# Sum of nucleus less than sum of neutrons & protons? [duplicate]

Some of the comments to the last question hint to the answer. The total energy of the bound system is $E~=~k~+~U$, for $K$ kinetic energy and $U$ potential energy. The kinetic energy is $K~=~(\gamma~-~1)mc^2$ which for semi-relativistic physics becomes $K~=~\frac{1}{2}mv^2~+~mc^2$ $+~O(v^4/c^4)$ with $v^2~<<~c^2$. This is then $N$ = number of nucleons times the mass of a nucleon $\sim~1840 MeV$. The potential energy part is negative, and is about a few $MeV$ per nucleon. Therefore the binding energy, or the negative potential energy, is sufficient to actually reduce the mass of the nucleus.