Discussion Forum

A fission  reaction is given by   $_{92}^{236}U\rightarrow_{54}^{140}Xe +_{38}^{94} Sr+x+y$  , where  x and y are two paricles.  Considering   $_{92}^{236}U$ to be at rest, the kinetic energies of the products are denoted bt K_xe , K_sr , K_x  (2MeV) and  K_y   (2 MeV), respectively, Let the binding energies per nucleon of  $_{92}^{236}U$ . $_{54}^{140}xe$  and  $_{38}^{94}sr$ be 7.5 MeV , 8.5 MeV and 8.5 MeV, respectively. Considering different vconservation laws , the correct options is/are

Answer:

Explanation:

From conservation laws of mass number and atomic number, we can say that x= n, y=n

         $(x=_0^1n, y=_0^1n)$

  $\therefore$   Only (a) and (d)  options may be correct

   From the conservation of momentum.

            $|p_{xe}|= |p_{sr}|$

    From $K=\frac{P^{2}}{2m}\Rightarrow K\propto\frac{1}{m}$

              $\frac{K_{sr}}{K_{xe}}=\frac{m_{xe}}{m_{sr}}$

$\therefore$           K_sr  =129MeV

                     K_xe      =86MeV

Note:  There is no need of finding the total energy released in the process.