One million electron volt $(1\,MeV)$ is equal to

As the mass number increases, the binding energy per nucleon .....

The dependence of binding energy per nucleon,$B_N$ on the mass number,$A$,is represented by

Find the binding energy $(BE)$ per nucleon of ${^{56}Fe}$,where $m({^{56}Fe}) = 55.936 \ u$,$m_{n} = 1.00866 \ u$,and $m_{p} = 1.00727 \ u$ (in $MeV$).

Find the binding energy per nucleon for $^{120}_{50}Sn$. Given: mass of proton $m_{p} = 1.00783 \, U$,mass of neutron $m_{n} = 1.00867 \, U$,and mass of tin nucleus $m_{Sn} = 119.902199 \, U$. (Take $1 \, U = 931 \, MeV$) (in $, MeV$)

Consider the following nuclear reaction : $^{238}\text{U} \rightarrow ^{234}\text{Th} + ^4\text{He}$. Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^4\text{He}$ as $238.050 \text{ u}$, $234.043 \text{ u}$ and $4.003 \text{ u}$, respectively. The $Q$ value for the reaction, in $\text{keV}$, is : [Given : $1 \text{ u} = 931.5 \text{ MeV/c}^2$]