Question

The energy released per fission of a 92U235 nucleus is nearly

A
200eV
B
20eV
C
200MeV
D
2000eV
Solution
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The fission process represented by the equation, 92U235+0n156Ba144+36Kr89+30n1
Masses of reactants =234.39+1.01=235.4amu
Masses of products =143.28+88.89+3(1.01)=235.2amu
Energy released = mass difference =235.4235.2=0.2amu=0.2×931200MeV

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