# Physics exercises_solution: Chapter 43

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## 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