# Mass of an Atom

So the mass deficit of an atom, denoted by $\Delta M(Z,A)$, is given by the following formula,

$\Delta M(Z,A) = M(Z,A) - Z(M_{p} + m_{e}) -NM_{n}$

However since the rest mass of an electron is a lot smaller than the rest mass of a proton, we generally tend to ignore it in calculations. So does this mean that the mass of an atom is approximately equal to the mass of the nucleus? If so then essentially when we are talking about the binding energy of an atom we mean the binding energy of the nucleus since the contribution from the electron is negligible anyways?

Have a look at this table, which shows the binding energy (i.e. the ionisation energy) of electrons in various atoms. The highest energies are a few hundred eV, and those are for the core electrons (though admittedly it doesn't show $1s$ energies for the heavy atoms) so the average electron binding energy will be lower.
• So if I understand correctly, in short we can assume that the mass deficit of an atom is approximately equal to $M(Z,A)-ZM_{p}-NM_{n}$ which basically implies that the mass deficit of an atom and nucleus are essentially the same – user1314 Apr 15 '15 at 15:40