1. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
Chapter 39. Nuclear Physics and the Nucleus
The Elements
208
39-1. How many neutrons are in the nucleus of 82 Pb ? How many protons? What is the ratio
N/Z? (N is the number of neutrons and Z is the number of protons.)
A
A = N + Z; N = A – Z = 126 neutrons; Z = 82 protons; = 1.54
Z
39-2. The nucleus of a certain isotope contains 143 neutrons and 92 protons. Write the symbol
for this nucleus.
235
A = N + Z = 143 + 92 = 235; Z = 92: 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?
N
Z = 55; = 1.49; N = 1.49(55) = 81.95; A = N + Z ; A = 137
Z
39-4. Most nuclei are nearly spherical in shape and have a radius that may be approximated by
1
r = r0 A 3 r0 = 1.2 x 10−15 m
197
What is the approximate radius of the nucleus of a gold atom 79 Au ?
r = (1.2 x10−15 m) 3 197 = 6.98 x 10-15 m ; 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.
N
Beryllium: A = 9; Z = 4; N = 9 – 4 = 5; = 1.25
Z
26
2. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
N
39-5. (Cont.) Copper: A = 64; Z = 29; N = 64 – 29 = 35; = 1.21
Z
N
Radium: A = 224; Z = 88; N = 224 – 88 = 136; = 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 ; m = 3.32 x 10-21 kg
1.00 u
39-7. Find the mass of a 2-kg copper cylinder in atomic mass units? In Mev? In joules?
1u
m = 2 kg -27 ; m = 1.20 x 1027 u
1.6606 x 10 kg
931 MeV
m = 1.204 x 1027 u ; m = 1.12 x 1030 MeV
1u
30 1.6 x 10-13 J
m = 1.12 x 10 MeV ; m = 1.79 x 1017 J
1 MeV
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 ; m = 0.00591 u
931 MeV
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); m = 107.816 u
27
3. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
*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?
E 120, 000 V/m
v= = ; v = 2 x 105 m/s
B 0.6 T
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
1u
mv 2 mv (3.32 x 10-27 kg)(2 x 105 m/s)
= qvB; R= ; R= ; R = 6.92 cm
R qB (1.6 x 10-19 C)(0.600 T)
Mass Defects and Binding Energy
(Refer to Table 39-4 for Nuclidic Masses.)
20
*39-12. Calculate the mass defect and binding energy for the neon-20 atom 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 ; E = 161 MeV
1u
3
*39-13. Calculate the binding energy and the binding energy per nucleon for tritium 1 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
28
4. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
931 MeV
39-13. (Cont.) mD = 0.009106 u; E = mD c 2 = (0.009106 u) = 8.48 MeV
1u
EB 8.48 MeV EB
E = 8.48 MeV; = ; = 2.83 MeV/nucleon
A 3 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 7 Li . What is the binding energy per nucleon?
3
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 ; E = 38.4 MeV
1u
EB 38.4 MeV EB
= ; = 5.48 MeV/nucleon
A 7 A
12
*39-15. Determine the binding energy per nucleon for carbon-12 ( 6 C ).
mD = [ 6(1.007825 u) + 6(1.008665 u) ] − 12.0000 u = 0.09894 u
931 MeV EB 92.1 MeV
E = mD c 2 = (0.09894 u) ; = = 7.68 MeV/nucleon
1u A 12
197
*39-16. What is the mass defect and the binding energy for a gold atom 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
29
5. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
120
*39-17. Determine the binding energy per nucleon for tin-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 EB
= ; = 8.50 MeV/nucleon
A 120 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); nuclei = 6.22 x 1012 nuclei
1 Ci
60
39-19. The cobalt nucleus 27 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 ; m = 0.00129 u
931 MeV
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?
t / T½ 1
1 t 4.3 h 1
R = R0 ; = = 1; R = (1 mCi) ; R = 0.5 mCi
2 T½ 4.3 h 2
t / T½ 2
1 t 8.6 h 1
R = R0 ; = = 2; R = (1 mCi) ; R = 0.25 mCi
2 T½ 4.3 h 2
t / T½ 3
1 t 12.9 h 1
R = R0 ; = = 3; R = (1 mCi) ; R = 0.125 mCi
2 T½ 4.3 h 2
30
6. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
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 ½
1 t 30 min 1
N = N0 ; = = 0.5; N = (7.7 x 1011 ) = 5.44 x 1011 nucl.
2 T½ 60 min 2
t / T½ ½
1 1
R = R0 ; R = (4 mCi) ; R = 2.83 mCi
2 2
*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 n
1 R 1 1
R = R0 ; = = ; n=2
2 R0 2 4
t t
n= = = 2; t = 2(28 yr); t = 56 yr
T½ 28 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 n
1 m 1.2 g 1
m = m0 ; = = 0.300; = 0.300
2 m0 4g 2
The solution is accomplished by taking the common log of both sides:
−0.523
n log(0.5) = log(0.300); − 0.301n = −0.523; n= = 1.74
−0.301
t t
n= = = 1.74; t = 1.74 (78 h) t = 135 h
T½ 78 h
31
7. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
*39-24. If one-fifth of a pure radioactive sample remains after 10 h, what is the half-life?
n
1 1
= ; 0.200 = (0.5) n ; n log(0.5) = log(0.2); − 0.301 n = −0.699
5 2
t 10 h 10 h
n= = = 2.32; T½ = ; T½ = 4.31 h
T½ T½ 2.32
Nuclear Reactions
*39-25. Determine the minimum energy released in the nuclear reaction
19
9 F +1 H →4 He + 16 O + energy
1
2 8
The atomic mass of 19
9 F is 18.998403 u, 4
2 He = 4.002603 u, 1
1H = 1.007824 u.
E= 19
9 F + 1H - 2 He - 16 O ; (Energy comes from mass defect)
1 4
8
E = 18.998403 u + 1.007825 u – 4.002603 – 15.994915 u = 0.0087 u
931 MeV
E = (0.00871 u) ; E = 8.11 MeV
1u
*39-26. Determine the approximate kinetic energy imparted to the alpha particle when
Radium-226 decays to form Radon-222. Neglect the energy imparted to the radon
nucleus.
226
88 Ra → 222
86 Rn + 4 He + Energy;
2
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 + 1H → 2 He + 2 He + energy
1 4 4
32
8. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
931 MeV
E = 7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u = 17.3 MeV
1u
*39-28. Compute the kinetic energy released in the beta minus decay of thorium-233.
233
90 Th → 233
91 Pa + 0 β + energy;
+1
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 ; E = 0.735 MeV
1u
*39-29. What must be the energy of an alpha particle as it bombards a Nitrogen-14 nucleus
17
producing 8 O and 1H? (17 O = 16.999130 u) .
1 8
4
2 He + 14 N + Energy →
7
17
8
1
O + 1H
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 ; m = 1.83 x 10-26 kg
1.00 u
*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 ; E = 76.2 MeV
1u
33
9. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
EB 76.2 MeV EB
= ; = 6.92 MeV/nucleon
A 11 A
*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 EB 1621 MeV
E = mD c 2 = (1.740846 u) ; = = 7.87 MeV/nucleon
1u 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) ; EB = 1610 MeV
1u
*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
= = ; 0.200 = (0.5) n ; n log(0.5) = log(0.2); − 0.301 n = −0.699
R0 5 2
0.699 t 6.8 h 6.8 h
n= = 2.32; n = = = 2.32; T½ = ; T½ = 2.93 h
0.301 T½ T½ 2.32
*39-35. How much energy is required to tear apart a deuterium atom? (1 H = 2.014102 u)
2
mD = [ 1(1.007825 u) + 1(1.008665 u) ] − 2.014102 u = 0.002388 u
931 MeV
E = mD c 2 = (0.002388 u) ; EB = 2.22 MeV
1u
34
10. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
*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?
t / T½ 8
1 t 4h 1
m = m0 ; = = 8.00; m = (4 g) ; m = 15.6 mg
2 T½ 0.5 h 2
232
94 Pu → 228
92 U + 4 He + energy
2
*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
= 9
= ; 0.0625 = (0.5) n ; n log(0.5) = log(0.0625); − 0.301 n = −1.204
N 0 32 x 10 2
1.204 t 48 h 48 h
n= = 4.00; n = = = 4.00; T½ = ; T½ = 12.0 h
0.301 T½ T½ 4.00
*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
N 0 10 2
1.00 t 4h 4h
n= = 3.32; n = = = 3.32; T½ = ; T½ = 72.3 min
0.301 T½ T½ 3.32
*39-39. When a 6 Li nucleus is struck by a proton, an alpha particle and a produce nucleus are
3
released. Write the equation for this reaction. What is the net energy transfer in this case?
6
3 Li + 1H → 2 He + 3 He + energy
1 4
2
35
11. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
931 MeV
E = 6.015126 u + 1.007825 u - 4.002603 u - 3.016030 = 0.00432 u = 4.02 MeV
1u
*39-40. Uranium-238 undergoes alpha decay. Write the equation for the reaction and calculate
the disintegration energy.
238
92 U→ 224
90
4
Th + 2 He + energy
E = 238.05079 u - 234.04363 u - 4.002603 u = 0.00456 u
931 MeV
E = 0.00456 u ; E = 4.24 MeV
1u
*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
= = ; 0.600 = (0.5) n ; n log(0.5) = log(0.6); − 0.301 n = −0.222
R0 5 Ci 2
0.222 t 40 min 40 min
n= = 0.737; n = = = 0.737; T½ = ; T½ = 54.3 min
0.301 T½ T½ 0.737
n 0.737
1 1
m = m0 = (9 g) m = 5.40 g
2 2
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 + 3 He →2 He +1 H+1H + energy
2
4 1 1
931 MeV
E = 2(3.016030 u) - 4.002603 u - 2(1.007825 u) = 0.013807 u
1u
36
12. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
E = 12.9 MeV
*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 = ; 0.400 = (0.5) n ; n log(0.5) = log(0.4); − 0.301 n = −0.399
R0 2
0.399 t t
n= = 0.737; n = = = 1.32; t = 1.32(5740 yr); t = 7590 yr
0.301 T½ 5740 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?
E 50, 000 V/m
v= = ; v = 2.5 x 105 m/s
B 0.2 T
mv 2 qBR (1.6 x 10-19 C)(0.2 T)(0.091 m) R
= qvB; m = ; m= ;
R v (2.5 x 105 m/s)
1u
m = 1.165 x 10-26 kg -27 ; m = 7.014 u B = 0.2 T into paper
1.66 x 10 kg
*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?
37
13. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition
2 x 106 J/s 1.25 x 1019 MeV/s
P= = 1.25 x 1019 MeV/s; = 6.25 x 10216 fissions/s
1.6 x 10-19 J/MeV 200 MeV/fission
14
*39-46. Consider an experiment which bombards 7 N with an alpha particle. One of the two
product nuclides is 1 H . The reaction is
1
4
2 He +14 N → Z X +1 H
7
A
1
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 N + enery →
7
17
8 O +1 H
1
E = 16.99913 u + 1.007825 u - 4.002603 u - 14.003074 u = 0.001282 u
931 MeV
E = 0.001282 u ; E = 1.19 MeV
1u
*39-47. When passing a stream of ionized lithium atoms through a mass spectrometer, the radius
of the path followed by 7 Li (7.0169 u) is 14.00 cm. A lighter line is formed by the 6 Li
3 3
6
(6.0151 u). What is the radius of the path followed by the 3 Li isotopes?
mv 2 mv
= qvB; R = ; (See Problems 39-10 and 39-11)
R qB
m1 R1 m2 R1 (6.0151 u)(14 cm)
= ; R2 = = ; R2 = 6.92 cm
m2 R2 m1 7.0169 u
38