I've just been taught about the effective nuclear charge and screening effect in school. I'd like to ask and confirm if the effective nuclear charge is only defined for valence electrons. If not then why don't we consider the repulsion from the electrons further away from the nucleus than the test electron, which by common sense should add to the attractive force by nucleus?
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$\begingroup$ In chemistry you often worry only about the outermost electrons. In atomic physics they worry about more of them (e.g. levels leading to x-ray generation). As you have noticed, trying to pack all the effects into one parameter (effective nuclear charge) doesn't work so well for all of the electrons. $\endgroup$– Jon CusterCommented May 12, 2021 at 12:41
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$Z=Z_{eff}+\sigma$, where $\sigma$ is shielding constant calculated according to Slater's rule. Since Slater's rule can be applied to any electron regardless of whether or not it's a valence electron, $Z_{eff}$ and $Z$ can be defined for any electron too, $Z$ in particular remaining constant. In his paper, which I have attached below, Slater states that the approximations to the wave functions that he gives have been obtained empirically, and do not involve any of the outer electrons.
Original paper by JC Slater ($1930$): https://web.archive.org/web/20120323031605/http://astrophysics.fic.uni.lodz.pl/100yrs/pdf/04/008.pdf
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$\begingroup$ Can you please elaborate how the Slater's rules explain the effect of outer electrons empirically? $\endgroup$ Commented Jun 14, 2021 at 3:18
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$\begingroup$ Explain, as in according to slater's rules the calculation of shielding constant only involves inner electrons. That is what I meant. $\endgroup$ Commented Jun 14, 2021 at 11:48
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$\begingroup$ This still doesn't answer the question. Why don't the outer electrons matter? $\endgroup$ Commented Jun 14, 2021 at 16:20