Question

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

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

Answer 1

Answer: (A)

$^{236}_{92}U \rightarrow ^{140}_{54}Xe+^{94}_{38}Sr+x+y$
$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 $^{140}_{54}Xe$ or $^{94}_{38}Sr=8.5\,MeV$
$Q$ value of reaction,
$Q =$ Net kinetic energy gained in the process
$=K_{xe}+K_{sr}+2+2-0=K_{xe}+K_{sr}+4 \ldots\left(i\right)$
As number of nucleons is conserved in a reaction, so $Q$ = Difference of binding energies of the nuclei
$=140\times8.5+94\times8.5-236\times7.5=219\,MeV \ldots\left(ii\right)$
From eqns. $\left(i\right)$ and $\left(ii\right)$
$K_{xe}+K_{sr}=219-4=215\,MeV$
$Xe$ and $Sr$ have momentum of same magnitude but in opposite directions
Hence, lighter body has larger kinetic energy.
So, from options,
$K_{sr}=129\,MeV$, and $K_{xe}=86\,MeV$