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The disintegration energy Q for the nuclear fission of 235U 140Ce + 94Zr + n is MeV. Given atomic masses of 235U : 235.0439 u; 140Ce : 139.9054 u, 94Zr : 93.9063 u; n : 1.0086 u, Value of c2 = 931 MeV/u.
208
Q = [m( 235U) - \m( 140Ce) + m( 94Zr) + mn\ ]c2
Mass of products = 139.9054 + 93.9063 + 1.0086 = 234.8203 u
m = 235.0439 - 234.8203 = 0.2236 u
Q = 0.2236 931 = 208.17 208 MeV
Answer: 208
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