Atoms & Nuclei

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Question 194 of 239

The disintegration energy QQ for the nuclear fission of 235U→140Ce+94Zr+n{ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+n is ______ MeV\mathrm{MeV}.
Given atomic masses of 235U:235.0439u;140Ce:139.9054u{ }^{235} \mathrm{U}: 235.0439 u ;{ }^{140} \mathrm{Ce}: 139.9054 u,
94Zr:93.9063u;n:1.0086u, { }^{94} \mathrm{Zr}: 93.9063 u ; n: 1.0086 u \text {, }
Value of c2=931MeV/uc^2=931 \mathrm{MeV} / \mathrm{u}

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Answer: 208

235U→140Ce+94Zr+n{ }^{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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