[AdiDex]

In one post, Enthalpy wrote

Electron fall. All s orbitals have their maximum of probability density per volume unit at the nucleus.

But electrons don't concentrate at the nucleus because they are waves and as such, have a volume. And in a more detailed manner: concentrating the electron in a smaller volume implies a higher kinetic energy, which counters the advantage of proximity to the positive nucleus. There is an optimum which is the atom's radius.

In the simple description of the Hydrogen atom, we consider the nucleus as a point particle but in reality, it is not. But what if we consider its size. Then volume element around the nucleus will also increase. From my fuzzy notion, The maxima of the radial probability curve should shift towards the nucleus, since there is a higher volume around the nucleus to start with.

My thought experiment,
What if instead of the nucleus, we take a hollow sphere whose electric field identical (eg. the electric potential at r from the nucleus = electric potential at x from the periphery of this hollow sphere) to the nucleus but has a bigger size. Now the initial volume element will be much bigger than the real case. Will maxima of radial probability will move towards the hollow sphere? Will there be any such case when maxima will be x 0 (like 0.000000001 Å) or r_max  r_o ? what if r_o > 0.59 Å ?

r_o = radius of hollow sphere, assuming potential inside the sphere is infinite so that electron can't go inside.
(Not asking about r, r= r_o + x )

[Borek]

Then volume element around the nucleus will also increase.

Please elaborate, sounds to me like you are juggling meaningless words here.

[Enthalpy]

Nuclei do have a finite volume, and this correction is made in accurate computations. It must be pretty useless for chemistry, but on the measures of the energy of last ionisation for varied elements, you can observe it. Or as well on the energy of a pseudo-hydrogen atom made of one proton and one muon, as this "atom" is smaller hence more sensitive to the proton size (...and figures don't fit, that's one open research question presently, and nobody knows the consequences).

The usual model is to stray the nucleus' charge uniformly over its volume rather than over the nucleus' surface. Though, we might argue that since they repel an other, protons should concentrate a bit more at the nucleus' surface. I ignore if someone tried to model that.

The effect is that:

• Outside the nucleus, the field and the potential are the same as for a point charge.
• Inside the nucleus, the field is less steep because a fraction of the charge is behind you as you dive to the centre, so the potential is less deep when the charge is strayed.
• For the same electron orbital, the attraction energy would be less strong, which means that the adapted orbital is wider.

This correction is smaller than the relativistic effect of the electron's kinetic energy.

And by the way, I still want to know why the energy of the electrostatic interaction between the electron and the nucleus has no inertia. It should be about as big as the relativistic correction, hence much bigger than other corrections which are experimentally confirmed, but this inertia is never included in successful models.