Question 314: Atoms & Nuclei

Question

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

Accepted Answer

Q. value
$$\begin{aligned}
& =\{(235.0439)-[39.9054+93.9063+1.0086]\} 	imes 931 \mathrm{~MeV} \
& =208 \mathrm{~MeV}
\end{aligned}$$

Atoms & Nuclei

Integer / Numeric Answer Type