A nucleus ${ }_Z^A X$ has mass represented by $m(A, Z)$. If $m_p$ and $m_n$ denote the mass of proton and neutron respectively and $B E$ the binding energy (in MeV ) then,

In the case of formation of nucleus the evolution of energy equals to the binding energy of the nucleus takes place due to disappearance of a fraction of total mass. If the quantity of mass disappearing is $\Delta m$, then the binding energy is $B E=\Delta m c^2$ cFrom the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write $\Delta m=Z m_p-N m_n-m(A, Z)$ where $m(A, Z)$ is the mass of the atom of mass number $A$ atomic number $Z$. Hence, the binding energy of nucleus is $\begin{aligned} & B E=\left[Z m_p+N m_n-m(A, Z)\right] c^2 \\ & B E=\left[Z m_p+(A-Z) m_n-m(A, Z)\right] c^2\end{aligned}$