CH 29 – Nuclear Physics
In the early part of the 1900’s Ernest Rutherford demonstrated that the atom consisted of
a small, positive nucleus that contained most of the mass surrounded by electrons. In the
previous chapter we discussed the configurations and energies of the electrons. In this
chapter we discuss the makeup of the nucleus.
The nucleus of an atom consists of protons and neutrons. The proton has a charge e =
1.602 x 10-19 C and the neutron has no net charge. The nucleus is characterized by
atomic number (Z), neutron number (N), and mass number (A = Z + N):



Atomic number Z: number of protons in nucleus
Neutron number N: number of neutrons in nucleus
Mass number A: number of nucleons (neutrons and protons) in nucleus
Symbolically, we describe a nucleus as ZA X , where X is the element (e.g., 11H , 24He , 37Li ,
etc.)
The Z number determines the element. Elements with different N and A numbers are
12 13 14
isotopes. For example, the isotopes of carbon are 11
6 C, 6C, 6C, 6C .
The mass number, A, which is given for the elements in the periodic table is generally not
an integer. The reason is that it represents an average of the naturally occurring isotopes
and it includes the mass equivalent of the binding energy of the nucleons.
Atomic mass unit (u)
The atomic mass unit (u) is defined as 1/12th the mass of 126 C . Since 12 grams of 126 C is
a mole, or Avogadro’s number of atoms, the atomic mass unit is
1u 
1  0.012kg / mole 

  1.6606 x10  27 kg
23
12  6.0221x10 / mole 
The rest mass energy of 1u is
E  mc 2  ( 1.6606 x10 27 kg )( 2.998 x108 m / s )2  1.492 x10 10 J
 ( 1.492 x10 10 J ) /( 1.602 x10 19 J / ev )  931.5Mev
From the equation E = mc2, we can write
1 u  931.5Mev / c 2
1
The masses of the proton and neutron are slightly larger than 1 u. This is because of the
binding energy of the nucleus, as will be discuss later. The proton, neutron and electron
masses are
Proton:
m = 1.6726 x 10-27 kg = 1.007276 u = 938.28 Mev/c2
Neutron:
m = 1.6750 x 10-27 kg = 1.008665 u = 938.57 Mev/c2
Electron:
m = 9.109 x 10-31 kg = 5.486 x 10-4 u = 0.511 Mev/c2
Size of Nucleus
The size of the nuclei can be determined by experiments in which charged particles such
as alpha particles (helium nuclei) are scattered from a material. The angular distribution
of the scattering depends on the distance of closest approach of the alpha particles to the
nuclei. The radii of the nuclei has been found to follow the relationship
r  r0 A1 / 3 ,
where r0 = 1.2 x 10-15 m and A is the atomic number. This equation suggests that the
densities of the nuclei of the different elements are all approximately the same.
The density of the nuclei can be estimated as follows:
V  4  r 3  4  r0 3 A ( nuclear volume )
3
3
m
A
( nuclear mass )
NA

A/ NA
m
1
1



3
3
26
4
4
4
V
( 6.02 x10 / kg )  ( 1.2 x10 15 m )3
 r0 A N A  r0
3
17
3
3
 2.3 x10 kg / m 3
Nuclear Stability
For small atomic numbers, the number of protons and neutrons in the nucleus is about the
35
same. For example, 24 He 126C , 17
Cl . But for Z greater than about 20, the number of
neutrons exceeds the number of protons (e.g., 198
79 Au ). The figure below is a plot of the
number of neutrons (N) versus the number of protons (Z). The solid points represent
stable nuclei while the shaded area represents radioactive nuclei. The stability of the
nucleus depends on competition between electrostatic repulsion between the protons and
the attractive nuclear force between nucleons. Without the nuclear force the nucleus
2
would fly apart due to the electrostatic repulsion. The nuclear force is a short-range force
that exists mainly between nearest-neighbor nucleons; whereas, the electrostatic force is
comparatively long range. If there were no neutrons in the nucleus, then the repulsive
electrostatic force would be too great for stability. In a sense, neutrons help to spread the
protons apart, reducing their repulsion. If there were no neutrons, then the electrostatic
force on the outer protons would increase in proportion to the nuclear radius. If Q is the
total nuclear charge and  the nuclear charge density, then the force on an outer proton
(e) would be
F k
Qe
r
2
k
 4  r 3e
3
r2
r
So, as the number of protons increases, there must be an increasing excess of neutrons to
reduce the nuclear charge density.
3
Binding Energy of Nuclei
The mass of a nucleus is less than the sum of the masses of the individual protons and
neutrons because of the binding energy of the nucleus. An amount of energy equal to the
binding energy must be supplied to the nucleus to break it apart into individual protons
and neutrons. The mass difference is related to the binding energy by the equation
Eb  m c 2
Example:
Calculate the binding energy of deuterium.
Solution:
The deuterium nucleus consists of a proton and a neutron. So we find the difference
between the sum of the masses of the protons and neutron and the mass of the deuterium
nucleus (deuteron),
m  ( m p  mn )  md
If we use the atomic mass for md, then this includes the mass of the nucleus and the mass
of the electron. If we also use the atomic mass of hydrogen in place of the mass of the
proton, then the mass of the electron will cancel out in the calculation. Then
m  ( 1.007825 u  1.008665 u )  2.014102 u  0.002388 u
 ( 0.002388 u )( 931.49 Mev / c 2 / u )  2.224 Mev / c 2
Eb  m c 2  2.224 Mev
Per nucleon, the binding energy is
Eb / A  2.224 Mev / 2  1.11 Mev
Example:
Calculate the binding energy of 126 C .
4
Solution:
m  ( 6m p  6mn )  mC 12
 6( 1.007825 u  1.008665 u )  12 u  0.09894 u
 ( 0.09894u )( 931.49 Mev / c 2 / u )  92.16 Mev / c 2
Eb  m c 2  92.16 Mev
Per nucleon,
Eb / A  92.16 Mev / 12  7.68 Mev
The binding energy per nucleon generally increases rapidly with increasing A up to about
A ~ 60 and then more slowly decreases with further increase in A. This means that the
nuclei with intermediate masses are the most stable compared with the light and heavy
nuclei. The binding energy per nucleon is shown below as a function of A. This figure
suggests that we can convert mass to energy by combining lighter nuclei to make nuclei
of intermediate size (fusion) or breaking apart heavy nuclei into nuclei of intermediate
size (fission).
5
Radioactive Decay
An unstable nucleus can transform to a more stable nucleus by emitting an alpha particle
or a beta particle (electron or positron). A nucleus in an excited state can also emit
gamma rays. The alpha particle is the nucleus of the helium-4 atom, 24 He . The positron
is the antiparticle of the electron. Its mass is the same as that of the electron but it has a
positive charge +e. A gamma ray is a high energy photon and has no mass or charge.
Alpha Decay
Alpha decay can be expresses symbolically as
A
A 4
4
Z X  Z  2Y  2 He
Note that both the total mass number A and the atomic number Z are the same before and
after the decay. An example of alpha-decay is the decay of radium-226 into radon-222:
226
222
4
88 Ra  86 Rn  2 He
Example:
Calculate the energy released in the alpha decay of radium.
Solution:
m  mRa  ( m Rn  mHe )
 226.025402 u  ( 222.017571u  4.002602 u )  0.005229 u
E  ( 0.005229 u )( 931.494 Mev / u )  4.97 Mev
Since the alpha particle is much lighter than the radon nucleus, most of the released
energy appears as the kinetic energy of the alpha particle.
Beta Decay
Beta decay can be expressed as
A
A
Z X  Z 1Y
 e
(negative beta decay)
A
A
Z X  Z 1Y
 e
(positive beta decay)
or
6
Note that the mass number doesn’t change, but the nuclear charge either increases (for
electron emission) or decreases (for positron emission).
The emission of an electron is equivalent to the decay inside the nucleus of a neutron into
a proton and an electron –
1
1

0 n1 p  e
Similarly, the emission of a positron is equivalent to the conversion of a proton in the
nucleus into a neutron and a positron.
A process related to beta decay is K-capture. In this process a nucleus ‘captures’ an
orbital electron in the K-shell which combines with a proton to form a neutron. This
process is expressed as
A
ZX
 e   Z A1Y
(K-capture)
Example:
Calculate the energy released in the beta decay of C-14 into N-14.
Solution:
The process is
14
14

6 C 7 N  e
We can neglect the mass of the electron on the right side of the expression if we use the
atomic masses, since atomic N has one more electron than atomic C. Then
m  mC 14  m N 14
 14.003242 u  14.003074 u  0.000168 u
E  ( 0.000168 u )( 931.494 Mev / u )  0.156 Mev
If the electron were the only particle emitted in this decay process, then essentially all of
this energy would be imparted to the electron since its mass is small compared to the
mass of the nitrogen nucleus. However, it is observed that most of the time the electron’s
kinetic energy is significantly less than this amount. The observed distribution of kinetic
energies is as shown in the figure on the next page.
This discrepancy led to the prediction by Pauli that a third particle must be involved to
carry away the unaccounted for energy. In addition, the third particle was required to
have a spin angular quantum number of ½ to account for conservation of angular
7
momentum. This particle was expected to have no charge and a very small mass and was
called the ‘neutrino’. It was eventually discovered in 1956. The complete decay process
would be given as
14
14

6 C 7 N  e 
 is an anti-neutrino. In + beta decay processes,
a neutrino ( ) is emitted. It has recently been
shown that there are several types of neutrinos
and that they have a finite, but small, mass.
Because of their small mass they travel very
near the speed of light. Neutrinos are extremely
abundant (the sun produces a large flux of
neutrinos); however, they are extremely difficult
to detect.
Gamma Decay
Quite often in alpha or beta decay or in nuclear reactions a nucleus is left in an excited
stated, somewhat like an excited electronic state in an atom. The nucleus then decays to
its ground state by emitting one or more high energy photons (gamma rays). This process
is described as
A *
A
Z X Z X

Below is an example of gamma decay:
60
60 *

27 Co 28 Ni  e  
60 *
60
28 Ni  28 Ni  
Penetrating power of radiation
The penetrating power of radiation depends on the type and energy. Gamma rays are
generally the most penetrating since they have no charge to interact with the electrons
and nuclear charges in matter. Several centimeters of lead would be required to provide
safe shielding from a high energy gamma emitter source. Alpha particles are the least
penetrating both because of their charge and their mass. They easily lose energy in air
and can be shielded by thick paper. Beta particles have intermediate penetrating capacity.
8
Radioactive Decay Rates
The number of nuclei N that will decay in a time interval t is proportional to the
number of radioactive nuclei N and the time interval,
N  Nt
By using calculus it can be shown that this leads to an exponential dependence of N and
the decay rate N/t on time –
N  N 0 e  t
R
N
 N 0 e   t  N
t
 is the decay constant and has units of inverse time. The unit for decay rate, R, is the
curie.
1 curie  1Ci  3.7 x1010 decay / s
The half-life T1/2 is the time for one-half of the
nuclei to decay. The relationship between the halflife and the decay constant can be obtained as
follows.
1 N  N e   T1 / 2
0
2 0
 T1 / 2
e
2
ln( e  T1 / 2 )  T1 / 2  ln( 2 )
Or,
T1 / 2 
ln( 2 )

The exponential decay formula is somewhat like that for the discharge of a capacitor
through a resistor, where  is analogous to the RC time constant.
9
Example:
Tritium (hydrogen-3) decays into helium-3 by beta decay with a half-life of 12.3 yrs:
3
3

1 H 2 He  e  
What is the activity (decay rate) of 1 mg of tritium?
Solution:
number moles  n 
0.001g
 3.33x10  4
3g
N  nN A  ( 3.33x10  4 mole )( 6.02 x10 23 / mole )  2.01x1019

ln( 2 )
0.693

 1.79 x10  9 / s
T1 / 2 ( 12.3 yr )( 3.15 x10 7 s / yr )
R  N  ( 1.79 x10  9 / s )( 2.01x1019 )  3.60 x1010 / s
 ( 3.60 x1010 / s )( 1 Ci / 3.7 x1010 / s )  0.97 Ci
Example:
A curie is a very large and dangerous amount of radioactivity. How long would one have
to wait for the tritium activity to reduce to 1 mCi?
Solution:
R  N 0 e  t  R0 e  t
0.001Ci  0.97Cie   t
e t 
0.97
 970
0.001
ln( 970 )
ln( 970 )
t

 9.9 T1 / 2  ( 9.9 )( 12.3 yr )  122 yr

ln( 2 ) / T1 / 2
Carbon Dating
CO2 in the earth’s atmosphere has a small amount of radioactive C-14. The C-14 is
produced from N-14 in the upper atmosphere by neutrons which are generated in cosmic
ray interactions :
1 14
14
1
0 n  7 N  6 C 1 H
10
C-14 is radioactive and decays back into N-14 through beta decay:
14
14

6 C 7 N  e 
The lifetime of C-14 is 5730 yr, and the fraction of C-14 in the atmosphere is about 1 part
in 1012. Living organisms ingest C-14 from the atmosphere and from ingesting other
organisms and contain this same fraction of C-14. The C-14 in a living organism
produces a decay rate of 15 decays/min·g.
When a living organism dies, it no longer ingests C-14 and its fraction of C-14 decreases
with time. By comparing the decay rate of a dead organism with that of a live organism,
the time of death can be determined.
Example:
20 g of carbon is extracted from a fossil and the C-14 activity is measured to be 100
decays/min. What is the age of the fossil?
Solution:
The activity of 20-g of ‘live’ carbon would be
R0  ( 15 decays / min g )( 20 g )  300 decays / min
So,
R  R0 e  t
100  300e   t
ln( 3 ) ln( 3 )
ln( 3 )
t

T1 / 2 
( 5730 yr )  9082 yr

ln( 2 )
ln( 2 )
Radioactive Series
All elements for which Z > 83 are radioactive. These elements decay through alpha or
beta decay into lower Z elements which may also be radioactive. Eventually, they decay
into an element which is stable. The various radioactive elements that are produced in
this process can be classified into four series depending on the starting and ending nuclei.
These four radioactive series are given listed as follows:
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Series
Uranium
Actinium
Thorium
Neptunium
Starting isotope
Half-life (yrs)
Stable end product
238
92 U
235
92 U
232
90Th
237
93 Np
4.47x109
206
82 Pb
207
82 Pb
208
82 Pb
209
83 Bi
7.04x108
1.41x1010
2.14x106
The figure below shows the Thorium Series.
Nuclear Reactions
In a nuclear reaction a target nucleus is struck by an energetic particle such as a neutron,
alpha particle or another nucleus, and the particles are transformed into one or more other
particles.
Examples:
4
21
24
1
2 He 10 Ne12 Mg  0 n
12
1
4
2
3
0 n 2 He1 H 1H
Nuclear reactions can be either exothermic or endothermic, depending on whether the
final kinetic energy is greater or smaller than the initial kinetic energy. The ‘Q-value’ of
the reaction is the energy released (or absorbed). For exothermic reactions Q > 0; for
endothermic Q < 0. The Q-value can be obtained from the difference in mass of the
initial and final products.
Q  mc 2  ( mi  m f )c 2
A reaction with a negative Q requires that the incident particle have a minimum amount
of kinetic energy in order for the reaction to occur. If the target particle is at rest, then the
threshold kinetic energy for the reaction must exceed |Q| in order to conserve momentum.
For example, if KEi = |Q|, then KEf = 0 and the final momentum is zero, so momentum is
not conserved. By requiring that both energy and momentum be conserved, it can be
shown that the threshold kinetic energy of the incident particle for an endothermic
reaction to occur is
m

KEmin  1 
|Q|
M

where m is the mass of the incident particle and M is the mass of the target particle.
Example:
Calculate the Q for the following reaction, and calculate the minimum energy of the
incident neutron for the reaction to occur.
1 14
11
4
0 n  7 N  5 B  2 He
Solution:
Q  (mn  m N14 )  (m B11  m He 4 )
 (1.008665  14.003074)  (11.009306  4.002603)  0.00017 u
 (0.00017 u )(931.494 Mev / u )  0.158 Mev

mn
KEmin  1 
 m N14

1
 | Q | 1  (0.158Mev )  0.17 Mev
 14 

13