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

• Correct answer
  A

  $x=n,\mathrm{ }y=n,\mathrm{ }{K}_{Sr}=129\mathrm{ }MeV,\mathrm{ }{K}_{Xe}=86\mathrm{ }MeV$
• B

  $x=p,\mathrm{ }y=e^{-},\mathrm{ }{K}_{Sr}=129\mathrm{ }MeV,\mathrm{ }{K}_{Xe}=86\mathrm{ }MeV$
• C

  $x=p,\mathrm{ }y=n,\mathrm{ }{K}_{Sr}=129\mathrm{ }MeV,\mathrm{ }{K}_{Xe}=86\mathrm{ }MeV$
• D

  $x=n,\mathrm{ }y=n,\mathrm{ }{K}_{Sr}=86\mathrm{ }MeV,\mathrm{ }{K}_{Xe}=129\mathrm{ }MeV$

Hence, the correct answer is option: 'A'.