# Why is Mosely screening in heavy atom K-shell 1 unit?

The Wikipedia article on Moseley's law seems to show that the screening of heavy atoms is by 1 electron charge exactly (in the limit of large Z, experimental precision, within nonrelativistic limits, and).

But why is this exactly one unit? The other K-shell electron is not screening exactly one unit, and this seems to be a conspiracy of other electrons. I suspect it is because of an unappreciated hole-picture of deep holes in heavy atoms (electrons missing in deep shells), and I will describe this theory briefly.

If you remove an electron from close to the nucleus, the electron-hole behaves as an object with positive charge and negative mass (this is why it orbits the nucleus that it is repelled by). The state is not a vacuum quite, because of the presence of other electrons, but the rigidity of the Fermi liquid near the nucleus of a heavy atom means that the hole behaves as a single particle. This single-particle behavior is in the potential background of the nucleus and the other electrons, and it is possible that the result can give an exact 1 unit screening. I developed the formalism a little bit to see what the form should be, but I did not see any reason for 1 unit screening. Perhaps there is none, but it looks to be more than a coincidence.

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I am confident that your explanation is at least on the right track. At any rate, even the Wikipedia article does say that $(Z-1)$ arises because of differences in the electron-electron interactions between the initial and final states and holes could be very useful to quantify this difference. It must be possible to justify the number 1 in some way. –  Luboš Motl Apr 13 '12 at 10:06
Ron, what about charge conservation? In truth there are Z-1 electrons going around and the hole must take its apparent charge from the Z of the nucleus to have charge conservation of the system from afar. Charge number is quantized after all in units of e.This might leave the nucleus at Z-1. –  anna v Apr 13 '12 at 10:38
@annav: This doesn't work--- the Z-1 is only for K-shell, and I don't see a reason this is linked to the number of electrons. I also am not sure this is nonrelativistically exact--- I just saw a convergence in the values at large Z by eye. –  Ron Maimon Apr 13 '12 at 19:32
Are you saying that by "remove" you mean the electron is on a higher energy shell but still attached to the atom? If it is off the atom then the atom is ionized and will have a charge +1 . –  anna v Apr 14 '12 at 3:10
look at "binding energy" here : en.wikipedia.org/wiki/Ionization_energy . –  anna v Apr 14 '12 at 3:14

## 1 Answer

I doubt that it is exactly 1. See Effective_nuclear_charge for references therein. The Clementi tables goes up to Radon and uses Slater Type orbitals. You can calculate it on your own with a quantum chemistry programm, defining the correct symmetry group for your exited wavefunction with an empty k-shell. The 1s electron coefficients are not integer but close to (Z-1).

Another important point is the inclusion of relativistic effects, most important for s-shell electrons in heavy elements. E.g. the hyperfine interaction (the Fermi contact term) is increased by about 20% for heavy elements

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I know it isn't exactly 1 for finite Z, the question is whether it asymtptotes to 1 for large Z. I see it pass 1, but I think for large atoms it is even closer to 1. It might be coincidental, but I don't think so. –  Ron Maimon Apr 13 '12 at 16:19
I don't want relativistic--- the question is whether a nonrelativistic enormous Z potential with Z electrons and one K-shell hole has energy given by Bohr model with (Z-1). It's a well defined mathematical question, and I don't see a clear "no". If you find the answer for Z=400 and it is significanty different from 1, it is good evidence that the asymptotic value is not exactly 1, but maybe 1.08 or something. I was trying to see if there is a reason for a near-1 value, and this limit of Z going to infinity is the only thing I could think of. But +1 for the link –  Ron Maimon Apr 13 '12 at 19:31
Also, you might be right, it might not be 1 exactly, and if you give a little evidence, I will accept. The problem is that I don't know how far you can get a good solution for the K-shell screening--- I never looked at the numerical methods. Experimentally, extremely heavy atoms could screw up the convergence due to relativistic corrections. –  Ron Maimon Apr 13 '12 at 19:46
Please wait for an anwer from another user, that is a specialist in x-ray analysis or a particle physicist. The Schrödinger equation is just analytically solvable for the hydrogen atom, for other you have to rely on numerical methods. I stumbled over this problem from the reverse - the effective potential of the valence electron for alkalis. There is always this assumption of an "light electron", the electron cloud screening Z-1 charge. But this picture is not right at all. –  Alex1167623 Apr 13 '12 at 19:52
Yes--- the valence picture is only a crude rough approximation. But the inner picture should be mathematically correct because of the large separation between inner shells and outer shells in energy and distance both. This means that in the limit of large Z, the inner shell transitions and orbits are exactly (nonrelativistically) described by a one-hole dynamical picture, where the x-ray transitions are the negative mass positive charge hole going up in "n" (down in energy--- it has negative mass) with matrix elements given by the hole dipole moment. This isn't in the literature, should be. –  Ron Maimon Apr 13 '12 at 20:14