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Recentely we were discussing / learning Moseley's law in class. We have successfully constructed the most basic version of it:

$$f=f_R*(Z-1)^2*(1-\frac{1}{n^2})$$

with $n$ being the main quantum number, $f_R=\frac{m_ee^4}{8\epsilon_0^2h^3}$ being the Rydberg freqency and $f$ being the frequency of the characteristic X-Ray line and of course $Z$ being the atomic number of the atom in question.
This is Moseley's law for the so-called $K_\alpha$ line, where the electron drops from the L-shell ($n=2$) to the K-shell ($n=1$).

Now we have discussed where the $Z-1$ term comes from and concluded that it is caused by the remaining electron in the K-shell which reduce the effective nuclear charge by one. No problems so far.

Next we thought of how the law would look like for the $L_\alpha$ line where an electron drops from the M-shell ($n=3$) to the L-shell ($n=2$). Somebody in class came up with the idea that $Z-1$ needed to be replaced by $Z-7$ then, which makes some sense as there are 7 electrons remaining in the L-shell. But we quickly found out that it should be $Z-9$ because the two electrons in the K-shell also reduce the effective nuclear charge.

So I looked this up on the internet and found at several places that it is in fact $Z-7.4$ for the $L_\alpha$ line. These places also told me that it is something called electron screening and quantum effects that cause it to be 7.4 rather than 9, but I don't really understand what this "screening" is and which effects are involved.

So here's my question:

Why is the effective nuclear charge for this $Z-7.4$ and not $Z-9$ and how can I construct this constant for other shells / lines like $M_\alpha$?

And if possible only assume the least possible quantum effects / mechanics knowledge for an answer. And please don't consider $f_R$ being wrong (= not corrected) part of the question.

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Screening means that from the view of the electron for which this is being calculated, the effect of the nuclear charge is lessened due to the other electrons which are orbiting the nucleus. In the first case, the K-shell electron is in the $1s$ orbital(which is spherically symmetric around the nucleus, and so it can effectively screen the nuclear charge making the EAN (Effective atomic number) = $Z-1$ . But in case of L-shell, only 2 electrons are in the $2s$ orbital and effectively screen the nucleus. The other 5 electrons are in $2p$ orbital which is dumbbell shaped, with the nucleus at the centre of the dumbbell. This is not so effecient in screening the charge and so the contribution of each electron will be lesser than 1, making the $EAN = Z-2-2-3.4 = Z-7.4$ (instead of $Z-2-2-5 = Z-9$).

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    $\begingroup$ Just to add to this, if you'd want to calculate the reduction in Effective Atomic Number you'd almost certainly need to run numerical simulations to crunch the numbers, or look up empirical results. There's likely no way you could work it out analytically $\endgroup$ – Chris Cundy Sep 20 '15 at 13:35
  • $\begingroup$ @ChrisCundy We can work them out using quantum mechanics. $\endgroup$ – sarat.kant Sep 20 '15 at 17:52

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