The disintegration energy $Q$ for the nuclear fission of $${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+n$$ is $\underline{\hspace{2cm}}\mathrm{MeV}$. 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 $c^2=931 \mathrm{~MeV} / \mathrm{u}$.