The disintegration energy Q for the nuclear fission of 235 U rightarrow 140 Ce + 94 Zr + n is MeV . Given atomic masses of 235 U : 235.0439 u , 140 Ce ; 139.9054 u , 94 Zr : 93.9063 u ; n : 1.0086 u Value of c 2 =931 MeV / u

Physics · Nuclear Physics

JEE Main 2024 — 4 April, Shift 2 — Question 54

The disintegration energy $Q$ for the nuclear fission of ${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$ is $\qquad$ MeV .

Given atomic masses of ${ }^{235} \mathrm{U}: 235.0439 \mathrm{u},{ }^{140} \mathrm{Ce} ; 139.9054 \mathrm{u}$ , ${ }^{94} \mathrm{Zr}: 93.9063 \mathrm{u} ; \mathrm{n}: 1.0086 \mathrm{u}$

Value of $\mathrm{c}^{2}=931 \mathrm{MeV} / \mathrm{u}$

Answer: 208

Numerical answer — enter this value.

Step-by-step solution

${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$

Disintegration energy

$\begin{array}{*{35}{r}}\text{Q} & ~=\left( {{\text{m}}_{\text{R}}}-{{\text{m}}_{\text{p}}} \right)\cdot {{\text{c}}^{2}} \\\text{ }\!\!~\!\!\text{ }{{\text{m}}_{\text{R}}}&~=235.0439\text{u}\\\text{}\!\!~\!\!\text{}{{\text{m}}_{\text{p}}}&~=139.9054\text{u}+93.9063\text{u}+1.0086\text{u} \\{} & ~=234.8203\text{u} \\\text{Q} & ~=\left( 235.0439\text{u}-234.8203\text{u} \right){{\text{c}}^{2}} \\{} & ~=0.2236{{\text{c}}^{2}} \\{} & ~=0.2236\times 931 \\\text{Q} & ~=208.1716 \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction