To explain the characteristic X-ray emission peaks for various elemental targets, a formula was developed which was similar in construct to the Rydberg equation for H-atom as derived by the Bohr-model.

$$\tilde\nu=R\left(\frac{1}{n_\mathrm{f}^2}-\frac{1}{n_1^2}\right)\left(Z-\sigma\right)^2$$ where $\tilde\nu$ is the wavenumber, $\sigma=1$ for the $K_\alpha$ line and $7.4$ for the $L_\alpha$ line. My question is:
Although Bohr model is wrong in light of the modern quantum mechanical concept of atomic structure, why does the energy for transition (and X-ray emission) of complicated atoms with complicated screening effect be represented by a simple adjustment of the $Z$ factor in the Rydberg equation with the value of $R$ used in case of H atom? Why isn’t there a need of complicated corrections (instead of simple subtraction of an experimentally obtained $\sigma$) to predict the X-ray wavelength?

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    $\begingroup$ Though the Bohr model is oversimplified, the Rydberg equation itself is a good approximation (I wonder if it can be derived using the more accurate orbital model). Either way the $K$ and $L$ shell electrons are very close to the nucleus and maybe they don't suffer as much from complicated electron-electron interactions (or rather they likely become less important relative to electron-nuclei attraction), so apparently they can be well approximated by just considering a screening effect which changes the effective nuclear charge. $\endgroup$ – Nicolau Saker Neto Nov 20 '13 at 13:08
  • $\begingroup$ I guess the equation is used as more of a fitting curve for experimental data points rather than an accurate theoretical equation, especially if the value of $\sigma$ is allowed to vary slightly. $\endgroup$ – Nicolau Saker Neto Nov 20 '13 at 13:09
  • $\begingroup$ Also perhaps part of the reason the Rydberg equation does not work well with transitions between valence electrons is that in the valence shell there is a significant energy difference between subshells. For core electrons, I think the degeneracy is close to restored, so it becomes less inaccurate to work with only the principal quantum number. $\endgroup$ – Nicolau Saker Neto Nov 20 '13 at 15:18
  • $\begingroup$ @NicolauSakerNeto I guess all these comments could be posted as a good answer. Though a bit speculative, They effectively answer my main concern. $\endgroup$ – stochastic13 Nov 22 '13 at 8:24
  • $\begingroup$ I appreciate it, but I would rather abstain. I hope someone can come forth with a more direct answer. My comments were mostly musings to get some ideas flowing. I don't know the explanation for certain myself. $\endgroup$ – Nicolau Saker Neto Nov 23 '13 at 18:33

The concept to understand here is that most theories in physics and chemistry are obtained by observing then theorizing, this results in the theories and models being made in such a way that they can explain most (if not all) observable phenomenon.

So, when Bohr made his model of the atom, it had to explain the the phenomenon of "X-ray emission" otherwise it wouldn't be accepted, because, X-ray emission is an observed phenomenon of atoms. Following on, it was discovered that the Bohr-model couldn't explain all observable phenomenon as equipment got better and more things could be seen (notably the splitting of spectral lines in magnetic fields). So any theory that followed on to explain this splitting of spectral lines had to explain X-ray emission as well, without disturbing the results of Bohr's theory.

A suitable analogy would be the discovery of special relativity by Einstein, whatever discovered had to explain why Newton's Law had stuck around for hundred's of years without people realizing it was wrong. (And it did, the correction at velocities we know are less than one part in a million and that's why we didn't think Newton was wrong)

So the deeper question is, even a one part in a million error is still an error and so we should know that the theory is wrong even early on... this is not the case as most instrumental errors on their own are much larger in this regard (at least in Newton's time), and the same applies here in that older instruments weren't usually sensitive enough to notice the change brought on by quantum mechanics on X-ray emission to that of Moseley's rough equation. (Though now it is usually possible to detect differences to a good degree.)

So to the final part of your question, there is a horrendous equation waiting if you want to plug in exact values to find the exact wavenumber, but most like the instrument barely notices a difference, so in a sense, it doesn't matter which theory we use. (It's very well possible a hundred years down the road, another genius comes up with an even smaller correction that can be observed only with futuristic devices)

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