Unit 19 Notes –Nuclear Physics
Chapter 32
Nuclear Structure
A= atomic mass number
Z= atomic number
N= number of neutrons
A=Z+N
Notation:
A
where X is an element symbol
Z X
proton: 11 H because the proton is the nucleus of a hydrogen atom
neutron: 01 n
electron:
0
1
e because electron has no nucleus A = 0 and Z=-1 because it has a negative charge
isotope: nuclei with same number of protons but different number of neutrons
nucleon: general name for particles in the nucleus, i.e. protons and neutrons
atomic mass unit: amu or u: 1u = 1.660540x10-27 kg
(define relative to 126C having mass of exactly 12u)
Mass Defect of nucleus and Nuclear Binding Energy
- binding energy – energy required to separate a stable nucleus into its constituent protons
and neutrons
- see example 32-6
- In Einstein’s theory of special relativity, mass and energy are equivalent:
o change Δm in mass of a system is equivalent to a change ΔE0 in the rest energy of
the system by an amount
ΔE0 = (Δm)c2
c, of course, is the speed of light; E is energy; m is mass
o the energy equivalent of 1 u is
1 eV


E = mc2 = (1.660540x10-27 kg)(2.99792x108 m/s)2 
19 
 1.6022x10 J 
= 931.5 MeV
(note that this is over 6 orders of magnitude larger than electron energies)
o Since mass and energy are equivalent, mass can be expressed in terms of energy:
1 u = 931.5 MeV/c2
- the binding energy used to disassemble the nucleus appears as extra mass of the separated
nucleons.
o in other words, the sum of the individual masses of the separated protons and
neutrons is greater by an amount Δm than the mass of the stable nucleus
o the difference in mass is known as the mass defect of the nucleus
1
Strong Nuclear Force
- The attractive force that holds the nucleus together – without it, the electrostatic force
would cause the nucleus to fly apart.
- Acts on short ranges - only a couple of fermis (1 fermi = 1 femtometer = 10-15 m)
- Always attractive
- Nearly equal strength between proton-proton, proton-neutron, and neutron-neutron
- Does not act on electrons
Stability
Stability is determined by competition between repulsive electrostatic force and attractive strong
nuclear force.
Radioactivity
α-rays – least penetrating, blocked by 0.01 mm of lead
β-rays – next most penetrating, blocked by 0.1 mm of lead
γ-rays – most penetrating, blocked by 100 mm of lead
2
The disintegration process that produces α-rays, β-rays, and γ-rays must obey these laws:
- conservation of mass/energy
- conservation of electric charge
- conservation of linear momentum
- conservation of angular momentum
- conservation of nucleon number
α decay
-
-
-
4
an α-ray consists of positively charged particles, each one being the 2 He nucleus of
helium. thus, an α particle has a charge of +2e and a nucleon number of A=4.
Example:
238
234
4
92 U 
90Th  2 He
parent nucleus (uranium)  daughter nucleus (thorium) + α particle (helium nucleus)
Note that the sum of the A numbers on each side of the equation are equal (238=234+4)
and the sum of the Z numbers on each side of the equation are equal (92=90+2). This is
conservation of nucleon number
Since the parent and daughter nuclei are different, this is called transmutation
when α decay occurs, the energy released appears as kinetic energy of the recoiling
daughter nucleus and the α particle
β- decay
- a β-ray consists of negatively charged particles, or β- particles
- experiments show β- particles are electrons( 10 e )
- Example:
234
234
0
90Th 
91 Pa  -1 e
parent nucleus (thorium)  daughter nucleus (protactinium) + β- particle (electron)
- note that law of conservation of nucleon is still followed
- the electron emitted does NOT actually exist within the parent nucleus and is NOT one of
the orbital electrons. the electron is created when a neutron decays into a proton and an
electron; when this occurs, the proton number of the parent nucleus increases from Z to Z
+1 and the nucleon number remains unchanged. the electron is usually fast-moving and
escapes from the atom, leaving behind a positively charged atom
- it turns out that the decay of a neutron emits not only a proton and an electron, but also a
neutrino (“little neutral one”, so named by Fermi after being hypothesized by Pauli).
this was discovered when it was found that electrons given off in beta decay have
energies less than that predicted by conservation of energy (and cons.of momentum). the
neutrino is very small in mass and has no charge. the neutrino given off in β- decay is
actually an antineutrino. the neutrino itself is given off in β+ decay.
β+ decay
- in some cases, the particle emitted by the nucleus is a positron, rather than an electron
- a positron, aka β+ particle, 10 e , has the same mass as an electron, but carries a charge of
+e instead of –e.
γ decay
- the nucleus, like electrons, only exists in discrete energy levels. when it goes from an
excited state to a lower state, a photon is emitted. The photon is very high energy and is
called a γ-ray.
- general formula:
3
P   AZ P  
excited energy state  lower energy state + γ-ray
Note that this is NOT transmutation. The identity of the element doesn’t change.
A
Z
-
See example 32-3, 32-4, active example 32-1
Radioactive Decay Series
- when an unstable parent nucleus decays, the resulting daughter nucleus is sometimes also
unstable. if so, the daughter then decays and produces its own daughter, and so on, until
a completely stable nucleus is produced.
- this sequential decay is called a radioactive decay series
- Example radioactive decay series for 238U
Induced Nuclear Reactions
- nuclear reaction – occurs whenever the incident nucleus, particle, or photon causes a
change to occur in a target nucleus:
4
14
17
1
2 He  7 N  8 O  1 H
incident α particle + nitrogen (target)  oxygen + proton
- in this case, the incident particle changes nitrogen to oxygen, so this is called induced
nuclear transmutation
Nuclear Fission
- nuclear fission – splitting of a massive nucleus into two less-massive fragments
- example:
1
235
236
141
92
1
0 n  92 U 
92 U  56 Ba  36 Kr  3 0 n
neutron + U-235  compound nucleus (unstable)  barium + krypton + 3 neutrons
- this is only one possibility for when U-235 undergoes fission.
- Those produced neutrons can the hit another U-235 and cause it to undergo fission.
- This is a chain reaction. read section in book on chain reactions p1071-1072
- See example 32-7
4
Nuclear Fusion
- nuclear fusion – two light nuclei combine to form a more massive nucleus
Particles to know:
Name
hydrogen
deuterium
tritium
1
1
2
1
3
1
H
H
H
deuteron – nucleus of deuterium
triton – nucleus of tritium
#protons
1
1
1
#neutrons
0
1
2
#electrons
1
1
1
1
1
1
2
0
0
positron – “positive electron” – is the antiparticle of an electron; same mass of electron, charge
of +e
Practice
1. Identify Z, N, and A for the following isotopes: (a)
2. What are the nuclear radii of (a) 197
79 Au and (b)
60
27
238
92
U, (b)
239
94
Pu, (c) 144
60 Nd.
Co?
3. Complete the following nuclear reaction:
4
7
1
 2 He  ?
3 Li  1 H 
4. Complete the following nuclear reaction:
230
234
88 Ra  ?
90 Th 
5. Complete the following nuclear reaction:
14
?
 7 N  e   
5
6. Complete the following nuclear reaction, and determine the amount of energy it releases:
3
3
2 He  ? ?
1H 
7. The atomic mass of gold-197 is 196.96654 u. How much energy is required to completely
separate the nucleons in a gold-197 nucleus?
8. The atomic mass of lithium-7 is 7.01435 u. How much energy is required to completely
separate the nucleons in a lithium-7 nucleus?
9. Calculate the average binding energy per nucleon of (a)
56
26
Fe and (b)
238
92
U.
10. Find the number of neutrons released by the following fission reaction:
1
235
101
132
0 n  92 U 
50 Sn  42 Mo  neutrons
11. Complete the following fission reaction and determine the amount of energy it releases:
1
235
1
133
0 n  92 U 
51 Sb  ?50 n
12. Find the energy released in the following fusion reaction:
1
2
 23 He  
1 H 1H 
6
Solutions
238
1. (a) 92 U
Z  92
N  A  Z  238  92  146
A  238
(b) 239
94 Pu
Z  94
N  A  Z  239  94  145
A  239
(c) 144
60 Nd
Z  60
N  A  Z  144  60  84
A  144
2. (a) 197
79 Au
r  (1.2 1015 m) A1/ 3  (1.2 1015 m)(197)1/ 3  7.0 1015 m
60
(b) 27
Co
r  (1.2  1015 m)(60)1/ 3  4.7  1015 m
7
Li  11H  24 He  24 He
3. 3
234
4
Th  230
88 Ra  2 He
4. 90
5.
14
6C
 147 N  e –  
3
3
–
6. 1H  2 He  e  
mi  3.016049 u
mf  3.016029 u
m  mf  mi  3.016029 u  3.016049 u  0.000020 u
 931.494 MeV

c2  2
E  m c 2  (0.000020 u) 
c  19 keV


1u


7.
197
79 Au
mi  196.96654 u
mf  (79)(1.007825 u)  (118)(1.008665 u)  198.640645 u
m  1.67411 u
 931.494 MeV

c2  2
E  mc 2  (1.67411 u) 
c  1559.42 MeV


1u


7
8. 37 Li
mi  7.01435 u
mf  (3)(1.007825 u)  (4)(1.008665 u)  7.058135 u
m  0.04379 u
 931.494 MeV

c2  2
E  mc  (0.04379 u) 
c  40.79 MeV


1u


2
9. (a) 56
26 Fe
mi  55.934939 u
mf  26(1.007825 u)  30(1.008665 u)  56.463400 u
m  56.463400 u  55.934939 u  0.528461 u
 931.494 MeV
 2
c2
(0.528461 u) 
c
1u
2


m c
E

  8.79033 MeV/nucleon


56
56
56
(b) 238
92 U
mi  238.050786 u
mf  92(1.007825 u)  146(1.008665 u)  239.984990 u
m  239.984990 u  238.050786 u  1.934204 u
 931.494 MeV
 2
c2
(1.934204 u) 
c
1
u
2


m c
E

  7.57017 MeV/nucleon


238
238
238
1
235
132
101
10. 0 n  92 U  50 Sn  42 Mo  neutrons
Sn neutrons = 132 50 = 82
Mo neutrons = 101  42 = 59
U neutrons = 235  92 = 143
neutrons released = 1 + 143  82  59 = 3
133
1
11. 01n  235
92 U  51Sb  ?  5 0 n
U neutrons = 235  92 = 143
Sb neutrons = 133  51 = 82
Z  92  51  41
N  1  143  82  5  57
A  41  57  98
The missing atom is
98
41 Nb
.
mi  1.008665 u  235.043925 u  236.052590 u
mf  132.915237 u  97.910331 u  5(1.008665 u)  235.868893 u
m  235.868893 u  236.052590 u  0.183697 u
 931.494 MeV

c2  2
E  m c 2  (0.183697 u) 
c  171.113 MeV
12.


1u


1
2
3
1 H  1 H  2 He  
mi  1.007825 u  2.014102 u  3.021927 u
mf  3.016029 u
m  3.016029 u  3.021927 u  0.005898 u
 931.494 MeV

c2  2
E  m c 2  (0.005898 u) 
c  5.494 MeV


1u


8