Question

A fission reaction is given by
${}_{92}^{236}U{\to }_{54}^{140}Xe{+}_{38}^{94}Sr+x+y$, where $x$ and $y$ are two particles. Considering ${}_{92}^{236}U$ Uto be at rest, the kinetic energies of the products are denoted by $K_{Xe}$, $K_{Sr}$, $K_X$ $(2MeV)$ and $K_Y(2MeV)$, respectively. Let the binding energies per nucleon of ${}_{92}^{236}U$ , ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$ be $7.5MeV$, $8.5MeV$ and $8.5MeV$ , respectively. Considering different conservation laws, the correct option(s) is(are)

• A. $x=n$ , $y=n$, $K_{Sr}=129MeV$, $K_{Xe}=86MeV$
• B. $x=p$ , $y=e^{-}$, $K_{Sr}=129MeV$, $K_{Xe}=86MeV$
• C. $x=p$ , $y=n$, $K_{Sr}=129MeV$, $K_{Xe}=86MeV$
• D. $x=n$ , $y=n$, $K_{Sr}=86MeV$, $K_{Xe}=129MeV$

Answer 1

Answer: (A)

${}_{92}^{236}U{\to }_{54}^{140}Xe{+}_{38}^{94}Sr+x+y$
$K_x=2MeV,K_y=2MeV,K_{xe}=?,K_{sr}=?$
By conservation of charge number and mass number,
$x\equiv y\equiv n$
$B.E$. per nucleon of ${}_{54}^{140}Xe$ or ${}_{38}^{94}Sr=8.5MeV$
$Q$ value of reaction,
$Q=$ Net kinetic energy gained in the process
$=K_{xe}+K_{sr}+2+2-0=K_{xe}+K_{sr}+4\dots \left(i\right)$
As number of nucleons is conserved in a reaction, so $Q$ = Difference of binding energies of the nuclei
$=140\times 8.5+94\times 8.5-236\times 7.5=219MeV\dots \left(ii\right)$
From eqns. $\left(i\right)$ and $\left(ii\right)$
$K_{xe}+K_{sr}=219-4=215MeV$
$Xe$ and $Sr$ have momentum of same magnitude but in opposite directions
Hence, lighter body has larger kinetic energy.
So, from options,
$K_{sr}=129MeV$, and $K_{xe}=86MeV$