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A fission reaction is given by \( { }_{92}^{236} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+x+y \), where \( x \) and \( y \) are two particles. Considering \( { }_{92}^{236} \mathrm{U} \) to be at rest, the kinetic energies of the products are denoted by \( K_{\mathrm{Xe}}, K_{\mathrm{S} r}, K_{x}(2 \mathrm{MeV}) \) and \( K_{y}(2 \mathrm{MeV}) \), respectively. Let the binding energies per nucleon of \( { }_{92}^{236} \mathrm{U} \), \( { }_{54}^{140} \mathrm{Xe} \) and \( { }_{38}^{94} \mathrm{Sr} \) be \( 7.5 \mathrm{MeV}, 8.5 \mathrm{MeV} \) and \( 8.5 \mathrm{MeV} \), respectively. Considering different conservation laws, the correct option is
(a) \( x=n, y=n, K_{\mathrm{Sr}}=129 \mathrm{MeV}, K_{\mathrm{Xe}}=86 \mathrm{MeV} \)
(b) \( x=p, y=e^{-}, K_{\mathrm{Sr}}=129 \mathrm{MeV}, K_{\mathrm{Xe}}=86 \mathrm{MeV} \)
(c) \( x=p, y=n, K_{\mathrm{Sr}}=129 \mathrm{MeV}, K_{\mathrm{Xe}}=86 \mathrm{MeV} \)
(d) \( x=n, y=n, K_{\mathrm{Sr}}=86 \mathrm{MeV}, K_{\mathrm{Xe}}=129 \mathrm{MeV} \)
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