Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
Chapter 39. Nuclear Physics and the Nucleus
The Elements
39-1. How many neutrons are in the nucleus of
N/Z?
208
82
Pb ? How many protons? What is the ratio
(N is the number of neutrons and Z is the number of protons.)
A = N + Z;
N = A – Z = 126 neutrons;
A
 1.54
Z
Z = 82 protons;
39-2. The nucleus of a certain isotope contains 143 neutrons and 92 protons. Write the symbol
for this nucleus.
235
92
A = N + Z = 143 + 92 = 235; Z = 92:
U
39-3. From a stability curve it is determined that the ratio of neutrons to protons for a cesium
nucleus is 1.49. What is the mass number for this isotope of cesium?
Z = 55;
N
 1.49;
Z
N  1.49(55)  81.95; A  N  Z ;
A = 137
39-4. Most nuclei are nearly spherical in shape and have a radius that may be approximated by
1
r  r0 A3
r0  1.2 x 1015 m
What is the approximate radius of the nucleus of a gold atom
r  (1.2 x1015 m) 3 197  6.98 x 10-15 m ;
197
79
Au ?
r = 6.98 x 10-15 m
39-5. Study Table 39-4 for information on the several nuclides. Determine the ratio of N/Z for
the following nuclides: Beryllium-9, Copper-64, and Radium 224.
Beryllium: A = 9; Z = 4; N = 9 – 4 = 5;
538
N
 1.25
Z
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
39-5. (Cont.)
Copper: A = 64; Z = 29; N = 64 – 29 = 35;
N
 1.21
Z
Radium: A = 224; Z = 88; N = 224 – 88 = 136;
N
 1.55
Z
The Atomic Mass Unit
39-6. Find the mass in grams of a gold particle containing two million atomic mass units?
 1.66 x 10-27 kg 
m  2 x 106 u 
;
1.00
u


m = 3.32 x 10-21 kg
39-7. Find the mass of a 2-kg copper cylinder in atomic mass units? In Mev? In joules?


1u
m  2 kg 
;
-27
 1.6606 x 10 kg 
 931 MeV 
m = 1.204 x 1027 u 
;
 1u

m = 1.20 x 1027 u
m = 1.12 x 1030 MeV
 1.6 x 10-13J 
m = 1.12 x 10 MeV 
;
 1 MeV 
30
m = 1.79 x 1017 J
39-8. A certain nuclear reaction releases an energy of 5.5 MeV. How much mass (in atomic
mass units) is required to produce this energy?
 1u

m  5.5 MeV 
;
 931 MeV 
m = 0.00591 u
39-9. The periodic table gives the average mass of a silver atom as 107.842 u. What is the
average mass of the silver nucleus?
(Recall that me = 0.00055 u )
For silver, Z = 47. Thus, the nuclear mass is reduced by the mass of 47 electrons.
m = 107.842 u – 47(0.00055 u);
539
m = 107.816 u
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-10. Consider the mass spectrometer as illustrated by Fig. 39-3. A uniform magnetic field of
0.6 T is placed across both upper and lower sections of the spectrometer., and the electric
field in the velocity selector is 120 V/m. A singly charged neon atom (+1.6 x 10-19 C) of
mass 19.992 u passes through the velocity selector and into the spectrometer. What is the
velocity of the neon atom as it emerges from the velocity selector?
v
E 120, 000 V/m

;
B
0.6 T
v = 2 x 105 m/s
R
*39-11. What is the radius of the circular path followed by the
neon atom of Problem 19-10?
B = 0.6 T into paper
 1.66 x 10-27 kg 
26
m  19.992 u 
  3.32 x10 kg
1
u


mv 2
 qvB;
R
mv
;
R
qB
(3.32 x 10-27 kg)(2 x 105m/s)
R
;
(1.6 x 10-19C)(0.600 T)
R = 6.92 cm
Mass Defects and Binding Energy
(Refer to Table 39-4 for Nuclidic Masses.)
*39-12. Calculate the mass defect and binding energy for the neon-20 atom
20
10
Ne .
mD  [(ZmH  Nmn )]  M  10(1.007825 u)  10(1.008665 u) 19.99244 u
mD = 20.016490 u –19.99244 u ;
mD = 0.17246 u
 931 MeV 
E  mD c 2  (0.17246 u) 
  160.6 MeV ;
 1u

E = 161 MeV
*39-13. Calculate the binding energy and the binding energy per nucleon for tritium 13 H . How
much energy in joules is required to tear the nucleus apart into its constituent nucleons?
mD  [(ZmH  Nmn )]  M  1(1.007825 u)  2(1.008665 u)  3.016049 u
540
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
39-13. (Cont.) mD = 0.009106 u;
E = 8.48 MeV;
 931 MeV 
E  mD c 2  (0.009106 u) 
  8.48 MeV
 1u

EB 8.48 MeV

;
A
3
EB
 2.83 MeV/nucleon
A
Energy to tear apart: EB = 8.48 MeV(1.6 x 10-13 J/MeV) = 1.36 x 10-12 J
*39-14. Calculate the mass defect of 73 Li . What is the binding energy per nucleon?
mD  [(ZmH  Nmn )]  M  3(1.007825 u)  4(1.008665 u)  7.016930 u
mD = 7.058135 u –7.016930 u ;
mD = 0.041205 u
 931 MeV 
E  mD c 2  (0.041205 u) 
  38.4 MeV ;
 1u

EB 38.4 MeV

;
A
7
E = 38.4 MeV
EB
 5.48 MeV/nucleon
A
*39-15. Determine the binding energy per nucleon for carbon-12 (
12
6
C ).
mD  6(1.007825 u)  6(1.008665 u) 12.0000 u  0.09894 u
 931 MeV 
E  mD c 2  (0.09894 u) 
;
 1u

EB 92.1 MeV

= 7.68 MeV/nucleon
A
12
*39-16. What is the mass defect and the binding energy for a gold atom
197
79
Au ?
mD  79(1.007825 u)  118(1.008665 u) 196.966541 u  1.674104 u
 931 MeV 
E  mD c 2  (1.674104 u) 
  1.56 GeV;
 1u

EB 1.56 GeV

= 7.91 MeV/nucleon
A
197
541
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-17. Determine the binding energy per nucleon for tin-120 (
120
50
Sn ).
mD  50(1.007825 u)  70(1.008665 u) 119.902108 u  1.09569 u
EB  (1.09569 u)(931 MeV/u 

;
A 
120

EB
 8.50 MeV/nucleon
A
Radioactivity and Nuclear Decay
39-18. The activity of a certain sample is rated as 2.8 Ci. How many nuclei will have
disintegrated in a time of one minute?
 3.7 x 1010s-1 
Nuclei  2.8 Ci 
 (60 s);
1 Ci


60
27
39-19. The cobalt nucleus
nuclei = 6.22 x 1012 nuclei
Co emits gamma rays of approximately 1.2 MeV. How much mass
is lost by the nucleus when it emits a gamma ray of this energy?
 1u

E  1.2 MeV 
;
 931 MeV 
m = 0.00129 u
39-20. The half-life of the radioactive isotope indium-109 is 4.30 h. If the activity of a sample is
1 mCi at the start, how much activity remains after 4.30, 8.60, and 12.9 h?
1
R  R0  
2
1
R  R0  
2
1
R  R0  
2
t / T½
1
t
4.3 h

 1;
T½ 4.3 h
1
R  (1 mCi)   ;
2
;
t
8.6 h

 2;
T½ 4.3 h
1
R  (1 mCi)   ;
2
;
t 12.9 h

 3;
T½ 4.3 h
1
R  (1 mCi)   ;
2
;
t / T½
t / T½
R = 0.5 mCi
2
542
R = 0.25 mCi
3
R = 0.125 mCi
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
39-21. The initial activity of a sample containing 7.7 x 1011 bismuth-212 nuclei is 4.0 mCi. The
half-life of this isotope is 60 min. How many bismuth-212 nuclei remain after 30 min?
What is the activity at the end of that time?
t / T½
1
N  N0  
2
½
t
30 min
1
;

 0.5; N  (7.7 x 1011 )   = 5.44 x 1011 nucl.
T½ 60 min
2
1
R  R0  
2
t / T½
½
1
R  (4 mCi)   ;
2
;
R = 2.83 mCi
*39-22. Strontium-90 is produced in appreciable quantities in the atmosphere during a nuclear
explosion. If this isotope has a half-life of 28 years, how long will it take for the initial
activity to drop to one-fourth of its original activity?
n
1
R  R0   ;
2
n
n
R 1 1
   ; n2
R0  2 
4
t
t

 2;
T½ 28 yr
t  2(28 yr); t = 56 yr
*39-23. Consider a pure, 4.0-g sample of radioactive Gallium-67. If the half-life is 78 h, how
much time is required for 2.8 g of this sample to decay?
When 2.8 g decay, that leaves 4 g – 2.8 g or 1.20 g remaining.
n
1
m  m0   ;
2
m 1.2 g

 0.300;
m0
4g
n
1
   0.300
2
The solution is accomplished by taking the common log of both sides:
n log(0.5)  log(0.300);
n
 0.301n  0.523;
t
t

 1.74;
T½ 78 h
t = 1.74 (78 h)
543
n
0.523
 1.74
0.301
t = 135 h
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-24. If one-fifth of a pure radioactive sample remains after 10 h, what is the half-life?
n
1 1
  ;
5 2
0.200 = (0.5) n ; n log(0.5)  log(0.2);
n
t 10 h

 2.32;
T½
T½
T½ 
10 h
;
2.32
 0.301 n  0.699
T½ = 4.31 h
Nuclear Reactions
*39-25. Determine the minimum energy released in the nuclear reaction
19
9
The atomic mass of
E
19
9
19
9
F 11 H 24 He + 16
8 O + energy
F is 18.998403 u,
4
2
He  4.002603 u,
1
1
H = 1.007824 u.
F + 11H - 42 He - 16
(Energy comes from mass defect)
8 O ;
E = 18.998403 u + 1.007825 u – 4.002603 – 15.994915 u = 0.0087 u
 931 MeV 
E  (0.00871 u) 
;
 1u

E = 8.11 MeV
*39-26. Determine the approximate kinetic energy imparted to the alpha particle when Radium226 decays to form Radon-222. Neglect the energy imparted to the radon nucleus.
226
88
Ra 
222
86
Rn + 24 He + Energy;
226
88
Ra = 226.02536
 931 MeV 
E  226.02536 u - 222.017531 u - 4.002603 u = 0.00523 u 
 = 4.87 MeV
 1u

*39-27. Find the energy involved in the production of two alpha particles in the reaction
7
3
Li  11H  42 He  42 He + energy
 931 MeV 
E  7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u 
 = 17.3 MeV
 1u

544
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-28. Compute the kinetic energy released in the beta minus decay of thorium-233.
233
90
Th 
233
91
Pa + 0+1  + energy;
233
90
Th = 233.041469 u;
233
91
Pa = 233.040130
E  233.041469 u - 233.040130 u - 0.00055 u = 0.000789 u
 931 MeV 
E  0.000789 u 
;
 1u

E = 0.735 MeV
*39-29. What must be the energy of an alpha particle as it bombards a Nitrogen-14 nucleus
producing
17
8
O and 11H? (17
8 O = 16.999130 u) .
4
2
He + 14
7 N + Energy 
17
8
O + 11H
 931 MeV 
E  16.999130 u + 1.007825 u - 14.003074 - 4.002603 u = 0.001278 u 

 1u

E = 1.19 MeV
This is the Threshold Energy for the reaction
Challenge Problems
*39-30. What is the average mass in kilograms of the nucleus of a boron-11 atom?
 1.66 x 10-27 kg 
m  11.009305 u 
;
1.00
u


m = 1.83 x 10-26 kg
*39-31. What are the mass defect and the binding energy per nucleon for boron-11?
mD  [(ZmH  Nmn )]  M  5(1.007825 u)  6(1.008665 u) 11.009305 u
mD = 11.09112 u –11.009305 u ;
mD = 0.08181 u
 931 MeV 
E  mD c 2  (0.08181 u) 
  76.2 MeV ;
 1u

EB 76.2 MeV

;
A
11
E = 76.2 MeV
EB
 6.92 MeV/nucleon
A
545
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-32. Find the binding energy per nucleon for Thallium-206.
mD  81(1.007825 u)  125(1.008665 u)  205.976104 u  1.740846 u
 931 MeV 
E  mD c 2  (1.740846 u) 
;
 1u

EB 1621 MeV

= 7.87 MeV/nucleon
A
206
*39-33. Calculate the energy required to separate the nucleons in mercury-204.
mD  80(1.007825 u)  124(1.008665 u)  203.973865 u  1.7266 u
 931 MeV 
E  mD c 2  (1.7266 u) 
;
 1u

EB = 1610 MeV
*39-34. The half-life of a radioactive sample is 6.8 h. How much time passes before the activity
drops to one-fifth of its initial value?
n
R 1 1
   ;
R0 5  2 
n
0.200 = (0.5) n ; n log(0.5)  log(0.2);
0.699
t
6.8 h
 2.32; n 

 2.32;
0.301
T½
T½
T½ 
6.8 h
;
2.32
 0.301 n  0.699
T½ = 2.93 h
*39-35. How much energy is required to tear apart a deuterium atom? (12 H  2.014102 u)
mD  1(1.007825 u)  1(1.008665 u)  2.014102 u  0.002388 u
 931 MeV 
E  mD c 2  (0.002388 u) 
;
 1u

546
EB = 2.22 MeV
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-36. Plutonium-232 decays by alpha emission with a half-life of 30 min. How much of this
substance remains after 4 h if the original sample had a mass of 4.0 g? Write the equation
for the decay?
1
m  m0  
2
t / T½
t
4h

 8.00;
T½ 0.5 h
;
232
94
Pu 
228
92
8
1
m  (4 g)   ;
2
m = 15.6 mg
U + 42 He + energy
*39-37. If 32 x 109 atoms of a radioactive isotope are reduced to only 2 x 109 atoms in a time of
48 h, what is the half-life of this material?
n
N
2 x 109  1 

   ; 0.0625 = (0.5) n ; n log(0.5)  log(0.0625);  0.301 n  1.204
9
N0 32 x 10  2 
n
1.204
t
48 h
 4.00; n 

 4.00;
0.301
T½
T½
T½ 
48 h
;
4.00
T½ = 12.0 h
*39-38. A certain radioactive isotope retains only 10 percent of its original activity after a time of
4 h. What is the half-life?
n
N
1 1
    ; 0.100 = (0.5) n ; n log(0.5)  log(0.100);  0.301 n  1.00
N0 10  2 
n
1.00
t
4h
 3.32; n 

 3.32;
0.301
T½ T½
T½ 
4h
;
3.32
T½ = 72.3 min
*39-39. When a 63 Li nucleus is struck by a proton, an alpha particle and a produce nucleus are
released. Write the equation for this reaction. What is the net energy transfer in this case?
6
3
Li  11H  42 He  32 He + energy
 931 MeV 
E  6.015126 u + 1.007825 u - 4.002603 u - 3.016030 = 0.00432 u 
 = 4.02 MeV
 1u

547
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-40. Uranium-238 undergoes alpha decay. Write the equation for the reaction and calculate
the disintegration energy.
238
92
U
224
90
Th + 24 He + energy
E  238.05079 u - 234.04363 u - 4.002603 u = 0.00456 u
 931 MeV 
E  0.00456 u 
;
 1u

E = 4.24 MeV
*39-41. A 9-g sample of radioactive material has an initial activity of 5.0 Ci. Forty minutes later,
the activity is only 3.0 Ci. What is the half-life? How much of the pure sample remains?
n
R 3 Ci  1 

  ;
R0 5 Ci  2 
n
0.600 = (0.5)n ; n log(0.5)  log(0.6);
0.222
t
40 min
 0.737; n 

 0.737;
0.301
T½
T½
n
1
1
m  m0    (9 g)  
2
2
T½ 
40 min
;
0.737
 0.301 n  0.222
T½ = 54.3 min
0.737
m = 5.40 g
Critical Thinking Problems
*39-42. Nuclear fusion is a process that can produce enormous energy without the harmful
byproducts of nuclear fission. Calculate the energy released in the following nuclear
fusion reaction:
3
2
H 32 He 24 He 11 H+11H  energy
 931 MeV 
E  2(3.016030 u) - 4.002603 u - 2(1.007825 u) = 0.013807 u 

 1u

E = 12.9 MeV
548
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-43. Carbon-14 decays very slowly with a half-life of 5740 years. Carbon dating can be
accomplished by seeing what fraction of carbon-14 remains, assuming the decay process
began with the death of a living organism. What would be the age of a chunk of charcoal
if it was determined that the radioactive C-14 remaining was only 40 percent of what
would be expected in a living organism?
n
R
1
 0.40    ;
R0
2
n
0.400 = (0.5) n ; n log(0.5)  log(0.4);
0.399
t
t
 0.737; n 

 1.32;
0.301
T½ 5740 yr
 0.301 n  0.399
t  1.32(5740 yr); t = 7590 yr
*39-44. The velocity selector in a mass spectrometer has a magnetic field of 0.2 T perpendicular
to an electric field of 50 kV/m. The same magnetic field is across the lower region. What
is the velocity of singly charged Lithium-7 atoms as they leave the selector? If the radius
of the path in the spectrometer is 9.10 cm, find the atomic mass of the lithium atom?
v
E 50, 000 V/m

;
B
0.2 T
v = 2.5 x 105 m/s
mv 2
qBR
(1.6 x 10-19C)(0.2 T)(0.091 m)
 qvB; m 
; m
;
R
v
(2.5 x 105m/s)


1u
m = 1.165 x 10-26 kg 
;
-27
 1.66 x 10 kg 
m = 7.014 u
R
B = 0.2 T into paper
*39-45. A nuclear reactor operates at a power level of 2.0 MW. Assuming that approximately
200 MeV of energy is released for a single fission of U-235, how many fission processes
are occurring each second in the reactor?
P
2 x 106 J/s
1.25 x 1019 MeV/s
19

1.25
x
10
MeV/s;
 6.25 x 10216 fissions/s
-19
1.6 x 10 J/MeV
200 MeV/fission
549
Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
*39-46. Consider an experiment which bombards
14
7
N with an alpha particle. One of the two
product nuclides is 11 H . The reaction is
4
2
A
1
He 14
7 N  Z X 1 H
What is the product nuclide indicated by the symbol X? How much kinetic energy must
the alpha particle have in order to produce the reaction?
Conservation of nucleons means:
4
2
He 14
7 N + enery 
17
8
O 11 H
E  16.99913 u + 1.007825 u - 4.002603 u - 14.003074 u = 0.001282 u
 931 MeV 
E  0.001282 u 
;
 1u

E = 1.19 MeV
*39-47. When passing a stream of ionized lithium atoms through a mass spectrometer, the radius
of the path followed by 73 Li (7.0169 u) is 14.00 cm. A lighter line is formed by the 63 Li
(6.0151 u). What is the radius of the path followed by the
6
3
Li isotopes?
mv 2
mv
 qvB; R 
; (See Problems 39-10 and 39-11)
R
qB
m1 R1
 ;
m2 R2
R2 
m2 R1 (6.0151 u)(14 cm)

;
m1
7.0169 u
550
R2 = 6.92 cm