2.6 Electro-Weak Interactions
10.93 The observation of the process νμe−
→ νμe−
, signifies the presence of a
neutral current interaction. Similarly, why does the process νee−
→ νee−
,
not indicate the presence of such an interaction?
10.94 From the data on the partial and full decay width of Z0
boson show that the
number of neutrino generations is 3 only.
Γz(total) = 2.534 GeV, Γ(z0
→ hadrons) = 1.797 GeV,
Γ(z0
→l+
l−
) = 0.084 GeV.Theoretical value for Γ(z0
→νlνl) = 0.166 GeV.
10.95 (a) What are the experimental signatures and with what detectors would one
measure (a) W → eν and W → μν (b) Z0
→ e+
e−
and Z0
→ μ+
μ−
(b) The weak force is due to W, Z exchange, mass ∼
= 100 GeV. Give the
range in meters.
10.96 Using the results of electro-weak theory of Salam and Weinberg, calculate
the masses of W and Z bosons. Take the fine-structure constant α = 1/128
and Fermi’s constant GF/3
c3
= 1.166 × 10−5
GeV−2
and Weinberg angle
θW = 28.17◦
10.2.7 Feynman Diagrams
10.97 Explain which force is responsible for the following particle interactions
and draw a Feynman diagram for each:
(i) τ+
→ μ+
+ νμ + ντ
(ii) K−
+p → Ω−
+K+
+K0
(K−
= su; K+
= us; K0
= ds; Ω−
= sss)
(iii) D0 → K+
+ π−
(D0 = uc; K+
= us; π−
= du)

10.2 Problems 557
10.98 Sketch the Feynman diagrams and replace the symbols l with the correct
leptons or anti-leptons in the following:
(i) l + n → e−
+ p
(ii) τ−
→ μ−
+ ℓ + l̄
(iii) B0
→ D−
+ μ+
+ l
[The quark content of B0
is bd and D−
is cd]
10.99 Draw the lowest order Feynman diagram for the following decays
(a) Δ0
→ pπ−
(b) Ω−
→ ΛK−
10.100 (a) Draw the lowest order Feynman diagram for e+
e−
→ νμνμ
(b) Draw the lowest order Feynman diagram for D0
→ K−
π+
and estimate
the ratio of the transition rates.
[Quark contents (masses in MeV/c2
):Δ+
= uud(1, 232), Ω−
= sss(1, 672),
Ξ0
= uss(1, 315), p = uud(938), n = udd(940), π0
= 1
√
2
,
uu + dd
-
(135),
π+
= ud(140), D0
= cu(1, 865), K−
= ūs(494)]
[Adapted from University of Cambridge, Tripos 2004]
10.101 Draw the lowest order Feynman diagram for the following processes:
(a) e−
− e−
elastic scattering (Moller scattering)
(b) e+
e−
→ e+
e−
(Bhabha scattering)
10.102 Draw the lowest order Feynman diagram at the quark level for the following
decays
(a) Λ → p + e−
+ νe
(b) D−
→ K0
+ π−
10.103 Draw the Feynman diagrams at the quark level for the reactions:
(a) π−
+ p → K0
+ Λ
(b) e+
+ e−
→ B0 + B0
, where B is a meson containing a b-quark.
10.104 Draw the lowest order Feynman diagram for the decay K−
→ μ−
+ νμ+γ
and hence deduce the form of the overall effective coupling.
10.105 Explain with the aid of Feynman diagrams, why the decay D0
→ K−
+π+
can occur as a charged-current weak interaction at lowest order, but the
decay D+
→ K0
+ π+
cannot.
10.106 Why is the mean lifetime of the charged pion much longer than that of the
neutral pion? Draw Feynman diagram to illustrate your answer.
10.107 Draw Feynman diagrams for
(a) Bremsstrahlung
(b) pair production.
10.108 Draw Feynman diagrams for
(a) photo electric effect
(b) Compton scattering

10.109 Draw Feynman diagrams for the processes
(a) e+
e−
→ qq
(b) νμ + N → νμ + X.
10.110 Draw Feynman diagrams for the decays
(a) Ξ−
→ Λ + π−
(b) K+
→ π+
π+
π−
10.111 (a) Draw the Feynman diagram for the semi leptonic decay of D+
→ K0
+
l + l̄.
(b) Draw Feynman diagram for the tauon decay τ−
→ π−
+ ντ
10.112 Draw Feynman diagrams for (a) Two-photon annihilation (b) Three-photon
annihilation of positronium.
10.113 Draw the Feynman diagram for the decay Λ → p + π−
10.3 Solutions
10.1 (i) Weak decay
Fig. 10.2(a)
(ii) Forbidden because Baryon number is violated
(iii) Charge current weak interaction
Fig. 10.2(b)
(iv) Forbidden because energy is not conserved

10.3 Solutions 559
(v)
Fig. 10.2(c) A EM
interaction
Fig. 10.2(d) Weak
interaction
10.2 (i) Weak interaction because neutrino is involved
(ii) Does not occur because lepton number is violated
(iii) Electromagnetic interaction as gamma ray is involved and ΔS = 0
(iv) Allowed as a weak decay if proton is bound but forbidden when proton
is free because proton is lighter than the sum of masses of the product
particles.
(v) Does not occur as a strong or electromagnetic interaction because
ΔS = 0
(vi) Strong interaction because ΔS = 0 and other quantum numbers are
conserved.
(vii) Weak interaction, because a lepton- antilepton pair is involved.
10.3 If P0 is the momentum of Kaon, p1 and p2 the momenta of the pions, then
momentum conservation requires
p0 = p1 − p2 = 2p2 − p2 = p2
Energy conservation requires
E2 + E1 = E0
or
$
p2
2 + m2
π +
$
p2
1 + m2
π =
$
p2
0 + m2
K (1)
But p2 = p0 and p1 = 2p2 = 2p0 (2)
Using (2) in (1) and solving the resultant equation
p0 =
mK
2

m2
K − 4m2
π
2m2
K + m2
π
1/2
(3)

Substituting mK = 498 MeV/c2
and mπ = 140 MeV/c2
, we get
p0 = 142.8 MeV/c2
10.4 (i) Does not occur because energy is not conserved
(ii) Weak interaction because neutrinos are involved
(iii) Does not occur because lepton number is not conserved
(iv) Weak interaction because leptons are involved
(v) Does not occur because of non-conservation of baryon number and
lepton number
(vi) Electromagnetic interaction because γ -rays are involved, charge and
c-parity are conserved.
(vii) Occurs as strong interaction because strangeness is conserved
(viii) Does not occur because strangeness is not conserved.
10.5 ρ0
has J p
= 1−
. The conservation of angular momentum requires that the
two π0
’s are in l = 1 state of orbital angular momentum. This state is anti-
symmetric and is therefore forbidden for identical bosons which require the
state to be symmetric with respect to the exchange of two bosons.
10.6 (a) This decay mode is allowed and is observed. ΔS = 1 as required for the
weak decays of strange particles.
(b) The decay is forbidden as ΔS = 2
(c) The decay is allowed as ΔS = 1. Further, the rest mass energy of Ω−
is
greater than the sum of the energies of decay products. Also Q/e and B
are conserved.
(d) The decay is forbidden although ΔS = 1 and Q/e and B are conserved.
But E is violated.
10.7 (a) Strong interaction
(b) Electromagnetic interaction
(c) Weak interaction
(d) Weak interaction
Relative strength:1 : 10−2
: 10−7
: 10−7
10.8 (a) r → −r under P- operation as x → −x, y → −y and z → −z but r → r
under T- operation.
(b) P reverses its sign under both P and T operation, P → −P. Both r and p
are known as polar vectors.
(c) σ or L are axial vectors. σ = r × p. Since both r and p change their
sign under P-operation, L does not. However under T-operation r does
not change sign but p does and so σ changes its sign.
(d) E = −∂V/∂r for the above argument changes sign under P-operation as r
changes its sign and does not under T-operation as r does not.
(e) The magnetic field like angular momentum is an axial vector B = i × r.
Under p-operation B → B because i → −i and r → −r but under
T-operation because r → r and i → −i so that B → −B

10.3 Solutions 561
(f) σ.E → −σ.E under both P and T-operations.
(g) Similarly σ.B → −σ.B under both p-operation and under T-operation.
10.9 The deuteron which is the nucleus of deuterium (heavy hydrogen) consists of
one proton and one neutron. Since the parity of neutron and proton are +1,
that of deuteron is also +1. Spin of deuteron is 1, and l = 0 mostly, with
4% admixture of l = 2 so that the parity determined by (−1)l
= +1 for
l = 0 or 2. The deuteron is in a state of total angular momentum J = 1.
Thus J p
= 1+
. Using the spectroscopic notation 2s+1
LJ, deuteron’s state is
described by 3
S1 and 3
D1.
10.10 (a) For ρ, η and π, the IG
values are 1+
, 0+
and 1−
respectively. There-
fore G-parity is violated. Therefore, the decay of ρ is forbidden by strong
interaction.
(b) For ω, η and π, the parities are −1. The overall parity of η and π will
be (−1)(−1) = +1, as they are emitted in the s-state of relative angular
momentum. Therefore, both strong and em interactions will be forbidden.
The strong interaction will be forbidden also because of non conservation
of isospin.
10.11 (a) In the decay ρ0
→ π+
π−
, all the quantum numbers (Q/e, B, I, G, π)
are conserved as required by a strong interaction. Therefore its lifetime
(∼10−23
s) is characteristic of a strong interaction. On the other hand the
decay K0
→ π+
π−
, violates strangeness. It is therefore a weak decay,
characterized by relatively long lifetime of the order of 10−10
s.
(b) Δ0
→ p + π−
→ n + π0
The Δ0
has T = 3/2 and T3 = −1/2. In both the decays T3 is conserved
(for the first one p + π−
system has T3 = +1/2 − 1 = −1/2, and for the
second one n + π0
system has T3 = −1/2 + 0 = −1/2). Therefore, the
decay proceeds through strong interaction with a characteristic life time
of ∼10−23
s.
In the case of Λ,
Λ → p + π−
→ n + π0
Λ has T3 = 0, and therefore T3 is violated. The decay being weak has the
characteristic life time of 10−10
s.
10.12 (a) For ρ0
(770 MeV), Jp
= 1−
and IG
= 1+
while for π (139 MeV),
Jp
= 0−
and IG
= 1−
. In the decay ρ0
→ π+
π−
, all the quantum
numbers (Q/e, B, I, G, π) are conserved. Note that the decay involves
a large Q-value (491 MeV) so that the pions will come off with relative
angular momentum of l = 1, contributing (−1) to overall parity. Thus,
the overall parity is conserved (each pion has intrinsic parity −1).
(b) Decay is forbidden because of Bose symmetry.

(c) Decay via strong or electromagnetic interaction is forbidden because of
violation of cp invariance. For ρ0
, cp = −1. But cη = cπ0 = +1 because
both decay to two gamma rays.
(d) The decay is allowed via electromagnetic interaction.
10.13 Parity of deuteron πd = πp πn. (−1)l
. As πp = πn = +1 for s or d state
l = 0 or 2.
πd = +1
The intrinsic parity of quarks is assumed to be positive because the intrinsic
parity of a nucleon (+) comes from the parities of three quarks and l = 0.
10.14 Decay (a) is forbidden by the ΔS = ΔQ rule for the semi – leptonic decays
and (b) is forbidden by the ΔS = 0, ±1 rule for the hadronic weak decays.
(c) and (d) are allowed and both have been experimentally observed.
10.15 Since strange particles are always produced in pairs, as in the reaction π−
+
p → Λ+K0
, the intrinsic parity of a strange particle can only be determined
relative to that of another. Thus, for example, one can determine the kaon
parity relative to that of Λ, which by convention is assigned a positive parity.
Consider the reaction
K−
+ He4
→ ΛHe4
+ π−
→ ΛH4
+ π0
These reactions are known to occur in a helium bubble chamber. Now, the
Λ is bound in an s-state relative to the nuclear core He3
or H3
which have
positive parity. Furthermore, all the participants in the reaction are spinless.
Linitial = Lfinal. The orbital angular momentum does not contribute to the
parity because of s-state. The only relevant parities in the above reaction are
PK = PΛ. Pπ− = −PΛ, as Pπ− = −1
The validity of the argument obviously hinges on the hyper-nuclei having
zero spin. If the spin were 1, for example, angular momentum conservation
would require l = 1 in the final state, thus reversing the conclusion. The
spin of ΛH4
has also been experimentally determined to be indeed zero. It is
concluded that the relative parity of K−
is negative.
10.16(a) (i) Hadron is an elementary particle which participates in strong interac-
tions, examples being neutron and pion.
(ii) Lepton participates in weak interaction and if charged in em interac-
tion as well, examples being electron and muon. Lepton number is
universally conserved.
(iii) Baryon comprises nucleons and hyperons which participate in strong
interaction, examples being proton and Σ-hyperon. Baryons are
fermions and baryon number is universally conserved.
(iv) Mesons are the carrier of strong forces, examples being pion and kaon.

10.3 Solutions 563
(b) Each photon will carry 67.5 MeV
λ =
1241
67.5 × 106
nm = 18.385 × 10−6
nm = 1.84 × 10−14
m
10.17 (a) τ =

Γ
=
c
Γc
=
197.3 × 10−15
(MeV − m)
(120MeV)(3 × 108 m/s)
= 5.5 × 10−24
s
γ = 1 +
200
1.236
∼
= 163
β ∼
= 1
d = βcγ τ = 1 × 3 × 108
× 163 × 5.5 × 10−24
m
∼
= 2.7 × 10−13
m
∼
= 0.0003 nm
a distance which is much less than the resolution obtainable by the avail-
able techniques. The best resolution obtained in photographic emulsions
is only 1 µm.
(b) Uncertainty Principle:
Γ.τ ≥
τ ≥

Γ
=
c
Γc
=
197.3 MeV − fm
6.5(MeV) × 3 × 108(m/s)
=
197.3 × 10−15
19.5 × 108
τ ≥ 10
−22
s
10.18 π+
+ p → π+
+ p (1)
π−
+ p → π−
+ p (2)
π−
+ p → π0
+ n (3)
The cross-section is proportional to the square of the matrix element Mif
connecting the initial and final states
Mif =
7
ψ f |H| ψi
8
where H is the isospin operator, and
σ ∝
%
%Mi f
%
%2
As pion has T = 1 and proton T = 1/2, the reactions can proceed either
through I = 1/2 or I = 3/2 channels. Designating the corresponding oper-
ators by H1 and H3 and the matrix elements for the reactions by M1 and M3,
we can write
M1 =
5
ψ f

1
2
%
%
%
% H1
%
%
%
%ψi

1
2
6
M3 =
5
ψ f

3
2
%
%
%
% H3
%
%
%
%ψi

3
2
6
As I-spin is conserved there is no operator connecting different isospin
states. Reaction (1) involves a pure state of I = 3/2, I3 = +3/2.

Therefore, σ1 = C |M3|2
where C = constant
For reaction (2) we have the mixture of I = 1/2 and I = 3/2 states.
Refering to the table for 3/2 × 1/2 C.G. coefficients we can write
|ψi =
%
%ψf
8
=
1
3
%
%
%
%φ

3
2
, −
1
2
6
−
2
3
%
%
%
%φ

1
2
, −
1
2
6
Therefore, σ2 = C
7
ψ f |H1 + H3| ψi
82
= C
%
%
%
%
1
3
M3 +
2
3
M1
%
%
%
%
2
For the reaction (3) we have
|ψi =
1
3
%
%
%
%φ

3
2
, −
1
2
6
−
2
3
%
%
%
%φ

1
2
, −
1
2
6
%
%ψ f
8
=
2
3
%
%
%
%φ

3
2
, −
1
2
6
+
1
3
%
%
%
%φ

1
2
, −
1
2
6
Therefore, σ3 = C
%
%
%
$
2
9
M3 −
$
2
9
M1
%
%
%
2
The ratio of cross-sections are
σ1 : σ2 : σ3 = |M3|2
:
1
9
|M3 + 2M1|2
:
2
9
|M3 − M1|2
If a1/2 a3/2, M1 M3, σ1 : σ2 : σ3 = 9 : 1 : 2
And if a1/2 a3/2, M1 M3, σ1 : σ2 : σ3 = 0 : 2 : 1
10.19 From the known π − N scattering cross-sections in terms of the amplitudes
a3/2 and a1/2 obtained in Problem 10.18 we can construct a diagram for the
amplitudes in the complex plane, Fig. 10.3
Fig. 10.3 Diagram of
amplitudes
√
σ+ = |a3|
√
σ− =
1
3
|a3 + 2a1|
√
2σ0 =
2
3
|a3 − a1|
where for brevity we have written a1 for a1/2 and a3 for a3/2
From the triangle the required inequality follows from the fact that the sum
of two sides is equal to or greater than the third side.

10.3 Solutions 565
10.20 The nucleon has T = 1/2 and pion T = 1. The Δ+
has T = 3/2 and
T3 = +1/2. Using the C.G.C. for 1 × 1/2 (Table of Chap. 3), we have
%
%
%
%
3
2
,
1
2
6
=
2
3
|1, 0
%
%
%
%
1
2
,
1
2
6
+
1
3
|1, 1
%
%
%
%
1
2
, −
1
2
6
%
%
%
%Nπ;
3
2
,
1
2
6
=
2
3
%
%Pπ0
8
+
1
3
%
%nπ+
8
∴
Γ(Δ+
→ π0
+ p)
Γ(Δ+ → π+ + n)
=

2/3
2
3

1/3
2
= 2
Note that for the charge states Δ++
, the strong decay is through only one
channel (Δ++
→ p + π+
), so also for Δ−
, viz Δ−
→ n + π−
.
10.21 X+
→ n + π+
→ p + π0
X+
can have either T = 3/2 or 1/2. For T = 3/2 the predicted ratio
Γ(X+
→ n π+
)/Γ(X+
→ pπ0
) = 1/2 (as in Problem 10.20) which is
in disagreement with the experimental ratio of 36/18 or 2.
If we assume the value T = 1/2 for X+
the state
%
%1
2
, 1
2
8
must be orthogonal
to the state
%
%3
2
, 1
2
8
. Therefore
X+
%
%
%
%Nπ,
1
2
,
1
2
6
= −
1
3
%
%pπ0
8
+
2
3
%
%nπ+
8
upto an overall phase factor.
The branching ratio
∴ Γ(X+
→ nπ+
)/Γ(X+
→ pπ0
) = 2/1, which is in agreement with
experimental ratio of 36/18 or 2/1.
10.22 τ =
Γ
= c
Γc
= 197.3 MeV−fm
158 (MeV)×3×108 (m/s)
= 197.3×10−15
(MeV−m)
474 (MeV)×108 (m/s)
= 4 × 10−24
s
a lifetime which is characteristic of a strong interaction. Therefore, the ρ-
meson decays via strong interaction.
10.23 We first write down the isospin for a pair of pions and then combine the
resultant with the third pion. (a) Each pion has T = 1, so that the π+
, π−
combination has I = 2, 1, 0. When π0
is combined, possible values are I =
3, 2, 1, 0 (b) Two π0
’s give I = 2. When the third π0
is added total I = 3
or 1.
10.24 From the conservation laws for strong interactions the quantum numbers for
X are B = +1, Q/e = −1, S = −2. It is Xi hyperon (Ξ−
) and decays
weakly with lifetime τ∼10−13
s. Its quark structure is dss; the quantum
numbers for Y are B = +1, Q/e = 0, S = −1. It is a sigma hyperon
(Σ0
) which decays electromagnetically with lifetime of the order of 10−20
s.
Its quark structure is uds.

10.25 (a) From Problem 10.18, we have the result
%
%a3/2
%
%2
∝ 190 mb

1
√
3
%
%a3/2 + 2a1/2
%
%
2
∝ 70 mb
Dividing one by the other, and solving we find a1/2 = 0.0256 a3/2. Thus
the amplitude a1/2 is negligible. The resonance is therefore characterized
by I = 3/2.
(b) The fact that π+
p scattering cross-section has a fairly high value at
p = 230 MeV/c implies that a3
2
amplitude is dominant.
10.26 From the analysis of π-N scattering (Problem 10. 18) the ratio
σ(π+
+ p → Δ)
σ(π− + p → Δ)
=
%
%
%a3
2
%
%
%
2
.
1
√
3
%
%
%a3
2
+ 2a1
2
%
%
%
/2
If we put a1/2 = 0, we get the ratio as 3.
10.27 The analysis is identical with that for πp reactions (problem 10.18). The
Σ’s are isospin triplet (T = 1) with the third components T3 = +1, 0, −1
for Σ+
, Σ0
, Σ−
similar to π+
, π0
, π−
. Further, the K+
and K0
form
a doublet (T = 1/2) with T3 = +1/2 and −1/2, analogous to p and n.
Therefore the result on the ratio of cross-section will be identical with that
for pion – nucleon reactions as in Problem 10.18
σ(π+
+ p → Σ+
+ K+
) : σ(π−
+ p → Σ−
+ K+
) :
σ(π−
+ p → Σ0
+ K0
)
=
%
%a3/2
%
%2
: 1
9
%
%a3/2 + 2a1/2
%
%2
: 2
9
%
%a3/2 − a1/2
%
%2
If a1/2 a3/2, then the ratio becomes 9:1:2.
10.28 Pions have T = 1 and nucleons T = 1/2, so that the resultant isospin both in
the initial state and the final state can be I = 1/2 or 3/2. Looking up the table
for Clebsch – Gordon Coefficients of 1 × 1/2 (Table 3.3) we can write
%
%π−
p
8
=
1
3
%
%
%
%
3
2
, −
1
2
6
−
2
3
%
%
%
%
1
2
, −
1
2
6
%
%π0
n
8
=
2
3
%
%
%
%
3
2
, −
1
2
6
+
1
3
%
%
%
%
1
2
, −
1
2
6
Therefore,
7
π−
p |H| π−
p
8
= 1
3
a3 + 2
3
a1
7
π0
n
%
% H
%
%π−
p
8
=
√
2
3
a3 −
√
2
3
a1

10.3 Solutions 567
The ratio of cross sections is
σ−
σ0
=
|a3 + 2a1|2
%
%
%
√
2 (a3 − a1)
%
%
%
2
Put a3 = a1eiΦ
with Φ = ±30◦
σ1
σ2
=
1
2
%
%eiφ
+ 2
%
%2
%
%eiφ − 1
%
%2
=
1
2

eiφ
+ 2

e−iφ
+ 2

eiφ − 1

e−iφ − 1
=
1
2
(5 + 4 cos φ)
(2 − 2 cos φ)
= 15.8
10.29 Deuteron has spin 1 and is in the s-state (l = 0) mostly, with an admixture
of l = 2. The total angular momentum J = 1 so that J p
= 1+
. Thus
deuteron’s state is described by 3
S1 and 3
D1. Next we consider the parity
arising from the angular momentum of the π−
–d system. The time for a π−
to reach the K− orbit in the deuterium mesic atom is estimated as ∼10−10
s.
Furthermore, direct capture of π−
from 2p level is negligible compared to the
transition from 2p to 1s level. It is therefore concluded that all π−
’s will be
captured from the s-state of the deuterium atom before they decay. Therefore,
the parity arising from the angular momentum is (−1)l
= (−1)0
= +1. If
pπ− is the parity of π−
then the parity of the initial state will be
π(initial) = pd.pπ− .1 = pπ−
In the final state the neutrons obey Fermi – Dirac statistics and must be
in the anti symmetric state. The two neutrons can have total spin S = zero
(singlet, antisymmetric) or S = 1 (triplet, symmetric). Now the total wave
function which is the product of space and spin parts must be antisymmetric.
ψ = ϕ(space)χ(spin)
Now the symmetry of the spin function is (−1)s+1
and that for the spatial
part it is (−1)L
. Hence the overall symmetry of the wavefunction ψ under
the interchange of both space and spin will be (−1)L+S+1
.
The initial state of π−
− d system is characterized by L = 0 and S = 1
(∵ the deuteron spin = 1 and the pion spin is zero) and L = 0. Hence the
total angular momentum of the initial state is 1. By conservation of angular
momentum, final state must also have J = 1. The requirement J = 1 implies
L = 0, S = 1 or L = 1, S = 1 or L = 1, S = 0 or 1, or L = 2, S = 1. The
possible final states are enumerated 3
S1(symmetrical); 3
P1(antisymmetrical);
3
D1(symmetrical); 1
P1(symmetrical). Of these combinations L = S = 1
alone gives L + S even.
Thus the only correct state is 3
P1 with parity (−1)l
= −1, which also
requires negative parity for the initial state as parity is conserved in strong
interactions. It follows that pπ− = −1.
It may be pointed out that the assumed parity of +1 for neutron and
deuteron is immaterial because the parities of baryon particles get cancelled
in any reaction due to conservation of baryon number.

10.30 In the initial state T (K) = 1/2 and T (p) = 1/2, so that I = 0 or 1. In the
first two reactions I = 2, 1 or 0; so that these reactions can proceed through
I = 1 or 0. In the third reaction T (Λ) = 0 and T (π0
) = 1 so that I = 1 only.
Since the resonance does not go through, the conclusion is that the resonance
has T = 0.
10.31 K−
+ p → Σ+
+ π−
(1)
K−
+ p → Σ0
+ π0
(2)
K−
+ p → Σ−
+ π+
(3)
K−
+ n → Σ−
+ π0
(4)
K−
+ n → Σ0
+ π−
(5)
The initial K−
+ p system of two particles of T = 1/2 has I3 = 0 and
consists equally of I = 0 and I = 1 state. The final Σ + π state will be
|ψ =
1
√
2
(a0 |φ (0, 0) + a1 |φ (1, 0) )
For I1 = I2 = 1, referring to the C.G. Coefficients (Table 3.3) we can write
|φ (0, 0) =
1
√
3
%
%Σ+
π−
8
−
%
%Σ0
π0
8
+
%
%Σ−
π+
8
|φ (1, 0) =
1
√
2
%
%Σ−
π+
8
−
%
%Σ+
π−
8
|ψ =
1
√
2

a0
√
3
−
a1
√
2

%
%Σ+
π−
8
−
a0
√
3
%
%Σ0
π0
8
+

a0
√
3
+
a1
√
2

%
%Σ−
π+
8

∴

Σ+
π−

:

Σ0
π0

:

Σ−
π+

=
1
2

a0
√
3
−
a1
√
2
2
:
a2
0
6
:
1
2

a0
√
3
+
a1
√
2
2
The reactions of K−
with n go through I = 1 only. Since I3 = −1 and the
final Σπ state is a1 |ψ (1, −1) = a1
√
2
,%
%Σ−
π0
8
−
%
%Σ0
π−
8-

Σ−
π0

:

Σ0
π−

=
a2
1
2
:
a2
1
2
If K−
is incident with equal frequency on p and n, then
Σ−
+ Σ+
=
1
2

a0
√
3
−
a1
√
2
2
+
1
2

a0
√
3
+
a1
√
2
2
+
a2
1
2
=
a2
0
3
+ a2
1
Σ0
=
a2
0
6
+
a2
1
2
It follows that Σ−
+ Σ+
= 2Σ0
10.32 B+
→ ω0
+ π+
Jp
1−
0−
I 0 1

10.3 Solutions 569
Assuming that the decay proceeds via strong interaction, the parity
PB = Pω Pπ = (−1)(−1)(−1)0
= +1
Jω = 1 + 0 = 1 (because l = 0)
I = 0 + 1 = 1
For the B+
meson, J p
= 1+
and I = 1
In case of weak decay, the spin would still be 1 but it would not be mean-
ingful to talk about I or P.
10.33 The nucleon has T = 1/2 and pion T = 1. The Δ0
has T = 3/2 and
T3 = −1/2. Using Clebsch – Gordon coefficients (C.G.C) for 1 × 1/2
(Table 3.3 of Chap. 3), we have
%
%
%
%
3
2
, −
1
2
6
=
2
3
|1, 0
%
%
%
%
1
2
, −
1
2
6
+
1
3
|1, −1
%
%
%
%
1
2
,
1
2
6
Δ0
π0
n π−
p
The ratio of the amplitudes for the decays Δ0
→ π−
p and Δ0
→ π0
n
is given by the ratio of the corresponding C.G.C. and the ratio of the cross
sections by the squares of the C.G.C. Thus, the branching ratio would be

2
3
#24
1
3
#2
= 2
10.34
Fig. 10.4 S − I3 plot for
pseudoscalar meson octet
10.35
Fig. 10.5 S − I3 plot for
vector meson nonet

10.36
Fig. 10.6 Y − I3 plot for
baryon octet
10.37
Fig. 10.7 Y − I3 plot for
baryon decuplet
10.38 (a) The reaction channels for the pion reactions π−
+ p → π−
+ p and
π−
+p → π0
+n are charge symmetric to the reactions, π+
+n → π+
+n
and π+
+n → π0
+ p. We can therefore expect the reaction cross sections
with pions to be nearly identical. On the other hand, the reactions with
kaons are not charge symmetric because K+
and K−
do not constitute an
iso spin doublet but are particle–antiparticle. Typical reactions with K+
are K+
+ n → K+
+ n and K+
+ n → K0
+ p
However for K−
many more channels are open.
K−
+ p → K−
+ p
→ K0
+ n
→ Σ+
+ π−
→ Σ0
+ π0
→ Σ−
+ π+
So, there is no reason for σ(K−
+ p) to be equal to σ(K+
+ n).
(b) Because of the CPT theorem, K0
and K0 which are particle and anti-
particle, cannot be distinguished by their decay modes, but can be identi-
fied through their distinct strong interaction process. Thus

10.3 Solutions 571
K0
+ p → K+
+ n allowed
K0
+ p → Σ+
+ π0
K0
+ n → K−
+ p

forbidden because of strangeness non-conservation
K0 + p → Σ+
+ π0
K0 + n → K−
+ p
2
allowed
K0 + p → K+
+ n forbidden
10.39 4
He is singlet and so T = 0, while K−
has T = 1/2. Therefore, the initial
state is a pure I = 1/2 with I3 = −1/2. In the final state 4
ΛHe and 4
ΛH form a
doublet, with T = 1/2. As pion has T = 1, the final state will be a mixture
of I = 3/2 and 1/2. Looking up the Clebsch – Gordon Coefficients 1 × 1/2
(given in Table 3.3 of Chap. 3), we find the
final state ∼
1
√
3
%
%π0
HΛ
8
−
2
3
%
%π−
HeΛ
8
σ(Λ
4
He)
σ(Λ
4H)
=
$
2

3
2
1
?√
3
2
=
2
1
10.40 As both K−
and π0
have zero spin, l is conserved. Thus the orbital angular
momentum l must be the same for the initial and final states. Since this is a
strong interaction, conservation of parity requires that
(−1)l
Pk = (−1)l
Pπ = (−1)l
(−1)
Therefore, Pk = −1
10.41 The deuteron spin Jd = 1 and the capture of pion is assumed to occur from
the s-state. Particles in the final state must have J = 1. As the Q- value is
small (0.5 MeV), the final nnπ0
must be an s-state. It follows that the two
neutrons must be in a triplet spin state (symmetric) which is forbidden by
Pauli’s principle.
10.42 Referring to the two nucleons by 1 and 2, + for proton and – for neutron,
and using the notation |I, I3 for the isotopic state and the Clebsch–Gordon
Coefficients for 1/2× 1/2 we can write the p–p state:
|p |p ≡ |1+ |2+ = |1, 1
The p–n state
|p |n ≡ |1+ |2− =
1
√
2
[|1, 0 + |0, 0 ]

While the p–p system is a pure I = 1 state the p-n system is a linear super-
position with equal statistical weight of I = 1 and I = 0 states. Therefore if
isospin is conserved only the I = 1 state can contribute since the final state
is a pure I = 1 state. The expected ratio of cross-sections at a given energy is
σ+
σ0
=
1
1
?√
2
2
= 2
The observed ratio 3.15/1.5 = 2.1 is consistent with the expected ratio of
2.0.
10.43 Pions obey Bose statistics and the system of pions must be symmetric upon
interchange of space and isospin coordinates of any two pions. For each pion
T = 1 and for a system of two pions, total isospin, I = 2, 1, 0. The corre-
sponding states are
|2, ±2 , |2, ±1 , |2, 0 (Symmetric)
|1, ±1 , |1, 0 (Anti symmetric)
|0, 0 (Symmetric)
Now ψ(total) = ψ (space)ψ (isospin)
Thus for the symmetric states like |2, ±1 or |0, 0 , ψ (space) must be
symmetric because of Bose symmetry. But for two pions the interchange of
the spatial coordinates introduces a factor (−1)l
so that only even l states
are allowed. Thus, (π0
, π0
), (π+
, π+
), (π−
, π−
) in (a), (b) and (f) will
be found in the states L = 0, 2, . . . However, the states |1, ±1 , |1, 0 are
antisymmetric and this requires, L = odd, that is L = 1, 3, . . . (c) being
antisymmetric can have odd values for L, that is L = 1, 3, . . . (d) and (e)
can exist in I = 1 regardless of the L values because of two particles in
these two cases are fermion and boson, so that the previous considerations
are inconsequential.
10.3.3 Quarks
10.44 High energy collisions can cause the quarks within hadrons, or newly created
quark–antiquark pairs, to fly apart from each other with very high energies.
As the distance from the origin increases the colour force also increases,
and it is energetically more favourable for the quarks to be fragmented,
that is transformed into a jet of hadrons, mostly pions. Gluons may also
radiate a separate jet. However, the quarks can not be dislodged as free
particles.

10.3 Solutions 573
10.45
Fig. 10.8 Chain decay of
B-meson
The decays are weak at each stage as the quantum numbers b, c and S
change by one unit and is evidenced by narrow decay widths characteristic
of weak interactions. The route for the chain decays is in accordance with the
Cabibbo scheme as explained in the Sect. 10.1.
10.46 The QED formula is
Γ(3γ ) =
(2π2
− 9)α6
mec2
9π
(1)
(a) By analogy the QCD formula for charmonium (cc bound state) would be
Γ(3g) =
(2π2
− 9)α6
s mcc2
9π
(2)
where αs is the coupling constant for strong interaction and mcc2
≈
1.55 GeV is the constituent mass of the c-quark. Substituting Γ (3g) =
8 × 10−5
GeV in (2), we find αs = 0.307
(b) For the radiative decay of the bb bound state the analogous formula
would be
Γ(ggγ ) =
2(π2
− 9)α4
s α2
mbc2
9π
Substituting Γ(ggγ ) = 1.6 × 10−7
GeV, α = 1/137, mbc2
= 4.72 GeV,
we find αs = 0.319
10.47 The ratio of the cross sections
R =
3σ(e+
e−
→ QQ)
σ(e+e− → μ+μ−)
=
3

i
(qi /e)2
1
(1)
where qi /e is the charge of the quark in terms of the charge of the electron.
The charges and masses for the quarks are tabulated below
q/e M (GeV/c2
)
u-quark +2/3 0.31
d-quark −1/3 0.31
s-quark −1/3 0.5
c-quark +2/3 1.6
b-quark −1/3 4.6
t-quark +2/3 175

The sum of the squares of the quark charges refer to those quark pairs
which can possibly contribute to the reactions at a given CMS energy (
√
s).
The factor 3 in (1) arises due to 3 colors. At the low s-values, below the c c
threshold, only u, d, s quarks are involved, the expected ratio being
R(
√
s 3 GeV) = 3 × [(2

3)2
+ (1

3)2
+ (1

3)2
] = 2
R(3.2
√
s 9.2 GeV) = 3 × [(2

3)2
+ (1

3)2
+ (1

3)2
+ (2

3)2
] = 10

3
R(9.2
√
s 350 GeV) = 3 × [(2

3)2
+ (1

3)2
+ (1

3)2
+ (2

3)2
+ (1

3)2
] = 11

3
R(
√
s 350 GeV) = 3 × [(2

3)2
+ (1

3)2
+ (1

3)2
+ (2

3)2
+ (1

3)2
+ (2

3)2
] = 5
10.48 In the quark model Σ−
= dds, P = uud, n = udd, K−
= su, π−
= du
σ(Σ−
n) = 9σ(qq)
σ(pp) = 9σ(qq)
σ(K−
p) = 3σ(qq) + 3σ(qq)
σ(π−
p) = 3σ(qq) + 3σ(qq)
It follows that
σ(Σ−
n) = σ(pp) + σ(K−
p) − σ(π−
p)
10.49 In the quark model, Λ = uds, p = uud, n = udd, K−
= su, π+
= ud
σ(Λp) = σ(uds)(uud)
= σ(uu + uu + ud + du + du + dd + su + su + sd) = 9σ(qq)
Similarly σ(pp) = 9σ(qq)
σ(K−
n) = 3σ(qq) + 3σ(qq)
σ(π+
p) = 3σ(qq) + 3σ(qq)
Therefore σ(Λp) = σ(pp) + σ(K−
n) − σ(π+
p)
where we have assumed that σ(qq) = σ(qq)
10.50 Substract 1 MeV from the rest energies of the charged particles and take the
difference in the masses.
n(940) − p(938 − 1) = 3 MeV
= udd − uud = d − u
Σ−
(1,197 − 1) − Σ0
(1,192) = 4 MeV
= dds − uds = d − u
Σ0
(1192) − Σ+
(1189 − 1) = 4 MeV
= uds − uus = d − u
K0
(498) − K+
(494 − 1) = 5 MeV
= ds − us = d − u
Thus the mean difference in the masses of d-quark and u-quark is 4 MeV.

10.3 Solutions 575
10.51 (a) Δ−
: ddd, Δ0
: ddu, Δ+
: uud, Δ++
: uuu
(b) The fundamental difficulty is that Pauli’s principle is violated. For exam-
ple consider the spin of Δ−
. All the three d-quarks have to be aligned with
the same jz = −1/2. The same difficulty arises for Δ++
. The difficulty
is removed by endowing a new intrinsic quantum number (colour) to the
quarks. Thus, the three d-quarks in Δ−
or the three u-quarks in Δ++
differ
in colour, green, blue, and red.
(c) The higher values of J are accounted for by endowing higher orbital angu-
lar momentum values to the quarks. The parity is determined by the value
of (−1)l
, where l is the orbital angular momentum quantum number.
10.52 (a) Σ+
: uus, Σ−
: dds, n : udd, p : uud
(b) K+
: us, K−
: ds, π+
: ud, π−
: ud
10.53 (a) Δ++
, Δ−
, Ω−
(b) The quantum numbers of the b-quark are spin = 1/2 , charge Q/e =
−1/3, mass ∼4.5 GeV/c2
, and Beauty = −1.
10.54 Figure 10.9a, b are the quark flow diagrams for the decays Φ → K+
K−
and ω → π+
π0
π−
, respectively. Figure 10.9 c shows that the decay Φ →
π+
π0
π−
is suppressed because of unconnected quark lines.
10.55 σ(π N) = σ(qq)(qqq) = 6 σ(qq)
σ(NN ) = σ(qqq)(qqq) = 9 σ(qq)
where we have assumed σ(qq) = σ(qq)
∴ σ(π N)/σ(NN ) = 6/9 = 2/3
which is in agreement with the ratio 25/38.
10.56 The cross section for this electromagnetic process is proportional to the
square of the quark charge. In the annihilation of π−
(= ud) with 12
C nucleus
(= 18u + 18d), 18u are involved, the cross section being proportional to
18Q2
U or 18(4/9) or 8. In the annihilation of π+
(= ud) with 12
C nucleus,
18d are involved, the cross section being proportional to 18Q2
d or 18(1/9)
or 2.
Therefore the cross section ratio σ(π−
C)/σ(π+
C) = 8/2 or 4:1
10.57 According to the quark model, the ratio
R =
σ

e+
e−
→ hadrons

σ (e+e− → μ+μ−)
=
3

Q2
i
1
where Qi is the charge of the quark and the factor 3 arises due to the three
colours in which the quarks can appear. Now the charges of the three quarks,
u, d, and s are +2/3, −1/3, and −1/3
Therefore ΣQ2
i =
2
3
2
+

−1
3
2
+

−1
3
2
= 2
3
and R = 3 × 2/3 = 2
Hence σ

e+
e−
→ hadrons

= 3 × 2
3
× σ

e+
e−
→ μ+
μ−

= 2 × 20 nb =
40 nb

Fig. 10.9
Actually the value of R will be slightly greater than 2 because of an
increased value of αs, the running coupling constant.
10.58 E = ψ |H| ψ

ψ | ψ =
∞

0
e−r/a

−
2
2μ

d2
dr2
+
2
r
d
dr

+ Br

e−r/a
.4πr2
dr
∞

0
e−2r/a.4πr2dr
where μ = mq

2 is the reduced mass
E =
2
2μ
1
a2
+
3
2
Ba (1)

10.3 Solutions 577
The ground state energy is obtained by minimizing E, ∂E
∂a
= 0;
−
2
μa3
+
3
2
B = 0 → a =

22
3μB
1/3
(2)
Substituting (2) in (1) and putting μ = mq

2
E = E0 = 2.45

B2
2
mq
1/3
10.3.4 Electromagnetic Interactions
10.59
σ =
dσ
dΩ
.dΩ =
dσ
dΩ
.2π sin θ dθ
=
2πα2
2
c2
4E2
cM
1

−1
(1 + cos2
θ)d cos θ
=
2π
4
×
1
1372
×
(197.3 MeV − fm)2
(2 × 104 MeV)2

cos θ +
cos3
θ
3
1
−1
= 2.17 × 10−8
fm2
= 0.217nB
The Feynman diagram is given in Fig. 10.10
Fig. 10.10 e+
e−
→ γ
→ μ+
μ−
10.60 (a) Γ =

τ
=
c
τc
=
197.3 MeV − fm
(7.4 × 10−20s)(3 × 108 × 1015fm/s)
= 8.89 × 10−3
MeV = 8.89 keV
Note that because of the inverse dependence of decay width on lifetime,
weak decays with relatively long lives ( 10−3
s) have widths of the order
of a small fraction of eV, while em decays (τ∼10−19
–10−20
s) have widths
of the oder a kV, and strong decays (τ∼10−23
− 10−24
s) have widths of
several MeV.
(b) Photon has spin 1 and pion zero and K+
meson is known to have zero
spin. Pion will be emitted with l = 0 as its energy is small. Therefore the
occurrence of the radiative decay would constitute the violation of angular
momentum conservation.

10.61 Neither e+
nor e−
is an eigen state of C. However, the system e+
e−
in a
definite (l, s) state is an eigen state of C. According to the generalized Pauli
Principle under the total exchange of particles consisting of changing Q, r
and s labels, the total function Ψ must change its sign.
Space exchange gives a factor (−1)l
as this involves parity operation.
Spin exchange gives a factor (−1)S+1
.
Charge exchange gives a factor C.
The condition becomes
(−1)l
(−1)S+1
C = −1 or C = (−1)l+S
where S is the total spin.
We shall now show how C-invariance restricts the number of photons into
which the positrinium annihilates. Let n be the number of photons in the final
state. Conservation of C-parity gives
(−1)l+S
= (−1)n
Two cases arise
(i) Singlet s state 1
S0; l = S = 0 (para – positronium)
e+
e−
→ 2γ allowed with lifetime 1.25 × 10−7
s.
→ 3γ forbidden
(ii) Triplet s state 3
S1; l = 0, S = 1 (ortho – positronium)
e+
e−
→ 3γ allowed with lifetime 1.5 × 10−7
s.
→ 2γ forbidden
Note that annihilation into a single photon is not possible as it would violate
the conservation of linear momentum.
10.62 (a) It is forbidden as electromagnetic interaction because ΔS = 0 and also
forbidden as weak interaction because there is no strangeness changing
current
(b) It is allowed as an electromagnetic process because ΔS = 0 (both Σ0
and
Λ have S = −1. S does not apply to e+
, and e−
)
10.3.5 Weak Interactions
10.63 The decay rate W is given by the inverse of lifetime multiplied by the branch-
ing fraction B, that is W = B/τ
W(D+
) =
0.19
10.6 × 10−13
= 1.8 × 1011
s−1
W(D0
) =
0.08
4.2 × 10−13
= 1.9 × 1011
s−1

10.3 Solutions 579
10.64 σ ∝
1
,
q2 + m2
-2
At low energies q → 0
σ
σ0
=
80
100
=

m2
302 + m2
2
m = 87 GeV/c2
10.65 Number of W+
→ e+
νe =
σ(pp−
→ W+
)
σ(pp− → anything)
× 109
=
1.8 × 10−9
70 × 10−3
× 109
= 26
10.66 The decay D+
(= cd) → K0(= sd) + μ+
+ νμ involves the transformation
of a c-quark to an s-quark which is Cabibbo favored. In the case of the decay
D+
(= cd) → π0
(=
uu − dd
√
2
) + μ+
+ νμ
the transformation of a c-quark to a u-quark is Cabibbo suppressed. This
explains why the former process is more likely than the latter.
10.67 The difference in the decay rates is due to two reasons (i) the Q - values in
the decays are different. For Σ−
→ n + e−
+ νe, Q1 = 257 MeV, while for
Σ−
→ Λ + e−
+ νe, Q2 = 81 MeV, so that by Sargent’s law, the decay rate
(ω) will be proportional to Q5
. (ii) The first decay involves |ΔS| = 1, so that
ω will be proportional to sin2
θc, where θc is the Cabibbo angle. The second
one involves |ΔS| = 1, so that ω is proportional to cos2
θc. In all
R =
ω

Σ−
→ n + e−
+ νe

ω (Σ− → Λ + e− + νe)
=
sin2
θc
cos2 θc

Q1
Q2
5
= 0.0533

257
81
5
= 17.14
where we have used θc = 13◦
. The experimental value for R is 17.8
10.68 The leptonic decays of τ−
are
τ−
→ e−
+ νe + ντ (1)
τ−
→ μ−
+ νμ + ντ (2)
which from lepton – quark symmetry have equal probability. In addition, the
possible hadronic decays are of the form τ−
→ ντ + X
where X can be du or sc so that X may have negative charge. However the
latter possibility is ruled out because mτ ms + mc. We are then left with
only one possibility for X so that the hadronic decay will be
τ−
→ ντ + du (3)

Reaction (3) has relative weightage of 3 because of three colours. Thus
the branching fraction of (1) is predicted to be 1/5 or 0.2, a value which is in
agreement with the experimental value of 0.18.
10.69 According to Sargent’s law, Γ ∝ Q5
Decay Q (MeV) τ (s) Q5
τ (MeV5
− s)
(a) μ+
→ e+
+ νe + νμ 105.15 2.197 × 10−6
2.82 × 104
(b) n → p + e−
+ νe 0.782 900 2.63 × 102
(c) τ+
→ e+
+ νe + ντ 1784 3.4 × 10−13
6.14 × 103
(d) π+
→ π0
+ e+
+ νe 4.08 2.56 2.89 × 103
(e) 14
O → 14
N∗
+ e+
+ νe 1.81 198 3.85 × 103
The numbers in the last column vary only over one order of magnitude
which is small compared to the lifetimes which span over 15 orders of mag-
nitude, thereby conforming to the Sargents’ law (τ ∝ 1/T 5
max) and therefore
favoring the same coupling constant for the weak decays.
10.70 (a) Forbidden because lepton number is not conserved, left side has Lμ = −1
while right side has Lμ = +1
(b) Forbidden because charge is not conserved.
(c) Allowed because B, Lμ, Q etc are conserved. Also the selection rule,
ΔQ = ΔS = ±1 is obeyed.
(d) Allowed for reason stated in (c).
10.71 Introduce the spurion, (Sp) a hypothetical particle of spin 1/2 and isospin 1/2
and convert the weak decay into a strong interaction
Sp + Ξ−
→ Λ + π−
I 1/2 1/2 0 1
I3 − 1/2 − 1/2 0 − 1
The reaction must occur in pure I = 1 state. Looking up the Clebsch –
Gordon coefficients for 1/2× 1/2, in Table 3.3,
%
%Λ, π−
8
= a1 |1, −1
For the second decay, we associate again a spurion and consider the reaction
Sp + Ξ0
→ Λ + π0
I 1/2 1/2 0 1
I3 − 1/2 + 1/2 0 0

10.3 Solutions 581
The initial state is a mixture of I = 0 and I = 1 states. The final state can
exist only in I = 1.
|Λ, π0
=
1
√
2
[a1|1, 0 +a0|0, 0 ]
λ(Ξ−
→ Λ + π−
)
λ(Ξ0 → Λ + π0)
=
1
(1/
√
2)2
= 2
10.72 Let us introduce a fictitious particle called spurion of I = 1/2, I3 = −1/2,
and neutrally charged and add it on the left hand side so that the weak decay
is converted into a strong interaction in which I is conserved. The reactions
with Σ+
can proceed in I = 1/2 and 3/2 channels while that with Σ−
through
pure I = 3/2 channel. We can then write down the amplitudes for the initial
state by referring to the C.G. coefficients for 1 × 1/2:
Σ+
+ s →
1
3
a3 +
2
3
a1
Σ−
+ s → a3
where a1 and a3 are the I = 1/2 and I = 3/2 contributions. Similarly, for
the final state the amplitudes are
nπ+
→
1
3
a3 +
2
3
a1
nπ−
→ a3
pπ0
→
2
3
a3 −
1
3
a1
Therefore,
a+ =
7
Σ+
%
% nπ+
8
= 1
3
a2
3 + 2
3
a2
1
a− =
7
Σ−
%
% nπ−
8
= a2
3
a0 =
7
Σ+
%
% pπ0
8
=
√
2
3
a2
3 −
√
2
3
a2
1
giving a+ +
√
2 a0 = a−
10.73 Converting the decay into a reaction
S + Λ → p + π−
→ n + π0
where s is the fictitious particle of isospin T = 1/2 and T3 = −1/2. Because
Λ has T = 0 and T3 = 0, the initial state will be a pure I = 1/2 state with
I3 = −1/2. In the final state the nucleon has T = 1/2 and pion has T = 1,
so that I = 3/2 or 1/2. I and I3 conservation require that the final state must
be characterized by I = 1/2 and I3 = −1/2. Looking up the table for C.G.
Coefficients for 1 × 1/2 (given in Table 3.3) we can write down

|N, π, 1/2, −1/2 = −
2
3
%
%pπ−
8
+
1
3
%
%nπ0
8
Therefore,
ω

Λ → pπ−

ω

Λ → nπ0
=

2
3
#24
1
3
#2
= 2
10.74 L⊙ = 3.83 × 1026
Js−1
=
3.83 × 1026
1.6 × 10−13
MeV s−1
= 2.39 × 1039
MeV s−1
Number of α′
s produced, Nα =
2.39 × 1039
26.72
= 8.94 × 1037
s−1
As the number of neutrinos produced is double the number of α’s,
Nν = 2Nα = 1.788 × 1038
s−1
Number of neutrinos received per square metre on earth’s surface, that is flux
φ =
Nυ
4πr2
=
1.788 × 1038
(4π)

1.5 × 1011
2
= 6.33 × 1014
m−2
s−1
10.75 Using the formula E = mc2
γ , the speed of antineutrinos of mass m
v = c

1 −
m2
c4
E2
1/2
The time taken to travel to earth
t =
d
v
=
d
c

1 −
m2
c4
E2
− 1/2
∼
=
d
c

1 +
1
2
m2
c4
E2

The difference in travel time for two neutrinos of energies E1 and E2 where
E2 E1 m,
Δt =
d
2c

mc2
2

1
E2
1
−
1
E2
2

Substituting Δt = 4 s, E1 = 5 MeV and E2 = 15 MeV, d
c
= 17 × 104
×
3.15×107
seconds, and solving for mc2
we find mc2
= 6.48×10−6
MeV =
6.5 eV
10.76 (a) Q/e = +1 ∵ ΔQ = 0
(b) B = 0 ∵ ΔB = 0
(c) lμ = 0 ∵ Δlμ = 0
(d) T = 1/2 ∵ ΔT3 = ±1/2
(e) s = ±1 ∵ ΔS = ±1
(f) spin = 0 or 1 ∵ spin of π0
, μ and ν are 0, 1/2 and 1/2 respectively
(g) Boson ∵ on right hand side there are two fermions (μ, ν) and
one boson (π)

10.3 Solutions 583
(h) Mass mx ≥

mπ + mμ + mυ

/c2
= (135 + 106 + 0) /c2
= 241 MeV/c2
(i) X is identified as K+
-meson.
10.77 The initial state 16
O∗
has odd parity, while the parity of 12
C is even and so
also that of α. If α is emitted with l = 0 the parity of the final state is even.
Clearly the decay goes through a weak interaction because parity is violated
and the observed width is consistent with a weak decay. On the other hand
the width of the electro-magnetic decay 16
O∗
→16
O + γ , is 3 × 10−3
eV.
10.78 ττ = B×τμ
mμ
mτ
5
= 17.7
100
×2.197×10−6
×
105.658
1784
5
= 0.28×10−12
s a value
which is in excellent agreement with the experimental value of 0.3 × 10−12
s
10.79 Pions have T = 1. A system of two pions can exist in the isospin state I = 0,
1, and 2 state. Therefore, isospin is conserved in all the three cases.
Now pions obey Bose statistics and so Ψtotal is symmetrical. As J = 0
for pions, the spatial part of pions system must be symmetrical. The intrinsic
parity of each pion being negative, does not contribute to the parity of the
state. The parity of the state is mainly determined by the L-value; p = (−1)l
.
Therefore, allowed states correspond to L = 0, 2, 4, . . . It follows that par-
ticles with J p
= 0+
, 2+
. . . can decay to two pions, so that only possible
decay is f 0
→ π+
+ π−
; the other two particles ω0
and η0
actually decay
into three pions.
10.80 The mass of η – meson is 549 Mev/c2
while the mass of four pions is
558 MeV/c2
if all the four pions are charged and 540 MeV/c2
if uncharged.
In the first state the decay cannot occur because of energy conservation and
in the second case the decay will be heavily suppressed because of a small
phase space on account of small Q-value. Apart from this the only possible
pionic decay is the 3-body mode, the strong decay into two pions or four
pions being forbidden by parity conservation.
10.81 The leptonic decays are assumed to proceed via exchange of a single virtual
photon, Fig. 10.11
Fig. 10.11 Leptonic decay of
vector meson
Apart from numerical factors the partial width Γ(V → l+
l−
) is propor-
tional to the squared sum of the charges of the quarks in the meson, that is
Γ ∝

|ai Qi |2
, where ai the amplitudes from all the quarks in the meson
are superposed. The approach is similar to the Rutherford scattering where
the cross-sections are assumed to be proportional to the sum of squares of
the charges of quarks. The quark wave functions and the quantity
%
%

ai Qi
%
%2
are tabulated below

Meson Quark wave function |Σai Qi |2
ρ0
(uu − dd)/
√
2
*
1
√
2
2
3
−

−1
3
+2
= 1
2
ω0
(uu + dd)/
√
2
*
1
√
2
2
3
− 1
3
+2
= 1
18
Φ0
ss
1
3
2
= 1
9
The expected leptonic widths are in the ratio
Γ(ρ0
) : Γ(ω0
) : Γ(ϕ0
) = 9 : 1 : 2
in agreement with the observed ratios.
10.82 For the decay of charmed particles, the selection rules are (i) ΔC = ΔS =
ΔQ = ±1, for Cabibbo allowed and (ii) ΔC = ΔQ = ±1, ΔS = 0, for
Cabibbo suppressed decays
(ΔQ applies to hadrons only). Using these rules (a) is Cabibbo allowed and
(b) is Cabibbo suppressed
10.83 For the charmed particles, the selection rule for the Cabibbo allowed decay
is ΔC = ΔS = ΔQ = ±1 and for the Cabibbo suppressed decays the
selection rule is ΔC = ΔQ = ±1, ΔS = 0. Using these rules we infer that
the decays are (a) Cabibbo allowed (b) forbidden
10.84 For semileptonic decays the rule is ΔS = ΔQ = ±1,. In (a) both Q and S
increase by one unit. Hence it is allowed and experimentally observed. (b) is
also a semileptonic decay in which Q decreases by one unit, but S increases
(from −2 to −1) by one unit. Therefore, the decay is forbidden. (c) is a
non-leptonic weak decay in which ΔS = +1 and the energy is conserved. It
is allowed and experimentally observed.
10.85 All the decays are semi-leptonic. For Cabibbo allowed decay,
ΔQ = ΔS = ±1.
In (a) ΔQ = −1 and ΔS = −1 (∵ S changes from +1 to 0) Therefore it is
allowed.
In (b) ΔQ = +1 while ΔS = −1. Therefore, forbidden.
In (c) ΔQ = +1 and ΔS = +1(∵ S = −1 → S = 0), therefore allowed.
In (d) ΔQ = −1 and ΔS = +1, therefore forbidden.
10.86 If lepton number is not absolutely conserved and neutrinos have finite masses,
then mixing may occur between different types of neutrinos (νe, νμ, ντ ) In
what follows consider only two types of neutrinos νe and νμ. The neutrino
states νμ and νe which couple to the muon and electron, respectively could
be linear combinations

10.3 Solutions 585
νμ = ν1 cos θ + ν2 sin θ (1)
and νe = −ν1 sin θ + ν2 cos θ (2)
which form a set of ortho normal states. νμ and νe are the sort of states which
are produced in charged pion decay (π → μ + νμ) and neutron decay (n →
p + e−
+ νe); ν1 and ν2 are the mass eigen states, corresponding to the
neutrino masses m1 and m2. In the matrix form

νμ
νe

=

cos θ sin θ
− sin θ cos θ

ν1
ν2

(3)
The difference in the masses leads to difference in the characteristic fre-
quencies with which the neutrinos are propagated. Here θ is known as the
mixing angle which is analogous to the Cabibbo angle θc. Using natural units
( = c = 1) at time t = 0,
ν1(t) = ν1(0)e−i E1 t
1
ν2(t) = ν2(0)e−i E2 t
2 (4)
where the exponentials are the usual oscillating time factors associated with
any quantum mechanical stationary state. Since the momentum is conserved
the states ν1(t) and ν2(t) must have the same momentum p. If the mass
mi Ei(i = 1, 2)
Ei = (p2
+ m2
i )1/2 ∼
= p +
m2
i
2p
(5)
Suppose at t = 0, we start with muon type of neutrinos so that, νμ(0) = 1
and νe(0) = 0 then by (3),
ν2(0) = νμ(0) sin θ (6)
ν1(0) = νμ(0) cos θ
and νμ(t) = cos θν1(t) + sin θν2(t) (7)
Thus at time t = 0, the muon–neutrino beam is no longer pure but devel-
ops an electron neutrino component. Similarly, an electron–neutrino beam
would develop a muon–neutrino beam component.
νμ(t)
νμ(0)
= cos2
θ.e−i E1t
+ sin2
θ.e−i E2t
and the intensity
Iμ(t)
Iμ(0)
=
%
%
%
%
νμ(t)
νμ(0)
%
%
%
%
2
= cos4
θ + sin4
θ + sin2
θ cos2
θ
,
ei(E2−E1)t
+ e−i(E2−E1)t
-
P(νμ → νe) = 1 − sin2
2θ sin2

(E2 − E1)
2
t

(8)
where we have used (cos4
θ + sin4
θ) = (cos2
θ + sin2
θ)2
− 2 cos2
θ sin2
θ
in simplifying the above equation.

10.87 (a) Pe(t) = 1 − Pμ(t) = sin2
2θ sin2
*
(E2−E1)
2
t
+
(b) Pμ(t) = Pe(t)
1 − sin2
2θ sin2

(E2 − E1)
2
t

= sin2
2θ sin2

(E2 − E1)
2
t

2 sin2
2θ sin2

(E2 − E1)
2
t

= 1
Restoring to practical units the above equation becomes
2 sin2
2θ sin2

Δm2
c4
2E
t

= 1
where Δm2
= m2
2
− m1
2
and θ = 340
If m1 and m2 are in eV/c2
, and E in MeV and L the distance from the source,
then the last equation becomes
2 sin2
2θ sin2

1.27Δm2
.L
E

= 1
Inserting θ = 340
, Δm2
= 52
− 32
= 16 and E = 1,000 MeV, we find
L = 426 m, giving t = L/C = 1.42 × 10−6
s.
10.88 τ(μ+
→ e+
νe ντ ) =
G2
(c)6
m5
μ
192π3
(1)
G2
∼ g2

M2
w , where Mw is the mass of W-boson.
From the τ lepton lifetime and formula (1) for the dependence of parent
particle mass, we can test the universality of the couplings gμ and gτ to the
W – boson

gτ
gμ
4
= B

τ+
→ e+
νeντ

mμ
mτ
5
τμ
ττ

Inserting B = 0.178, mμ = 105.658 MeV/c2
, mτ = 1777.0 MeV/c2
, τμ =
2.197 × 10−6
s and ττ = 2.91 × 10−13
s, we find
gτ
gμ
= 0.987
Comment: From the branching fractions for τ+
→ e+
νeντ and τ+
→
μ+
νμντ the ratio gμ/ge = 1.001. A similar result is obtained from the
branching ratio of π → eν and π → μe, proving thereby different flavours
of leptons have identical couplings to the W±
bosons.
The principle of universality is equally valid for the Z0
coupling. Thus,
the branching fractions are predicted as
Z0
→ e+
e−
: μ+
μ−
: τ+
τ−
= 1 : 1 : 1
in agreement with the experimental ratios. Formula (1) affords the most accu-
rate determination of G, the Fermi constant because the mass and lifetime of
muon are precisely known by experiment.

10.3 Solutions 587
Fig. 10.12
10.89 The reaction (a) can go via the mechanism of the diagram shown in Fig. 10.12a.
However, no diagram with single W exchange can be drawn for the reaction
(b) which at the quark level implies the transformation
uus → udd + e+
+ νe
as in Fig. 10.12b and would require two separate quark transitions which
involve the emission and absorption of two W bosons – a mechanism which
is of higher order and therefore negligibly small.
The above conclusion can also be reached by invoking for the selection
rule for semi leptonic decays. Reaction (a) obeys the rule ΔS = ΔQ = +1
and is therefore allowed, while in reaction (b) we have ΔS = +1, but ΔQ =
−1, and therefore forbidden.
10.90 The difference in the two decay rates is due to two factors (i) The decay
D+
→ K0μ+
νμ involves |Δs| = 1 and hence proportional to sin2
θc, where
θc = 12.90
is the Cabibbo angle, while D+
→ π0
μ+
νμ involves |Δs| = 0
and is proportional to sin2
θc (ii) The Q-values are different for these decays.
For the first one Q1 = 1870 − (498 + 106) = 1,266 MeV and for the second
one Q2 = 1,870 − (135 + 106) = 1629MeV. Thus using Sargent’s rule
R ≈
sin2
θc
cos2 θc

Q1
Q2
5
= tan2
12.90

1266
1629
5
∼
= 0.015

10.91 Let a K0
beam be formed through a strong interaction like π−
+p → K0
+Λ.
Neither
%
%K0
8
nor
%
%
%K0
@
is an eigen state of |cp . However, linear combinations
can be formed.
|Ks =
1
√
2
%
%K0
8
+
%
%
%K0
@
cp |Ks =
1
√
2
%
%
%K0
@
+
%
%K0
8
= |Ks
cp = +1
|KL =
1
√
2
%
%K0
8
−
%
%
%K0
@
cp |KL =
1
√
2
%
%
%K0
@
−
%
%K0
8
= − |KL
cp = −1
while K0
and K0 are distinguished by their mode of production, Ks and
KL are distinguished by the mode of decay. Typical decays are Ks →
π0
π0
, π+
π−
, KL → π+
π−
π0
, πμν.
At t = 0, the wave function of the system will have the form
ψ(0) =
%
%K0
8
=
1
√
2
(|Ks + |KL )
As time develops Ks and KL amplitudes decay with their characteristic life-
times. The intensity of Ks or KL components can be obtained by squaring
the appropriate coefficient in Ψ(t). The amplitudes therefore contain a factor
e−iEt/
which describes the time dependence of an energy eigen function in
quantum mechanics.
In the rest frame of the K0
we can write the factor e−iEt/
as e−imc2
t/
,
where m is the mass. The complete wavefunction for the system can therefore
be written as
ψ(t) =
1
√
2

|Ks e
−t 1
2τs
+ ims c2

+ |KL e
−t

1
2τL
+
imL c2

'
=
1
√
2
e−imsc2
t/
*
|Ks e− t
2τs + |KL eiΔmc2
t/
+
where Δm = ms − mL and we have neglected the factor e−t/2τL
which varies
slowly (τL
∼
= 70τs). Reexpressing |K1 and |K2 in terms of
%
%K0
8
and
%
%
%K0
@

10.3 Solutions 589
|Ks =
1
√
2
%
%K0
8
+
%
%
%K0
@
|KL =
1
√
2
%
%K0
8
−
%
%
%K0
@
ψ(t) =
1
2
e−imsc2
t/
*
e−t/2τs
%
%K0
8
+
%
%
%K0
@
+ eiΔmc2
t/
%
%K0
8
−
%
%
%K0
@+
=
1
2
e−imsc2
t/
*%
%K0
8
e−t/2τs
+ eiΔmc2
t/

+
%
%
%K0
@
e−t/2τs
− eiΔmc2
t/
+
The intensity of the component is obtained by taking the absolute square of
the coefficient of
%
%
%K0
@
I
%
%K0
8
=
1
4
,
e−t/τs
+ 1 + 2e−t/2τs
cos

Δmc2
t/
-
Similarly,
I
%
%
%(K0
@
=
1
4
,
e−t/τs
+ 1 − 2e−t/2τs
cos

Δmc2
t/
-
10.92 Refering to Problem 10.91, the KL state can be written as
|KL =
1
√
2
%
%K0
8
−
%
%
%K0
@
(1)
When KL enters the absorber, strong interactions would occur with K0
(S =
+1) and
%
%
%K0
@
(S = −1) components of the beam of the original K0
beam
intensity, 50% has disappeared by KS-decay. The remaining KL component
consists of 50% K0. Upon traversing the material the existence of K0 with
S = −1 is revealed by the production of hyperons in a typical reaction,
K0 + p → Λ + π+
While K0
components can undergo elastic and charge-exchange scattering
only, the K0 component can in addition participate in absorption processes
resulting in the hyperon production. The emergent beam from the slab will
then have the K0
amplitude f
%
%K0
8
and K0 amplitude f
%
%
%K0
@
with f
f 1. The composition of the emergent beam from the slab is given by
modifying (1).
1
√
2
f
%
%K0
8
− f
%
%
%K0
@
=

f + f

2
√
2
%
%K0
8
−
%
%
%K0
@
+

f − f

2
√
2
%
%K0
8
+
%
%
%K0
@
=
1
2

f + f

|KL +
1
2

f − f

|KS

Since f = f , we conclude that some of the KS-state has regenerated. The
regeneration of the short lived K1-component in a long-lived KL beam was
experimentally confirmed from observation of two-pion decay mode (1956).
10.3.6 Electro-weak Interactions
10.93
Fig. 10.13(a)
νµ+e– →z0→νµ+e–
A charged current weak interaction is mediated by the exchange of W±
boson, as for the decay μ−
→ e−
+ νe + νμ. Neutral current interaction
is mediated by the exchange of Z0
, as in the scattering, νμ + p → νμ + p.
For νμ + e−
→ νμ + e−
, the Feynman diagram for weak neutral current
shown in Fig. 10.13a is unambiguous.
νe+e– →z0→νe+e–
νe+e– →w→νe+e–
Fig. 10.13(b), (c)

10.3 Solutions 591
However, for νe+e−
→ νe +e−
, there are two diagrams shown in Fig. 10.13b
and Fig. 10.13c.
Thus the reaction can be described by neutral or charged current and there-
fore does not constitute an unequivocal evidence for neutral current.
10.94 If Nν is the number of neutrino types in the sequence νe, νμ, ντ and assum-
ing that there are only three charged leptons e, μ and τ, the balance equation
for the decay rate can be written as
Γz(total) = Γ(Z0
→ hadrons) + 3 Γ(Z0
→ l+
l−
) + NνΓ(Z0
→ νlνl )
The factor 3 is for the three charged leptons. Substituting the given data
2.534 = 1.797 + 3 × 0.084 + 0.166Nν
and solving, we get Nν = 2.92 or 3.
10.95 (a) W±
and Z0
bosons are produced in the annihilation of p−
− p at high
energies (Fig. 10.14). The elementary production and decay processes are
u + d → W+
→ e+
+ νe, μ+
+ νμ
u + d → W−
→ e−
+ νe, μ−
+ νμ
u + u
d + d

→ Z0
→ e+
e−
, μ+
μ−
Detector: It consists of calorimeter detectors, the central tracking cham-
ber to detect individual secondary particles, surrounded by an electro-
magnetic calorimeter to detect electron – photon showers, a much larger
calorimeter to detect and measure hadron jets and an outside muon detec-
tor.
Signature for W → eν event:
(i)An isolated single electron track with high transverse momentum (PT)
in the central track detector.
(ii)The electron track points to a shower in the electromagnetic calorime-
ter with appreciable energy deposition in the nearby hadronic calorime-
ter in the neighborhood
(iii)There should be missing PT overall when summation is made over all
the secondaries. The missing PT is attributed to the unseen neutrino
from the W-decay.
Signature of Z0
→ e+
e−
events:
Two isolated tracks with large PT values and invariant mass Me+e−
50 GeV, pointing to localized track clusters in the electromagnetic
calorimeter.
For muonic decays, W±
→ μν, Z0
→ μμ are observed by imposing
high PT requirements on the PT values for muons which are able to pene-
trate hadron calorimeter and observed in the external muon chambers.
Fig. 10.14 Annihilation
process

(b) R ∼
c
mc2
=
197 MeV − fm
100 × 103 MeV
= 2 × 10−3
fm
10.96 MW
2
=
4πα
8 sin2
θW
·
√
2
(c)3
GF
=
(4π)
8 sin2
28.170
·
√
2
128
×
1
1.166 × 10−5
= 6, 670
MWc2
= 81.67 GeV
the experimental value being 81 GeV
MW/MZ = cos θW → MZ = MW/ cos θW
MZ = 81.67/ cos 28.170
= 92.64 GeV
the experimental value being 94 GeV.
10.97 (i) Weak interaction
Fig. 10.15(i) τ+
decay
τ+
→ μ+
+ νμ + ντ
(ii) Strong interaction
Fig. 10.15(ii) Ω−
production
in K−
interaction
K−
+ p → Ω−
+ K+
+ K0
(iii)
Fig. 10.15(iii) Decay of D0
D0 → K+
+ π−
10.98 (i) l = νe

10.3 Solutions 593
Diagram is the same as in Problem 10.2(iii)
(ii) τ−
→ µ−
+ l + l
Fig. 10.16(a) Decay of τ−
τ−
→ ντ + μ−
+ νμ
(iii) l = νμ
Fig. 10.16(b) Decay of B0
meson
10.99 (a) Δ0
→ pπ−
The decay Δ0
→ p + π−
proceeds via strong interaction in which a
gluon is involved. Δ0
= udd; p = udu; π−
= du
Fig. 10.17 (a)

(b) Ω−
→ ΛK−
Fig. 10.17 (b)
10.100 (a) e+
e−
→ νμνμ
Fig. 10.18 (a)
(b) D0
→ K−
π+
Fig. 10.18 (b)

10.3 Solutions 595
10.101 (a)
Fig. 10.19(a) Moller
scattering
(b)
Fig. 10.19(b) Bhabha
scattering
10.102 (a) Λ → p + e−
+ νe
Fig. 10.20(a)

(b)
Fig. 10.20(b)
10.103 (a)
Fig. 10.21(a)
(b)
Fig. 10.21(b)
10.104 The Feynman diagram is
Fig. 10.22

10.3 Solutions 597
The two vertices 1 and 2 where the W boson couples pertain to weak
interactions and have strengths
√
αw. Vertex 3 is electromagnetic and has
strength
√
αem. The overall strength of the diagram is
√
αw.
√
αw.
√
αem or
αw
√
αem
10.105 The decay of D0
→ K−
+ π+
is accomplished by the exchange of W+
boson as illustrated by the Feynman diagram Fig. 10.23.
Fig. 10.23
The quark composition (D0
= cu, K−
= su, π+
= ud) is also indicated.
The favoured route for the decay is via c → s. Hence the decay occurs via
lowest order charge current weak interaction. However, for D+
→ K0
+
π+
the c-quark is required to decay to d-quark via W emission and the
subsequent decay of W+
to π+
. This would mean that the d quark in the
D+
decay to an s quark in the K0
which is not possible as they both have
the same charge. Furthermore the transition c → d is not favoured in the
Cabibbo scheme.
10.106 Consider the Feynman diagrams for the decays π+
→ μ+
+ νμ and π0
→
γ + γ
Fig. 10.24
For charged pion there are two vertices of strength
√
αw, and a propagator
1
Q2 + M2
wc2
∼
=
1
M2
wc2
In the limit Q Mwc momentum transfer squared Q2
carried by W
boson is quite small. Therefore, the decay rate will be proportional to
√
αw
√
αw
M2
w
2
=
α2
w
M4
w

For the neutral pion there are two vertices of strength
√
αem but the propa-
gator term is absent because photon has zero rest mass. Thus the decay rate
will be proportional to α2
em. But αw
∼
= αem according to the electro-weak
unified theory. Therefore, the ratio of the decay rates
ω(π±
→μν)
ω(π0→2γ )
≈ 1
M4
W
which is quite small and so charged pion lifetime (∼10−8
s) is much larger
than that of neutral pion (∼10−17
s).
10.107 (a)
Fig. 10.25(a) Dominant Feynman diagrams for the Bremsstrahlung process e−
+ (Z, A) → e−
+
γ + (Z, A)
(b) The pair production process γ + (Z, A) → e−
+ e+
+ (Z, A)
Fig. 10.25(b)

10.3 Solutions 599
10.108 (a)
Fig. 10.26(a) Photo electric
effect
(b)
Fig. 10.26(b) Lowest order
Feynman diagrams for the
Compton scattering
10.109
Fig. 10.27(a)
Fig. 10.27(b)

10.110
Fig. 10.28(a)
Fig. 10.28(b)
The decay K+
→ π+
π+
π−
. In all hadrons, gluon interactions keep the
gluon bound. In this example a dd̄ pair is created.
10.111 (a) There are five basic mechanisms for the D+
decay
Fig. 10.29(a)
D+
→ K0
+ l + l

10.3 Solutions 601
where the combinations (1,2) are (e+
, νe) or (μ+
, νμ) or (d̄, u), the last
one three times for three colors. The combinations (τ+
, ντ ) and (s̄, c) are
ruled out as energy is violated.
(b) Hadronic decay of τ−
τ−
→ π−
ντ
Fig. 10.29(b)
10.112 (a)
Fig. 10.30(a)
Two-photon annihilation; two ordered diagrams
(b)
Fig. 10.30(b)

10.113
Fig. 10.31
Λ → pπ−
Λ = uds; p = udu; π−
= ūd

Appendix: Problem Index
Chapter 1 Mathematical Physics
If ϕ = 1/r, ∇ϕ = r/r3
1.1
Unit vector to the given surface at a point 1.2
Divergence of 1/r2
is zero 1.3
If A and B are irrotational, A × B is solenoidal 1.4
(a) For central field F, Curl F = 0 (b) For a solenoidal field
function F(r),
1.5

c A.dr between two points along the curve r, where A is defined 1.6
!
c A.dr around closed curve c defined by two equations where A is
defined
1.7
(a) Given field F is conservative (b) To find scalar potential
(c) Work done in moving a unit mass between two points
1.8
To verify Green’s theorem in the given plane, the bonded region
being defined by two equations
1.9
!
s A.ds, where S is sphere of radius R and A is defined. 1.10

A.dr around a circle in xy-plane where A is defined 1.11
(a) Curl of gradient is zero (b) divergence of curl is zero 1.12
Gradient and Laplacian of given function ϕ 1.13
(a) A unit vector normal to given surface at the given point
(b) Directional derivative at the given point in the given direction 1.14
The divergence of inverse square force is zero 1.15
Angle between two surfaces at given point 1.16
Fourier series expansion for saw-tooth wave 1.17
Fourier series expansion for square wave 1.18
To express π/4 as a series 1.19
Fourier transform of a square wave 1.20
To use Fourier integral to prove the given definite integral 1.21
Fourier transform of a Gaussian function is another Gaussian
function
1.22
603

604 Appendix: Problem Index
1.2.3 Gamma and Beta Function
(a) Γ(Z + 1) = ZΓ(Z) (b) Γ(n + 1) = n! for positive integer 1.23
β(m, n) = Γ(m)Γ(n)/Γ(m + n) 1.24
To evaluate a definite integral using β functions 1.25
(a) Γ(n)Γ(1 − n) = π/ sin(nπ); 0 n 1
(b) |Γ(in)|2
= π/n sinh(nπ)
1.26
Characteristic roots of Hermitian matrix are real 1.27
Characteristic equation and eigen values of the given matrix 1.28
Effect of a set of matrices on position vector 1.29
Eigen values of the given matrix 1.30
Diagnalization of a matrix and its trace 1.31
Eigen values, eigen vector, modal matrix C and its inverse C−1
,
product C−1
AC
1.32
Solution of a cubic equation by Newton’s method 1.33
(a) Turning points of f(x) (b) Whether f(x) is odd or even or neither 1.34
1.2.6 Series
Interval of convergence of series 1.35
Expansion of log x by Taylor’s series 1.36
Expansion of cos x into infinite power series 1.37
Expansion of sin(a + x) by Taylor’s series 1.38
Sum of series 1 + 2x + 3x2
+ 4x3
+ . . . , |x| 1 1.39
1.2.7 Integration
(a)

sin3
x cos6
x dx (b)

sin4
x cos2
x dx 1.40

(2x2
− 3x − 2)−1
dx 1.41
(a) Sketch of curve r2
= a2
sin 2θ (b) area within the curve between
θ = 0 and θ = π/2
1.42

(x3
+ x2
+ 2)(x2
+ 2)−2
dx 1.43
∞
0 4a3
(x2
+ 4a2
)−1
dx 1.44
(a)

tan6
x sec4
x dx (b)

tan5
x sec3
x dx 1.45
4
2 (2x + 4)(x2
− 4x + 8)−1
dx = ln 2 + π 1.46
Area included between curve y2
= x3
and line x = 4 1.47
Surface of revolution of hypocycloid about x-axis 1.48
a
0
√
a2−x2
0 (x + y)dydx 1.49
Area enclosed between curves y = 1/x and y = −1/x and lines
x = 1 and x = 2
1.50

(x2
− 18x + 34)−1
dx 1.51
1
0 x2
tan−1
x dx 1.52
(a) Area bounded by curves y = x2
+ 2 and y = x − 1 and lines
x = −1 and x = 2
(b) Volume of solid of revolution obtained by rotating area enclosed
by lines x = 0, y = 0, x = 2 and 2x + y = 5 through 2π rad about
y-axis
1.53

Appendix: Problem Index 605
(a) Area enclosed by curve y = x sin x and x-axis
(b) Volume generated when curve rotates about x-axis 1.54
dy/dx = (x3
+ y3
)/3xy2
1.55
d3
y/dx3
− 3d2
y/dx2
+ 4y = 0 1.56
d4
y/dx4
− 4d3
y/dx3
+ 10d2
y/dx2
− 12dy/dx + 5y = 0 1.57
d2
y/dx2
+ m2
y = cos ax 1.58
d2
y/dx2
− 5dy/dx + 6y = x 1.59
Equation of motion for damped oscillator 1.60
Modes of oscillation of coupled springs 1.61
SHM of a rolling cylinder with a spring attached to it 1.62
d2
y/dx2
− 8dy/dx = −16y 1.63
x2
dy/dx + y(x + 1)x = 9x2
1.64
d2
y/dx2
+ dy/dx − 2y = 2 cosh (2x) 1.65
xdy/dx − y = x2
1.66
(a) y′
− 2y/x = 1/x2
(b) y′′
+ 5y′
+ 4y = 0 1.67
(a) dy/dx + y = e−x
(b) d2
y/dx2
+ 4y = 2 cos (2x) 1.68
dy/dx + 3y/(x + 2) = x + 2 with boundary conditions 1.69
(a) d2
y/dx2
− 4dy/dx + 4y = 8x2
− 4x − 4, with boundary
conditions
(b) d2
y/dx2
+ 4y = sin x 1.70
d3
y/dx3
− d2
y/dx2
+ dy/dx − y = 0 1.71
Solution of radioactive chain decay 1.72, 73
(a) £(eax
) = 1/(s − a) (b) £(cos ax) = s/(s2
+ a2
)
(c) £(sin ax) = a/(s2
+ a2
)
1.74
For Hermite polynomials (a) H′
n = 2nHn−1
(b) Hn+1 = 2ξHn − 2nHn−1
1.75
For Bessel function (a) d
dx
[xn
Jn(x)] = xn
Jn−1(x)
(b) d
dx
[x−n
Jn(x)] = −x−n
Jn+1(x)
1.76
(a) Jn−1(x) − Jn+1(x) = 2 d
dx
Jn(x)
(b) Jn−1(x) + Jn+1(x) = 2(n/x)Jn(x)
1.77
(a) J1/2(x) =
$
2
πx
sin x (b) J−1/2(x) =
$
2
πx
cos x 1.78
For Legendre polynomials
1
−1 pn(x)pm(x)dx = 2
2n+1
, if m = n
and = 0 if m = n
1.79
For large n and small θ, pn(cos θ) ≈ J0(nθ) 1.80
(a) (l + 1)pl+1 = (2l + 1)xpl − lpl−1
(b) Pl (x) + 2x pl
′
(x) = pl
′
+1(x) + pl
′
−1(x) 1.81
For Laguerre’s polynomials Ln(0) = n! 1.82
!
c dZ/(Z − 2) where C is (a) Circle |Z| = 1 (b) Circle |Z + i| = 3 1.83
!
c(4Z2
− 3Z + 1)(Z − 1)−3
dZ, C encloses Z = 1 1.84

Location of singularities of
(4Z3
− 2Z + 1)(Z − 3)−2
(Z − i)−1
(Z + 1 − 2i)−1
1.85
Residues of (Z − 1)−1
(Z + 2)−2
at poles Z = 1 and Z = −2 1.86
Laurent series about singularity for ex
(Z − 2)−2
1.87
∞
0 (x4
+ 1)−1
dx 1.88
Curve which has shortest length between two points 1.89
Brachistochrome (curve of quickest descent) 1.90
Equation to the surface of revolution of a soap film stretched
between two parallel circular wires.
1.91
Sphere is the solid of revolution which for a given surface area
has maximum volume
1.92
Poisson(px ) (a) Normalization of px (b) x= m (c) σ =
√
m
(d) pm−1 = pm (e) px−1 = (x/m)pm px+1 = mpx /(x + 1)
1.93
Binomial (Bx ) (a) Normalization of Bx (b) x = Np (c)
σ =
√
Npq
1.94
S.D. with net counts due to source, given total count rate and
background rate
1.95
(a) When p is fixed B(x) → Normal as N → ∞
(b) When Np is fixed B(x) → P(x) as N → ∞ 1.96
If b is background counting rate and g background plus source,
for minimum statistical error tb/tg =
$
b
g
1.97
Radioactivity equation, decay constant and half-life from data on
activity measured after each hour
1.98
Interval distribution and exponential law of radioactivity 1.99
Carbon dating experiment 1.100
Best fit for a parabolic curve 1.101
Relative precision of capacitance, given errors in the radii of
concentric cylinder
1.102
Mean and most probable value of x for the probability
distribution f (x) = x exp(−x/λ)
1.103
10
1 x2
dx by Trapezoidal rule 1.104
10
1 x2
dx by Simpson’s rule 1.105
Chapter 2 Quantum Mechanics-1
Photon energy and wavelength 2.1
Wavelength of electrons accelerated by P.D 2.2
Wavelength of neutron 2.3
For electron λ = (150/V )1/2
2.4

Thermal Neutrons wavelength 2.5
K.E of neutrons bound in nucleus 2.6
2.2.2 Hydrogen Atom
En from Bohr model.Quantum numbers in Carbon atom. 2.7
Ground state energy of Carbon atom if electrons are replaced
by pions
2.8
Atomic units 2.9
Condition for potential energy to be negative. Internal
Energy of molecule formed in atomic collision 2.10
Bohr’s condition of quantization from de Broglie relation
and to find ionization energy 2.11
Radii of mesic atom and the resulting photons 2.12
Mu-mesic atom in which meson in lowest orbit just
touches nuclear surface 2.13
Wavelengths of Lyman series of Positronium 2.14
En for positronium, radii and transition energies 2.15
For force f (r) = −kr, to find P.E, V 2
,r2
, En, λ 2.16
(En − En−1) ∝ 1
n3 for n → ∞ 2.17
Transitions in which given wavelength occurs and the series to
which it belongs
2.18
For n → ∞, fn = νn, νn = (En − En−1)/h 2.19
Identification of a hydrogen-like ion from observations of Balmer
and Lyman series
2.20
The series to which a spectral line belongs having wave number
equal to the difference between two known lines of Balmer series
2.21
2.2.3 X-rays
Screening parameter of the K-shell electrons 2.22
Kα wavelength given ionization energy of H-atom 2.23
K-absorption edges of constituents of Cobalt alloy 2.24
λmin in X-ray tube 2.25
Planck’s constant from λmin in X-ray tube 2.26, 27
λα in copper from that of iron 2.28
To find applied voltage from difference (λα − λc) 2.29
Allowed transitions for Lα line under selection rule
Δl = ±1, Δ j = 0, ±1
2.30
Identification of target from study of λα and λc with
increasing voltage
2.31
Number of elements with Kα lines between 241 and 180 pm 2.32
Moseley’s law 2.33, 34
2.2.4 Spin and µ and Quantum Numbers-Stern–Gerlah’s Experiment
Key features of Stern–Gerlah’s experiment 2.35
Allowed values of j 2.36
Magnetic dipole moment of electron in H-atom 2.37
Beam separation in Stern–Gerlah’s experiment 42, 43 2.38,

Electron configuration and μ for H and Na 2.39
Magnetic moment, Stern–Gerlah’s experiment 2.40
Transition element, rare-earth element, electronic structure 2.41
Bohr magneton 2.44
2.2.5 Spectroscopy
Allowed values of l and m 2.45
Forbidden transitions in dipole transitions 2.46
Hyperfine quantum number for 9
Be+
2.47
Doppler line width temperature broadening 2.48
Zeeman effect in weak field in alkali atom 2.49
Calcium triplet-Fine structure 2.50
Zeeman effect in Sun spot 2.51
Normal Zeeman effect 2.52
Sketch for Zeeman splitting 2.53
Energy levels of mercury spectrum 2.54
Life times of 2p → 1s and 2s − 1s transitions 2.55
Population of states in Helium-Neon laser 2.56
2.2.6 Molecules
Modes of motion of a diatomic molecule 2.57
Rotational spectral lines in H-D molecule 2.58
Alternate intensities of rotational spectrum 2.59
Excited state of CO molecule 2.60
Boltzman distribution of rotational states 2.61
Vibrational states of NO molecule 2.62
Rotational states of CO molecule 2.63
H2 molecule as a rigid molecule 2.64
Mass number of unknown carbon isotope 2.65
Force constant of H2 molecule 2.66
2.2.7 Commutator
To show eipα/
xe−ipα/
= x + α, Hermicity of operators 2.67
If A is Hermicity to show that ei A
is unitary operator
To distinguish between eikx
and e−ikx
and sin ax and cos ax 2.68
To show (a) exp(iσ xθ) = cos θ + iσ x sin θ
(b)
d
dx
†
= −d/dx 2.69
To show (a) [x, px ] = i etc (b) [x2
, px ] = 2ix 2.70
Linearity of hermitian operator 2.71
Hermicity of momentum operator 2.72
Necessary condition for commuting operators 2.73
(a) Hermitian adjoint operator (b) Commutators
[Â, x̂], [Â, Â], [Â, p̂]
2.74
(a) Eigen value (b) Eigen state (c) Observable 2.75
(a) [x, H] = ip/μ (b) [[x, H], x] = 2
/μ 2.76
[A2
, B] = A[A, B] + [A, B]A 2.77
(σ.A)(σ.B) = A · B + iσ · (A × B) 2.78

(a) σy is real (b) Eigen value and Eigen vectors (c) projector
operators
2.79
(i) σ2
x = 1 (ii) [σx , σy] = 2iσz 2.80
Condition for two operators to commute 2.81
Ground state energy of a linear oscillator 2.82
Uncertainty for energy and momentum 2.83
Uncertainty for position and momentum 2.84
KE of electron in H-atom 2.85
Two uncertainty principles, KE of neutron in nucleus. 2.86
Chapter 3 Quantum Mechanics 2
3.2.1 Wave Function
Infinitely deep potential well-photon energy 3.1
Variance of Gaussian function 3.2
Normalization constant 3.3
Flux of particles 3.4
Klein–Gordon equation-probability density 3.5
Normalized wave functions for square well 3.6
Thomas-Reich-Kuhn sum rule 3.7
Laporte rule 3.8
Eigen values of a hermitian operator 3.9
Rectangular distribution of ψ, Normalization constant, Constant
probability, x, σ2
, momentum probability distribution
3.10, 11
Probability for exponential ψ 3.12
Solution of radial equation for n = 2 3.13
Ehrenfest’s theorem 3.14
Separation of equation into r, θ and ø parts 3.15
Derivation of quantum number m 3.16
Derivation of quantum number l 3.17
Particle trapped in potential well of infinite depth. Wave functions
and Eigen values
3.18, 48
V0 R2
= constant for deuteron 3.19
Expectation value of potential energy of deuteron 3.20
Average distance of separation and interaction of n and P in
deuteron
3.21
V0 for deuteron 3.22, 29
Root mean square separation of n p in deuteron 3.23
Photon wavelength for transition of electron trapped in a potential
well
3.24

Class I and II solutions of Eigen values for particle in a potential
well of finite depth.
3.25
Fraction of time that neutron and proton in deuteron are outside
range of nuclear forces
3.26
Results of energy levels for infinite well follow from finite well 3.27
Impossibility of excited states of deuteron 3.28
Transmission of particles through potential barrier 3.30, 31
Expected energy value for particle in a harmonic potential well 3.32
Infinitely deep potential well, momentum distribution 3.33
Quantum mechanical tunneling of alpha particle 3.34
One dimensional potential well, condition for one bound state and
two bound states
3.35
To express normalization constant A in terms of α, β and a in
Problem 3.25
3.36
Mean position and variance for particle in an infinitely deep well 3.37
Energy eigen values for given Hamiltonian 3.38
Eigen value for 3-D rectangular well 3.39
Degeneracy of energy levels in (3.39) 3.40
Potential step 3.41, 42
A 1-D potential well has ψ(x) = A cos
3πx
L

for −L/2 ≤ x ≤ L/2
and zero elsewhere. To find probability and Eigen value
3.43
“Top hat” potential; transmission coefficient. Numerical problem 3.44
Energy eigen functions and eigen values for 2D potential well 3.45
Number of states with energy less than E in 3-D infinite
potential well
3.46
Force exerted on the walls of a hollow sphere by a particle 3.47
Transmission amplitude of a beam of particles through
a rectangular well
3.49
Real and virtual particles. Klein–Gordon equation
and Yukawa’s potential
3.50
3.2.4 Simple Harmonic Oscillator (SHO)
Hermite equation for SHO 3.51
Probability in classical forbidden region 3.52
Energy of a 3-D SHO 3.53
Zero point energy and uncertainty principle 3.54
For n → ∞, Q.M. SHO→classical SHO 3.55
Probability distribution for classical SHO 3.56
E and V(x) for potential well, ψ ∼ exp(−x2
/2a2
) 3.57
SHO and uncertainty principle 3.58
Vibrational or rotational transitions from a set of wave numbers 3.59
Degeneracy of energy levels of isotropic oscillator 3.60
Oscillations of probability density of SHO state 3.61
Given the eigen functions of SHO, to find the expectation
value of energy
3.62
Eigen functions and eigen values for the lowest two states 3.63

3.2.5 Hydrogen Atom
Expectation value of U and E 3.64
Maximum electron density and mean radius 3.65
To show that 3d functions are orthogonal to each other 3.67
Degree of degeneracy for n = 1, 2, 3, 4 3.68
Parity of 1s, 2p, 3d states 3.69
To show that 3d functions are spherically symmetric 3.70
Probability for electron to lie within a sphere of radius R 3.71, 76
Maximum and minimum electrons density in 2s orbit 3.72
In the phosphorous mesic atom photon wavelength in the transition
3d → 2p is calculated and lifetime in the 3d state estimated 3.73
Momentum probability distribution of electron 3.74
Most probable momentum and mean value of electron 3.75
[Lx , Ly] = iLz 3.77
Eigen value of S1.S2 3.78
σp.σn = −3 for singlet state and = 1 for triplet state 3.79
Lz = −i ∂
∂ϕ
3.80
Expressions for Lx and Ly in spherical polar coordinates 3.81
[L2
, Lx ] = 0 etc 3.82
Expression for L2
/(i)2
in spherical polar coordinates 3.83
Angular momentum matrices for j = 1/2 and j2
3.84
Angular momentum matrices for j = 1 and j2
3.85
Clebsch-Gordon coefficients for j1 = 1 and j2 = 1/2 3.86
For given wavefunction to show the probability is zero for l = 0
and l = 1, and is unity for l = 2
3.87
To show that a set of wavefunctions belong to Lz and to obtain
eigen values
3.88
For given stationary state wavefunction to find Lz and L2
3.89
Given a wavefunction for H-atom to find Lz and parity 3.90
To apply L+ and L− to 2p eigen functions of H-atom 3.91
Nuclear spin from rotational spectra of diatomic molecules 3.92
To re-express given angular wave function as combination of
spherical harmonics
3.93
Given a wavefunction of a hydrogen-like atom, to find Lz 3.94
Given the wavefunction for a state, to get values of Lz and to find
the corresponding probabilities
3.95
To prove commutation rules involving Jx , Jy, Jz and J+ 3.96
To calculate correction to the potential of hydrogen atom 3.97
Stationary energy levels of a charged particle oscillating with a
given frequency in an electric field
3.98
To work out the perturbed levels 3.99
To find the ionization energy of helium atom by variation method 3.100

Stark effect of H-atom 3.101
Application of variation method to SHO 3.102
Application of first order perturbation to 2-D potential 3.103
3.3.8 Scattering (Phase-Shift Analysis)
To derive partial-wave expansion 3.104
α-He scattering, classical and quantum mechanical 3.105
σ(E) from δl 3.106
Energy at which p-wave is important in n-p scattering 3.107
To find δ0 from known σ 3.108
Hard sphere scattering 3.109
σel and σtotal for scattering from a black sphere 3.110
Ramsauer effect 3.111
Explanation for low energy n-p cross-sections 3.112
Form factor of proton and characteristic radius 3.113
Simplification of elastic scattering amplitude under the assumption
of spherically symmetric potential
3.114
Optical theorem 3.115
Reduction of Mott scattering due to finite size effects 3.116
Given the scattering amplitude, to obtain the form factor 3.117
Scattering from a shielded Coulomb potential for point
charged nucleus
3.118
Form factor for scattering from extended nucleus of constant
charge density
3.119
Root mean square radius from scattering data 3.120
Form factor for Gaussian charge distribution and mean square radius 3.121
Scattering amplitude for spherically symmetric potential 3.122
To obtain form of scattering amplitude for Yukawa’s potential 3.123
Chapter 4 Thermodynamics and Statistical Physics
Maxwellian law
Velocity distribution 4.1
Energy distribution 4.2
Mean speed 4.3
vrms 4.4
Most probable speed 4.5
vp :v:vrms 4.5
vrms for H2 4.6
vrms given ρ p 4.7
1/v 4.8
N(α)dα, α = v/vp 4.9

Δv/v for O2 4.10
vrms of H2 at T2 given vrms at T1 4.11
vrelative 4.12
vrms(N2) = vescape 4.13
Fraction of molecules with MFP λ and 2λ 4.14
Viscosity 4.15
MFP and collision frequency 4.16
Degrees of freedom 4.17
T from E 4.18
γ from degrees of freedom 4.19
K, η, Cv γ 4.20
4.2.2 Thermodynamic Relations
4.21,
4.22,
4.23
Clausius-Clapeyron equation 4.24
Latent heat of vaporization of water 4.25
Stefan-Boltzmann law 4.26
Cp − CV = R 4.27,
Cp − CV = R(1 + 2a/RT V ) 4.28,
Cp − CV = T Eα5
V 4.29
Tds equations 4.30,
4.31,
4.32
Joule Thompson effect 4.33,
4.34,
4.35
ES/ET 4.36
Ratio of adiabatic to Isobaric pressure coefficient 4.37
Ratio of adiabatic to Isochoric pressure coefficient 4.38
Internal energy of ideal gas 4.39
Application of Clapeyron equation 4.40,
4.41,
4.42
Blackbody radiation 4.43
Dulong Petit law 4.44
Rotational states of H2 4.45
Normal modes of frequency 4.46
Sterling’s approximation 4.47
Rotational states 4.48,
49, 50
Hydrogen atoms in chromosphere with n = 1, 2, 3 and 4 4.51

Fermi level 4.52,
53, 54
Distribution of distinguishable and indistinguishable particles 4.55
Vibrational states 4.56, 59
Accessible states 4.57
Standard deviation 4.58
Boltzmann formula 4.60
Temperature of wire 4.61
Sun’s temperature 4.62, 65
Radiation pressure 4.63
Stefan’s law → Newton’s law 4.64
Loss of sun’s mass 4.66
Temperature fall of spheres 4.67
Wien’s law 4.68
Planck’s law 4.69,
70, 71,
72, 73,
74, 75
Temperature of earth 4.76
Solar constant 4.77
Chapter 5 Solid State Physics
Volume in cubic structure 5.1
Volume in body-centered 5.2
Separation of planes 5.3
Diffraction angle for (111) plane 5.4
Higher order reflection 5.5
Davisson and Germer experiment 5.6
Atomic radii, FCC, BCC, diamond 5.7
Madelung constant 5.8, 9
Equilibrium of atoms 5.10, 11,
12
Melting point of lead 5.13
5.2.3 Metals
Fermi energy/level 5.14, 21,
22, 25, 26,
27, 28, 32,
33, 34, 35

Drift velocity 5.15, 17
Flow of current 5.16
Wiedemann–Franz constant 5.18
Hall effect 5.19, 20
λ/d 5.23
Average amplitude of vibration 5.24
Debye’s model 5.29, 31
Einstein’s model 5.30
pn junction 5.36
ne and nh 5.37
Effective mass of e 5.38
Doping 5.39
Conduction bands 5.40, 41
Forward bias 5.42
Depletion layer width 5.43
Shockley equation 5.44
Carrier concentration 5.45
Cooper pair 5.46, 47
Josephson junction 5.48, 51
BCS theory 5.49, 50
Quantum Interference device 5.52
Chapter 6 Special theory of relativity
Simultaneity of two events 6.1
Lorentz transformation corresponds to rotation iα 6.2
Velocity of muon in lab system 6.3, 4, 5
Maxwell’s equations for em are invariant 6.6
KE of π+
from decay K0
→ π+
π−
in flight 6.7
Emission angle of muon in lab from π → µν, neutrinio being
emitted at 900
to pion velocity
6.8
Reference frame velocity when the particle velocity components in
S and S′
are specified
6.9
In the two body decay energy carried by the particles of known
masses and Q-value
6.10
Maximum lab energy of muon and neutrino from decay of pion of
known energy in flight
6.11
CMS velocity of e+
− e−
pair 6.12
Angular relation in relativistic elastic scattering between identical
particles
6.13

Maximum and minimum muon energy from the decay sequence
K → π → μ
6.14
Mass of a particle decaying into three pions of known energies
and emission angles
6.15
Contraction of a rod traveling with known velocity 6.16
Speed of a rod shrunk by known length 6.17
Shrinking of earth’s diameter 6.18
Speed of muons from lifetime measurement 6.19
Lifetime of muons of known velocity from observed
lifetime
6.20
Average distance of muons before decaying in lab system and
muon system
6.21
Velocity with which a person must travel from the centre to the edge
of our galaxy so that the trip may last 40 years
6.22
A pion produced in earth’s atmosphere 1 km above sea level travels
vertically down with speed of 0.99 c and decays. To locate the
decay point
6.23
Relative velocity of two particles approaching each other with
velocity 0.5 c each
6.24
Observed length of an evacuated tube when the observer travels
with the speed of that of 100 MeV electron
6.25
Speed of a spacecraft when the mass of a 100 kg man in the
spacecraft is registered as 101 kg from earth
6.26
When a spaceship moving with speed of 0.5 c passes near Mars a
radio signal is sent from Mars which is received on earth 1,125 s
later. Time taken by spaceship to reach earth according to observers
on earth and spaceship crew
6.27, 28
A spaceship is moving away from earth with speed 0.6 c. A radio
signal is sent from the earth when the ship is 5 × 108
km from earth.
To find the time for the signal to reach the ship according to earth’s
observers and ship’s crew
6.29, 30
Mean lifetime of pion of known energy from track length up to
decay point
6.31
Intensity of neutrino flux at the end of flight path of pions. Given E,
I0, L and τπ
6.32
Distance traveled by pion of KE = mπ c2
6.33
Technique for obtaining a neutrino beam free from contamination
of pions and muons
6.34
Mean lifetime of pions of known energy and initial intensity from
counting rates in three counters
6.35
Velocity with which an observer must move such that the velocities
of two objects appear to have equal and opposite velocities
6.36
Lifetime measurement from observations of counting rate of a beam
of particles of given velocity recorded in two counters in tandem
6.37

A sphere of mass m and radius r has speed
√
3c/2. To find E, x
before decay and shape as seen in lab frame
6.38
Photon has velocity c in all frames of reference 6.39
Effective mass, KE and momentum of muons, given observed and
proper lifetimes
6.40
Energy from annihilation of 1g of mass 6.41
Speed of proton whose KE = mc2
6.42
Speed of P and e when accelerated to 1GeV 6.43
Energy needed to break up 12
C into (a) 6p + 6n (b) 3α (c) 3α into p
and n
6.44
Mass of a particle from T and p measurements 6.45
Rest mass energy of electron 6.46
P.D. required for electron to accelerate from rest to 0.6 c 6.47
Velocity at which T differs classical value by (a) 1% (b) 10% 6.48
For v ≪ c, T ≪ mc2
, T (relativistic)→ T (classical) 6.49
Effective mass of photon for (a) visible light (b) X-rays 6.50
1 amu = 931.5 MeV/c2
6.51
Energy from explosion of 5.0 kg fissionable material 6.52
Energy imparted to electron emitted at angle θ in collision of proton
of velocity βc with electron at rest
6.53
Energy carried by muon and neutrino in the decay of pion at rest 6.54
In the complete inelastic collision of a body with identical body at
rest to find the speed of resulting body and its rest mass
6.55
Mass of a body decaying into kaon and pion with equal and
opposite momenta
6.56
Speed of muon in π → μν at rest, in terms of masses involved 6.57
EB in A → B + C at rest in terms of masses of A, B and C 6.58
Tmax of electron in muon decay 6.59
In a symmetric elastic collision between a particle with an identical
particle at rest, angle between outgoing particles
6.60
An electron has T = mc2
. A photon with Pγ = Pe has
Eγ =
√
3mc2
6.61
In the decay of muon at rest if the energy released is divided equally
among the final leptons whose mass is neglected angle between the
paths of any two leptons is 1200
6.62
Energy imparted to an electron which is emitted at 30
in the
collision of energetic proton
6.63
(a) mπ from Tμ in pion decay at rest (b) Te (max) in muon decay 6.64
Eπ0 in p−
+ d → n + π0
6.65
Mass of K∗
particle deduced from momenta of kaon and pion into
which it decays and angle between the tracks
6.66
A proton of KE 940 MeV makes an elastic symmetric collision with
a stationary proton. To calculate the angle between the outgoing
protons
6.67

Available energy for the incident proton of known momentum
on a target proton of Fermi momentum when it is (a) parallel
(b) antiparallel (c) orthogonal
6.68
Elastic scattering of an antiproton with a stationary proton 6.69
Formula for the KE acquired by recoiling target nucleus when
elastically scattered by ultra-relativistic electron
6.70
Formula for particle mass (M≫me) of momentum p when
elastically scattered by electron
6.71
Pmax transferred to electron in high energy neutrino-electron
collision
6.72
In the relativistic elastic collision of mass m1 with m2(m1 m2)θmax
for m2 depends only on particle masses.
6.73
Formula for 4-momentum transfer squared in collision with electron 6.74
Momenta and velocity of electron in photon–electron collision 6.75
Compton scattering wavelength shift using 4-vectors 6.76
Squared four-momentum transfer in high energy electron scattering
at given angle
6.77
Mass of a particle which decays into two pions with equal
momentum at right angles to each other
6.78
Formula for mass of a particle decaying into two particles of known
mass, p, E and θ
6.79
Relations for Mandelstam variables 6.80, 81
Angle between two γ -rays from π0
decay and θmin at Eπ0 = 10 GeV 6.82
Rest energy of ω0
in p−
+ p → π+
+ π−
+ ω0
6.83
Mass M of a composite particle formed in the collision of m1 of
velocity v with m2 at rest
6.84
Q-value of Λ-hyperon from the energy and angular measurements
of decay products
6.85
Mass, velocity and direction of a particle decaying into two particles
of momenta p1 and p2 at right angles and energy E1 and E2
6.86
Analysis of a V-type of event observed in a bubble chamber 6.87
Derivation of formula for the angle between two γ -rays in π0
decay
from invariance principle
6.88
Maximum 4-momentum transfer in neutron decay 6.89
Doppler wavelength shift of a receding star 6.90
Doppler wavelength shift of a nebula 6.91
For slow speed approximate formula for Doppler shift 6.92
Speed of a motorcycle at which red light would appear green 6.93
Color of light seen by a spaceship passenger when it appears orange
from earth
6.94
Doppler shift of sodium light moving in a circle 6.95
Emission angle of electron with energy E in scattering of neutrino
of energy E0 with electron
6.96

Mass of kaon from its decay characteristics 6.97
Neutral pions of fixed energy decay into two γ -rays, velocity and
rest mass energy of pion, energy and angular distribution of
γ -rays angle between two γ -rays and minimum angle
6.98, 99,
100, 101,
103, 104
In problem 6.98, locus of the tip of momentum vector and the
disparity for the ratio of γ-ray energy
6.102, 105
Maximum lab angle of K∗
produced in pion-proton collision 6.106
Relativistic transformations of velocity and angle 6.107
Relative Doppler shift of wavelength of light from excited
carbon atoms
6.108
Fractional change of frequency of a spectral line of a star 6.109
Angle between the two photons in the lab from π0
decay when
emitted at right angles in the pion system
6.110
Velocity of a moving object from aberration of light 6.111
The product of the longitudinal component of the length of a rod
and the tangent of its orientation
6.112
Threshold energy for p−
p pair production in p–p collision 6.113
Threshold energy for e−
e+
pair production in γ − e−
collision 6.114
Threshold energy for pion collision in N–N collision 6.115
The threshold for antiproton production with Fermi energy 6.116
Threshold energy for γ + p → K+∗
+ Λ 6.117
Threshold energy for two pion production in pion-hydrogen
collision
6.118
Threshold energy for π−
+ p → K0
+ Λ 6.119
Threshold energy for nπ production in p–p collision 6.120
Threshold energy for γ + p → p + π0
6.121
Threshold energy for π−
+ p → Ξ−
+ K+
+ K0
6.122
Lower limit for the mass of W+
boson in neutrino-proton
interaction
6.123
Threshold energy and invariant mass of the system in
p + p → p + Λ + K+
at threshold energy
6.124
Minimum K−
momentum for Ω−
production in
K−
+ P → K0
+ K+
+ Ω−
6.125
Feasibility of P + P → d + π, in Cu target, with Fermi
momentum
6.126
Chapter 7 Nuclear Physics I
Elastic scattering 7.1, 4
Maximum scattering angle 7.2, 5
Quasi elastic scattering 7.3
Scattering in CMS 7.6
vmax imparted 7.7

σ(θ) versus σ(θ∗
) 7.8
dσp/dEp 7.9
σ(ϕ) 7.10
Scattering from sphere 7.11
σ(ϕ) 7.12
N(600
− 900
)/N(900
− 1200
) 7.13
Scattering probability 7.14, 20
Field of force 7.15
θ∗
for b = R0/2 7.16
Breakdown of Rutherford scattering 7.17, 26
N(θ 900
)/N 7.18
Scattering with L = 7.19
Z from Rutherford scattering 7.21
dσ/dW for electron 7.22
σ = σg(1 − R0/R) 7.23
rmin for α scattering 7.24, 28
Emin to reach nucleus 7.25
θ for b = R0 7.27
Scattering for θ 300
7.29
Scattering from brass 7.30
Scattering from gold 7.31
Darwin’s formula 7.32
Radiation loss of D and e 7.33
Muon penetration 7.34
Range-energy for D and P 7.35
Range of P, D and α 7.36
Range of α and P in Al and air 7.37
Straggling of 3
He and 4
He 7.38
Mass estimation from ranges 7.39
Range of D and α 7.40
dE/dx of α 7.41
Stopping power of P and D 7.42
Specific ionization of α and P 7.43
Stopping power in air and Al 7.44
Bragg–Kleeman rule 7.45
Geiger’s rule 7.46
Geiger–Nuttal law 7.47
Production of ion pairs 7.48
Energy loss of P and D 7.49, 50
Radiation energy loss of e 7.51
Multiple scattering and angular distribution in radiation processes 7.52

Kinematics of Compton scattering 7.53
(Δν)max 7.54
Recoil velocity of electron 7.55
Ph.elec, Comp, Pair 7.56
λ0 7.57
ΔE for various situations 7.58
Attenuation of Eγ 7.59
(Δλ)max 7.60
hν0 given hν θ 7.61
Range of Eγ from annihilation radiation 7.62
υe 7.63
T and hν, given Br and BE 7.64
V0, given λ0 and λ 7.65
Photo effect not possible with free electron 7.66
Attenuation of γ rays 7.67
Absorption edges 7.68
h, λ0 and W 7.69
μph and μc 7.70
Photo effect from hydrogen 7.71
h (Planck’s constant) 7.72
Photoelectric current 7.73
T and λe 7.74
Eγ (threshold) 7.75
Eγ from e+
e−
annihilation 7.76
γ → e+
e−
, not possible in vacuum 7.77
Range of n 7.78
KE of proton 7.79
No. of photons emitted 7.80
Mosbauer experiment 7.81
ΔW when M∗
→ M 7.82
Thermal broadening 7.83
Red shift 7.84
T1 and T2 7.85
Heating effect 7.86
Radioactive equilibrium 7.87, 92,
93, 95

Age of radioactive alloy 7.88
Source strength 7.89
Electricity generation 7.90
Carbon dating 7.91
Half life time of 55
Co 7.94
Partial half lives of 242
Pu 7.96
Chain decay 7.97
Age of earth 7.98
Activity in Curies 7.99
Rad 7.100
Diagnostic 7.101
7.2.10 Alpha-Decay
Gamow’s formula 7.102,
105
Minimum α energy to force it into nucleus 7.103
Gieger–Nuttal rule 7.104
7.2.11 Beta-Decay
Beta-decay transitions 7.106
Mean lifetime 7.107, 108
Evolution of heat 7.109
Kurie plot 7.110
Energy of e, ν recoiling nucleus 7.111
Beta ray spectrum 7.112
Chapter 8 Nuclear Physics II
Mass spectroscopy 8.1, 2, 3
Mirror nuclei 8.4
Uncertainty relation 8.5
Mass of 14
O 8.6
Electrostatic energy of nucleus 8.7
Potential (rR) 8.8
Rotating proton 8.9
Hyperfine structure 8.10
Magnetic resonance 8.11
NMR 8.12

8.2.4 Electric Quadrupole Moment (eqm)
Calculation of eqm 8.13, 16
Major to minor axes ratio 8.14
Condition for vanishing of eqm 8.15
Condition for α decay 8.17
Radius from separation energies 8.18
Decay via α or β emission 8.19
Competitive decays of 64
Cu 8.20
γ-ray energy in 28
Al → 28
Si + β−
+ γ 8.21
22
Na mass in 22
Na →22
Ne + β+
+ γ 8.22
Whether 7
Be → 7
Li + β+
+ ν possible 8.23
To show p = 2ρn Ef /5 8.24
Pf , Ef and V 8.25
8.2.7 Shell Model
J and l values 8.26
Spin and parities 8.27, 30,
31, 32
Gap between neutron shells 8.28
Quadrupole moment 8.29
Zmin/A for light and heavy nuclei 8.33
Most stable isobar 8.34, 37
Coulomb coefficient 8.35
Radii of nuclei 8.36
Separation energy 8.38
Binding energy of nuclei 8.39
Binding energy/mass number curve 8.40
Beta stability of isobars 8.41
Beta stability of 27
Mg 8.42
8.2.9 Optical Model
Imaginary part of potential 8.43
λ and absorption of neutrons 8.44
Reaction threshold 8.45, 48
Energy of products in K capture 8.46
Inelastic scattering 8.47
Difference in B.E. of 3
H and 3
He 8.49
Q value and KE of α 8.50
Q value of 27
Al (p, n) 27
Si 8.51
Ethreshold for 3
H (p, n) 3
He 8.52

Q-value in 30
Si (d, p) 31
Si 8.53
Q-value for inelastic scattering 8.54
Eγ for producing 5 MeV protons in Compton like scattering 8.55
Elastic and inelastic scattering of protons from 10
B 8.56
Ep at 900
from 27
Al (d, p) 28
Al 8.57
Fusion reaction 8.58
Which is unstable 17
O or 17
F? 8.59
Energy level diagram 8.60
Neutron and hydrogen atom mass difference 8.61
Range of neutron energy in 3
H (d, n) 4
He 8.62
Forward and backward reactions 8.63
Ethreshold for endoergic reaction 8.64
8.2.11 Cross-Sections
Neutron absorption in indium foil 8.65
60
Co production in reactor 8.66, 67
σ(θ) parameters 8.68
σt for 48
Ca (α, p) 8.69
Contribution of scattering to resonance at 0.178 eV 8.70
σ(n, γ ) σ(n, α) given Γn, Γγ , Γα and ER 8.71
Γel given ER, Γ and JN = 0 8.72
Spin value from detailed balance 8.73
E, ΔE θ in D-stripping 8.74
Production and activity of 24
Na 8.75
Production and activity of 198
Au 8.76
Energy released in fission of 238
U 8.77
Spatial distribution of neutrons 8.78, 87,
88, 91
Thermal utilization factor 8.79
Critical size of bare reactor 8.80
235
U consumption 8.81
Fission processes/second 8.82
Annual electricity consumption in Britian 8.83
Neutron flux near reactor centre 8.83
To show ln(E1/E2)= 1 for neutron scattering 8.84
Neutron moderation 8.85
Criticality of homogeneous reactor 8.86
Thermal diffusion time 8.89
Generation time 8.90
Number of collisions required for thermalization of fission
neutrons
8.92

8.2.15 Fusion
Range of neutrino energies in p + p → D + e+
+ ν 8.93
Neutron energy in d(d, n) 3
He 8.94
For d–d reaction, required temperature modified due to tunneling
reaction
8.95
Lawson criterion 8.96
Chapter 9 Particle Physics I
Conversion : kg → GeV/c2
9.1
m → GeV−1
, GeV−2
→ mb, second → GeV−1
9.2
Compton wavelength, Bohr radius, νe in H-atom 9.3
Natural units → practical units 9.4
Mean life time of muon 9.5
9.2.2 Production
Emission angles of photons in LS and CMS 9.6
For fixed target experiments ECM ∝
√
Elab 9.7
Interaction length for σ(e+
e−
→ μ+
μ−
) in Pb 9.8
Ethreshold for e+
e−
→ pp−
9.9
9.2.3 Interaction
Possible production of particles in p–p collisions 9.10
Neutrino interactions 9.11
Muon interactions 9.12
Proton interactions 9.13
Number of photons in π−
p → π0
n, π0
→ 2γ 9.14
Attenuation of pions muons in iron 9.15
Non-occurrence of reactions 9.17
Hyper-fragment mass 9.18
4-momentum transfer 9.19
9.2.4 Decay
Pion energy in decay K+
→ π+
π0
at rest 9.20
Pion lifetime from attenuation of beam intensity 9.21
Tmax of pion in decay K+
→ π+
π+
π−
9.22
To express E′
ν in terms of mπ and mμ 9.23
Tn(max) in decay Σ+
→ nμ+
νμ 9.24
To sketch decay configuration and value for Eν(max) in K → μν 9.25
Lifetime of kaon from track measurements 9.26
Difference in travel time of muon and neutrino 9.27
Tmax Tmin of µ’s from pion decays in flight 9.28
Drop in intensity in transport of kaons and Λ beams 9.29
Relation of d, p, τ and m 9.30

In decay π → μν, relation for Eν in terms of γ , Eπ , mμ, mπ and θ 9.31
Transport of sigma hyperon beam 9.32
Total decay width of muon from electron energy spectrum,
Ee(max) and muon helicity
9.33
Disparity in lifetimes of π+
and π0
, equality of masses of π+
and
π−
but difference in masses of
Σ+
and Σ−
, lifetime of Σ0
much smaller than Λ and Ξ0
9.34
Identification of a neutral unstable particle from momenta and
angular measurements of decay products
9.35
Dead time of counting system 9.36, 45
Energy of α source and ion pairs in ionization chamber 9.37
Rate of entry of beta particles in ionization chamber 9.38
Gas multiplication in proportional counter 9.39
Pulse height in proportional counter 9.40
Plateau of G.M. counter 9.41
Maximum radial field of a G.M. tube and its lifetime 9.42
Distance from anode at which electron gains enough energy to
ionize
Argon in one MFP. 9.43
Difference in efficiency of G.M. and scintillation counters 9.44
Leakage resistance in a dosimeter 9.46
Maximum voltage for avalanche 9.47
Potential of an electrometer 9.48
Time of flight method 9.49
Discovery of antiproton 9.50
Pulse height 9.51
Average size and standard deviation 9.52
Dead time of multichannel pulse height analyser 9.53
Electron multiplication per stage 9.54
Gamma ray spectrum of 22
Na 9.55
Half-width at half-maximum of photopeak 9.56
Standard deviation of energy and coefficient of variation of energy 9.57
Identification of kaons and pions of same energy 9.58, 61
Speed of electron 9.59
mc2
in terms of p and θ 9.60
Number of Cerenkov photons produced 9.62
Length of gas counter 9.63
Threshold Cerenkov counter 9.64
Potential from alpha particle absorption 9.65

9.2.9 Emulsions
Range-energy graph for 2
H and 3
He 9.66
p = 0.3Br 9.67
An energetic proton heading toward earth’s centre just
misses it
9.68
Null deflection in crossed E and B 9.69
Mass of particle from E and r 9.70
Deflection of a particle in magnetic field 9.71
An electron recrossing x-axis in electric field 9.72
Radius of curvature of helical path 9.73
A proton circling around magnetic equator 9.74
Power delivered to earth by cosmic rays 9.75
Mass spectroscopy of kaons 9.76
9.2.11 Betatron
Electron energy 9.77
Number of revolutions, Emax and E 9.78
9.2.12 Cyclotron
KE in terms of r and B 9.79
Field strength and acceleration time 9.80
ΔB for cyclotron resonance 9.81
Resonance condition for P, D, α 9.82
Range of frequency modulation in synchrocyclotron 9.83
Calculation of B and r for given Emax and f 9.84
Drop of B for cyclotron resonance when alphas are accelerated
instead of deuterons
9.85
Tmax imparted to electron from proton collision 9.86
To calculate final frequency from given initial frequency and Emax
for protons
9.87
9.2.13 Synchrotron
R of protons for given B and E 9.88, 89
Percentage depth of modulation 9.90
Energy loss in electron synchrotron 9.91
Derivation of R = m0c(n2
+ 2n)1/2
/qB, n = T/m0c2
9.92
Proton energy when B rises to known value 9.93
Acceleration of D and 3
He when protons are replaced 9.94
Given B, V, φs and Tmax for protons, to find r, t to calculate Δf
(initial), frequency modulation range
9.95
Feasibility of 500 GeV electron synchrotron 9.96
KE of proton in known orbit making 1 revolution/µs 9.97
For an electron synchrotron to find initial RF, total time and
distance traveled
9.98

Length and number of drift tubes 9.99
Average beam current and power output in SLA 9.100
Output energy after the fifth drift tube and total length 9.101
Length of longest tube operating at known frequency and
maximum energy
9.102
9.2.15 Colliders
Luminosity calculation 9.103
μ+
μ−
pair production 9.104
E2
CM = 4E1 E2 9.105
Comparision of available energy in fixed target machines and
colliders
9.106, 107
Calculation of CMS energy for HERA accelerator
(electron–proton collider)
9.108
CMS energy in a collider with a crossing angle 9.109
Chapter 10 Particle Physics 2
Interactions allowed and force involved 10.1
Strong, em or weak interactions 10.2, 4
Kaon momentum from known values of pion momenta 10.3
ρ0
→ π+
π−
allowed but not ρ0
→ π0
π0
10.5
Allowed decays of omega minus 10.6
Interactions and relative coupling strength 10.7
Transform of various physical quantities under P and T operation 10.8
Deuteron exists in 3
S1 and 3
D1 states 10.9
ρ → η + π and ω → η + π are forbidden 10.10
Disparity in lifetimes of ρ0
K0
and Δ0
Λ 10.11
Decay modes of ρ0
by strong em reactions 10.12
Parity of u d-quarks 10.13
Hyperon decays by weak interactions 10.14
Parity of K−
meson 10.15, 40
Definition of hadron, Lepton, baryon, Meson, wavelength of
photon in decay of π0
10.16
Distance of travel of Δ++
; lifetime of W-boson 10.17
Analysis of pion-proton scattering via isospin amplitudes 10.18
Inequality in cross-sections for pion-proton scattering 10.19
Branching ratios for Δ+
→ pπ0
Δ+
→ nπ+
10.20
Isospin of X+
(1520) 10.21
Decay of ρ-meson 10.22
Isospin states of three pions 10.23

Identification of particles and quark content produced in K−
p
interactions
10.24
Resonance produced in πp interactions 10.25, 26
Ratio of cross-section for πp → ΣK 10.27
Ratio of cross-section for π−
p → π−
p π−
p → π0
n 10.28
Parity of negative pion 10.29
Isospin of resonance in K−
p interactions 10.30
Number of charged and neutral sigma hyperons in K−
p and
K−
n interactions
10.31
Spin, parity and isospin of B-meson 10.32
Branching ratio of the decay modes of Δ0
(1232) 10.33
Pseudoscalar mesons octet 10.34
Vector mesons nonet 10.35
Baryon octet 10.36
Baryon decuplet 10.37
σ(π−
p) = σ(π+
n) but σ(K−
p) = σ(K+
n)
Distinction between K0
and K̄0
mesons 10.38
Ratio of σ for K−4
He → 4
HeΛ K−4
He → 4
HΛ 10.39
π−
d → nnπ0
cannot occur at rest 10.41
Isospin predictions for σ(pp → dπ+
) σ(pn → dπ0
) 10.42
Pairs of particles with I = 1 10.43
10.2.3 Quarks
Phenomena when a quark is struck by energetic electron 10.44
Decay of B−
meson 10.45
Effective strong interaction coupling constant αs 10.46
σ(e+
e−
→ hadrons/σ(e+
e−
→ μ+
μ−
) 10.47
Cross-section relations using quark model 10.48, 49
Mass difference of u- and d-quark 10.50
Quark constituents of Δ-states and fundamental difficulty.
Higher excited states of nucleons 10.51
Quark composition of hyperons, nucleons, kaons and pions 10.52
Particles with three identical quarks.Properties of b quark 10.53
Quark flow diagrams for the decays of ϕ and ω0
10.54
σ(πN)/σ(NN) explained by quark model 10.55
Ratio σ(π−
C)/σ(π+
C) explained by quark model 10.56
Estimation of σ(e+
e−
→ hadrons) from σ(e+
e−
→ μ+
μ−
) 10.57
Ground state energy of meson by variation method 10.58
10.2.4 Electromagnetic Interaction
Feynman diagram and σ for e+
e−
→ μ+
μ−
10.59
Width for Σ0
decay. Spin zero for K+
10.60
C-invariance and positronium decay 10.61
Three body decays of K+
and Σ0
10.62
10.2.5 Weak Interaction
Decay rates of D+
and D0
10.63

Mass of exchanged boson from variation of σ(νe−
→ νe−
) 10.64
Number of W+
→ e+
νe events in pp−
annihilation 10.65
Cabibbo’s theory to explain D+
→ K̄0
μ+
νμ and
D+
→ π0
μ+
νμ
10.66
Ratio Γ(Σ−
→ nēνe)/Γ(Σ−
→ Λē ν̄e) 10.67
Branching fraction for τ−
→ ēνeντ, from lepton universality 10.68
Sergent’s law of beta decay 10.69
Allowed and forbidden weak decays 10.70
Use of spurion for hyperon decays 10.71, 73
ΔI = 1/2 rule for Σ hyperon decays 10.72
Flux of neutrinos received on earth 10.74
Neutrino mass from observations of supernova 10.75
Properties of particle X from its decay 10.76
Parity violation in alpha decay of 16
O 10.77
Tauon lifetime from muon lifetime 10.78
Feasibility of two-pion decay of ω0
, η0
, f 0
10.79
Non-existence of η → 4π decay 10.80
Ratios Γ(ρ0
) : Γ(ω0
) : Γ(ϕ0
) from Van Royen–Weisskopf
formula
10.81
Classification of D+
decay as Cabibbo
allowed/suppressed/forbidden interaction
10.82, 83
Allowed decays for K−
, Ξ0
, Ω−
10.84
Allowed and forbidden decays of K0
and K
0
10.85
Oscillation of muon neutrino 10.86, 87
Universality of weak coupling constant 10.88
Non-existence of decay Σ+
→ ne+
νe 10.89
Relative decay rates of D+
→ K
0
μ+
νμ D+
→ π0
μ+
νμ 10.90
Starting with pure beam of K0
’s at t = 0 to find I(K0
) I(K
0
)
at time t
10.91
Regeneration phenomenon of Ks 10.92
10.2.6 Electroweak Interactions
Criterion for the existence of neutral current interaction 10.93
Number of neutrino generations 10.94
Experimental signatures for W and Z0
bosons 10.95
Masses of W and Z bosons from Salam–Weinberg theory 10.96
τ+
→ μ+
νμντ, K−
+ p → Ω−
K+
K0
, D
0
→ K+
π−
10.97
l + n → e−
+ p, τ−
→ μ−
ll̄, B0
→ D−
μ+
l 10.98
Δ0
→ pπ−
, Ω−
→ ΛK−
10.99
e+
e−
→ νµνµ, D0
→ K−
π+
10.100
Moller scattering, Bhabha scattering 10.101
Λ → pe−
νe, D−
→ K0
π−
10.102
π−
p → K0
Λ, e+
e−
→ B
0
B0
10.103
Effective coupling in K−
→ μ−
ν̄μγ 10.104

D0
→ K−
π+
, charge current interaction but not D+
→ K0
π+
10.105
τ(π±
) ≫ τ(π0
) via Feynman diagram 10.106
Bremsstrahlung, pair production 10.107
Photoelectric effect, Compton scattering 10.108
e+
e−
→ qq, νμ + N → νμ + X 10.109
Ξ−
→ Λπ−
, K+
→ π+
π+
π−
10.110
D+
→ K0
ll̄, τ−
→ π−
ντ 10.111
e+
e−
→ 2γ , e+
e−
→ 3γ 10.112
Λ → pπ−
10.113

Index
A
Alpha decay, 393
Angular momentum, 149
Approximate methods, 152
Atomic masses, 434
Atomic radii, 434
B
Beta decay, 393
Beta function, 23
Betatron, 498
Blackbody radiation, 256
Born approximation, 154
C
Calculus of variation, 31
Cerenkov counter, 496
Cerenkov radiation, 390
Colliders, 502
Commutators, 100
Complex variables, 30
Compound nucleus, 443
Compton scattering, 387
Conservation laws, 544
Cross-sections, 442
Crystal properties, 294
Crystal structure, 294
Cyclotron, 499
D
de Broglie waves, 92
Differential equations, 26
Direct reactions, 443
Doppler effect, 328
E
Electric energy, 435
Electric potential, 435
Electric quadrupole moment, 435
Electromagnetic interactions, 577
Electro-weak interactions, 590
Emulsions, 497
F
Fermi gas model, 437
Feynman diagrams, 592
Fission, 444
Fourier series, 22
Fourier transform, 22
Fusion, 447
G
Gamma function, 23
G.M. Counter, 493
H
Hydrogen atom, 92, 147
I
Integration, 25
Ionization, 385
chamber, 493
K
Kinematics, 382
Kinetic theory, 258
L
Laplace transforms, 29
Linear accelerator, 501
Liquid drop model, 438
Lorentz transformations, 319
angles, 328
length, time, velocity, 320
M
Magnetic field, motion of charged particles,
497
633

634 Index
Magnetic moment, 96
Matrix algebra, 24
Maxima, 24
Metals, 295
Minima, 24
Molecules, 99
N
Nuclear magnetic moment, 435
Nuclear reactions, 440
Nuclear reactors, 444
Nuclear resonance, 390
Nuclear spin, 435
Nuclear stability, 436
Numerical integration, 33
O
Optical model, 439
P
Pair production, 390
Particle interaction, 491
Particle production, 489
Phase-shift analysis, 153
Photoelectric effect, 411
Potential barriers, 140
Potential wells, 140
Proportional counters, 493
Q
Quadrupole moment, 435
Quarks, 550
R
Radioactivity, 391
Range, 385
Rutherford scattering, 383
S
Scintillation counter, 495
Semiconductors, 297
Series, 25
Shell model, 437
Simple harmonic oscillator, 146
Solid state detector, 497
Special functions, 29
Spectroscopy, 97
Spin, 96
Statistical distributions, 32, 255
Stern–Gerlah experiment, 96
Straggling, 385
Strong interactions, 546
Superconductor, 298
Synchrotron, 500
System of units, 488
T
Thermodynamic relations, 253
Threshold of particle production, 330
U
Uncertainty principle, 101
V
Vector calculus, 21
W
Wave function, 137
Weak interactions, 578
X
X-rays, 95

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