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

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?

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.

• I still don't get it. Rydberg formula is used for atoms that have a similar structure to hydrogen (only 1 electron). Therefore it makes sense that Z is squared as the electron is pulled by the same amount of charges (Z) no matter the level. But in Moseley's law, it applies to atoms with more electrons. Due to the shielding effect, the sigma value should be different for different level of the electron. So how is the (Z-sigma) squared in the Moseley's law? – boboboy131 Oct 3 '18 at 13:40
• Added more detail to the answer. Hopefully that helps. // User Farcher made an excellent suggestion to read the Wikipedia article on Mosely's Law. – MaxW Oct 3 '18 at 14:02
• Mathjax hint : If you use \left( and \right) for brackets they will be displayed at a larger size to match the contents of the brackets. – StephenG Oct 3 '18 at 15:21
• @MaxW So my understanding now is that the Moseley's law is empirical. The $\sigma$ value comes from the experiment results. It can only be used to describe certain de-excitation of atoms. The part I am confused about is that my professor said the $\sigma$ value in Moseley's law theoretically should be the no. of electrons under the jumping electron. Just like the example I gave in the post. Is he correct? – boboboy131 Oct 3 '18 at 16:39
• @boboboy131 - Yes, Moseley's law is empirical, though it mimics a form of the Rydberg formula which has a much stronger theoretical foundation. In the formula I showed I think of $\sigma$ more or less as being the shielding of the most bound electron. So for the $K_{\alpha}$ lines $\sigma$ would be, more or less, for the 1s electrons - like from Salter's rules. Not sure what a "best fit" would be for what I have as $k_{nm}$ and $\sigma$ over all the $K_{\alpha1}$ lines for example. – MaxW Oct 3 '18 at 17:14