# Ch43Word - Ch 43 Page 1 CHAPTER 43 Nuclear Energy Effects...

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Ch. 43 Page 1 CHAPTER 43 – Nuclear Energy; Effects and Uses of Radiation Note: A factor that appears in the analysis of energies is e 2 /4p Å 0 = (1.60 × 10 –19 C) 2 /4p(8.85 × 10 –12 C 2 /N · m 2 )= 2.30 × 10 –28 J · m = 1.44 MeV · fm. 1. We find the product nucleus by balancing the mass and charge numbers: Z (X) = Z ( 27 Al) + Z (n) = 13 + 0 = 13; A (X) = A ( 27 Al) + A (n) = 27 + 1 = 28, so the product nucleus is Al 13 28 . If Al 13 28 were a β + emitter, the resulting nucleus would be Mg 12 28 , which has too many neutrons relative to the number of protons to be stable. Thus we have a β emitter . The decay is Al 13 28 Si 14 28 + e – 1 0 + ν , so the product is Si 14 28 . 2. For the reaction H 1 2 (d,n) He 2 3 , we find the difference of the initial and the final masses: ? M = M ( 2 H) + M ( 2 H) – m (n) – M ( 3 He) = 2(2.014102 u) – (1.008665 u) – (3.016029 u) = + 0.003510 u. Thus no threshold energy is required. 3. For the reaction U 92 238 (n, γ ) U 92 239 with slow neutrons, whose kinetic energy is negligible, we find the difference of the initial and the final masses: ? M = M ( 238 U) + m (n) – M ( 239 U) = (238.050782 u) + (1.008665 u) – (239.054287 u) = + 0.005160 u. Thus no threshold energy is required, so the reaction is possible . 4. For the reaction Li 3 7 (p, α ) He 2 4 , we determine the Q -value: Q = [ M ( 7 Li) + M ( 1 H) – M ( 4 He) – M ( 4 He)] c 2 = [(7.016004 u) + (1.007825 u) – 2(4.002603 u)] c 2 (931.5 MeV/u c 2 ) = + 17.35 MeV. Thus 17.35 MeV is released . 5. For the reaction Be 4 9 ( α ,n) C 6 12 , we determine the Q -value: Q = [ M ( 9 Be) + M ( 4 He) – m (n) – M ( 12 C)] c 2 = [(9.012182 u) + (4.002603 u) – (1.008665 u) – (12.000000 u)] c 2 (931.5 MeV/u c 2 ) = + 5.701 MeV. Thus 5.701 MeV is released . 6. ( a ) For the reaction Mg 12 24 (n, d) Na 11 23 , we determine the Q -value: Q = [ M ( 24 Mg) + m (n) – M ( 2 H) – M ( 23 Na)] c 2 = [(23.985042 u) + (1.008665 u) – (2.014102 u) – (22.989770 u)] c 2 (931.5 MeV/u c 2 ) = – 9.469 MeV. Because ( K + Q ) > 0, the reaction can occur . ( b ) The energy released is K + Q = 10.00 MeV – 9.469 MeV = 0.53 MeV . 7. ( a ) For the reaction Li 3 7 (p, α ) He 2 4 , we determine the Q -value: Q = [ M ( 7 Li) + M ( 1 H) – M ( 4 He) – M ( 4 He)] c 2 = [(7.016004 u) + (1.007825 u) – 2(4.002603 u)] c 2 (931.5 MeV/u c 2 ) = + 17.35 MeV. Because Q > 0, the reaction can occur . ( b ) The kinetic energy of the products is K = K i + Q = 2.500 MeV + 17.35 MeV = 19.85 MeV .

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Ch. 43 Page 2 8. ( a ) For the reaction N 7 14 ( α , p) O 8 17 , we determine the Q -value: Q = [ M ( 14 N) + M ( 4 He) – M ( 1 H) – M ( 17 O)] c 2 = [(14.003074 u) + (4.002603 u) – (1.007825 u) – (16.999131 u)] c 2 (931.5 MeV/u c 2 ) = – 1.191 MeV. Because ( K + Q ) > 0, the reaction can occur . ( b ) The kinetic energy of the products is K = K i + Q = 7.68 MeV – 1.191 MeV = 6.49 MeV . 9. For the reaction O 8 16 ( α , γ ) Ne 10 20 , we determine the Q -value: Q = [ M ( 16 O) + M ( 4 He) – M ( 20 Ne)] c 2 = [(15.994915 u) + (4.002603 u) – (19.992440 u)] c 2 (931.5 MeV/u c 2 ) = + 4.730 MeV . 10. For the reaction C 6 13 (d, n) N 7 14 , we determine the Q -value: Q = [ M ( 13 C) + M ( 2 H) – m (n) – M ( 14 N)] c 2 = [(13.003355 u) + (2.014102 u) – (1.008665 u) – (14.003074 u)] c 2 (931.5 MeV/u c 2 ) = + 5.326 MeV. The kinetic energy of the products is K = K i + Q = 36.3 MeV + 5.326 MeV = 41.6 MeV . 11. ( a ) We find the product nucleus by balancing the mass and charge numbers: Z (X) = Z ( 6 Li) + Z ( 2 H) – Z ( 1 H) = 3 + 1 – 1 = 3; A (X) = A ( 6 Li) + A ( 2 H) – A ( 1 H) = 6 + 2 – 1 = 7, so the product nucleus is Li 3 7 . ( b ) It is a “stripping” reaction because a neutron is stripped from the deuteron .
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