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

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Unformatted text preview: 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 = (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 Use Word 6.0c or later to view Macintosh picture. were a + emitter, the resulting nucleus would be Use Word 6.0c or later to view Macintosh picture. , which has too many neutrons relative to the number of protons to be stable. Thus we have a emitter . The decay is Use Word 6.0c or later to view Macintosh picture. , so the product is Si 14 28 . 2. For the reaction Use Word 6.0c or later to view Macintosh picture. , 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 Use Word 6.0c or later to view Macintosh picture. 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 Use Word 6.0c or later to view Macintosh picture. , 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 Use Word 6.0c or later to view Macintosh picture. , 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 Use Word 6.0c or later to view Macintosh picture. , 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....
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This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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

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