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CHAPTER 43 - 11

CHAPTER 43 - 11 - Z(X)=Z Be Z He)Z(n)=4 20=6 9 4 A(X)=A Be...

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Be 4 9 + He 2 4 C 6 12 + n 0 1 Z (X) =  Z ( 9 Be) +  Z ( 4 He) –  Z (n) = 4 + 2 – 0 = 6; A (X) =  A ( 9 Be) +  A ( 4 He) –  A (n) = 9 + 4 – 1  = 12, so the other nucleus is      . ( b ) For the reaction  , we determine the  Q -value: Q   = [ M ( 9 Be) +  M ( 4 He) –  M ( 12 C) –  m (n)] c 2   = [(9.012182 u) + (4.002603 u) – (12.000000 u) – (1.008665 u)] c 2 (931.5 MeV/u c 2 ) =     + 5.70 MeV . 56. We find the conversion from K  =  kT   by converting the units for  k : 1.38 × 10 –23  J/K = (1.38 × 10 –23  J/K)/(1.60 × 10 –16  J/keV) =       8.63 × 10 –8  keV/K . 57. The average kinetic energy is a function of the temperature: K  =  ! mv 2  =  * kT . When we form the ratio for the molecules of the two isotopes at the same temperature, we get v 235 / v 238  = ( m 238 / m 235 ) 1/2  =  { [238 u + 6(19 u)]/[235 u + 6(19 u)] } 1/2  =      1.004 . 58. ( a ) We find the number of fissions from n  = (20 kt)(5 × 10 12  J/kt)/(200 MeV/fission)(1.60 × 10 –13  J/MeV) = 3.13 ×

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