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.

By contrast, the average binding energies of nucleons is around 8 MeV. So in general the binding energy of the nucleons will be much greater than the binding energy of the electrons, and the mass deficit will be dominated by the nucleus.

But I'm not sure it makes much sense to just add together electron and nucleon binding energies to get a total binding energy for the atom. I suppose it does have a meaning in that it is the energy necessary to dismantle the atom into separate electrons, protons and neutrons. In practice it's hard to see how this ever be useful.

It is possible to work out the mass deficit due to the electron binding energy in the hydrogen atom. Obviously in this case the nucleus is just a proton so there is no nuclear binding energy. The masses of the proton, electron and hydrogen atom are precisely enough known to (just) be able to calculate the mass deficit. The mass deficit is of course 13.6eV.

  • $\begingroup$ 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 $\endgroup$ – user1314 Apr 15 '15 at 15:40
  • $\begingroup$ @user1314: yes. It's normally safe to ignore any mass deficit for the electrons. $\endgroup$ – John Rennie Apr 15 '15 at 15:41

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