JEE Advanced 2015 Paper 2 Physics Q16

Physics · Nuclei & Radioactivity

Problem

A fission reaction is given by

{}^{236}_{92}\mathrm U\rightarrow{}^{140}_{54}\mathrm{Xe}+{}^{94}_{38}\mathrm{Sr}+x+y

, where

x

and

y

are two particles. Considering

{}^{236}_{92}\mathrm U

to be at rest, the kinetic energies of the products are denoted by

K_{\mathrm{Xe}}

K_{\mathrm{Sr}}

K_x

(

2\,\mathrm{MeV}

) and

K_y

(

2\,\mathrm{MeV}

), respectively. Let the binding energies per nucleon of

{}^{236}_{92}\mathrm U

{}^{140}_{54}\mathrm{Xe}

and

{}^{94}_{38}\mathrm{Sr}

7.5\,\mathrm{MeV}

8.5\,\mathrm{MeV}

and

8.5\,\mathrm{MeV}

, respectively. Considering different conservation laws, the correct option(s) is (are)