 # Physics exercises_solution: Chapter 43

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8 ## Physics exercises_solution: Chapter 43

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 43

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## Nội dung Text: Physics exercises_solution: Chapter 43

1. 28 43.1: a) 14 Si has 14 protons and 14 neutrons. b) 85 Rb has 37 protons and 48 neutrons. 37 205 c) 81 Tl has 81 protons and 124 neutrons. 43.2: a) Using R  (1.2 fm) A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. b) Using 4R 2 for each of the radii in part (a), the areas are 163 fm 2 , 353 fm 2 and 633 fm 2 . 4 c) R 3 gives 195 fm 3 , 624 fm 3 and 1499 fm 3 . 3 d) The density is the same, since the volume and the mass are both proportional to A: 2.3  1017 kg m 3 (see Example 43.1). e) Dividing the result of part (d) by the mass of a nucleon, the number density is 3 0.14 fm 3  1.40  10 44 m . 43.3: E  μz B  ( μz B)  2 μz B hf But E  hf , so B  2 z (6.63  10 34 J  s)(2.27  10 7 Hz) B  0.533 T 2(2.7928)(5.051  10 27 J T) 43.4: a) As in Example 43.2, E  2(1.9130)(3.15245  10 8 eV T)(2.30 T)  2.77  10 7 eV.   Since N and S are in opposite directions for a neutron, the antiparallel configuration is lower energy. This result is smaller than but comparable to that found in the example for protons. E c b) f   66.9 MHz, λ   4.48 m. h f
2.     43.5: a) U  μ  B    z B. N and S point in the same direction for a proton. So if the spin magnetic moment of the proton is parallel to the magnetic field, U  0, and if they are antiparallel, U  0. So the parallel case has lower energy. The frequency of an emitted photon has a transition of the protons between the two states given by: E E  E 2 μ z B f    h h h 2(2.7928)(5.051  10 27 J T)(1.65 T)   7.02  10 7 Hz. (6.63  10 34 J  s) c 3.00  108 m s λ   4.27 m. This is a radio wave. f 7.02  107 Hz b) For electrons, the negative charge means that the argument from part (a) leads to 1   the m s   state (antiparallel) having the lowest energy, since N and S point in 2 opposite directions. So an emitted photon in a transition from one state to the other has a frequency E E  12  E  12  2 z B f    h h h But from Eq. (41.22), e  (2.00232)e μz  (2.00232) Sz  2me 4me (2.00232)eB (2.00232)(1.60  1019 C)(1.65 T)  f   4πme 4π (9.11  10 31 kg)  f  4.62  1010 Hz c 3.00  108 m s so λ    6.49  10 3 m. f 4.62  10 Hz 10 This is a microwave. 5 a) (13.6 eV) (0.511  10 eV)  2.66  10  0.0027%. 6 43.6: 3 b) (8.795 MeV) (938.3 MeV)  9.37  10  0.937%. 43.7: The binding energy of a deuteron is 2.224  10 6 eV. The photon with this energy has wavelength equal to hc (6.626  10 34 J  s)(2.998  108 m s) λ  19  5.576  10 13 m. E (2.224  10 eV)(1.602  10 J eV) 6
3. 43.8: a) 7(mn  mH )  m N  0.112 u, which is 105 MeV, or 7.48 MeV per nucleon. b) Similarly, 2(mH  mn )  mHe  0.03038 u  28.3 MeV, or 7.07 MeV per nucleon, slightly lower (compare to Fig. (43.2)). 11 43.9: a) For 5 B the mass defect is: m  5mp  6mn  5me  M( 11 B) 5  5(1.007277 u )  6(1.008665 u )  5(0.000549 u )  11.009305 u  0.081815 u  The binding energy EB  (0.081815 u)(931.5 MeV u)  76.21 MeV Z ( Z  1) ( A  2Z ) 2 b) From Eq. (43.11): E B  C1 A  C 2 A 3  C 3  C4 2 1 and there is no A3 A fifth term since Z is odd and A is even. 2 5(4)  E B  (15.75 MeV)11  (17.80 MeV)(11) 3  (0.7100 MeV) 1 11 3 (11  10) 2  (23.69 MeV) (11)  E B  76.68 MeV. 76.68 MeV  76.21 MeV So the percentage difference is  100  0.62% 76.21 MeV 62 Eq. (43.11) has a greater percentage accuracy for Ni.
4. 43.10: a) 34mn  29mH  mCu  34(1.008665) u  29(1.007825) u  62.929601 u  MeV 0.592 u, which is 551 MeV, or 8.75 MeV per nucleon (using 931.5 and u 63 nucleons). b) In Eq. (43.11), Z = 29 and N = 34, so the fifth term is zero. The predicted binding energy is 2 (29)(28) EB  (15.75 MeV)(63)  (17.80 MeV)(63) 3  (0.7100 MeV) 1 (63) 3 (5) 2  (23.69 MeV) (63)  556 MeV. (The fifth term is zero since the number of neutrons is even while the number of protons is odd, making the pairing term zero.) This result differs from the binding energy found from the mass deficit by 0.86%, a very good agreement comparable to that found in Example 43.4. 43.11: Z is a magic number of the elements helium (Z = 2), oxygen (Z = 8), calcium (Z = 20), nickel (Z =28), tin (Z = 50) and lead (Z = 82). The elements are especially stable, with large energy jumps to the next allowed energy level. The binding energy for these elements is also large. The protons’ net magnetic moments are zero. 43.12: a) 146mn  92mH  m U  1.93 u, which is b) 1.80  10 3 MeV, or c) 7.56 MeV per nucleon (using 931.5 MeV and 238 nucleons). u 43.13: a) α decay : Z decreases by 2, A decreases by 4  239 Pb  235 U   94 92 b) β  decay : Z decreases by 1, A remains the same :  24 Na  24 Mg    11 12 c)   decay : Z decreases by 1, A remains the same : 15 O15 N    8 7
5. 43.14: a)The energy released is the energy equivalent of mn  mp  me  8.40  10 4 u, or 783 keV. b) m n  m p , and the decay is not possible. 43.15: m  2M ( 4 He)  M ( 4 Be) 2 8  2(4.002603 u)  8.005305 u  m  9.9  10 5 u 43.16: a) A proton changes to a neutron, so the emitted particle is a positron (   ). b) The number of nucleons in the nucleus decreases by 4 and the number of protons by 2, so the emitted particle is an alpha-particle. c) A neutron changes to a proton, so the emitted particle is an electron (   ). 43.17: If   decay 14 C is possible, then we are considering the decay 14 6 C14 N    . 7 m  M (14 C)  M (14 N)  me 6 7  (14.003242 u  6(0.000549 u))  (14.003074 u  7(0.000549 u))  0.0005491  1.68  10  4 u : So E  (1.68  10  4 u)(931.5 MeV u )  0.156 MeV  156 keV 43.18: a) As in the example, (0.000898 u)(931.5 MeV u)  0.836 MeV. b) 0.836 MeV  0.122 MeV  0.014 MeV  0.700 MeV. 43.19: a) If tritium is to be unstable with respect to   decay, then the mass of the products of the decay must be less than the parent nucleus. M (3 H  )  3.016049 u  0.00054858 u  3.015500 u 1 M ( 2 He 2  )  3.016029 u  2(0.00054858 u)  3.014932 u 3  m  M ( 3 H  )  M ( 2 He  )  me  2.0  10 5 u, 1 3 so the decay is possible. b) The energy of the products is just E  (2.0  10 5 u )(931.5 MeV u)  0.019 MeV  19 keV.
6. 43.20: Note that Eq. 43.17 can be written as follows  t / T1 2 N  N0 2 . The amount of elapsed time since the source was created is roughly 2.5 years. Thus, we expect the current activity to be 2.6years N  (5000 Ci)2  5.271years  3600 Ci. ln (2) The source is barely usable. Alternatively, we could calculate λ   0.132(years) 1 T1 2 and use the Eq. 43.17 directly to obtain the same answer. 43.21: For 14 C, T1 2  5730 y  ( ln 2 ) t T1 A  A0e  λt ; λ  ln 2 T 1 2 so A  A0e 2 ; A0  180.0 decays min a) t  1000 y, A  159 decays min b) t  50,000 y, A  0.43 decays min 43.22: (a) 90 39 Sr     90 X 39 90 X has 39 protons and 90 protons plus neutrons, so it must be Y. (b) Use base 2 because we know the half life. t T 1 A  A0 2 2 t T 1 2 0.01A0  A0 2 T1 2 log 0.01 t log 2 (28 yr ) log 0.01   190 yr log 2 43.23: a) 3 1 H 0 e 2 He 1 3 b) N  N 0e  λt , N  0.100 N 0 and λ  (ln 2) T1 2  t ( ln 2 ) T1 2  ln(0.100)T1 2 0.100  e ;  t (ln 2) T1 2  ln (0.100); t  40.9 y ln 2
7. dN 43.24: a)  500Ci  (500  10 6 )(3.70  1010 s 1 ) dt dN dt  1.85  10 7 dec s ln 2 ln 2 ln 2 T1 2  λ   6.69  10 7 s λ T1 2 12 d (86,400 s d) dN dN dt 1.85  107 dec / s  λN  N    7 1  2.77  1013 nuclei dt λ 6.69  10 s 131 The mass of this many Ba nuclei is m  2.77  1013 nuclei  (131  1.66  1027 kg nucleus )  6.0  10 12 kg  6.0  10 9 g  6.0 ng (b) A  A0e  λt 1μ Ci  (500 μ Ci) e  λt ln (1/ 500)  λt ln (1 500) ln (1 500) t  λ 6.69  10 7 s 1  1d   9.29  106 s  86,400 s   108 days     t ( ln 2 ) / T1 / 2 43.25: A  A0e  λt  A0e (ln 2)t   ln ( A A0 ) T 12 (ln 2)t (ln 2)(4.00 days) T12     2.80 days ln( A A0 ) ln(3091 8318)
8. dN 43.26:  λN dt ln 2 ln 2 λ  T12  3.15  10 7 s 1620 yr      1yr  λ  1.36  10 11 s 1  6.022  10 23 atoms  N  1 g   2.665  10 25 atoms   226 g  dN dec  λN  (2.665  1025 )(1.36  10 11 s 1 )  3.62  1010 dt s  3.62  10 Bq 10 Convert to Ci:  1 Ci  3.62  1010 Bq  3.70  1010 Bq   0.98 Ci    43.27: Find the total number of carbon atoms in the sample. n  m M; N tot  nN A  m N A M  (12.0  10 3 kg ) (6.022  1023 atoms mol) (12.011  10  3 kg mol N tot  6.016  1023 atoms, so (1.3  1012 ) (6.016  1023 )  7.82  1011 carbon - 14 atoms N t  180 decays min  3.00 decays s  (N t ) N t   λN ; λ   3.836  1012 (s) 1 N T1 / 2  (ln 2) λ  1.807  1011 s  5730 y 43.28: a) Solving Eq. (43.19) for λ , ln 2 ln 2 λ   4.17  10 9 s 1. T1 2 (5.27 y) (365 days year)(24 hrs day) (3600 sec hr ) m 3.60  10 5 g b) N    3.61  1017. Au (60) (1.66  10  24 g ) dN c)  λN  1.51  109 Bq, which is d) 0.0408 Ci. The same calculation for radium, dt with larger A and longer half-life (lower λ ) gives T A   (5.27 yrs) (60)  λ RA N Ra  λ Co N Co  1/2Co Co   0.0408 Ci  T   (1.600 yrs) (226)   3.57  10 Ci.  5  1/2Ra ARa   
9. dN (0) 43.29: a)  7.56  1011 Bq  7.56  1011 decays s dt and 0.693 0.693 λ   3.75  10 4 s 1. T1 2 (30.8 min)(60 s min) so 1 dN (0) 7.56  1011 decays s N (0)    4 1  2.02  1015 nuclei. λ dt 3.75  10 s N (0) b) The number of nuclei left after one half-life is  1.01  1015 nuclei, and the 2 dN activity is half:  3.78  1011 decays s . dt c) After three half lives (92.4 minutes) there is an eighth of the original amount N  2.53  1014 nuclei, and an eighth of the activity:  dN    9.45  10 decays s . 10   dt  3070 decays min 43.30: The activity of the sample is  102 Bq kg , while the (60 sec min) (0.500 kg) activity of atmospheric carbon is 255 Bq kg (see Example 43.9). The age of the sample is then ln (102 255 ) ln (102 255 ) t   7573 y. λ 1.21  10  4 y 0.693 0.693 43.31: a)    T1 2 (1.28  10 y)(3.156  10 7 s y) 9    1.72  10 17 s 1 . In m  1.63  10 6 g of 40 K there are 1.63  10 9 kg N  2.45  1016 nuclei. 40(1.66  10  27 kg) dN So  λN  (1.72  1017 s 1 )(2.45  1016 nuclei)  0.421 decays s . dt dN 0.421 Bq b)   1.14  10 11 Ci. dt 3.70  10 Bq Ci 10 360  106 decays 43.32:  4.17  103 Bq  1.13  10 7 Ci  0.113 μCi. 86,400 s
11. 43.37: a) We need to know how many decays per second occur. 0.693 0.693 λ   1.79  10 9 s 1. T1 2 (12.3 y) (3.156  10 s y) 7 1 dN (0.35 Ci) (3.70  1010 Bq Ci) The number of tritium atoms is N (0)   λ dt 1.79  10 9 s 1  N (0)  7.2540  1018 nuclei. The number of remaining nuclei after one week is just 9 1 N (1 week )  N (0)e  λt  (7.25  1018 )e  (1.79 10 s ) ( 7 ) ( 24 ) (3600s )  N (1 week)  7.2462  1018 nuclei  N  N (0)  N (1 week)  7.8  1015 decays. So the energy absorbed is Etotal  N  Eγ  (7.8  1015 ) (5000 eV) (1.60  1019 J eV)  6.24 J. So the absorbed dose (6.24 J) is  0.125 J kg  12.5 rad. Since RBE = 1, then the equivalent dose is (50 kg) 12.5 rem. b) In the decay, antinentrinos are also emitted. These are not absorbed by the body, and so some of the energy of the decay is lost (about 12 keV ). 43.38: a) From Table (43.3), the absorbed dose is 0.0900 rad. b) The energy absorbed is (9.00  10 4 J kg) (0.150 kg)  1.35  10 4 J; each proton has energy 1.282  10 13 J, so the number absorbed is 1.05  10 9  c) The RBE for alpha particles is twice that for protons, so only half as many, 5.27  10 8 , would be absorbed. Nhc (6.50  1010 ) (6.63  10 34 J  s) (3.00  10 8 m s) 43.39: a) E total  NE     2.00  1011 m 4  E total  6.46  10 J. b) The absorbed dose is the energy divided by tissue mass: 6.46  10 4 J  100 rad  ` dose   (1.08  10 3 J kg )   J kg   0.108 rad.  0.600 kg   The rem dose for x rays (RBE = 1) is just 0.108 rem. 43.40: (0.72  106 Ci) (3.7  1010 Bq Ci) (3.156  107 s)  8.41  1011 α particles. The absorbed dose is (8.41  1011 ) (4.0  10 6 eV) (1.602  10 19 J eV)  1.08 Gy  108 rad. (0.50 kg) The equivalent dose is (20) (108 rad) = 2160 rem.
12. 43.41: a) 1 H  9 Be  Z X  4 He. So 2  9  A  4  A  7 and 1  4  Z  2  2 4 A 2 Z  3, so X is 3 Li. 7 b) m  M (1 H)  M ( 9 Be)  M ( 3 Li)  M ( 4 He) 2 4 7 2  2.014102 u  9.012182 u  7.016003 u  4.002603 u  7.678  10 3 u. So E  (m)c 2  (7.678  10 3 u) (931.5 MeV u )  7.152 MeV. c) The threshold energy is taken to be the potential energy of the two reactants when they just “touch.” So we need to know their radii: r2H  (1.2  10 15 m) (2)1 3  1.5  10 15 m r9Be  (1.2  10 15 m) (9)1 3  2.5  10 15 m So the centers’ separation is r  4.0  10 15 m 1 qH qBe 4(1.60  1019 C) 2 Thus U    15  2.3  1013 J 4πε0 r 4πε0 (4.0  10 m)  U  1.4  10 6 eV  1.4 MeV. 43.42: m 3 He  m 2 H  m 4 He  m 1 H  1.97  10 2 u, so the energy released is 18.4 MeV. 2 1 2 1 43.43: a) As in Ex. (43.41a ), 2  14  A  10  A  6, and 1  7  Z  5 Z  3, so X 3 Li. 6 b) As in Ex. (43.41b), using M (1 H)  2.014102 u , M (14 N)  14.003074 u , M ( 3 Li) , 2 7 6 6.01521 u, and M (10 B)  10.012937 u , m  0.010882 u, so energy is absorbed in the 5 reaction.  Q  (0.010882 u) (931.5 MeV u)  10.14 MeV. M c) From Eq. (43.24): K cm  K M m so M m 14.0 u  2.01 u K   K cm  (10.14 MeV)  11.6 MeV  M  14.0 u 43.44: (200  106 eV) (1.602  1019 J eV) (6.023  1023 molecules mol)  1.93  1013 J mol , which is far higher than typical heats of combustion. 43.45: The mass defect is m  M ( 92 U)  mn  M ( 92 U * ) 235 236  m  235.043923 u  1.008665 u  236.045562 u  0.007025 u So the internal excitation of the nucleus is: Q  (m)c 2  (0.007025 u) (931.5 MeV u )  6.544 MeV
13. 43 .46: a) Z  3  2  0  5 and A  4  7  1  10. b) The nuclide is a boron nucleus, 3 and mHe  mLi  mn  mB  3.00  10 u, and so 2.79 MeV of energy is absorbed. 43.47: The energy liberated will be M ( 3 He)  M ( 4 He)  M ( 7 Be)  (3.016029 v  4.002603 v  7.016929 v) 2 2 4  (931.5 MeV v)  1.586 MeV. 43.48: a) 14 Si    12 Mg  Z X. A  24  28 so A  4. Z  12  14 so Z  2. X is an 28 24 A  particle. b) KEγ  mc 2  (23.985042 v  4.002603 v  27.976927 v) (931.5 MeV v)  9.984 MeV 43.49: Nuclei: Z X z   Z  4 Y ( Z 2 )  4 He 2 A A 2 2 Add the mass of Z electrons to each side and we find: m  M ( Z X)  M ( Z 4 Y)  A A 2 4 M ( 2 He), where now we have the mass of the neutral atoms. So as long as the mass of the original neutral atom is greater than the sum of the neutral products masses, the decay can happen. 43.50: Denote the reaction as A  Z X  Z 1 Y  e . A The mass defect is related to the change in the neutral atomic masses by [mX  Zme ]  [mY  ( Z  1)me ]  me  (mX  mY ), where mX and mY are the masses as tabulated in, for instance, Table (43.2). 43.51: Z X z   Z  1 Y ( Z 1)     A A Adding (Z –1) electron to both sides yields Z X  Z 1 Y   A  A  So in terms of masses:  m  M  Z X    M Z 1 Y  me A A    X  m   M  Y  m  M A Z e A Z 1 e  M  X   M  Y   2m . A A Z Z 1 e So the decay will occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses. A 43.52: Denote the reaction as Z X  e   Z  1 Y. A The mass defect is related to the change in the neutral atomic masses by [mX  Zme ]  me  [mY  ( Z  1)me ]  (mX  mY ), where mX and mY are the masses as tabulated in, for instance, Table (43.2).
14. 25 25 43.53: a) Only the heavier one (13 Al) can decay into the lighter one (12 Mg). b) (13 Al)  (12 Mg )  Z X  A  0, Z  1  X is a positron 25 25 A    decay or (13 Al)  A X  12 Mg  A  0, Z  1  X  is an electron 25 Z 25  electron capture c) Using the nuclear masses, we calculate the mass defect for   decay: m  ( M (13 Al)  13me )  ( M (12 Mg )  12me )  me 25 25  24.990429 u  24.985837 u  2(0.00054858 u )  3.495  10 3 u  Q  (m)c 2  (3.495  10 3 u ) (931.5 Me V u )  3.255 MeV. For electron capture: m  M (13 Al)  M (12 Mg)  24.990429 u  24.985837 u 25 25  4.592  10 3 u  Q  (m)c 2  (4.592  10 3 u) (931.5 Me V u)  4.277 MeV. 43.54: a) m 210 Po  m 206 Pb  m 4 He  5.81  10 3 u, or Q  5.41 MeV. The energy of the 84 82 2 alpha particle is (206 210) times this, or 5.30 MeV (see Example 43.5) b) m 210 Po  m 209 Bi  m1 H  5.35  10 3 u  0, so the decay is not possible. 84 83 1 c) m 210 Po  m 209 Po  mn  8.22  10 3 u  0, so the decay is not possible. 84 84 d) m 210 At  m 210 Po , so the decay is not possible (see Problem (43.50)). 85 84 e) m 210 Bi  2me  m 210 Po , so the decay is not possible (see Problem (43.51)). 83 84 43.55: Using Eq: (43.12): Z M  ZM H  Nmn  EB c 2  M (11 Na)  11M H  A 24 (11)(10) 13mn  E B c 2 . But E B  15.75 MeV (24)  (17.80 MeV)(24) 2 3  (0.7100 MeV  (24)1 3 (24  2(11)) 2 (23.69 MeV)  (39 MeV)(24)  4 3  198.31 MeV. 24 (198.31 MeV)  M (11 Na)  11(1.007825 u)  13(1 .008665 u)  24  23.9858 u 931.5 MeV u 23.990963  23.9858 % error   100  0.022%. 23.990963 If the binding energy term is neglected, M (11 Na)  24.1987 u and so the percentage error 24 24.1987  23.990963 would be  100  0.87%. 23.990963
15. 226 43.56: The  -particle will have of the mass energy (see Example 45.5) 230 226 (mTh  mRa  m )  5.032  10 3 u or 4.69 MeV. 230 43.57: 198 Au 198 Hg    m  M (198 Au )  M (198 Hg)  197.968225 u  197.966752 u  79 80 79 80 1.473  10 3 u. And the total energy available was Q  (m)c 2  Q  (1.473  10 3 u)(931.5 MeV u  1.372 MeV. The emitted photon has energy 0.412 MeV, so the emitted electron must have kinetic energy equal to 1.372 MeV  0.412 MeV  0.960 MeV. 43.58: (See Problem (43.51)) m11 C  m11 B  2me  1.03  10 3 u. Decay is energetically 6 5 possible. 43.59: 13 N 13 C    As in Problem 43.51, β  decay has a mass defect in terms of 7 6 neutral atoms of m  M (13 N)  M (13 C)  2me 7 6  13.005739 u  13.003355  2(0.00054858 u)  1.287  10 3 u Therefore the decay is possible because the initial mass is greater than the final mass. 43.60: a) A least-squares fit to log of the activity vs. time gives a slope of ln 2 λ  0.5995 hr 1 , for a half-life of  1.16 hr. b) The initial activity is N 0 λ, so λ (2.00  10 4 Bq) N0 1  1.20  10 8. (0.5995 hr )(1 hr 3600 s) c) N 0e  λt  1.81  106. dN (t ) dN (t ) dN (0) 43.61: The activity A(t )  but  λN (t ) so   λN (0)  dt dt dt dN (t ) dN (0)  λt  λN 0  A0 . Taking the derivative of N (t )  N 0e  λt    λN 0e  λt  e dt dt or A(t )  A0e  λt .  t   ( ln 2)    λt  λt  T1 2  43.62: From Eq.43.17 N (t )  N 0e but N 0e  N 0e    t   t       ln ( 1 )  n    T1 2   T1 2  (ln 2)     1 t  N0 e  N 0 e 2  . So N (t )  N 0   where n  .    2 T1 2 Recall a ln x  ln( x ), e  (e ) , and e  x. a ax ax a ln x
16. 0.693 0.693 43.63:     4.62  10 19 s 1 T1 2 (4.75  10 y)(3.156  10 s y) 10 7 N 87  N 087 e  t  N 087  N 87 e  t 19 s 1 ) ( 4.6109 y ) ( 3.156107 s y)  N 087  N 087 e ( 4.6210  N 0 87  1.0694 N 87 N 87 0.2783 N 85 But we also know that  0.2783  N 87   0.3856 N 85  N 85  N 87 (1  0.2783) N 087 1.0694(0.3856) 0.3856 N 085 . So   0.2920. N 085  N 087 (1  1.0694(0.3856)) So the original percentage of 87 Rb is 29%. ( N 085  N 85 since it doesn’t decay.) 43.64: a) (6.25  10 )(4.77  106 MeV)(1.602  1019 J eV) (70.0 kg)  0.0682 Gy  12 m ln(2) 6.82 rad. b) (20)(6.82 rad)=136 rem (c) Nλ   1.17  109 Bq  Au T1 2 6.25  1012 31.6 mCi. d)  5.34  10 3 s, about an hour and a half. Note that this time is 1.17  10 9 Bq so small in comparison with the half-life that the decrease in activity of the source may be neglected. dN 43.65: a)  2.6  10  4 Ci(3.70  1010 decays s  Ci)  9.6  10 6 decays s so in one dt second there is an energy delivered of 1  dN  1 6 1 19 E    t  Eγ  (9.6  10 s )(1.00 s)(1.25  10 eV)(1.60  10 J eV) 6 2  dt  2  9.62  10 7 J s. E 9.6  10 7 J s b) Absorbed dose   m 0.500 kg  rad   1.9  10 6 J kg  s100    1.9  10  4 rad.  J kg  s   c) Equivalent dose  0.7(1.9  10 )rad  1.3  10 4 rem. 4 200 rem d)  1.5  10 6 s  17 days. 1.3  10  4 rem s 43.66: a) After 4.0 min = 240 s, the ratio of the number of nuclei is  1 1  2 240 122.2 ( 240 )     240 26.9  2  26.9 122.2   124. 2 b) After 15.0 min = 900 s, the ratio is 7.15  10 7.
17. N 43.67:  0.21  e  λt N0 ln(0.21) 5730 y t    ln(0.21)   13000 y λ 0.693 43.68: The activity of the sample will have decreased by a factor of (4.2  10 6 Ci)(3.70  1010 Bq Ci)  1.097  10 6  2 20.06 ; (8.5 counts min)(1 min 60s) this corresponds to 20.06 half-lifes, and the elapsed time is 40.1 h. Note the retention of extra figures in the exponent to avoid roundoff error. To the given two figures the time is 40 h. 43.69: For deuterium: 1 e2 (1.60  1019 C) 2 a) U    7.61  1014 J  4 0 r 4 0 2(1.2  10 m)(2) 15 13   0.48 MeV b) m  2M (1 H)  M ( 2 He)  mn 2 3  2(2.014102 u)  3.016029 u  1.008665 u  3.51  10 3 u  E  (m)c 2  (3.51  10  3 u)(931.5 MeV u )  3.270 MeV  5.231  10 13 J c) A mole of deuterium has 6.022  10 23 molecules, so the energy per mole is (6.022  10 23 )(5.231  10 13 J)  3.150  1011 J. This is over a million times more than the heat of combustion. 43.70: a) m 16 O  m 15 N  m 1 H  1.30  10 2 u, so the proton separation energy is 8 7 1 12.1 MeV. b) m 16 O  m 15 O  mn  1.68  10 2 u, so the neutron separation energy is 8 8 15.7 MeV. c) It takes less energy to remove a proton. 43.71: Mass of 40 K atoms in 1.00 kg is (2.1  10 3 )(1.2  10 4 )kg  2.52  10 7 kg. 2.52  10 7 kg Number of atoms N   27  3.793  1018. 40 u(1.661  10 kg u ) dN (0.693)(3.793  1018 )  λN   2.054  109 decays y . dt 1.28  10 y 9 So in 50 years the energy absorbed is: E  (0.50 MeV decay)(50 y)(2.054  109 decay y) 5.14  1010 MeV  8.22  10 3 J. So the absorbed dose is (8.22  10 3 J)(100 J rad)  0.82 rad and since the RBE = 1.0, the equivalent dose is 0.82 rem.
18. 43.72: In terms of the number N of cesium atoms that decay in one week and the mass m  1.0 kg, the equivalent dose is N 3.5 Sv  ((RBE)γ E γ  (RBE)e E e ) m N  ((1)(0.66 MeV)  (1.5)(0.51 MeV)) m N  (2.283  1013 J), so m (1.0 kg)(3.5 Sv) N  1.535  1013. (2.283  1013 J) The number N 0 of atoms present is related to N by N 0  Ne λt , so 0.693 λ  7.30  10 10 sec 1  N 0  Ne λt  (1.535  (30.07 yr)(3.156  10 7 sec yr ) 10 sec 1 ) ( 7 days) (8.64  104 sec days) 1013 )e ( 7.30 10  1.536  1013. m 43.73: a) vcm  v  mM  m  M   vm vm  v  v  v vm  mM mM  mM 1 1 1 mM 2 1 Mm 2 K   mvm2  Mv2   M v2  v2 2 2 2 (m  M ) 2 2 (m  M ) 2 1 M  mM  2 m2   v 2 (m  M )  m  M m  M    M 1 2 M   mv   K   K  K cm mM 2  mM b) For an endoenergetic reaction K cm  QQ  0 at threshold. Putting this into part M  M  m  (a) gives  Q  K th  K th  Q M m M M 43.74: K  K  , where K  is the energy that the  -particle would have if the M  m nucleus were infinitely massive. Then, M  M Os  M   K   M Os  M   186 182  2.76 MeV c 2  181.94821 u   43.75: m  M 92 U  M 140 Xe  M 94 Sr  mn 235 54  38      235.043923 u  139.921636 u  93.915360 u  1.008665 u  0.1983 u  E  m c 2  0.1983 u  931.5 MeV u   185 MeV.
19. 43.76: a) A least-squares fit of the log of the activity vs. time for the times later than 4.0 hr   gives a fit with correlation  1  2  10 6 and decay constant of 0.361 hr 1 , corresponding to a half-life of 1.92 hr. Extrapolating this back to time 0 gives a contribution to the rate of about 2500/s for this longer-lived species. A least-squares fit of the log of the activity vs. time for times earlier than 2.0 hr gives a fit with correlation = 0.994, indicating the presence of only two species. b) By trial and error, the data is fit by a decay rate modeled by R  5000 Bq e t 1.733 hr   2500 Bq e t 0.361 hr  . This would correspond to half-lives of 0.400 hr and 1.92 hr. c) In this model, there are 1.04  10 7 of the shorter-lived species and 2.49  10 7 of the longer-lived species. d) After 5.0 hr, there would be 1.80  10 3 of the shorter-lived species and 4.10  10 6 of the longer-lived species.
20. 43.77: (a) There are two processes occurring: the creation of 128 I by the neutron dN irradiation, and they decay of the newly produced 128 I . So  K  λN where K is the dt N dN  t rate of production by the neutron irradiation. Then    dt. 0 K  λN  0  ln K  λN 0   λt N  ln K  λN   ln K  λt  N t    K 1  e  λt. λ    b) The activity of the sample is λN t   K 1  e  λt  1.5  106 decays s      0.693   . So the activity is 1.5  10 6 decays s 1  e 0.02772 t  , with t in minutes. So   25 min  t 1  e          dN   the activity   at various times is:  dt   dN   dN  (t  10 min) (t  1 min)  4.1  104 Bq;  3.6  105 Bq; dt dt  dN   dN  (t  50 min) (t  25 min)  7.5  105 Bq;  1.1  106 Bq; dt dt  dN   dN  (t  180 min) (t  75 min)  1.3  106 Bq;  1.5  106 Bq; dt dt K (1.5  10 ) (60) 6 c) N max    3.2  109 atoms . λ 0.02772 d) The maximum activity is at saturation, when the rate being produced equals that decaying and so it equals 1.5  10 6 decays s . 