CHAPTER 43 - 1 -...

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 Al 13 28  Mg 12 28   Al 13 28 Si 14 28 + e – 1 0 + ν   H 1 2 (d, n) He 2 3   U 92 238 (n, γ ) U 92 239   Li 3 7 (p, α ) He 2 4   Be 4 9 ( , n) C 6 12   Mg 12 24 (n, d) Na 11 23 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      . If   were a  β +  emitter, the resulting nucleus would be  , which has too many neutrons relative to the number of protons to  be stable.   Thus we have a        emitter . The decay is 
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at University of California, Berkeley.

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CHAPTER 43 - 1 -...

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