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

K_x = 2 MeV, K_y = 2 MeV, K_Xe = ?, K_Sr = ?

By conservation of charge number and mass number, x $$\equiv$$ y $$\equiv$$ n

B.E. per nucleon of $${}_{92}^{236}U = 7.5$$ MeV

B.E. per nucleon of $${}_{54}^{140}Xe$$ or $${}_{38}^{94}Sr = 8.5$$ MeV

Q value of reaction,

Q = Net kinetic energy gained in the process

$$ = {K_{Xe}} + {K_{Sr}} + 2 + 2 - 0 = {K_{Xe}} + {K_{Sr}} + 4$$ ..... (1)

As number of nucleons is conserved in a reaction, so Q = Difference of binding energies of the nuclei

$$ = 140 \times 8.5 + 94 \times 8.5 - 236 \times 7.5 = 219$$ MeV ...... (2)

From eqns. (i) and (ii)

$${K_{Xe}} + {K_{Sr}} = 219 - 4 = 215$$ MeV ....... (3)

The linear momentum of a particle of mass m and kinetic energy K is given by $$p = \sqrt {2mK} $$. Since the masses and kinetic energies of x and y are very small in comparison to that of $${}_{54}^{140}Xe$$ and $${}_{38}^{94}Sr$$, we can neglect the linear momentum of these particles. Initially, the linear momentum of $${}_{92}^{236}U$$ is zero (at rest). Finally, the products $${}_{54}^{140}Xe$$ and $${}_{38}^{94}Sr$$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus,

$$\sqrt {2{M_{Xe}}{K_{Xe}}} = \sqrt {2{M_{Sr}}{K_{Sr}}} $$, i.e.,

$$140{K_{Xe}} = 94{K_{Sr}}$$. ...... (4)

Solve equations (3) and (4) to get K_Xe = 86 MeV and K_Sr = 129 MeV.