The fundamental physics Atomic models theory

Bohr’s Atom
electrons in orbits
nucleus
Atomic Models
Atomic Models
Prof. Walid Tawfik
Prof. Walid Tawfik

2
2
INTRODUCTION
INTRODUCTION
 The purpose of this chapter is to build a simplest
The purpose of this chapter is to build a simplest
atomic model that will help us to understand the
atomic model that will help us to understand the
structure of atoms
structure of atoms
 This is attained by referring to some basic
This is attained by referring to some basic
experimental facts that have been gathered since
experimental facts that have been gathered since
1900’s (e.g. Rutherford scattering experiment,
1900’s (e.g. Rutherford scattering experiment,
atomic spectral lines etc.)
atomic spectral lines etc.)
 In order to build a model that well describes the
In order to build a model that well describes the
atoms which are consistent with the experimental
atoms which are consistent with the experimental
facts, we need to take into account the wave
facts, we need to take into account the wave
nature of electron
nature of electron
 This is one of the purpose we explore the wave
This is one of the purpose we explore the wave
nature of particles in previous chapters
nature of particles in previous chapters

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3
Basic properties of atoms
Basic properties of atoms
 1)
1) Atoms are of microscopic size, ~ 10
Atoms are of microscopic size, ~ 10-10
-10
m. Visible
m. Visible
light is not enough to resolve (see) the detail structure of
light is not enough to resolve (see) the detail structure of
an atom as its wavelength is only of the order of 100 nm.
an atom as its wavelength is only of the order of 100 nm.
 2)
2) Atoms are stable
Atoms are stable
 3)
3) Atoms contain negatively charges, electrons, but
Atoms contain negatively charges, electrons, but
are electrically neutral. An atom with
are electrically neutral. An atom with Z
Z electrons must also
electrons must also
contain a net positive charge of +
contain a net positive charge of +Ze
Ze.
.
 4)
4) Atoms emit and absorb EM radiation (in other
Atoms emit and absorb EM radiation (in other
words, atoms interact with light quite readily)
words, atoms interact with light quite readily)
Because atoms interacts with EM radiation quite strongly,
Because atoms interacts with EM radiation quite strongly,
it is usually used to probe the structure of an atom. The
it is usually used to probe the structure of an atom. The
typical of such EM probe can be found in the atomic
typical of such EM probe can be found in the atomic
spectrum as we will see now
spectrum as we will see now

HELIUM ATOM
+
N
N
+
-
-
proton
electron neutron
Shell
What do these particles consist of?

ATOMIC STRUCTURE
ATOMIC STRUCTURE
the number of protons in an atom
the number of protons and
neutrons in an atom
He
He
2
2
4
4 Atomic mass
Atomic number
number of electrons = number of protons

ATOMIC STRUCTURE
ATOMIC STRUCTURE
Electrons are arranged in Energy Levels or
Shells around the nucleus of an atom.
• first shell a maximum of 2 electrons
• second shell a maximum of 8 electrons
• third shell a maximum of 8 electrons

ATOMIC STRUCTURE
ATOMIC STRUCTURE
Particle
proton
neutron
electron
Charge
+ ve charge
-ve charge
No charge
1.6726219 ×
10-27
kilograms
1.6726219 ×
10-27
kilograms
9.10938356 ×
10-31
kilograms
Mass

ATOMIC STRUCTURE
ATOMIC STRUCTURE
There are two ways to represent the atomic
structure of an element or compound;
1. Electronic Configuration
2. Dot & Cross Diagrams

ELECTRONIC CONFIGURATION
With electronic configuration elements are represented
numerically by the number of electrons in their shells
and number of shells. For example;
N
Nitrogen
7
14
2 in 1st
shell
5 in 2nd
shell
configuration = 2 , 5
2 + 5 = 7

Write the electronic configuration for the following
elements;
Ca O
Cl Si
Na
20
40
11
23
8
17
16
35
14
28
B 11
5
a) b) c)
d) e) f)
2,8,8,2 2,8,1
2,8,7 2,8,4 2,3
2,6

DOT & CROSS DIAGRAMS
With Dot & Cross diagrams elements and compounds
are represented by Dots or Crosses to show electrons,
and circles to show the shells. For example;
Nitrogen N X
X X
X
X
X
X
N
7
14

Draw the Dot & Cross diagrams for the following
elements;
O Cl
8 17
16 35
a) b)
O
X
X
X
X
X
X
X
X
Cl
X
X
X
X X
X
X
X
X
X
X
X
X
X
X
X
X
X

 There are 3 types of subatomic particles:
Electrons (e–) Protons (p+) and Neutrons
(n0).
Neutrons have no charge and a mass similar
to protons
Elements are often symbolized with their
mass number and atomic number
E.g. Oxygen: O
16
8
These values are given on the periodic table.
For now, round the mass # to a whole number.
These numbers tell you a lot about atoms.
# of protons = # of electrons = atomic number
# of neutrons = mass number – atomic number

Atomic Information
1. The Atomic Number of an atom
= number of protons in the nucleus.
2. The Atomic Mass of an atom
= number of Protons + Neutrons in the nucleus
3. The number of Protons
= Number of Electrons.
4. Electrons orbit the nucleus in shells.
5. Each shell can only carry a set number of
electrons.

Isotopes
• Atoms with the same number of protons but
different numbers of neutrons
• Another way to say – atoms of the same element
with different numbers of neutrons
• An elements mass number is the number of
protons plus the number of neutrons
• REVIEW: to determine the number of neutrons
subtract the atomic number from the mass
number
• Mass # – atomic # = # of neutrons

Isotopic symbols
NOTE: Unlike on the periodic
table where the atomic number
is at the top of the box and the
average atomic mass is at the
bottom in isotopic symbols the
mass number is at the top and
the atomic number is at the
bottom.
If X is Hydrogen and its mass number is 2 then the isotopic symbol
would be
2
H
1

Isotopic symbols
If X is Fluorine and its mass
number is 20 then the correct
isotopic symbol is
Fluorine – 20.

If the atom has a charge the charge is written to
the upper right side of the symbol. If the charge
is either negative or positive 1 the 1 is not
written but understood to be 1.

Determine the number of neutron in each isotope
1. 238 – 92 = 146 neutrons
2. 84 – 36 = 48 neutrons
3. 35 – 17 = 18 neutrons
4. 14 – 6 = 8 neutrons

e–
Mass
Atomic
n0
p+
Review
Ca 20 40 20 20 20
Ar 18 40 18 22 18
Br 35 80 35 45 35

SUMMARY
SUMMARY
1. The Atomic Number of an atom = number of
protons in the nucleus.
2. The Atomic Mass of an atom = number of
Protons + Neutrons in the nucleus.
3. The number of Protons = Number of Electrons.
4. Electrons orbit the nucleus in shells.
5. Each shell can only carry a set number of electrons.

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Emission spectral lines
Emission spectral lines
 Experimental fact:
Experimental fact: A single atom or molecule in a very
A single atom or molecule in a very
diluted sample of gas emits radiation characteristic of the
diluted sample of gas emits radiation characteristic of the
particular atom/molecule species
particular atom/molecule species
 The emission is due to the de-excitation of the atoms
The emission is due to the de-excitation of the atoms
from their excited states
from their excited states
 e.g. if heating or passing electric current through the gas
e.g. if heating or passing electric current through the gas
sample, the atoms get excited into higher energy states
sample, the atoms get excited into higher energy states
 When a excited electron in the atom falls back to the
When a excited electron in the atom falls back to the
lower energy states (de-excites), EM wave is emitted
lower energy states (de-excites), EM wave is emitted
 The spectral lines are analysed with
The spectral lines are analysed with spectrometer
spectrometer,
,
which give important physical information of the
which give important physical information of the
atom/molecules by analysing the wavelengths
atom/molecules by analysing the wavelengths
composition and pattern of these lines.
composition and pattern of these lines.

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Line spectrum of an atom
Line spectrum of an atom
 The light given off by individual atoms, as in a
The light given off by individual atoms, as in a
low-pressure gas, consist of a series of discrete
low-pressure gas, consist of a series of discrete
wavelengths corresponding to different colour.
wavelengths corresponding to different colour.

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27
Comparing continuous and line
Comparing continuous and line
spectrum
spectrum
 (a) continuous
(a) continuous
spectrum produced
spectrum produced
by a glowing light-
by a glowing light-
bulb
bulb
 (b) Emission line
(b) Emission line
spectrum by lamp
spectrum by lamp
containing heated gas
containing heated gas

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Absorption line spectrum
Absorption line spectrum
 We also have absorption spectral line, in
We also have absorption spectral line, in
which white light is passed through a gas.
which white light is passed through a gas.
The absorption line spectrum consists of a
The absorption line spectrum consists of a
bright background crossed by dark lines
bright background crossed by dark lines
that correspond to the absorbed
that correspond to the absorbed
wavelengths by the gas atom/molecules.
wavelengths by the gas atom/molecules.

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29
Experimental arrangement for the
Experimental arrangement for the
observation of the absorptions lines
observation of the absorptions lines
of a sodium vapour
of a sodium vapour

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Comparing emission and
Comparing emission and
absorption spectrum
absorption spectrum
The emitted and absorption radiation displays
The emitted and absorption radiation displays
characteristic discrete sets of spectrum which
characteristic discrete sets of spectrum which
contains certain discrete wavelengths only
contains certain discrete wavelengths only
(a) shows ‘finger print’ emission spectral lines of H,
(a) shows ‘finger print’ emission spectral lines of H,
Hg and Ne. (b) shows absorption lines for H
Hg and Ne. (b) shows absorption lines for H

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A successful atomic model must be
A successful atomic model must be
able to explain the observed
able to explain the observed
discrete atomic spectrum
discrete atomic spectrum
We are going to study two attempts
We are going to study two attempts
to built model that describes the
to built model that describes the
atoms: the Thompson Plum-
atoms: the Thompson Plum-
pudding model (which fails) and the
pudding model (which fails) and the
Rutherford-Bohr model (which
Rutherford-Bohr model (which
succeeds)
succeeds)

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The Thompson model – Plum-
The Thompson model – Plum-
pudding model
pudding model
Sir J. J. Thompson (1856-
1940) is the Cavandish
professor in Cambridge who
discovered electron in cathode
rays. He was awarded Nobel
prize in 1906 for his research
on the conduction of electricity
by bases at low pressure.
He is the first person to
He is the first person to
establish the particle nature of
establish the particle nature of
electron. Ironically his son,
electron. Ironically his son,
another renown physicist
another renown physicist
proves experimentally electron
proves experimentally electron
behaves like wave…
behaves like wave…

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Plum-pudding model
Plum-pudding model
 An atom consists of
An atom consists of Z
Z electrons is embedded in
electrons is embedded in
a cloud of positive charges that exactly
a cloud of positive charges that exactly
neutralise that of the electrons’
neutralise that of the electrons’
 The positive cloud is heavy and comprising most
The positive cloud is heavy and comprising most
of the atom’s mass
of the atom’s mass
 Inside a stable atom, the electrons sit at their
Inside a stable atom, the electrons sit at their
respective equilibrium position where the
respective equilibrium position where the
attraction of the positive cloud on the electrons
attraction of the positive cloud on the electrons
balances the electron’s mutual repulsion
balances the electron’s mutual repulsion

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34
One can treat the
One can treat the
electron in the
electron in the
pudding like a point
pudding like a point
mass stressed by
mass stressed by
two springs
two springs
SHM

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35
The “electron plum” stuck on the
The “electron plum” stuck on the
pudding vibrates and executes
pudding vibrates and executes
SHM
SHM
 The electron at the EQ position shall vibrate like a
The electron at the EQ position shall vibrate like a
simple harmonic oscillator with a frequency
simple harmonic oscillator with a frequency
 Where ,
Where , R
R radius of the atom,
radius of the atom, m
m mass of
mass of
the electron
the electron
 From classical EM theory, we know that an
From classical EM theory, we know that an
oscillating charge will emit radiation with frequency
oscillating charge will emit radiation with frequency
identical to the oscillation frequency
identical to the oscillation frequency 
 as given
as given
above
above
m
k









2
1
3
2
4 R
Ze
k
o



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36
The plum-pudding model predicts
The plum-pudding model predicts
unique oscillation frequency
unique oscillation frequency
 Radiation with frequency identical to the
Radiation with frequency identical to the
oscillation frequency.
oscillation frequency.
 Hence light emitted from the atom in the plum-
Hence light emitted from the atom in the plum-
pudding model is predicted to have exactly
pudding model is predicted to have exactly one
one
unique
unique frequency as given in the previous
frequency as given in the previous
slide.
slide.
 This prediction has been falsified because
This prediction has been falsified because
observationally, light spectra from all atoms
observationally, light spectra from all atoms
(such as the simplest atom, hydrogen,) have
(such as the simplest atom, hydrogen,) have
sets of discrete spectral lines correspond to
sets of discrete spectral lines correspond to
many different frequencies (already discussed
many different frequencies (already discussed
earlier).
earlier).

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Experimental verdict on the plum
Experimental verdict on the plum
pudding model
pudding model
 Theoretically one expect the deviation angle of a scattered
Theoretically one expect the deviation angle of a scattered
particle by the plum-pudding atom to be small:
particle by the plum-pudding atom to be small:
 This is a prediction of the model that can be checked
This is a prediction of the model that can be checked
experimentally
experimentally
 Rutherford was the first one to carry out such experiment
Rutherford was the first one to carry out such experiment

1
~
ave
N



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38
Ernest Rutherford
Ernest Rutherford
British physicist Ernest Rutherford, winner of the
1908 Nobel Prize in chemistry, pioneered the field
of nuclear physics with his research and
development of the nuclear theory of atomic
structure
Born in New Zealand, teachers to many physicists
who later become Nobel prize laureates
Rutherford stated that an atom consists largely of
empty space, with an electrically positive nucleus
in the center and electrically negative electrons
orbiting the nucleus. By bombarding nitrogen gas
with alpha particles (nuclear particles emitted
through radioactivity), Rutherford engineered the
transformation of an atom of nitrogen into both
an atom of oxygen and an atom of hydrogen.
This experiment was an early stimulus to the
development of nuclear energy, a form of energy
in which nuclear transformation and
disintegration release extraordinary power.

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Rutherford experimental setup
Rutherford experimental setup
 Alpha particles from
Alpha particles from
source is used to be
source is used to be
scattered by atoms
scattered by atoms
from the thin foil
from the thin foil
made of gold
made of gold
 The scattered alpha
The scattered alpha
particles are detected
particles are detected
by the background
by the background
screen
screen

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41
“…
“…fire a 15 inch artillery shell at a
fire a 15 inch artillery shell at a
tissue paper and it came back and
tissue paper and it came back and
hit you”
hit you”
 In the scattering experiment Rutherford
In the scattering experiment Rutherford
saw some electrons being bounced back
saw some electrons being bounced back
at 180 degree.
at 180 degree.
 He said this is like firing “a 15-inch shell at
He said this is like firing “a 15-inch shell at
a piece of a tissue paper and it came back
a piece of a tissue paper and it came back
and hit you”
and hit you”
 Hence Thompson plum-pudding model
fails in the light of these experimental
result

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So, is the plum pudding model
So, is the plum pudding model
utterly useless?
utterly useless?
 So the plum pudding model does not work as its
So the plum pudding model does not work as its
predictions fail to fit the experimental data as well as
predictions fail to fit the experimental data as well as
other observations
other observations
 Nevertheless it’s a perfectly sensible scientific theory
Nevertheless it’s a perfectly sensible scientific theory
because:
because:
 It is a mathematical model built on sound and rigorous
It is a mathematical model built on sound and rigorous
physical arguments
physical arguments
 It predicts some physical phenomenon with definiteness
It predicts some physical phenomenon with definiteness
 It can be verified or falsified with experiments
It can be verified or falsified with experiments
 It also serves as a prototype to the next model which is
It also serves as a prototype to the next model which is
built on the experience gained from the failure of this
built on the experience gained from the failure of this
model
model

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How to interpret the Rutherford
How to interpret the Rutherford
scattering experiment?
scattering experiment?
 The large deflection of
The large deflection of
alpha particle as seen in
alpha particle as seen in
the scattering experiment
the scattering experiment
with a thin gold foil must
with a thin gold foil must
be produced by a close
be produced by a close
encounter between the
encounter between the
alpha particle and a very
alpha particle and a very
small but massive kernel
small but massive kernel
inside the atom
inside the atom
 In contrast, a diffused
In contrast, a diffused
distribution of the positive
distribution of the positive
charge as assumed in
charge as assumed in
plum-pudding model
plum-pudding model
cannot do the job
cannot do the job

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44
Comparing model with nucleus
Comparing model with nucleus
concentrated at a point-like nucleus
concentrated at a point-like nucleus
and model with nucleus that has
and model with nucleus that has
large size
large size

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45
Recap
Recap
the atomic model building story
the atomic model building story
 Plum-pudding model by
Thompson
 It fails to explain the
emission and absorption
line spectrum from atoms
because it predicts only a
single emission frequency
 Most importantly it fails to
explain the back-scattering
of alpha particle seen in
Rutherford’s scattering
experiment because the
model predicts only 
1
~
ave
N


m
k









2
1

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46
 Rutherford put forward an
Rutherford put forward an
model to explain the result
model to explain the result
of the scattering
of the scattering
experiment: the Rutherford
experiment: the Rutherford
model
model
 An atom consists of a very
An atom consists of a very
small nucleus of charge +
small nucleus of charge +Ze
Ze
containing almost all of the
containing almost all of the
mass of the atom; this
mass of the atom; this
nucleus is surrounded by a
nucleus is surrounded by a
swarm of
swarm of Z
Z electrons
electrons
 The atom is largely
The atom is largely
comprised of empty space
comprised of empty space
 R
Ratom
atom ~
~ 10
10-10
-10
m
m
 R
Rnucleus
nucleus ~
~ 10
10-13
-13
- 10
- 10-15
-15
m
m
The Rutherford model
The Rutherford model
(planetary model)
(planetary model)

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Infrared catastrophe: insufficiency
Infrared catastrophe: insufficiency
of the Rutherford model
of the Rutherford model
 According to classical
According to classical
EM, the Rutherford model
EM, the Rutherford model
for atom (a classical
for atom (a classical
model) has a fatal flaw: it
model) has a fatal flaw: it
predicts the collapse of
predicts the collapse of
the atom within 10
the atom within 10-10
-10
s
s
 A accelerated electron
A accelerated electron
will radiate EM radiation,
will radiate EM radiation,
hence causing the
hence causing the
orbiting electron to loss
orbiting electron to loss
energy and consequently
energy and consequently
spiral inward and impact
spiral inward and impact
on the nucleus
on the nucleus

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49
Rutherford model also can’t explain
Rutherford model also can’t explain
the discrete spectrum
the discrete spectrum
 The Rutherford model also cannot
The Rutherford model also cannot
explain the pattern of discrete
explain the pattern of discrete
spectral lines as the radiation
spectral lines as the radiation
predicted by Rutherford model is a
predicted by Rutherford model is a
continuous burst.
continuous burst.

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50
So how to fix up the problem?
So how to fix up the problem?
NEILS BOHR COMES TO
NEILS BOHR COMES TO
THE RESCUE
THE RESCUE
 Niels Bohr
Niels Bohr (1885 to
(1885 to
1962) is best known for
1962) is best known for
the investigations of
the investigations of
atomic structure and also
atomic structure and also
for work on radiation,
for work on radiation,
which won him the 1922
which won him the 1922
Nobel Prize for physics
Nobel Prize for physics
 He was sometimes
He was sometimes
dubbed “the God Father”
dubbed “the God Father”
in the physicist
in the physicist
community
community
 http://www-gap.dcs.st-
http://www-gap.dcs.st-
and.ac.uk/~history/
and.ac.uk/~history/
Mathematicians/Bohr_Niels.html
Mathematicians/Bohr_Niels.html

51
51
To fix up the infrared catastrophe …
To fix up the infrared catastrophe …
Neils Bohr put forward a model which is
Neils Bohr put forward a model which is
a
a hybrid of the Rutherford model with the
hybrid of the Rutherford model with the
wave nature of electron taken into
wave nature of electron taken into
account
account

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52
Bohr’s model of hydrogen-like atom
Bohr’s model of hydrogen-like atom
 We shall consider a simple atom
We shall consider a simple atom
consists of a nucleus with charge
consists of a nucleus with charge
Ze
Ze and mass of
and mass of M
Mnucleus
nucleus >> m
>> me
e
 The nucleus is surrounded by
The nucleus is surrounded by
only
only a single electron
a single electron
 We will assume the centre of the
We will assume the centre of the
circular motion of the electron
circular motion of the electron
coincides with the centre of the
coincides with the centre of the
nucleus
nucleus
 We term such type of simple
We term such type of simple
system: hydrogen-like atoms
system: hydrogen-like atoms
 For example, hydrogen atom
For example, hydrogen atom
corresponds to
corresponds to Z =
Z = 1; a singly
1; a singly
ionised Helium atom He
ionised Helium atom He+
+
corresponds to
corresponds to Z =
Z = 2 etc
2 etc
Diagram representing
the model of a
hydrogen-like atom
+Ze
M >>m

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53
Bohr’s postulate, 1913
Bohr’s postulate, 1913
 
r
v
m
r
e
Ze e
2
2
0
4
1


Coulomb’s attraction = centripetal force
Assumption: the mass of the
nucleus is infinitely heavy
compared to the electron’s
 Postulate No.1:
Postulate No.1: Mechanical
Mechanical
stability (classical mechanics)
stability (classical mechanics)
 An electron in an atom
An electron in an atom
moves in a circular orbit
moves in a circular orbit
about the nucleus under
about the nucleus under
Coulomb attraction obeying
Coulomb attraction obeying
the law of classical
the law of classical
mechanics
mechanics

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54
Postulate 2: condition for orbit
Postulate 2: condition for orbit
stability
stability
 Instead of the infinite orbit which could
Instead of the infinite orbit which could
be possible in classical mechanics (c.f
be possible in classical mechanics (c.f
the orbits of satellites), it is only possible
the orbits of satellites), it is only possible
for an electron to move in an orbit that
for an electron to move in an orbit that
contains an integral number of de Broglie
contains an integral number of de Broglie
wavelengths,
wavelengths,
 n
n
n
n =
= 2
2
r
rn
n,
, n
n = 1,2,3...
= 1,2,3...

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55
Bohr’s 2
Bohr’s 2nd
nd
postulate means that
postulate means that n
n
de Broglie wavelengths must fit into
de Broglie wavelengths must fit into
the circumference of an orbit
the circumference of an orbit

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56
Electron that don’t form standing
Electron that don’t form standing
wave
wave
 Since the electron must form
Since the electron must form
standing waves in the orbits,
standing waves in the orbits,
the the orbits of the electron
the the orbits of the electron
for each
for each n
n is quantised
is quantised
 Orbits with the perimeter that
Orbits with the perimeter that
do not conform to the
do not conform to the
quantisation condition cannot
quantisation condition cannot
persist
persist
 All this simply means: all
All this simply means: all
orbits of the electron in the
orbits of the electron in the
atom must be quantised, and
atom must be quantised, and
orbit that is not quantised is
orbit that is not quantised is
not allowed (hence can’t
not allowed (hence can’t
exist)
exist)

57
57
Quantisation of angular momentum
Quantisation of angular momentum
 As a result of the orbit
As a result of the orbit
quantisation, the angular
quantisation, the angular
momentum of the orbiting
momentum of the orbiting
electron is also
electron is also
quantised:
quantised:
 L =
L = (
(m
me
ev
v)
) r = pr
r = pr
(definition)
(definition)
 n
n
=
= 2
2
 r
r (orbit
(orbit
quantisation)
quantisation)
 Combining both:
Combining both:
 p= h/
p= h/
 = nh/
= nh/ 2
2
 r
r
 L = m
L = me
evr = p r = nh/
vr = p r = nh/ 2
2

p = mv
Angular momentum of the electron,
L = p x r. It is a vector quantity with
its direction pointing to the
direction perpendicular to the
plane defined by p and r

58
58
Third postulate
Third postulate
 Despite the fact that it is constantly
Despite the fact that it is constantly
accelerating, an electron moving in such an
accelerating, an electron moving in such an
allowed orbit does not radiate EM energy
allowed orbit does not radiate EM energy
(hence total energy remains constant)
(hence total energy remains constant)
 As far as the stability of atoms is concerned,
As far as the stability of atoms is concerned,
classical physics is invalid here
classical physics is invalid here
 My Comment: At the quantum scale (inside the
My Comment: At the quantum scale (inside the
atoms) some of the classical EM predictions
atoms) some of the classical EM predictions
fail (e.g. an accelerating charge radiates EM
fail (e.g. an accelerating charge radiates EM
wave)
wave)

59
59
Quantisation of velocity and radius
Quantisation of velocity and radius
 Combining the quantisation of angular
Combining the quantisation of angular
momentum and the equation of
momentum and the equation of
mechanical stability we arrive at the result
mechanical stability we arrive at the result
that:
that:
 the allowed radius and velocity at a given
the allowed radius and velocity at a given
orbit are also quantised:
orbit are also quantised:
2
2
2
0
4
Ze
m
n
r
e
n




n
Ze
vn
2
0
4
1



60
60
Some mathematical steps leading
Some mathematical steps leading
to quantisation of orbits,
to quantisation of orbits, 2
2
2
0
4
Ze
m
n
r
e
n



  2 2
2
2
0 0
1 1
(Eq. 2)
4 4
e
e
Ze e m v Ze
v
r r m r
 
  
(Eq.2)
(Eq.2) 
 (Eq.1)
(Eq.1)2
2
,
,
(
(m
me
evr
vr)
)2
2
=
= (
(nh/
nh/ 2
2
)
)2
2
LHS:
LHS: m
me
e
2
2
r
r2
2
v
v2
2
=
= m
me
e
2
2
r
r2
2
(
(Ze
Ze2
2
/ 4
/ 4

m
me
e r
r)
)
=
= m
me
e r Ze
r Ze2
2
/ 4
/ 4

=
= RHS
RHS =
= (
(nh/
nh/ 2
2
)
)2
2
r = n
r = n2
2
(
(h/
h/ 2
2
)
)2
2
4
4

 /
/ Ze
Ze2
2
m
me
e ≡
≡ r
rn
n ,
,
n
n = 1,2,3…
= 1,2,3… 


(Eq.1)
2
e
nh
m vr



61
61
Prove it yourself the quantisation of
Prove it yourself the quantisation of
the electron velocity
the electron velocity
using Eq.(1) and Eq.(2)
using Eq.(1) and Eq.(2)

n
Ze
vn
2
0
4
1



62
62
Important comments
Important comments
 The smallest orbit charaterised by
The smallest orbit charaterised by
 Z
Z = 1,
= 1, n
n=1 is the ground state orbit of the hydrogen
=1 is the ground state orbit of the hydrogen
 It’s called the Bohr’s radius = the typical size of an atom
It’s called the Bohr’s radius = the typical size of an atom
 In general, the radius of an hydrogen-like ion/atom with
In general, the radius of an hydrogen-like ion/atom with
charge
charge Ze
Ze in the nucleus is expressed in terms of the
in the nucleus is expressed in terms of the
Bohr’s radius as
Bohr’s radius as
 Note also that the ground state velocity of the electron in
Note also that the ground state velocity of the electron in
the hydrogen atom is m/s <<
the hydrogen atom is m/s << c
c
 non-relativistic
non-relativistic
0
2
2
0
0 5
.
0
4
A
e
m
r
e




Z
r
n
rn
0
2

6
0 10
2
.
2 

v

63
63
PYQ 7 Test II 2003/04
 In Bohr’s model for hydrogen-like atoms, an electron
In Bohr’s model for hydrogen-like atoms, an electron
(mass
(mass m
m) revolves in a circle around a nucleus with
) revolves in a circle around a nucleus with
positive charges
positive charges Ze
Ze. How is the electron’s velocity
. How is the electron’s velocity
related to the radius
related to the radius r
r of its orbit?
of its orbit?
 A.
A. B.
B. C.
C.
 D.
D. E.
E. Non of the above
Non of the above
 Solution: I expect your to be able to derive it from scratch without
Solution: I expect your to be able to derive it from scratch without
memorisation
memorisation
 ANS: D, Schaum’s series 3000 solved problems, Q39.13, pg 722
ANS: D, Schaum’s series 3000 solved problems, Q39.13, pg 722
modified
modified
mr
Ze
v
2
0
4
1

 2
2
0
4
1
mr
Ze
v

 2
0
4
1
mr
Ze
v


mr
Ze
v
2
0
2
4
1



64
64
The quantised orbits of hydrogen-
The quantised orbits of hydrogen-
like atom (not to scale)
like atom (not to scale)
+Ze
r0
r2
r3
r4
Z
r
n
rn
0
2


65
65
Strongly recommending the
Strongly recommending the
Physics 2000 interactive physics
Physics 2000 interactive physics
webpage by the University of
webpage by the University of
Colorado
Colorado
For example the page
http://www.colorado.edu/physics/2000/quantu
mzone/bohr.html
provides a very interesting explanation and
simulation on atom and Bohr model in
particular.
Please visit this page if you go online

66
66
Recap
Recap
 The hydrogen-like atom’s radii are quantised
The hydrogen-like atom’s radii are quantised
according to:
according to:
 The quantisation is a direct consequence of the
The quantisation is a direct consequence of the
postulate that electron wave forms stationary states
postulate that electron wave forms stationary states
(standing waves) at the allowed orbits
(standing waves) at the allowed orbits
 The smallest orbit or hydrogen, the Bohr’s radius
The smallest orbit or hydrogen, the Bohr’s radius
0
2
2
0
0 5
.
0
4
A
e
m
r
e




Z
r
n
rn
0
2

+Ze

67
67
Postulate 4
Postulate 4
 Similar to Einstein’s
Similar to Einstein’s
postulate of the energy of a
postulate of the energy of a
photon
photon
EM radiation is emitted if
an electron initially moving
in an orbit of total energy
Ei
, discontinuously changes
it motion so that it moves in
an orbit of total energy Ef
.
The frequency of the
emitted radiation,
  = (Ef
- Ei
)/h

68
68
Energies in the hydrogen-like atom
Energies in the hydrogen-like atom
 Potential energy
Potential energy of the electron at a distance
of the electron at a distance r
r from
from
the nucleus is, as we learned from standard
the nucleus is, as we learned from standard
electrostatics, ZCT 102, form 6, matriculation etc. is
electrostatics, ZCT 102, form 6, matriculation etc. is
simply
simply






r r
Ze
dr
r
Ze
V
0
2
2
0
2
4
4 

 -ve means that the EM force is attractive
-ve means that the EM force is attractive
Fe
-e
+Ze

69
69
Kinetic energy in the hydrogen-like
Kinetic energy in the hydrogen-like
atom
atom
 According to definition, the KE of the
According to definition, the KE of the
electron is
electron is
r
Ze
v
m
K e
0
2
2
8
2 


 Adding up KE + V, we obtain the total mechanical energy of
Adding up KE + V, we obtain the total mechanical energy of
the atom:
the atom:
The last step follows from the equation 2
0
2
2
4 r
Ze
r
v
me


  n
e
e
E
n
e
Z
m
n
Ze
m
Ze
r
Ze
r
Ze
r
Ze
V
K
E
































2
2
2
0
4
2
2
2
0
2
0
2
0
2
0
2
0
2
1
2
4
4
8
1
8
4
8









70
70
The ground state energy
The ground state energy
 For the hydrogen atom (
For the hydrogen atom (Z
Z = 1), the ground
= 1), the ground
state energy (which is characterised by
state energy (which is characterised by n
n =1)
=1)
 
eV
6
.
13
2
4
)
1
( 2
2
0
4
0 







e
m
n
E
E e
n
In general the energy level of a hydrogen
like atom with Ze nucleus charges can be
expressed in terms of
eV
6
.
13
2
2
2
0
2
n
Z
n
E
Z
En 



71
71
Quantization of H energy levels
Quantization of H energy levels
 The energy level of the
The energy level of the
electrons in the atomic
electrons in the atomic
orbit is quantised
orbit is quantised
 The quantum number,
The quantum number, n
n,
,
that characterises the
that characterises the
electronic states is called
electronic states is called
principle quantum
principle quantum
number
number
 Note that the energy state
Note that the energy state
is –ve (because it’s a
is –ve (because it’s a
bounded system)
bounded system)
eV
6
.
13
2
2
0
n
n
E
En 



72
72
Energy of the electron at very large
Energy of the electron at very large n
n
 An electron occupying an orbit
An electron occupying an orbit
with very large
with very large n
n is “almost free”
is “almost free”
because its energy approaches
because its energy approaches
zero:
zero:
 E
E = 0 means the electron is free
= 0 means the electron is free
from the bondage of the nucleus’
from the bondage of the nucleus’
potential field
potential field
 Electron at high
Electron at high n
n is not tightly
is not tightly
bounded to the nucleus by the
bounded to the nucleus by the
EM force
EM force
 Energy levels at high
Energy levels at high n
n
approaches to that of a
approaches to that of a
continuum, as the energy gap
continuum, as the energy gap
between adjacent energy levels
between adjacent energy levels
become infinitesimal in the large
become infinitesimal in the large
n
n limit
limit
  0



n
En

73
73
Ionisation energy of the hydrogen
Ionisation energy of the hydrogen
atom
atom
 The energy input required to remove the
The energy input required to remove the
electron from its ground state to infinity (ie.
electron from its ground state to infinity (ie.
to totally remove the electron from the
to totally remove the electron from the
bound of the nucleus) is simply
bound of the nucleus) is simply
 this is the ionisation energy of hydrogen
this is the ionisation energy of hydrogen
eV
6
.
13
0
0 



  E
E
E
Eionisation
Fe
-e
+Ze
Ionisation energy to pull
the electron off from the
attraction of the +ve
nucleus
+Ze
-e
Free electron (= free from the attraction of
the +ve nuclear charge, E= 0)
E=E0=-13.6 eV
+

74
74
Two important quantities to
Two important quantities to
remember
remember
 As a practical rule, it is strongly advisable
As a practical rule, it is strongly advisable
to remember the two very important values
to remember the two very important values
 (i) the Bohr radius,
(i) the Bohr radius, r
r0
0 = 0.53A and
= 0.53A and
 (ii) the ground state energy of the
(ii) the ground state energy of the
hydrogen atom,
hydrogen atom, E
E0
0 = -13.6 eV
= -13.6 eV

75
75
Bohr’s 4th postulate explains the
line spectrum of Hydrogen series
line spectrum of Hydrogen series

76
76
line spectrum
line spectrum
 When atoms are excited to an
When atoms are excited to an
energy state above its ground state,
energy state above its ground state,
they shall radiate out energy (in
they shall radiate out energy (in
forms of photon) within at the time
forms of photon) within at the time
scale of ~10
scale of ~10-8
-8
s upon their de-
s upon their de-
excitations to lower energy states –
excitations to lower energy states –
emission spectrum explained
emission spectrum explained
 When a beam of light with a range of
When a beam of light with a range of
wavelength from sees an atom, the few
wavelength from sees an atom, the few
particular wavelengths that matches the
particular wavelengths that matches the
allowed energy gaps of the atom will be
allowed energy gaps of the atom will be
absorbed, leaving behind other unabsorbed
absorbed, leaving behind other unabsorbed
wavelengthsto become the bright
wavelengthsto become the bright
background in the absorption spectrum.
background in the absorption spectrum.
Hence absorption spectrum explained
Hence absorption spectrum explained

77
77
Balmer series and the empirical
Balmer series and the empirical
emission spectrum equation
emission spectrum equation
 Since 1860 – 1898 Balmer have found an
Since 1860 – 1898 Balmer have found an
empirical
empirical formula that correctly predicted the
formula that correctly predicted the
wavelength of four
wavelength of four visible lines
visible lines of hydrogen:
of hydrogen:
where n = 3,4,5,….RH is called the Rydberg constant,
experimentally measured to be RH = 1.0973732 x 107
m-1
H
H

H
H

H
H

H
H










 2
2
1
2
1
1
n
RH

n = 6
n = 5 n = 4
n = 3
H
H

n  


78
78
Example
Example
 For example, for the H (486.1 nm) line, n = 4 in the
empirical formula
 According to the empirical formula the wavelength of
the hydrogen beta line is
 which is consistent with the observed value









 2
2
1
2
1
1
n
RH

nm
486
16
)
m
10
1.0973732
(
3
16
3
4
1
2
1
1 1
-
7
2
2

























H
H R
R

79
79
Other spectra series
Other spectra series
 Apart from the Balmer series others
Apart from the Balmer series others
spectral series are also discovered:
spectral series are also discovered:
Lyman, Paschen and Brackett series
Lyman, Paschen and Brackett series
 The wavelengths of these spectral lines
The wavelengths of these spectral lines
are also given by the similar empirical
are also given by the similar empirical
equation as
equation as
,...
7
,
6
,
5
,
1
4
1
1
,...
6
,
5
,
4
,
1
3
1
1
,...
4
,
3
,
2
,
1
1
1
1
2
2
2
2
2

































n
n
R
n
n
R
n
n
R
H
H
H



Lyman series, ultraviolet region
Paschen series, infrared region
Brackett series, infrared region

80
80
These are experimentally measured
These are experimentally measured
spectral line
spectral line
...
4
,
3
,
2
,
1
1
1
1
2










 n
n
RH

...
6
,
5
,
4
,
1
3
1
1
2
2










 n
n
RH

,
6
,
5
,
4
,
3
,
1
2
1
1
2
2










 n
n
RH


81
81
2 2
1 1 1
For Lyman series, 1, 2,3,4,...
For Balmer series, 2, 3,4,5...
For Paschen series, 3, 4,5,6...
For Brackett series, 4, 5,6,7...
For Pfund series, 5,
H
f i
f i
f i
f i
f i
f i
R
n n
n n
n n
n n
n n
n n

 
 
 
 
 
 
 
 
 
 6,7,8...


82
82
The empirical formula needs a
The empirical formula needs a
theoretical explanation
theoretical explanation









 2
2
1
1
1
i
f
H
n
n
R

is an empirical formula with RH measured
to be RH = 1.0973732 x 107
m-1
.
Can the Bohr model provide a sound
theoretical explanation to the form of this
formula and the numerical value of RH in
terms of known physical constants?
The answer is: YES

83
83
Theoretical derivation of the
Theoretical derivation of the
empirical formula from Bohr’s
empirical formula from Bohr’s
model
model
 According to the 4
According to the 4th
th
postulate:
postulate:
 
E = E
E = Ei
i – E
– Ef
f = h
= h
 = h
= hc/
c/
, and
, and
 E
Ek
k = E
= E0
0 /
/ n
nk
k
2
2
 =
= -
-13.6 eV /
13.6 eV / n
nk
k
2
2
 where
where k
k =
= i
i or
or j
j
 Hence we can easily obtain
Hence we can easily obtain
the theoretical expression for
the theoretical expression for
the emission line spectrum of
the emission line spectrum of
hydrogen-like atom
hydrogen-like atom

 
0
2 2
4
2 2 2 2 2
3
0
1 1 1
1 1 1 1
4 4
i f
i f
e
f i f i
E E E
c ch ch n n
m e
R
n n n n
c


 

 

   
 
 
 
   
   
   
   
   


84
84
The theoretical Rydberg constant
The theoretical Rydberg constant
 The theoretical Rydberg constant,
The theoretical Rydberg constant, R
R∞
∞, agrees with the
, agrees with the
experimental one up a precision of less than 1%
experimental one up a precision of less than 1%
 
1
-
7
2
0
3
4
m
10
0984119
.
1
4
4






c
e
m
R e
-1
7
m
10
0973732
.
1 

H
R
This is a remarkable experimental verification of
the correctness of the Bohr model

85
85
Real life example of atomic
Real life example of atomic
emission
emission
 AURORA are caused
AURORA are caused
by streams of fast
by streams of fast
photons and electrons
photons and electrons
from the sun that
from the sun that
excite atoms in the
excite atoms in the
upper atmosphere.
upper atmosphere.
The green hues of an
The green hues of an
auroral display come
auroral display come
from oxygen
from oxygen

86
86
Example
Example
 Suppose that, as a result of a collision, the
Suppose that, as a result of a collision, the
electron in a hydrogen atom is raised to the
electron in a hydrogen atom is raised to the
second excited state (
second excited state (n
n = 3).
= 3).
 What is (i) the energy and (ii) wavelength of the
What is (i) the energy and (ii) wavelength of the
photon emitted if the electron makes a direct
photon emitted if the electron makes a direct
transition to the ground state?
transition to the ground state?
 What are the energies and the wavelengths of
What are the energies and the wavelengths of
the two photons emitted if, instead, the electron
the two photons emitted if, instead, the electron
makes a transition to the first excited state (
makes a transition to the first excited state (n
n=2)
=2)
and from there a subsequent transition to the
and from there a subsequent transition to the
ground state?
ground state?

87
87
n = 3
n = 2
n = 1, ground state
E = E3 - E1
E = E3 - E2
E = E2 - E1
The energy of the proton emitted in the
transition from the n = 3 to the n = 1 state is
12.1eV
eV
1
1
3
1
6
.
13 2
2
1
3 











 E
E
E
the wavelength of this photon is
nm
102
1
.
12
1242






eV
nm
eV
E
ch
c


Likewise the energies of the two photons
emitted in the transitions from n = 3  n = 2
and n = 2  n = 1 are, respectively,
eV
89
.
1
2
1
3
1
6
.
13 2
2
2
3 











 E
E
E with wavelength nm
657
89
.
1
1242





eV
nm
eV
E
ch

eV
2
.
10
1
1
2
1
6
.
13 2
2
1
2 











 E
E
E with wavelength nm
121
2
.
10
1242





eV
nm
eV
E
ch

Make use of
Make use of E
Ek
k = E
= E0
0 /
/ n
nk
k
2
2
=
= -
-13.6 eV /
13.6 eV / n
nk
k
2
2

88
88
Example
Example
 The series limit of the Paschen (
The series limit of the Paschen (n
nf
f = 3)
= 3) is
is
820.1 nm (The series limit of a spectral
820.1 nm (The series limit of a spectral
series is the wavelength corresponds to
series is the wavelength corresponds to
n
ni
i
∞
∞).
).
 What are two longest wavelengths of the
What are two longest wavelengths of the
Paschen series?
Paschen series?

89
89
Solution
Solution
 Note that the Rydberg constant is not provided
Note that the Rydberg constant is not provided
 But by definition the series limit and the Rydberg constant
But by definition the series limit and the Rydberg constant
is closely related
is closely related
 We got to make use of the series limit to solve that problem
We got to make use of the series limit to solve that problem
 By referring to the definition of the series limit,
By referring to the definition of the series limit,
2
2
2
1
1
1
1
f
H
n
i
f
H
n
R
n
n
R i


 
















 Hence we can substitute
Hence we can substitute R
RH
H =
= n
nf
f
2
2
/
/ 
∞
∞ into
into
 and express it in terms of the series limit as
and express it in terms of the series limit as
 n = 4,5,6…
n = 4,5,6…









 2
2
1
1
1
i
f
H
n
n
R












2
2
1
1
1
i
f
n
n



90
90
 For Paschen series,
For Paschen series, n
nf
f = 3,
= 3,

 = 820.1 nm
= 820.1 nm









 2
2
3
1
nm
1
.
820
1
1
i
n

 The two longest wavelengths
The two longest wavelengths
correspond to transitions of the two
correspond to transitions of the two
smallest energy gaps from the energy
smallest energy gaps from the energy
levels closest to
levels closest to n
n = 3 state (i.e the
= 3 state (i.e the n
n
=
= 4,
4, n =
n = 5 states) to the
5 states) to the n
n = 3 state
= 3 state
nm
1875
9
4
4
nm
1
.
820
9
nm
1
.
820
:
4 2
2
2
2






















i
i
i
n
n
n 
nm
1281
9
5
5
nm
1
.
820
9
nm
1
.
820
:
5 2
2
2
2






















i
i
i
n
n
n 
nf=3
ni=4
ni=5

91
91
Example
Example
 Given the ground state energy of hydrogen
Given the ground state energy of hydrogen
atom -13.6 eV, what is the longest
atom -13.6 eV, what is the longest
wavelength in the hydrogen’s Balmer series?
wavelength in the hydrogen’s Balmer series?
 Solution:
Solution:
 
E = E
E = Ei
i – E
– Ef
f =
= -13.6 eV
-13.6 eV (1/
(1/n
ni
i
2
2
- 1/
- 1/n
nf
f
2
2
) =
) = hc
hc/
/

 Balmer series:
Balmer series: n
nf
f = 2. Hence, in terms of 13.6
= 2. Hence, in terms of 13.6
eV the wavelengths in Balmer series is given
eV the wavelengths in Balmer series is given
by
by
...
5
,
4
,
3
,
1
4
1
1nm
9
1
4
1
eV
6
.
13
nm
eV
1240
1
4
1
eV
6
.
13 2
2
2































 i
i
i
i
Balmer n
n
n
n
hc


92
92
 longest wavelength corresponds to the transition
longest wavelength corresponds to the transition
from the
from the n
ni
i = 3 states to the
= 3 states to the n
nf
f = 2 states
= 2 states
 Hence
Hence
...
5
,
4
,
3
,
1
4
1
1nm
9
2










 i
i
Balmer n
n

nm
2
.
655
3
1
4
1
1nm
9
2
max
, 










Balmer

 This is the red
This is the red H
H
 line in the hydrogen’s Balmer
line in the hydrogen’s Balmer
series
series
 Can you calculate the shortest wavelength (the
Can you calculate the shortest wavelength (the
series limit) for the Balmer series? Ans = 364 nm
series limit) for the Balmer series? Ans = 364 nm

93
93
PYQ 2.18 Final Exam 2003/04
PYQ 2.18 Final Exam 2003/04
 Which of the following statements are true?
Which of the following statements are true?
 I.
I..
. the ground states are states with lowest energy
the ground states are states with lowest energy
 II.
II. ionisation energy is the energy required to raise an
ionisation energy is the energy required to raise an
electron from ground state to free state
electron from ground state to free state
 III
III.
. Balmer series is the lines in the spectrum of atomic
Balmer series is the lines in the spectrum of atomic
hydrogen that corresponds to the transitions to the
hydrogen that corresponds to the transitions to the n
n = 1
= 1
state from higher energy states
state from higher energy states
 A.
A. I,IV
I,IV B.
B. I,II, IV
I,II, IV C.
C. I, III,IV
I, III,IV
 D.
D. I, II
I, II E.
E. II,III
II,III
 ANS: D, My own question
ANS: D, My own question
 (note: this is an obvious typo error with the statement
(note: this is an obvious typo error with the statement
IV missing. In any case, only statement I, II are true.)
IV missing. In any case, only statement I, II are true.)

94
94
PYQ 1.15 KSCP 2003/04
PYQ 1.15 KSCP 2003/04
Which of the following statement(s) is (are) true?
Which of the following statement(s) is (are) true?
 I.
I. The
The experimental proof for which electron posses a
experimental proof for which electron posses a
wavelength was first verified by Davisson and
wavelength was first verified by Davisson and
Germer
Germer
 II.
II. The experimental proof of the existence of discrete
The experimental proof of the existence of discrete
energy levels in atoms involving their excitation by
energy levels in atoms involving their excitation by
collision with low-energy electron was confirmed in
collision with low-energy electron was confirmed in
the Frank-Hertz experiment
the Frank-Hertz experiment
 III.
III. Compton scattering experiment establishes that light
Compton scattering experiment establishes that light
behave like particles
behave like particles
 IV.
IV. Photoelectric experiment establishes that electrons
Photoelectric experiment establishes that electrons
behave like wave
behave like wave

 A.
A. I,II
I,II B.
B. I,II,III,IV
I,II,III,IV C. I, II, III
C. I, II, III
 D.
D. III,IV
III,IV E.
E. Non of the above
Non of the above
 Ans: C
Ans: C Serway and Moses, pg. 127 (for I), pg. 133
Serway and Moses, pg. 127 (for I), pg. 133
(for II), own options (for III,IV)
(for II), own options (for III,IV)

95
95
PYQ 1.5 KSCP 2003/04
PYQ 1.5 KSCP 2003/04
 An electron collides with a hydrogen atom
An electron collides with a hydrogen atom
in its ground state and excites it to a state
in its ground state and excites it to a state
of
of n
n =3. How much energy was given to
=3. How much energy was given to
the hydrogen atom in this collision?
the hydrogen atom in this collision?
 A.
A. -12.1 eV
-12.1 eV B.
B. 12.1 eV
12.1 eV C.
C. -13.6 eV
-13.6 eV
 D.
D. 13.6 eV
13.6 eV E.
E. Non of the above
Non of the above
 Solution:
 ANS: B, Modern Technical Physics, Beiser, Example
25.6, pg. 786
 
0
3 0 0
2 2
13.6eV
( 13.6eV) 12.1eV
3 3
E
E E E E

        

96
96
PYQ 1.6 KSCP 2003/04
PYQ 1.6 KSCP 2003/04
 Which of the following transitions in a hydrogen
Which of the following transitions in a hydrogen
atom emits the photon of lowest frequency?
atom emits the photon of lowest frequency?
 A.
A. n
n = 3 to
= 3 to n
n = 4
= 4 B.
B. n
n = 2 to
= 2 to n
n = 1
= 1
 C.
C. n
n = 8 to
= 8 to n
n = 2
= 2 D.
D. n
n = 6 to
= 6 to n
n = 2
= 2
 E.
E. Non of the above
Non of the above
 ANS: D, Modern Technical Physics, Beiser, Q40, pg. 802,
ANS: D, Modern Technical Physics, Beiser, Q40, pg. 802,
modified
modified
0 0
2 2 2 2
2 2
Lowest frequency means lowest energy:
1 1
13.6eV
1 1
The pair { =6, =2} happens to give smallest
of 0.22.
i f
i f f i
i f
f i
E E
E E E
n n n n
n n
n n
 
      
 
 
 
 

 
 
 

97
97
Frank-Hertz experiment
Frank-Hertz experiment
 The famous experiment that shows the excitation of atoms to
The famous experiment that shows the excitation of atoms to
discrete energy levels and is consistent with the results
discrete energy levels and is consistent with the results
suggested by line spectra
suggested by line spectra
 Mercury vapour is bombarded with electron accelerated
Mercury vapour is bombarded with electron accelerated
under the potential
under the potential V
V (between the grid and the filament)
(between the grid and the filament)
 A small potential
A small potential V
V0
0 between the grid and collecting plate
between the grid and collecting plate
prevents electrons having energies less than a certain
prevents electrons having energies less than a certain
minimum from contributing to the current measured by
minimum from contributing to the current measured by
ammeter
ammeter

98
98
The electrons that arrive at the
The electrons that arrive at the
anode peaks at equal voltage
anode peaks at equal voltage
intervals of 4.9 V
intervals of 4.9 V
 As
As V
V increases, the
increases, the
current measured
current measured
also increases
also increases
 The measured current
The measured current
drops at multiples of a
drops at multiples of a
critical potential
critical potential
 V
V = 4.9 V, 9.8V,
= 4.9 V, 9.8V,
14.7V
14.7V

99
99
Interpretation
Interpretation
 As a result of inelastic collisions between the accelerated
As a result of inelastic collisions between the accelerated
electrons of KE 4.9 eV with the the Hg atom, the Hg atoms
electrons of KE 4.9 eV with the the Hg atom, the Hg atoms
are excited to an energy level above its ground state
are excited to an energy level above its ground state
 At this critical point, the energy of the accelerating electron
At this critical point, the energy of the accelerating electron
equals to that of the energy gap between the ground state and
equals to that of the energy gap between the ground state and
the excited state
the excited state
 This is a resonance phenomena, hence current increases
This is a resonance phenomena, hence current increases
abruptly
abruptly
 After inelastically exciting the atom, the original (the
After inelastically exciting the atom, the original (the
bombarding) electron move off with too little energy to
bombarding) electron move off with too little energy to
overcome the small retarding potential and reach the plate
overcome the small retarding potential and reach the plate
 As the accelerating potential is raised further, the plate current
As the accelerating potential is raised further, the plate current
again increases, since the electrons now have enough energy
again increases, since the electrons now have enough energy
to reach the plate
to reach the plate
 Eventually another sharp drop (at 9.8 V) in the current occurs
Eventually another sharp drop (at 9.8 V) in the current occurs
because, again, the electron has collected just the same
because, again, the electron has collected just the same
energy to excite the same energy level in the other atoms
energy to excite the same energy level in the other atoms

100
100
V = 14.7V
Ke= 4.9eV
Ke= 0
First
resonance
at 4.9 eV
Hg
Plate C Plate P
Ke= 0 after first
resonance
Electron continue to
be accelerated by the
external potential until
the second resonance
occurs
Hg
Ke reaches 4.9 eV again
Hg
First excitation
energy of Hg
atom E1 = 4.9eV
Hg
Ke= 0 after second
resonance
Electron continue to
be accelerated by the
external potential until
the next (third)
resonance occurs
Ke reaches 4.9 eV again
here
second resonance
initiated
Third
resonance
initiated
If bombared by electron with Ke = 4.9 eV excitation
of the Hg atom will occur. This is a resonance
phenomena
electron is
accelerated
under the
external
potential

101
101
 The higher critical potentials result from
The higher critical potentials result from
two or more inelastic collisions and are
two or more inelastic collisions and are
multiple of the lowest (4.9 V)
multiple of the lowest (4.9 V)
 The excited mercury atom will then de-
The excited mercury atom will then de-
excite by radiating out a photon of exactly
excite by radiating out a photon of exactly
the energy (4.9 eV) which is also detected
the energy (4.9 eV) which is also detected
in the Frank-Hertz experiment
in the Frank-Hertz experiment
 The critical potential verifies the existence
The critical potential verifies the existence
of atomic levels
of atomic levels

102
102
Bohr’s correspondence principle
Bohr’s correspondence principle
 The predictions of the quantum theory for the
The predictions of the quantum theory for the
behaviour of any physical system must
behaviour of any physical system must
correspond to the prediction of classical physics
correspond to the prediction of classical physics
in the limit in which the quantum number
in the limit in which the quantum number
specifying the state of the system becomes very
specifying the state of the system becomes very
large:
large:
 quantum theory = classical theory
quantum theory = classical theory
 At large
At large n
n limit, the Bohr model must reduce to a
limit, the Bohr model must reduce to a
“classical atom” which obeys classical theory
“classical atom” which obeys classical theory


n
lim

103
103
In other words…
In other words…
 The laws of quantum physics are valid in
The laws of quantum physics are valid in
the atomic domain; while the laws of
the atomic domain; while the laws of
classical physics is valid in the classical
classical physics is valid in the classical
domain; where the two domains overlaps,
domain; where the two domains overlaps,
both sets of laws must give the same
both sets of laws must give the same
result.
result.

104
104
 Which of the following statements are correct?
Which of the following statements are correct?
 I
I Frank-Hertz experiment shows that atoms are
Frank-Hertz experiment shows that atoms are
excited
excited to discrete energy levels
to discrete energy levels
 II
II Frank-Hertz experimental result is consistent with the
Frank-Hertz experimental result is consistent with the
results suggested by the line spectra
results suggested by the line spectra
 III
III The predictions of the quantum theory for the
The predictions of the quantum theory for the
behaviour of any physical system must correspond to
behaviour of any physical system must correspond to the
the
prediction of classical physics in the limit in which
prediction of classical physics in the limit in which the
the
quantum number specifying the state of the
quantum number specifying the state of the system
system
becomes very large
becomes very large
 IV
IV The structure of atoms can be probed by using
The structure of atoms can be probed by using
electromagnetic radiation
electromagnetic radiation
 A.
A. II,III
II,IIIB.
B. I, II,IV
I, II,IV C. II, III, IV
C. II, III, IV
 D.
D. I,II, III, IV
I,II, III, IV E. Non of the above
E. Non of the above
 ANS:D, My own questions
ANS:D, My own questions

105
105
Example
Example
(Read it yourself)
(Read it yourself)
 Classical EM predicts that an electron in a circular motion
Classical EM predicts that an electron in a circular motion
will radiate EM wave at the same frequency
will radiate EM wave at the same frequency
 According to the correspondence principle,
According to the correspondence principle,
the Bohr model must also reproduce this result in the
the Bohr model must also reproduce this result in the
large
large n
n limit
limit
More quantitatively
More quantitatively
 In the limit,
In the limit, n
n = 10
= 103
3
- 10
- 104
4
, the Bohr atom would have a size
, the Bohr atom would have a size
of 10
of 10-3
-3
m
m
 This is a large quantum atom which
This is a large quantum atom which is in classical domain
is in classical domain
 The prediction for the photon emitted during transition
The prediction for the photon emitted during transition
around the
around the n
n = 10
= 103
3
- 10
- 104
4
states should equals to that
states should equals to that
predicted by classical EM theory.
predicted by classical EM theory.

106
106
n (Bohr)
n  large
=  (classical theory)

107
107
Classical physics calculation
Classical physics calculation
 The period of a circulating electron is
T = 2r/(2K/m)1/2
= r(2m)1/2
(8e0
r)1/2
/e
 This result can be easily derived from the
mechanical stability of the atom as per
 Substitute the quantised atomic radius rn
= n2
r0
into T, we obtain the frequency as per
 n
= 1/T =me4
/323

2
3
n3
 
r
v
m
r
e
Ze e
2
2
0
4
1



108
108
Based on Bohr’s theory
Based on Bohr’s theory
 Now, for an electron in the Bohr atom at energy
level n = 103
- 104
, the frequency of an radiated
photon when electron make a transition from the n
state to n–1 state is given by
 n
= (me4
/643

2
3
)[(n-1)-2
- n-2
]
= (me4
/643

2
3
)[(2n-1)/n2
(n-1)2
]
In the limit of large n,
  (me4
/643

2
3
)[2n/n4
]
=me4
/323

2
3
)[1/n3
]

109
109
Classical result and Quantum
Classical result and Quantum
calculation meets at
calculation meets at n
n


 Hence, in the region of large
Hence, in the region of large n
n, where classical
, where classical
and quantum physics overlap, the classical
and quantum physics overlap, the classical
prediction and that of the quantum one is
prediction and that of the quantum one is
identical
identical
classical = Bohr = (me4
/323

2
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