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A fission reaction is given by $_{92}^{236} U \to_{54}^{140} X e +_{38}^{94} S r + x + y,$ where x and y are two particles. Considering $_{92}^{236} U$ to be at rest, the kinetic energies of the products are denoted by $K_{X e}, K_{S r}, K_{x} (2 M e V)$ and $K_{y} (2 M e V)$ respectively. Let the binding energies per nucleon of $_{92}^{236} U,_{54}^{140} X e$ and $_{38}^{94} S r$ be $7.5 M e V, 8.5 M e V$ and $8.5 M e V$ , respectively. Considering different conservation laws, the correct option(s) is (are)

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$x = p, y = n, K_{S r} = 129 M e V, K_{X e} = 86 M e V$

$x = n, y = n, K_{S r} = 86 M e V, K_{X e} = 129 M e V$

$x = n, y = n, K_{S r} = 129 M e V, K_{X e} = 86 M e V$

$x = p, y = e^{-}, K_{S r} = 129 M e V, K_{X e} = 86 M e V$

answer is A.

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Detailed Solution

From conservation laws of mass number and atomic number, we can say that x=n, y=n

$(x =_{0}^{1} n, y =_{0}^{1} n)$

From conservation of momentum

$\begin{array}{l} |P_{x e}| = |P_{s r}| \\ F r o m K = \frac{P^{2}}{2 m} \\ \Rightarrow K \propto \frac{1}{m} \\ S i n c e, m_{s r} 〈m_{x e} \Rightarrow K_{s r}〉 K_{x e} \\ ∴ K_{s r} = 129 M e V, \\ K_{x e} = 86 M e V \end{array}$

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