IIT JEE Physics (1978-2016: 39 Years) Topic-wise Complete Solutions

Fission and Fusion Processes

Problems from IIT JEE

Problem (IIT JEE 2015): A fission reaction is given by $\sideset{_{\ 92}^{236}}{}{U}\to \sideset{_{\ 54}^{140}}{}{Xe} + \sideset{_{38}^{94}}{}{Sr}+x+y$, where $x$ and $y$ are two particles. Considering $\sideset{_{\ 92}^{236}}{}{U}$ to be at rest, the kinetic energies of the products are denoted by $K_\text{Xe}$, $K_\text{Sr}$, $K_\text{x}({2}\;\mathrm{MeV})$ and $K_\text{y}({2}\;\mathrm{MeV})$, respectively. Let the binding energies per nucleon of $\sideset{_{\ 92}^{236}}{}{U}$, $\sideset{_{\ 54}^{140}}{}{Xe}$, and $\sideset{_{38}^{94}}{}{Sr}$ be ${7.5}\;\mathrm{MeV}$, ${8.5}\;\mathrm{MeV}$ and ${8.5}\;\mathrm{MeV}$, respectively. Considering different conservation laws, the correct option(s) is (are)

1. $x=n$, $y=n$, $K_\text{Sr}={129}\;\mathrm{MeV}$, $K_\text{Xe}={86}\;\mathrm{MeV}$
2. $x=p$, $y=e^\text{-}$, $K_\text{Sr}={129}\;\mathrm{MeV}$, $K_\text{Xe}={86}\;\mathrm{MeV}$
3. $x=p$, $y=n$, $K_\text{Sr}={129}\;\mathrm{MeV}$, $K_\text{Xe}={86}\;\mathrm{MeV}$
4. $x=n$, $y=n$, $K_\text{Sr}={86}\;\mathrm{MeV}$, $K_\text{Xe}={129}\;\mathrm{MeV}$

Solution: Given fission reaction is, \begin{align} \label{fzb:eqn:1} \sideset{_{\ 92}^{236}}{}{U}\to \sideset{_{\ 54}^{140}}{}{Xe} + \sideset{_{38}^{94}}{}{Sr}+\sideset{_{Z_x}^{A_x}}{}{x}+\sideset{_{Z_y}^{A_y}}{}{y}. \end{align} Apply conservation of charge and mass to get, \begin{align} \label{fzb:eqn:2} &92=54+38+Z_x+Z_y \\ \label{fzb:eqn:3} &236=140+94+A_x+A_y. \end{align} These equations gives $Z_x+Z_y=0$ and $A_x+A_y=2$. From the given options, these conditions are satisfied if $\sideset{_{Z_x}^{A_x}}{}{x}=\sideset{_{0}^{1}}{}{n}$ and $\sideset{_{Z_y}^{A_y}}{}{y}=\sideset{_{0}^{1}}{}{n}$.

The Q value of the given reaction is, \begin{align} \label{fzb:eqn:4} Q&=\mathrm{BE}_\text{products}-\mathrm{BE}_\text{reactants}\nonumber\\ &=(140\times 8.5+94\times8.5)-236\times7.5={219}\;\mathrm{MeV}. \end{align} The energy released in the reaction (Q value) is equal to the kinetic energy of the products i.e., $K_\mathrm{Xe}+K_\mathrm{Sr}+K_\mathrm{x}+K_\mathrm{y}=Q$, which gives, \begin{align} \label{fzb:eqn:5} K_\mathrm{Xe}+K_\mathrm{Sr}\!=\!219-(2+2)\!=\!{215}\;\mathrm{MeV}.\quad(\because K_\mathrm{x}\!=\!K_\mathrm{y}\!=\!{2}\;\mathrm{MeV}). \end{align} The linear momentum of a particle of mass $m$ and kinetic energy $K$ is given by $p=\sqrt{2mK}$. Since masses and kinetic energies of $x$ and $y$ are very small in comparison to that of $\sideset{_{\ 54}^{140}}{}{Xe}$ and $\sideset{_{38}^{94}}{}{Sr}$, we can neglect the linear momentum of these particles. Initially, the linear momentum of the $\sideset{_{\ 92}^{236}}{}{U}$ is zero (at rest). Finally, the products $\sideset{_{\ 54}^{140}}{}{Xe}$ and $\sideset{_{38}^{94}}{}{Sr}$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus, \begin{align} \label{fzb:eqn:6} &\sqrt{2M_\mathrm{X_e} K_\mathrm{Xe}}=\sqrt{2M_\mathrm{Sr} K_\mathrm{Sr}},&&\text{i.e.,}&& 140 K_\mathrm{Xe}=94 K_\mathrm{Sr}. \end{align} Solve above equations to get $K_\mathrm{Xe}={86}\;\mathrm{MeV}$ and $K_\mathrm{Sr}={129}\;\mathrm{MeV}$.