Question
In the options given below, let $$E$$ denote the rest mass energy of a nucleus and $$n$$ a neutron. The correct option is
A.
\[E\left( {\begin{array}{*{20}{c}}
{236}\\
{92}
\end{array}U} \right) > E\left( {\begin{array}{*{20}{c}}
{137}\\
{53}
\end{array}I} \right) + E\left( {\begin{array}{*{20}{c}}
{97}\\
{39}
\end{array}Y} \right) + 2E\left( n \right)\]
B.
\[E\left( {\begin{array}{*{20}{c}}
{236}\\
{92}
\end{array}U} \right) < E\left( {\begin{array}{*{20}{c}}
{137}\\
{53}
\end{array}I} \right) + E\left( {\begin{array}{*{20}{c}}
{97}\\
{39}
\end{array}Y} \right) + 2E\left( n \right)\]
C.
\[E\left( {\begin{array}{*{20}{c}}
{236}\\
{92}
\end{array}U} \right) < E\left( {\begin{array}{*{20}{c}}
{140}\\
{56}
\end{array}Ba} \right) + E\left( {\begin{array}{*{20}{c}}
{94}\\
{36}
\end{array}Kr} \right) + 2E\left( n \right)\]
D.
\[E\left( {\begin{array}{*{20}{c}}
{236}\\
{92}
\end{array}U} \right) = E\left( {\begin{array}{*{20}{c}}
{140}\\
{56}
\end{array}Ba} \right) + E\left( {\begin{array}{*{20}{c}}
{94}\\
{36}
\end{array}Kr} \right) + 2E\left( n \right)\]
Answer :
\[E\left( {\begin{array}{*{20}{c}}
{236}\\
{92}
\end{array}U} \right) > E\left( {\begin{array}{*{20}{c}}
{137}\\
{53}
\end{array}I} \right) + E\left( {\begin{array}{*{20}{c}}
{97}\\
{39}
\end{array}Y} \right) + 2E\left( n \right)\]
Solution :
Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to Uranium which has a big nuclei and less B.E/nucleon.
In other words, Iodine and Yttrium are more stable and therefore possess less energy and less rest mass. Also when Uranium nuclei explodes, it will convert into $$I$$ and $$Y$$ nuclei having kinetic energies.