JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 25)
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
208
Explanation
Q. value
$$\begin{aligned} & =\{(235.0439)-[39.9054+93.9063+1.0086]\} \times 931 \mathrm{~MeV} \\ & =208 \mathrm{~MeV} \end{aligned}$$
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