JEE Main 2024 Physics question
The disintegration energy Q for the nuclear fission of U → Ce + Zr + n is MeV. Given atomic masses of U: 235.0439u, Ce: 139.905u, Zr: 93.9063u, n: 1.0086u, Value of c = 931 MeV/u.
Answer: 208
Solution
Δm = 235.0439 − (139.905 + 93.9063 + 1.0086) = 0.2240 u; Q = 0.2240 × 931 ≈ 208 MeV.