CHAPTER 42 - 3

# CHAPTER 42 - 3 - 15(a . Bindingenergy =[3M H 3m n)M Li]c 1...

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15. ( a ) 6 Li consists of three protons and three neutrons.  We find the binding energy from the masses: Binding energy  = [3 M ( 1 H) + 3 m ( 1 n) –  M ( 6 Li)] c 2 = [3(1.007825 u) + 3(1.008665 u) – (6.015122 u)] c 2 (931.5 MeV/u c 2 ) =       32.0 MeV . Thus the binding energy per nucleon is (32.0 MeV)/6 =       5.33 MeV . ( b ) 208 Pb consists of 82 protons and 126 neutrons.  We find the binding energy from the masses: Binding energy  = [82 M ( 1 H) + 126 m ( 1 n) –  M ( 208 Pb)] c 2 = [82(1.007825 u) + 126(1.008665 u) – (207.976635 u)] c 2 (931.5 MeV/u c 2 =       1636 MeV . Thus the binding energy per nucleon is (1636 MeV)/208 =       7.87 MeV . 16. ( a ) 12 C less a proton becomes  11 B.  We find the binding energy of the last proton from the masses: Binding energy  = [ M ( 11 B) +  M ( 1 H) –  M ( 12 C)] c 2 = [(11.009305 u) + (1.007825 u) – (12.000000 u)] c 2 (931.5 MeV/u c 2 ) =       16.0 MeV . ( b ) 12 C less a neutron becomes  11 C.  We find the binding energy of the last neutron from the masses:
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