QJEEAdv15Phd906: JEE Advanced 2015 Physics - Class 12 JEEAdv15Phd906 Solution: A fission reaction is given by , where x and y are two…

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

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.