Why does the nuclear charge remain the same in Moseley's law?

Question

So Today I was taught about the Moseley's law and its relation with Bohr's formula :

$$\frac{hc}{\lambda} = R \left(\frac{(Z-\sigma)^2}{n^2}-\frac{(Z-\sigma)^2}{m^2}\right)$$

My understanding is that in $(Z-\sigma)$, Z is the number of protons and $\sigma$ is related to (but not exactly) the number of electrons that is in lower shells.

So my question is that if the electron jumps to a lower level, how could value of $\sigma$ not change as stated in the Moseley's formula?

For example, let's say that an electron in L shell is knocked out. Then one electron jumps from M shell to L shell to fill the hole, the electrons under the jumping electron will change from 9 to 2. Therefore, the value of $\sigma$ should at least change and not remain constant, right?
However my professor taught me that in the example that I just mentioned, the value of $\sigma$ would remain as 9 (theoretically). How is that possible?

Answer 1

The particular formulation of Moseley's law is a mashup with the Rydberg formula

$$\frac{1}{\lambda} = RZ^2\left( \frac{1}{n_1^2}-\frac{1}{n_2^2} \right)$$

and the relationship becomes clearer if you write it as:

$$\frac{hc}{\lambda} = R(Z-\sigma)^2 \left(\frac{1}{n^2}-\frac{1}{m^2}\right)$$

It would probably be even better to write the formula as

$$\frac{hc}{\lambda} = Rk_{nm}(Z-\sigma)^2 \left(\frac{1}{n^2}-\frac{1}{m^2}\right)$$

where $k_{nm}$ represents a correction factor for the particular line such as $K_{\alpha1}$, or $K_{\beta1}$, etc. Moseley's law was an attempt to explain how the energy/wavelength of a particular x-ray line varied as a function of atomic number, not how all the lines in a particular atom varied.

So all in all you're right. The energy difference between the two orbitals would be dependent on the shielding in both orbitals. See for example Slater's rules on shielding.