$A$ fission reaction is given by ${ }_{92}^{236} U \rightarrow{ }_{54}^{140} Xe +{ }_{38}^{94} Sr + x + y$,where $x$ and $y$ are two particles. Considering ${ }_{92}^{236} U$ to be at rest,the kinetic energies of the products are denoted by $K_{Xe}, K_{Sr}, K_x (2 \ MeV)$ and $K_y (2 \ MeV)$,respectively. Let the binding energies per nucleon of ${ }_{92}^{236} U, { }_{54}^{140} Xe$ 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):

• A
  $x = n, y = n, K_{Sr} = 129 \ MeV, K_{Xe} = 86 \ MeV$
• B
  $x = p, y = e^-, K_{Sr} = 129 \ MeV, K_{Xe} = 86 \ MeV$
• C
  $x = p, y = n, K_{Sr} = 129 \ MeV, K_{Xe} = 86 \ MeV$
• D
  $x = n, y = n, K_{Sr} = 86 \ MeV, K_{Xe} = 129 \ MeV$