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…

Question

The disintegration energy $Q$ for the nuclear fission of ${ }^{235} \mathrm{U} ightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+n$ is ______ $\mathrm{MeV}$. Given atomic masses of ${ }^{235} \mathrm{U}: 235.0439 u ;{ }^{140} \mathrm{Ce}: 139.9054 u$, ${ }^{94} \mathrm{Zr}: 93.9063 u ; n: 1.0086 u 	ext {, }$ Value of $c^2=931 \mathrm{MeV} / \mathrm{u}$

Aicharya · Accepted Answer

${ }^{235} \mathrm{U} ightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$ Disintegration energy $\begin{aligned} \mathrm{Q} & =\left(\mathrm{m}_{\mathrm{R}}-\mathrm{m}_{\mathrm{p}}ight) \mathrm{c}^2 \ \mathrm{~m}_{\mathrm{R}} & =235.0439 \mathrm{u} \ \mathrm{m}_{\mathrm{p}} & =139.9054 \mathrm{u}+93.9063 \mathrm{u}+1.0086 \mathrm{u} \ & =234.8203 \mathrm{u} \ \mathrm{Q} & =(235.0439 \mathrm{u}-234.8203 \mathrm{u}) \mathrm{c}^2 \ & =0.2236 \mathrm{c}^2 \ & =0.2236 	imes 931 \ \mathrm{Q} & =208.1716 \end{aligned}$

The disintegration energy $Q$ for the nuclear fission of ${ }^{235} \mathrm{U} \rightarrow{ }^{140}…

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