A fission reaction is given by ${}_{92}{}^{236}\mathrm{U}\to {}_{54}{}^{140}\mathrm{Xe}+{}_{38}{}^{94}\mathrm{Sr}+x+y$, 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 $K_{\mathrm{Xe}}$, $K_{\mathrm{Sr}}$, $K_{\mathrm{x}} \left(2 \mathrm{MeV}\right)$ and $K_{\mathrm{y}} \left(2 \mathrm{MeV}\right)$, 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(s) is (are)

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of the radius $R$ is given by $E=\frac{3}{5}\frac{Z\left(Z-1\right)e^2}{4\mathrm{\pi }{\epsilon }_{0}R}$.

The measured masses of the neutron, ${}_1{}^1\mathrm{H}$, ${}_7{}^{15}\mathrm{N}$ and ${}_8{}^{15}\mathrm{O}$ are $1.008665 \mathrm{u}$, $1.007825 \mathrm{u}$, $15.000109 \mathrm{u}$ and $15.003065 \mathrm{u}$, respectively. Given that the radii of both the ${}_7{}^{15}\mathrm{N}$ and ${}_8{}^{15}\mathrm{O}$ nuclei are the same, $1\mathrm{u}=931.5 \mathrm{MeV} {\mathrm{c}}^{-2}$ ($c$ is the speed of light) and ${\mathrm{e}}^2/\left(4\mathrm{\pi }{\epsilon }_0\right)=1.44 \mathrm{MeV} \mathrm{fm}$. Assuming that the difference between the binding energies of ${}_7{}^{15}\mathrm{N}$ and ${}_8{}^{15}\mathrm{O}$ is purely due to the electrostatic energy, the radius of either of the nuclei is $\left(1 \mathrm{fm}={10}^{–15} \mathrm{m}\right)$