Tài liệu Physics exercises_solution: Chapter 43 pdf

28 43.1: a) 14 Si has 14 protons and 14 neutrons b) 85 Rb has 37 protons and 48 neutrons 37 205 81 c) Tl has 81 protons and 124 neutrons 43.2: a) Using R  (1.2 fm) A1 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm b) Using 4R for each of the radii in part (a), the areas are 163 fm , 353 fm and 633 fm c) R gives 195 fm , 624 fm and 1499 fm d) The density is the same, since the volume and the mass are both proportional to A: 2.3  1017 kg m (see Example 43.1) e) Dividing the result of part (d) by the mass of a nucleon, the number density is 0.14 fm  1.40  10 44 m E  μz B  ( μz B)  μz B hf But E  hf , so B  2 z 43.3: B (6.63  10 34 J  s)(2.27  10 Hz)  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 c E  66.9 MHz, λ   4.48 m b) f  h f     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  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 μ z B  f   h h h 2(2.7928)(5.051  10 27 J T)(1.65 T)   7.02  10 Hz (6.63  10 34 J  s) c 3.00  108 m s   4.27 m This is a radio wave 7.02  107 Hz f b) For electrons, the negative charge means that the argument from part (a) leads to   the m s   state (antiparallel) having the lowest energy, since N and S point in 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 Sz  μz  (2.00232) 2me 4me λ  f  (2.00232)eB (2.00232)(1.60  1019 C)(1.65 T)  4πme 4π (9.11  10 31 kg)  f  4.62  1010 Hz so λ  c 3.00  108 m s   6.49  10 m 10 f 4.62  10 Hz This is a microwave 43.6: a) (13.6 eV) (0.511  10 eV)  2.66  10 b) (8.795 MeV) (938.3 MeV)  9.37  10 3 5  0.0027%  0.937% 43.7: The binding energy of a deuteron is 2.224  10 eV The photon with this energy has wavelength equal to λ hc (6.626  10 34 J  s)(2.998  108 m s)   5.576  10 13 m 19 E (2.224  10 eV)(1.602  10 J eV) 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)) 43.9: a) For 11 B the mass defect is: m  5mp  6mn  5me  M( 11 B)  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 b) From Eq (43.11): E B  C1 A  C A  C Z ( Z  1) ( A  2Z )  C4 and there is no A A3 fifth term since Z is odd and A is even  E B  (15.75 MeV)11  (17.80 MeV)(11)  (0.7100 MeV)  (23.69 MeV)  E B  76.68 MeV So the percentage difference is 5(4) 11 (11  10) (11) 76.68 MeV  76.21 MeV  100  0.62% 76.21 MeV Eq (43.11) has a greater percentage accuracy for 62 Ni 43.10: a) 34mn  29mH  mCu  34(1.008665) u  29(1.007825) u  62.929601 u  0.592 u, which is 551 MeV, or 8.75 MeV per nucleon (using 931.5 MeV 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 EB  (15.75 MeV)(63)  (17.80 MeV)(63)  (0.7100 MeV)  (23.69 MeV) (29)(28) (63) (5) (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 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  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    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 m  2M ( He)  M ( Be) 43.15:  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 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 C14 N    m  M (14 C)  M (14 N)  me  (14.003242 u  6(0.000549 u))  (14.003074 u  7(0.000549 u))  0.0005491  1.68  10  u : So E  (1.68  10  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 M ( He  )  3.016029 u  2(0.00054858 u)  3.014932 u  m  M ( H  )  M ( He  )  me  2.0  10 5 u, 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 43.20: Note that Eq 43.17 can be written as follows N  N0  t / T1 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 The source is barely usable Alternatively, we could calculate λ  ln (2)  0.132(years) 1 T1 and use the Eq 43.17 directly to obtain the same answer 43.21: For 14 C, T1  5730 y A  A0e  λt ; λ  ln T so A  A0e  ( ln ) t T1 ; A0  180.0 decays a) t  1000 y, A  159 decays b) t  50,000 y, A  0.43 decays 43.22: (a) 90 39 Sr     90 X 39 X has 39 protons and 90 protons plus neutrons, so it must be 90 Y (b) Use base because we know the half life A  A0 0.01A0  A0 t t T t T T1 log 0.01 log (28 yr ) log 0.01   190 yr log 43.23: a) 3 H 0 e He b) N  N 0e  λt , N  0.100 N and λ  (ln 2) T1 0.100  e  t ( ln ) T1 ;  t (ln 2) T1  ln (0.100); t  ln(0.100)T1 ln  40.9 y dN  500Ci  (500  10 6 )(3.70  1010 s 1 ) dt dN dt  1.85  10 dec s ln ln ln λ   6.69  10 s T1  λ T1 12 d (86,400 s d) 43.24: a) dN dN dt 1.85  107 dec / s  λN  N    2.77  1013 nuclei  1 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 6.69  10 s 1 λ  1d   9.29  106 s  86,400 s   108 days     t ( ln ) / T1 / 43.25: A  A0e  λt  A0e (ln 2)t   ln ( A A0 ) T 12 T12   (ln 2)t (ln 2)(4.00 days)   2.80 days ln( A A0 ) ln(3091 8318) 43.26: dN  λN dt ln  λ T12 ln  3.15  10 1620 yr   1yr  λ  1.36  10 11 s 1 s     6.022  10 23 atoms    2.665  10 25 atoms N  g   226 g   dN dec  λN  (2.665  1025 )(1.36  10 11 s 1 )  3.62  1010 dt s 10  3.62  10 Bq Convert to Ci:   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 kg ) (6.022  1023 atoms mol) (12.011  10  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  3.00 decays s  (N t ) N t   λN ; λ   3.836  1012 (s) 1 N T1 /  (ln 2) λ  1.807  1011 s  5730 y 43.28: a) Solving Eq (43.19) for λ , ln ln λ   4.17  10 s 1 T1 (5.27 y) (365 days year)(24 hrs day) (3600 sec hr ) b) N  3.60  10 5 g m   3.61  1017 Au (60) (1.66  10  24 g ) dN  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  (5.27 yrs) (60)  A  5 λ RA N Ra  λ Co N Co  1/2Co Co   0.0408 Ci   (1.600 yrs) (226)   3.57  10 Ci  T     1/2Ra ARa  c) 43.29: a) dN (0)  7.56  1011 Bq  7.56  1011 decays s dt and λ 0.693 0.693   3.75  10 s 1 (30.8 min)(60 s min) T1 so dN (0) 7.56  1011 decays s  N (0)   2.02  1015 nuclei  1 3.75  10 s λ dt N (0)  1.01  1015 nuclei, and the b) The number of nuclei left after one half-life is dN  3.78  1011 decays s activity is half: 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  10    9.45  10 decays s dt   3070 decays  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 1.21  10  y λ 43.30: The activity of the sample is 43.31: a)   0.693 0.693  T1 (1.28  10 y)(3.156  10 s y)    1.72  10 17 s 1 In m  1.63  10 6 g of 40 K there are N So 43.32: 1.63  10 9 kg  2.45  1016 nuclei 40(1.66  10  27 kg) dN  λN  (1.72  1017 s 1 )(2.45  1016 nuclei)  0.421 decays s dt 0.421 Bq dN b)   1.14  10 11 Ci 10 dt 3.70  10 Bq Ci 360  106 decays  4.17  103 Bq  1.13  10 Ci  0.113 μCi 86,400 s 43.33: a) λ  0.693 0.693   4.91  1018 s 1 (4.47  10 y)(3.156  10 s y) T1 decays   b) 1.20  10 5 Ci  (1.20  10 5 Ci)  3.70  1010   4.44  10 decays/s s  Ci   But dN dN 4.44  105 decays s  N  N   dt 4.91  10 18 s 1  dt  N  9.04  10 22 nuclei  m  (238 u) N  m  (238)(1.66  10 27 kg)(9.04  10 22 )  0.0357 kg c) Each decay emits one alpha particle In 60.0 g of uranium there are 0.0600 kg N  1.52  10 23 nuclei  27 238(1.66  10 kg ) dN  λN  (4.91  10 18 s 1 )(1.52  10 23 nuclei)  7.46  105 dt alpha particles emitted each second  43.34: (a) rem = rad  RBE 200 = x(10) x = 20 rad (b) rad deposits 0.010 J kg , so 20 rad deposit 0.20 J kg This radiation affects 25 g (0.025 kg) of tissue, so the total energy is (0.025 kg)(0.20 J kg)  5.0  10 3 J  5.0 mJ (c) Since RBE = for  -rays, so rem = rad Therefore 20 rad = 20 rem 43.35: rad = 10 2 Gy, so Gy = 100 rad and the dose was 500 rad rem=(rad) (RBE) = (500 rad) (4.0) = 2000 rem Gy  J kg , so 5.0 J kg 43.36: a) 5.4 Sv (100 rem Sv)  540 rem b) The RBE of gives an absorbed dose of 540 rad c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4 Gy) (65 kg)  351 J The energy required to raise the temperature of 65 kg 0.010 o C is (65 kg) (4190 J kg  K) (0.01 C)  kJ 43.37: a) We need to know how many decays per second occur 0.693 0.693  λ  1.79  10 s 1 (12.3 y) (3.156  10 s y) T1 dN (0.35 Ci) (3.70  1010 Bq Ci)  λ dt 1.79  10 s 1  N (0)  7.2540  1018 nuclei The number of remaining nuclei after one week is just The number of tritium atoms is N (0)  9 1 N (1 week )  N (0)e  λt  (7.25  1018 )e  (1.79 10 s ) ( ) ( 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)  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 ) is 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  c) The RBE for alpha particles is twice that for protons, so only half as many, 5.27  10 , would be absorbed Nhc (6.50  1010 ) (6.63  10 34 J  s) (3.00  10 m s)   2.00  1011 m 4  6.46  10 J 43.39: a) E total  NE    E total ` b) The absorbed dose is the energy divided by tissue mass:  100 rad  6.46  10 4 J 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 eV) (1.602  10 19 J eV)  1.08 Gy  108 rad (0.50 kg) The equivalent dose is (20) (108 rad) = 2160 rem 2 A 43.41: a) H  Be  Z X  He So   A   A  and   Z   Z  3, so X is Li b) m  M (1 H)  M ( Be)  M ( Li)  M ( He)  2.014102 u  9.012182 u  7.016003 u  4.002603 u  7.678  10 3 u So E  (m)c  (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  1.5  10 15 m r9Be  (1.2  10 15 m) (9)1  2.5  10 15 m So the centers’ separation is r  4.0  10 15 m qH qBe 4(1.60  1019 C) Thus U     2.3  1013 J 15 4πε0 4πε0 (4.0  10 m) r 43.42: m He  U  1.4  10 eV  1.4 MeV  m H  m He  m H  1.97  10 2 u, so the energy released is 18.4 MeV 43.43: a) As in Ex (43.41a ),  14  A  10  A  6, and   Z  Z  3, so X 3 Li b) As in Ex (43.41b), using M (1 H)  2.014102 u , M (14 N)  14.003074 u , M ( Li) , 6.01521 u, and M (10 B)  10.012937 u , m  0.010882 u, so energy is absorbed in the reaction  Q  (0.010882 u) (931.5 MeV u)  10.14 MeV M c) From Eq (43.24): K cm  K M m so 14.0 u  2.01 u M m K  (10.14 MeV)  11.6 MeV  K cm  14.0 u  M  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 235 236 43.45: The mass defect is m  M ( 92 U)  mn  M ( 92 U * )  m  235.043923 u  1.008665 u  236.045562 u  0.007025 u So the internal excitation of the nucleus is: Q  (m)c  (0.007025 u) (931.5 MeV u )  6.544 MeV 43 46: a) Z     and A     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 ( He)  M ( He)  M ( Be)  (3.016029 v  4.002603 v  7.016929 v) 2  (931.5 MeV v)  1.586 MeV 28 24 A 43.48: a) 14 Si    12 Mg  Z X A  24  28 so A  Z  12  14 so Z  X is an  particle b) KEγ  mc  (23.985042 v  4.002603 v  27.976927 v) (931.5 MeV v)  9.984 MeV A A 43.49: Nuclei: Z X z   Z  Y ( Z 2 )  He 2 2 A A Add the mass of Z electrons to each side and we find: m  M ( Z X)  M ( Z 4 Y)  2 M ( 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 A  Z X  Z 1 Y  e 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) A 43.51: Z X z   Z  Y ( Z 1)     Adding (Z –1) electron to both sides yields A A   Z X  Z 1 Y   So in terms of masses: A A m  M  Z X    M Z 1 Y  me A     X  m   M  Y  m  M  X   M  Y   2m  M A Z A Z A Z 1 e A Z 1 e 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 A 43.52: Denote the reaction as Z X  e   Z  Y 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) 25 25 43.53: a) Only the heavier one (13 Al) can decay into the lighter one (12 Mg) 25 25 A b) (13 Al)  (12 Mg )  Z X  A  0, Z  1  X is a positron    decay 25 25 or (13 Al)  A X  12 Mg  A  0, Z  1  X  is an electron Z  electron capture c) Using the nuclear masses, we calculate the mass defect for   decay: 25 25 m  ( M (13 Al)  13me )  ( M (12 Mg )  12me )  me  24.990429 u  24.985837 u  2(0.00054858 u )  3.495  10 3 u  Q  (m)c  (3.495  10 3 u ) (931.5 Me V u )  3.255 MeV For electron capture: 25 25 m  M (13 Al)  M (12 Mg)  24.990429 u  24.985837 u  4.592  10 3 u  Q  (m)c  (4.592  10 3 u) (931.5 Me V u)  4.277 MeV 43.54: a) m 210 Po  m 206 Pb  m He  5.81  10 3 u, or Q  5.41 MeV The energy of the 84 82 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 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 24 A 43.55: Using Eq: (43.12): Z M  ZM H  Nmn  EB c  M (11 Na)  11M H  13mn  E B c But E B  15.75 MeV (24)  (17.80 MeV)(24)  (0.7100 MeV (11)(10)  (24)1 (24  2(11))  (39 MeV)(24)   198.31 MeV 24 (198.31 MeV) 24  M (11 Na)  11(1.007825 u)  13(1 008665 u)   23.9858 u 931.5 MeV u 23.990963  23.9858 % error   100  0.022% 23.990963 (23.69 MeV) 24 If the binding energy term is neglected, M (11 Na)  24.1987 u and so the percentage error 24.1987  23.990963  100  0.87% would be 23.990963 43.56: The  -particle will have 226 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  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 possible 43.59: 13 N 13 C    As in Problem 43.51, β  decay has a mass defect in terms of neutral atoms of m  M (13 N)  M (13 C)  2me  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 λ  0.5995 hr 1 , for a half-life of  1.16 hr b) The initial activity is N λ, so λ (2.00  10 Bq) N0  1.20  10 1 (0.5995 hr )(1 hr 3600 s) c) N 0e  λt  1.81  106 dN (t ) dN (t ) dN (0) but  λN (t ) so   λN (0)  dt dt dt dN (t ) dN (0)  λt  λN  A0 Taking the derivative of N (t )  N 0e  λt    λN 0e  λt  e dt dt or A(t )  A0e  λt 43.61: The activity A(t )  43.62: From Eq.43.17 N (t )  N 0e  t   T1       λt but N 0e  t   T1       λt  N 0e  t  ( ln 2)   T1      n  ln ( )  t 1  N0 e  N e So N (t )  N   where n   T1  2   a ax ax a ln x Recall a ln x  ln( x ), e  (e ) , and e  x  (ln 2)  43.63:   0.693 0.693   4.62  10 19 s 1 10 T1 (4.75  10 y)(3.156  10 s y) N 87  N 087 e  t  N 087  N 87 e  t  N 087  N 087 e ( 4.6210 19 s 1 ) ( 4.6109 y ) ( 3.156107 s y) N 87 0.2783 N 85  0.2783  N 87   0.3856 N 85  N 85  N 87 (1  0.2783) But we also know that 0.3856 N 085 So N 087 N 085  N 087 So the original percentage of  N 87  1.0694 N 87 87  1.0694(0.3856)  0.2920 (1  1.0694(0.3856)) 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  m ln(2) 6.82 rad b) (20)(6.82 rad)=136 rem (c) Nλ   1.17  109 Bq  Au T1 12 6.25  1012  5.34  10 s, about an hour and a half Note that this time is 1.17  10 Bq so small in comparison with the half-life that the decrease in activity of the source may be neglected dN  2.6  10  Ci(3.70  1010 decays s  Ci)  9.6  10 decays s so in one 43.65: a) dt second there is an energy delivered of  dN  1 19 E    t  Eγ  (9.6  10 s )(1.00 s)(1.25  10 eV)(1.60  10 J eV)  dt  31.6 mCi d)  9.62  10 7 J s b) Absorbed dose  E 9.6  10 7 J s  m 0.500 kg  rad    1.9  10  rad  1.9  10 6 J kg  s100  J kg  s    4 c) Equivalent dose  0.7(1.9  10 )rad  1.3  10 4 rem 200 rem d)  1.5  10 s  17 days 1.3  10  rem s 43.66: a) After 4.0 = 240 s, the ratio of the number of nuclei is  1   ( 240 )   240 122.2   26.9 122.2   124  240 26.9 b) After 15.0 = 900 s, the ratio is 7.15  10 N  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  20.06 ; (8.5 counts min)(1 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: e2 (1.60  1019 C) a) U    7.61  1014 J 15 13 4 r 4 2(1.2  10 m)(2)  0.48 MeV b) m  2M (1 H)  M ( He)  mn  2(2.014102 u)  3.016029 u  1.008665 u 43.67:    3.51  10 u  E  (m)c  (3.51  10  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 H  1.30  10 2 u, so the proton separation energy is 12.1 MeV b) m 16 O  m 15 O  mn  1.68  10 2 u, so the neutron separation energy is 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 Number of atoms N  2.52  10 7 kg  3.793  1018  27 40 u(1.661  10 kg u ) dN (0.693)(3.793  1018 )  λN   2.054  109 decays y dt 1.28  10 y 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 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 of atoms present is related to N by N  Ne λt , so 0.693  7.30  10 10 sec 1  N  Ne λt  (1.535  λ (30.07 yr)(3.156  10 sec yr ) 1013 )e ( 7.30 10 10 sec 1 ) ( days) (8.64  104 sec days)  1.536  1013 m mM m vm  M     vm  v  v vm  v mM mM  mM 1 mM Mm  K   mvm2  Mv2  v2  v2 M 2 2 (m  M ) (m  M ) 43.73: a) vcm  v  m2 M  mM   (m  M )  m  M m  M   v   M 1 2 M  K  K cm  mv   K   mM 2 mM  b) For an endoenergetic reaction K cm  QQ  0 at threshold Putting this into part M  M  m  K th  K th  (a) gives  Q  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 2.76 MeV c  181.94821 u 182 235 43.75: m  M 92 U  M 140 Xe  M 94 Sr  mn 54 38  235.043923 u  139.921636 u  93.915360 u  1.008665 u           0.1983 u  E  m c  0.1983 u  931.5 MeV u   185 MeV 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    10 6 and decay constant of 0.361 hr 1 , corresponding to a half-life of 1.92 hr Extrapolating this back to time 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 of the shorter-lived species and 2.49  10 of the longer-lived species d) After 5.0 hr, there would be 1.80  10 of the shorter-lived species and 4.10  10 of the longer-lived species   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 t dN    dt rate of production by the neutron irradiation Then  K  λN  N  ln K  λN 0   λt  ln K  λN   ln K  λt   K  e  λt  N t   λ     b) The activity of the sample is λN t   K  e  λt  1.5  106 decays s   1  e     So the activity is 1.5  10 decays s 1  e 0.02772 t  , with t in minutes So     dN   the activity   at various times is:  dt   dN   dN  (t  10 min)  3.6  105 Bq; (t  min)  4.1  104 Bq; dt dt  dN   dN  (t  50 min)  1.1  106 Bq; (t  25 min)  7.5  105 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)  3.2  109 atoms c) N max   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 decays s  0.693    25  t    43.78: The activity of the original iron, after 1000 hours of operation, would be (9.4  106 Ci) (3.7  1010 Bq Ci)2 (1000 hr) ( 45 d  24 hr d )  1.8306  105 Bq The activity of the oil is 84 Bq, or 4.5886  10 4 of the total iron activity, and this must be the fraction of the mass worn, or mass of 4.59  10 2 g The rate at which the piston rings lost their mass is then 4.59  10 5 g hr ... 15 4πε0 4πε0 (4.0  10 m) r 43. 42: m He  U  1.4  10 eV  1.4 MeV  m H  m He  m H  1.97  10 2 u, so the energy released is 18.4 MeV 43. 43: a) As in Ex (43. 41a ),  14  A  10  A ... 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 24 A 43. 55: Using Eq: (43. 12): Z M  ZM H  Nmn  EB c ... 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 possible 43. 59: 13 N 13 C    As in Problem 43. 51, β  decay has a mass