The disintegration energy Q for the nuclear fission of ^ 235 U arrow ^ 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

Correct answer: 0 ^ 235 U arrow ^ 140 Ce+ ^ 94 Zr+n Disintegration energy aligned Q & =(m_ R -m_ p ) c^2 ~m_ R & =235.0439 u m_ p & =139.9054 u+93.9063 u+1.0086 u & =234.8203 u Q & =(235.0439 u-234.8203 u) c^2 & =0.2236 c^2 & =0.2236 × 931 Q & =208.1716 aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2024 · 4 April · Evening Shift · Physics · Nuclear Physics

The disintegration energy Q for the nuclear fission of ^ 235 U...

The disintegration energy

Q

for the nuclear fission of

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

is ______

\mathrm{MeV}

. Given atomic masses of

{ }^{235} \mathrm{U}: 235.0439 u ;{ }^{140} \mathrm{Ce}: 139.9054 u

{ }^{94} \mathrm{Zr}: 93.9063 u ; n: 1.0086 u \text{, }

Value of

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

Correct answer

Step-by-step solution

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

Disintegration energy

\begin{aligned} \mathrm{Q} & =\left(\mathrm{m}_{\mathrm{R}}-\mathrm{m}_{\mathrm{p}}\right) \mathrm{c}^2 \\ \mathrm{~m}_{\mathrm{R}} & =235.0439 \mathrm{u} \\ \mathrm{m}_{\mathrm{p}} & =139.9054 \mathrm{u}+93.9063 \mathrm{u}+1.0086 \mathrm{u} \\ & =234.8203 \mathrm{u} \\ \mathrm{Q} & =(235.0439 \mathrm{u}-234.8203 \mathrm{u}) \mathrm{c}^2 \\ & =0.2236 \mathrm{c}^2 \\ & =0.2236 \times 931 \\ \mathrm{Q} & =208.1716 \end{aligned}

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