Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition
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 A 3
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;
26
N = 1.25 Z
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition
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 10-13 J 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);
27
m = 107.816 u
Chapter 39. Nuclear Physics and the NucleusPhysics, 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? 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 105 m/s) R= ; (1.6 x 10-19 C)(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
28
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition
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
29
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *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
39-19. The cobalt nucleus
60 27
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
R = 0.25 mCi
3
30
R = 0.125 mCi
Chapter 39. Nuclear Physics and the NucleusPhysics, 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? 1 N = N0 2
t / T½
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)
31
n=
−0.523 = 1.74 −0.301
t = 135 h
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *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 →42 He + 16 8 O + energy
F is 18.998403 u,
4 2
He = 4.002603 u,
1 1
H = 1.007824 u.
F + 11H - 24 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 Radium-226 decays to form Radon-222. Neglect the energy imparted to the radon nucleus. 226 88
Ra →
222 86
Rn + 42 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 → 24 He + 24 He + energy
931 MeV E = 7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u = 17.3 MeV 1u
32
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *39-28. Compute the kinetic energy released in the beta minus decay of thorium-233. 233 90
Th →
233 91
Pa + 0+1 β + 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 33
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *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
34
EB = 2.22 MeV
Chapter 39. Nuclear Physics and the NucleusPhysics, 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? 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 N 0 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 N 0 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 → 24 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
35
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *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 →42 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
36
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *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
(1.6 x 10-19 C)(0.2 T)(0.091 m) mv 2 qBR m = ; ; = qvB; m = (2.5 x 105 m/s) R v 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
37
Chapter 39. Nuclear Physics and the NucleusPhysics, 6th Edition *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
38
R2 = 6.92 cm
