Nuclear Physics.ppt

Shri Govindrao Munghate Arts and Science College ,
Kurkheda
Department of Physics
Topic –Nuclear Physics
Prof. B.V. Tupte
Head, Department of Physics
Shri Govindrao Munghate Arts and
Science College Kurkheda.

Composition of Matter
All of matter is composed of at least three
fundamental particles (approximations):
Particle Fig. Sym Mass Charge Size
The mass of the proton and neutron are close, but
they are about 1840 times the mass of an electron.
Electron e- 9.11 x 10-31 kg -1.6 x 10-19 C 
Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm
Neutron n 1.675 x 10-31 kg 0 3 fm

The Atomic Nucleus
Beryllium Atom
Compacted nucleus:
4 protons
5 neutrons
Since atom is electri-
cally neutral, there
must be 4 electrons.
4 electrons

Modern Atomic Theory
The Bohr atom, which is
sometimes shown with
electrons as planetary
particles, is no longer a valid
representation of an atom, but
it is used here to simplify our
discussion of energy levels.
The uncertain position of an
electron is now described as a
probability distribution—loosely
referred to as an electron cloud.

Definitions
A nucleon is a general term to denote a nuclear
particle - that is, either a proton or a neutron.
The atomic number Z of an element is equal to the
number of protons in the nucleus of that element.
The mass number A of an element is equal to the
total number of nucleons (protons + neutrons).
The mass number A of any element is equal to
the sum of the atomic number Z and the number
of neutrons N :
A = N + Z

Symbol Notation
A convenient way of describing an element is by
giving its mass number and its atomic number,
along with the chemical symbol for that element.
 
A Mass number
Z Atomic number
X Symbol

For example, consider beryllium (Be):
9
4 Be

Example 1: Describe the nucleus of a lithium atom
which has a mass number of 7 and an atomic number
of 3.
Lithium Atom
N = A – Z = 7 - 3
A = 7; Z = 3; N = ?
Protons: Z = 3
neutrons: N = 4
Electrons: Same as Z
7
3 Li

Isotopes of Elements
Isotopes are atoms that have the same number
of protons (Z1= Z2), but a different number of
neutrons (N). (A1  A2)
Helium - 4
4
2 He
Helium - 3
3
2 He Isotopes
of helium

Nuclides
Because of the existence of so many
isotopes, the term element is sometimes
confusing. The term nuclide is better.
A nuclide is an atom that has a definite
mass number A and Z-number. A list of
nuclides will include isotopes.
The following are best described as nuclides:
3
2 He 4
2 He 12
6 C 13
6 C

Atomic Mass Unit, u
One atomic mass unit (1 u) is equal to one-
twelfth of the mass of the most abundant
form of the carbon atom--carbon-12.
Atomic mass unit: 1 u = 1.6606 x 10-27 kg
Common atomic masses:
Proton: 1.007276 u Neutron: 1.008665 u
Electron: 0.00055 u Hydrogen: 1.007825 u

Exampe 2: The average atomic mass of Boron-11 is 11.009305
u. What is the mass of the nucleus of one boron atom in kg?
Electron: 0.00055 u
11
5 B = 11.009305
The mass of the nucleus is the atomic mass
less the mass of Z = 5 electrons:
Mass = 11.009305 u – 5(0.00055 u)
1 boron nucleus = 11.00656 u
-27
1.6606 x 10 kg
11.00656 u
1 u
m
 
  
 
m = 1.83 x 10-26 kg

2 8
; 3 x 10 m/s
E mc c
 
Mass and Energy
Recall Einstein’s equivalency formula for m and E:
The energy of a mass of 1 u can be found:
E = (1 u)c2 = (1.66 x 10-27 kg)(3 x 108 m/s)2
E = 1.49 x 10-10 J Or E = 931.5 MeV
When converting
amu to energy:
2 MeV
u
931.5
c 

Example 3: What is the rest mass energy of a proton
(1.007276 u)?
E = mc2 = (1.00726 u)(931.5 MeV/u)
Proton: E = 938.3 MeV
Similar conversions show other
rest mass energies:
Electron: E = 0.511 MeV
Neutron: E = 939.6 MeV

The Mass Defect
The mass defect is the difference between
the rest mass of a nucleus and the sum of
the rest masses of its constituent nucleons.
The whole is less than the sum of the parts!
Consider the carbon-12 atom (12.00000 u):
Nuclear mass = Mass of atom – Electron masses
= 12.00000 u – 6(0.00055 u)
= 11.996706 u
The nucleus of the carbon-12 atom has this mass.
(Continued . . .)

Mass Defect (Continued)
Mass of carbon-12 nucleus: 11.996706
Proton: 1.007276 u Neutron: 1.008665 u
The nucleus contains 6 protons and 6 neutrons:
6 p = 6(1.007276 u) = 6.043656 u
6 n = 6(1.008665 u) = 6.051990 u
Total mass of parts: = 12.095646 u
Mass defect mD = 12.095646 u – 11.996706 u
mD = 0.098940 u

The Binding Energy
The binding energy EB of a nucleus is the
energy required to separate a nucleus into
its constituent parts.
EB = mDc2 where c2 = 931.5 MeV/u
The binding energy for the carbon-12 example is:
EB = (0.098940 u)(931.5 MeV/u)
EB = 92.2 MeV
Binding EB for C-12:

Binding Energy per Nucleon
An important way of comparing the nuclei of
atoms is finding their binding energy per nucleon:
Binding energy
per nucleon
MeV
=
nucleon
B
E
A
 
 
 
For our C-12 example A = 12 and:
MeV
nucleon
92.2 MeV
7.68
12
B
E
A
 

Formula for Mass Defect
The following formula is useful for mass defect:
 
D H n
m Zm Nm M
  
 
 
Mass defect
mD
mH = 1.007825 u; mn = 1.008665 u
Z is atomic number; N is neutron number;
M is mass of atom (including electrons).
By using the mass of the hydrogen atom, you avoid
the necessity of subtracting electron masses.

Example 4: Find the mass defect for the nucleus of
helium-4. (M = 4.002603 u)
 
D H n
m Zm Nm M
  
 
 
Mass defect
mD
ZmH = (2)(1.007825 u) = 2.015650 u
Nmn = (2)(1.008665 u) = 2.017330 u
M = 4.002603 u (From nuclide tables)
mD = (2.015650 u + 2.017330 u) - 4.002603 u
mD = 0.030377 u
4
2 He

Example 4 (Cont.) Find the binding energy per
nucleon for helium-4. (mD = 0.030377 u)
EB = (0.030377 u)(931.5 MeV/u) = 28.3 MeV
MeV
nucleon
28.3 MeV
7.07
4
B
E
A
 
A total of 28.3 MeV is required To tear apart
the nucleons from the He-4 atom.
Since there are four nucleons, we find that

Binding Energy Vs. Mass Number
Mass number A
Binding
Energy
per
nucleon
50 100 150 250
200
2
6
8
4
Curve shows that
EB increases with
A and peaks at
A = 60. Heavier
nuclei are less
stable.
Green region is for
most stable atoms.
For heavier nuclei, energy is released when they
break up (fission). For lighter nuclei, energy is
released when they fuse together (fusion).

Stability Curve
Atomic number Z
Neutron
number
N
Stable
nuclei
Z = N
20 40 60 80 100
40
100
140
20
60
80
120
Nuclear particles are
held together by a
nuclear strong force.
A stable nucleus remains
forever, but as the ratio
of N/Z gets larger, the
atoms decay.
Elements with Z > 82
are all unstable.

Radioactivity
As the heavier atoms become
more unstable, particles and
photons are emitted from the
nucleus and it is said to be
radioactive. All elements with
A > 82 are radioactive.
Examples are:
Alpha particles a b particles (electrons)
b particles (positrons)
Gamma rays g
b
g
b
a

The Alpha Particle
An alpha particle a is the nucleus of a helium
atom consisting of two protons and two
neutrons tightly bound.
Charge = +2e- = 3.2 x 10-19 C
Mass = 4.001506 u
Relatively low speeds ( 0.1c )
Not very penetrating

The Beta-minus Particle
A beta-minus particle b is simply an electron
that has been expelled from the nucleus.
Charge = e- = -1.6 x 10-19 C
-
High speeds (near c)
-
Mass = 0.00055 u
-
Very penetrating
-

The Positron
A beta positive particle b is essentially an
electron with positive charge. The mass and
speeds are similar.
Charge = +e- = 1.6 x 10-19 C
+
High speeds (near c)
+
Mass = 0.00055 u
+
Very penetrating
+

The Gamma Photon
A gamma ray g has very high electromagnetic
radiation carrying energy away from the
nucleus.
Charge = Zero (0)
g
Mass = zero (0)
g
Speed = c (3 x 108 m/s)
g
Most penetrating radiation
g

Radioactive Decay
As discussed, when the ratio of N/Z gets very
large, the nucleus becomes unstable and often
particles and/or photons are emitted.
Alpha decay results in the loss of two
protons and two neutrons from the nucleus.
4
2a
4 4
2 2
A A
Z Z
X Y energy
a


  
X is parent atom and Y is daughter atom
The energy is carried away primarily
by the K.E. of the alpha particle.

Example 5: Write the reaction that occurs when
radium-226 decays by alpha emission.
4 4
2 2
A A
Z Z
X Y energy
a


  
226 226 4 4
88 88 2 2
Ra Y energy
a


  
From tables, we find Z and A for nuclides.
The daughter atom: Z = 86, A = 222
226 222 4
88 86 2
Ra Rn energy
a
  
Radium-226 decays into radon-222.

Beta-minus Decay
Beta-minus b decay results when a neutron
decays into a proton and an electron. Thus,
the Z-number increases by one.
0
1 1
A A
Z Z
X Y energy
b
 
  
by the K.E. of the electron.
-

Beta-plus Decay
Beta-plus b decay results when a proton
decays into a neutron and a positron. Thus,
the Z-number decreases by one.
0
1 1
A A
Z Z
X Y energy
b
 
  
by the K.E. of the positron.
+

Radioactive Materials
The rate of decay for radioactive substances is
expressed in terms of the activity R, given by:
N
R
t



Activity
N = Number of
undecayed nuclei
One curie (Ci) is the activity of a radioactive
material that decays at the rate of 3.7 x 1010 Bq
or 3.7 x 1010 disintegrations per second.
One becquerel (Bq) is an activity equal to one
disintegration per second (1 s-1).

The Half-Life
The half-life T1/2 of
an isotope is the
time in which one-
half of its unstable
nuclei will decay.
No
0
2
N
0
4
N
Number of Half-lives
Number
Undecayed
Nuclei
1 4
3
2
0
1
2
n
N N
 
  
 
Where n is number
of half-lives

Half-Life (Cont.)
The same reasoning will apply to activity R or to
amount of material. In general, the following
three equations can be applied to radioactivity:
Nuclei Remaining
0
1
2
n
N N
 
  
 
Activity R
0
1
2
n
R R
 
  
 
Mass Remaining
0
1
2
n
m m
 
  
 
Number of Half-lives:
1
2
t
n
T


Example 6: A sample of iodine-131 has an initial activity of 5
mCi. The half-life of I-131 is 8 days. What is the activity of the
sample 32 days later?
First we determine the number of half-lives:
1/ 2
32 d
8 d
t
n
T
  n = 4 half-lives
4
0
1 1
5 mCi
2 2
n
R R
   
 
   
   
R = 0.313 mCi
There would also be 1/16 remaining of the
mass and 1/16 of the number of nuclei.

Nuclear Reactions
It is possible to alter the structure of a nucleus
by bombarding it with small particles. Such
events are called nuclear reactions:
General Reaction: x + X  Y + y
For example, if an alpha particle bombards
a nitrogen-14 nucleus it produces a
hydrogen atom and oxygen-17:
4 14 1 17
2 7 1 8
N H O
a   

Conservation Laws
For any nuclear reaction, there are three
conservation laws which must be obeyed:
Conservation of Charge: The total charge of a
system can neither be increased nor decreased.
Conservation of Nucleons: The total number of
nucleons in a reaction must be unchanged.
Conservation of Mass Energy: The total mass-
energy of a system must not change in a
nuclear reaction.

Example 7: Use conservation criteria to determine the
unknown element in the following nuclear reaction:
1 7 4
1 3 2
A
Z
H Li He X energy
   
Charge before = +1 + 3 = +4
Charge after = +2 + Z = +4
Z = 4 – 2 = 2
Nucleons before = 1 + 7 = 8
Nucleons after = 4 + A = 8
(Helium has Z = 2)
(Thus, A = 4)
1 7 4 4
1 3 2 2
H Li He He energy
   

Conservation of Mass-Energy
There is always mass-energy associated with any
nuclear reaction. The energy released or absorbed
is called the Q-value and can be found if the
atomic masses are known before and after.
1 7 4 4
1 3 2 2
H Li He He Q
   
   
1 7 4 4
1 3 2 2
Q H Li He He
   
Q is the energy released in the reaction. If Q
is positive, it is exothermic. If Q is negative, it
is endothermic.

Example 8: Calculate the energy released in the
bombardment of lithium-7 with hydrogen-1.
1 7 4 4
1 3 2 2
H Li He He Q
   
   
1 7 4 4
1 3 2 2
Q H Li He He
   
7
3 7.016003 u
Li 
4
2 4.002603 u
He 
1
1 1.007825 u
H 
Substitution of these masses gives:
Q = 0.018622 u(931.5 MeV/u) Q =17.3 MeV
4
2 4.002603 u
He 
The positive Q means the reaction is exothermic.

Summary
Particle Fig. Sym Mass Charge Size
Electron e 9.11 x 10-31 kg -1.6 x 10-19 C 
Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm
Neutron n 1.675 x 10-31 kg 0 3 fm
Fundamental atomic and nuclear particles
The mass number A of any element is equal to
the sum of the protons (atomic number Z) and
the number of neutrons N : A = N + Z

Summary Definitions:
A nucleon is a general term to denote a nuclear
particle - that is, either a proton or a neutron.
The mass number A of an element is equal to the
total number of nucleons (protons + neutrons).
Isotopes are atoms that have the same number
of protons (Z1= Z2), but a different number of
neutrons (N). (A1  A2)
A nuclide is an atom that has a definite mass
number A and Z-number. A list of nuclides will
include isotopes.

Summary (Cont.)
 
A Mass number
Z Atomic number
X Symbol

Symbolic notation
for atoms
 
D H n
m Zm Nm M
  
 
 
Mass defect
mD
Binding Energy
per nucleon
MeV
=
nucleon
B
E
A
 
 
 
Binding
energy

Summary (Decay Particles)
An alpha particle a is the nucleus of a helium
atom consisting of two protons and two tightly
bound neutrons.
A beta-minus particle b is simply an electron
that has been expelled from the nucleus.
A beta positive particle b is essentially an
electron with positive charge. The mass and
speeds are similar.
A gamma ray g has very high electromagnetic
radiation carrying energy away from the
nucleus.

Summary (Cont.)
4 4
2 2
A A
Z Z
X Y energy
a


  
Alpha Decay:
0
1 1
A A
Z Z
X Y energy
b
 
  
Beta-minus Decay:
0
1 1
A A
Z Z
X Y energy
b
 
  
Beta-plus Decay:

Summary (Radioactivity)
Nuclei Remaining
0
1
2
n
N N
 
  
 
Activity R
0
1
2
n
R R
 
  
 
Mass Remaining
0
1
2
n
m m
 
  
 
Number of Half-lives:
1
2
t
n
T

The half-life T1/2 of an isotope is the time in
which one-half of its unstable nuclei will decay.

Summary (Cont.)
Conservation of Charge: The total charge of a
system can neither be increased nor decreased.
Conservation of Nucleons: The total number of
nucleons in a reaction must be unchanged.
Conservation of Mass Energy: The total mass-
energy of a system must not change in a
nuclear reaction. (Q-value = energy released)
Nuclear Reaction: x + X  Y + y + Q

Nuclear Physics.ppt

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