Physics

Problem

A fission reaction is given by

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

, where

x

and

y

are two particles. Considering

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

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

K_{\text{Xe}}

K_{\text{Sr}}

K_x

(2 MeV) and

K_y

(2 MeV), respectively. Let the binding energies per nucleon of

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

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

and

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

be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is(are) [Adv. 2015] a.

x = n, y = n, K_{\text{Sr}} = 129 \text{ MeV}, K_{\text{Xe}} = 86 \text{ MeV}

x = p, y = e^-, K_{\text{Sr}} = 129 \text{ MeV}, K_{\text{Xe}} = 86 \text{ MeV}

x = p, y = n, K_{\text{Sr}} = 129 \text{ MeV}, K_{\text{Xe}} = 86 \text{ MeV}

x = n, y = n, K_{\text{Sr}} = 86 \text{ MeV}, K_{\text{Xe}} = 129 \text{ MeV}