# Understanding Moseley’s law from the Rydberg-type equation?

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?

• 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. – Nicolau Saker Neto Nov 20 '13 at 13:08
• 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. – Nicolau Saker Neto Nov 20 '13 at 13:09
• 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. – Nicolau Saker Neto Nov 20 '13 at 15:18
• @NicolauSakerNeto I guess all these comments could be posted as a good answer. Though a bit speculative, They effectively answer my main concern. – Satwik Pasani Nov 22 '13 at 8:24
• 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. – Nicolau Saker Neto Nov 23 '13 at 18:33