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Chapter 29
NUCLEAR PHYSICS
Problems
1. Strategy Nucleons have masses of approximately 1 u, so dividing the mass of the person by 1 u will give an
estimate of the total number of nucleons in the person’s body.
Solution Estimate the number of nucleons.
75 kg
75 kg

 4.5 1028
1u
1.660 539 1027 kg
9. Strategy A  107 for technetium-107. Find the radius of the nucleus using Eq. (29-4). Use the volume of a sphere
to find the volume of the nucleus.
Solution Find the radius.
r  r0 A1/ 3  (1.2 fm)(107)1/ 3  5.7 fm
Find the volume.
4
4
4
V   r 3   r03 A   (1.2 1015 m)3 (107)  7.7 10 43 m3
3
3
3
17. Strategy The nucleon number A is the sum of the total number of protons Z and neutrons N. Use Eq. (29-7) to
find the mass defect.
Solution The
defect.
14
N atom has 7 protons, 7 neutrons, 7 electrons, and a mass of 14.003 074 0 u. Find the mass
m  (mass of 7 1H atoms  mass of 7 neutrons)  (mass of 14 N atom)
 7 1.007 825 0 u  7 1.008 664 9 u  14.003 074 0 u  0.112 355 3 u
29. Strategy In Problem 21, the daughter nuclide in this decay was found to be 40
20 Ca. The maximum kinetic energy
of the beta particle is equal to the disintegration energy.
Solution The reaction is
40
19 K
40
 20
Ca  01 e  00
40
The atomic masses of 19
K and 40
20 Ca are 39.963 998 7 u and 39.962 591 2 u, respectively. To get the masses of
the nuclei, we subtract Zme from each. The mass of the electron is 0.000 548 6 u, and the neutrino’s mass is
negligible. The mass difference is
m  [(M Ca  20me )  me ]  (M K  19me )  M Ca  M K  39.962 591 2 u  39.963 998 7 u  0.001 407 5 u.
The disintegration energy is E  m c 2  0.001 407 5 u  931.494 MeV u  1.3111 MeV.
The maximum kinetic energy of the   particle is 1.3111 MeV .
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College Physics
Chapter 29: Nuclear Physics
39. Strategy The ratio of C-14 to C-12 in the bone is 1/4 as much as in a living sample. The ratio is reduced by a
factor of 1/2 for each half-life.
Solution Since 22  1 4, we conclude that the age of the bone is 2 half-lives, or 2  5730 yr  11,500 yr .
53. (a) Strategy The mass numbers on the two sides of the reaction must be equal. Let x be the number of neutrons.
Solution Find the number of neutrons.
235  1  141  93  x, so x  2 .
(b) Strategy From Figure 29.2, the binding energies per nucleon of 235 U, 141Cs, and 93Rb are approximately
7.6 MeV, 8.35 MeV, and 8.7 MeV, respectively. The energy released is equal to the increase in the binding
energy.
Solution The binding energies of the three nuclides are as follows:
U: 235  7.6 MeV  1786 MeV, 141Cs: 141 8.35 MeV  1177 MeV,
Find the binding energy.
1177 MeV  809 MeV 1786 MeV  200 MeV
The energy released is approximately 200 MeV.
235
93
Rb: 93  8.7 MeV  809 MeV.
141
93
(c) Strategy The atomic masses of 235
92 U, 55Cs, and 37Rb are 235.043 923 1 u, 140.920 044 0 u, and
92.922 032 8 u, respectively. Atomic masses can be used, since both sides include the same number of
electrons (92).
Solution Find the change in mass.
m  140.920 044 0 u  92.922 032 8 u  2 1.008 664 9 u  235.043 9231 u 1.008 664 9 u  0.193181 4 u
The energy released is E  m c2  0.193181 4 u  931.494 MeV u  179.947 MeV .
(d) Strategy Divide the energy released by the rest energy.
Solution
E
m
0.193181 4 u


 0.000 822
E0
m
235.043 9231 u
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Chapter 29: Nuclear Physics
College Physics
69. Strategy The original number of samarium-147 nuclei in the rock is equal to the current number plus the number
of neodymium-143. The number of nuclei is related to the mass by N  mN A M , where N A is Avogadro’s
number and M is the molar mass. Use Eq. (29-23).
Solution Find the age of the rock.
t T
N  N0  2 1 2
t T
NSm  ( NSm  N Nd )  2 1 2
NSm
log 2
NSm  N Nd
NSm
NSm  N Nd
2
t T1 2
 log 2 2
t T1 2
t  T1 2 log 2
t  T1 2 log 2

t
T1 2
NSm  N Nd
NSm
mSm N A M Sm  mNd N A M Nd
mSm N A M Sm
 m M 
 (0.150)(146.914 897 9) 
9
t  T1 2 log 2 1  Nd Sm   (1.06  1011 yr) log 2 1 
  7.67 10 yr
m
M
(3.00)(142.909
814
3)


Sm Nd 

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