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In several computational software dealing with electronic calculations, protons and neutrons are lumped together into a point particle. This is done to simplify the problem, but I am wondering what gets left out by using this approximation. I am also wondering about what would be the magnitude of what is left out. Is the effect small enough such that we do not have to worry about it ever? What are the exceptions? This question is for electronic structure calculations.

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  • $\begingroup$ +1. Nice question. I am glad to know that I have not to ask this myself now. $\endgroup$ – user104909 Nov 17 '16 at 8:56
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The lowest-energy excitations in an atomic nucleus are typically thousands or millions of electron-volts, while chemical excitations are typically a few or a few tens of electron-volts. So treating the nucleus as inert during chemical reactions at thermal energies is an excellent approximation.

The electronic effect is negligible. The nuclear radius is typically $\sim 10^{-15}\rm\,m$, while the atomic radius is closer to $\sim 10^{-10}\rm\,m$. The nucleus therefore occupies roughly $10^{-15}$ of the atomic volume, and treating it as a simple point is also a reasonable approximation.

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    $\begingroup$ We should not compare sizes, but difference in the potential energy. In a nucleus of a finite size the potential energy is finite, not $1/r$. The diagonal matrix element of the potential "difference" compared to the ground state energy is what is needed. $\endgroup$ – Vladimir Kalitvianski Nov 17 '16 at 9:16
  • $\begingroup$ But the integrated energy of a finite charge density (as in an electron cloud) interacting with an ideal $1/r$ potential is finite, because the total charge which overlaps the zero-size singularity is zero. $\endgroup$ – rob Nov 17 '16 at 15:50
  • $\begingroup$ Yes, it is finite, I know and this is what I proposed as a meaningful comparison. $\endgroup$ – Vladimir Kalitvianski Nov 17 '16 at 16:09
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    $\begingroup$ But one way to prove that the integrated energy is finite is to take the limit of smaller and smaller volumes. That's why I suggested "the volume is already very small" as an argument that the point-like approximation is good. (I think that we are agreeing with each other, but I'm having trouble expressing it clearly.) $\endgroup$ – rob Nov 17 '16 at 16:12
  • $\begingroup$ Apart from the volume element $r^2dr$, there is a wave function squared $\psi(r)^2$, whose behavior should "support" your "volume estimation". $\endgroup$ – Vladimir Kalitvianski Nov 17 '16 at 17:15

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