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A fission reaction is given by U 236 92 → Xe 140 54 + Sr 94 38 + x + y, where x and y are two particles. Considering U 236 92 to be at rest, the kinetic energies of the products are denoted by K Xe K Sr , K x (2MeV) and K y (2MeV), respectively. Let the binding energies per nucleon of U 236 92 , Xe 140 54 and Sr 94 38 be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option (s) is (are)

1. x = n , y = n, K Sr = 129 MeV, K Xe = 86 MeV
2. x = p , y = e – , K Sr = 129 MeV, K Xe = 86MeV
3. x = p , y = n, K Sr = 129 MeV, K Xe = 86 MeV
4. x = n , y = n, K Sr = 86 MeV, K Xe = 129 MeV

Correct answer: x = n , y = n, K Sr = 129 MeV, K Xe = 86 MeV

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