CHAPTER 45 - 3

# CHAPTER 45 - 3 - (c K=*kT(4.7MeV(1.60 10 17(a...

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O 8 16 + O 8 16 Si 14 28 + He 2 4 ( c ) We find the temperature from K  =  * kT ; (4.7 MeV)(1.60 × 10 –13  J/MeV) =  * (1.38 × 10 –23  J/K) T , which gives  T  =       4 × 10 10  K . 17. ( a ) For the reaction  , we determine the  Q -value: Q   = [ M ( 16 O) +  M ( 16 O) –  M ( 28 Si) –  M ( 4 He)] c 2   = [2(15.994915 u) – (27.976927 u) – (4.002603 u)] c 2 (931.5 MeV/u c 2 ) = 9.594 MeV. Thus       9.594 MeV is released . ( b ) We find the radius of the oxygen nucleus from r  = (1.2 fm) A 1/3  = (1.2 fm)(16) 1/3  = 3.02 fm. If the two nuclei are just touching, the Coulomb potential energy must be the initial kinetic  energies of the two nuclei: 2 K  =  U  =  Z O Z O e 2 /4p Å 0 2 r   = (8)(8)(1.44 MeV     fm)/2(3.02 fm) = 15.25 MeV. Thus each oxygen nucleus has a kinetic energy of       7.6 MeV . ( c

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CHAPTER 45 - 3 - (c K=*kT(4.7MeV(1.60 10 17(a...

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