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 $${ }^{235} \mathrm{U}: 235.0439 u ;{ }^{140} \mathrm{Ce}: 139.9054 u, { }^{94} \mathrm{Zr}: 93.9063 u ; n: 1.0086 u$$, Value of $c^2=931 \mathrm{~MeV} / \mathrm{u}$.
Answer (integer)
208
Solution
<p>Q. value</p>
<p>$$\begin{aligned}
& =\{(235.0439)-[39.9054+93.9063+1.0086]\} \times 931 \mathrm{~MeV} \\
& =208 \mathrm{~MeV}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Fission and Fusion
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