Why is the screening coefficient, $\sigma_K$, in Moseley's law only constant for high $Z$ and how can it be anything other than unity?
In trying to understand the screening coefficient, $\sigma_K$ and the limitations of Moseley's law (why only valid for high $Z$) I came across a section of this lab script I found online:
Moseley’s Law and the Determination of the Rydberg Constant
X-rays can also be absorbed. This is essentially due to the ionization of atoms, which release an electron from an inner shell, e.g. the $K$-shell, when an x-ray photon is absorbed. The transmission of a material is defined as
$$T=\frac{I}{I_0}\tag{1}$$ where $I_0$ and $I$ are the x-ray intensities incident on and transmitted through the material respectively.
In 1913, the English physicist <PERSON> measured the $K$-absorption edges for various elements and formulated the law that bears his name: $$\sqrt{\frac{1}{\lambda_K}}=\sqrt{R}(Z-\sigma_K)\tag{2}$$ where is $R$ is the Rydberg constant, $Z$ is the atomic number of the absorbing elements and $\sigma_K$ is the screening coefficient. This equation can be brought into agreement with the predictions of <PERSON>’s model of the atom by considering the following: The nuclear charge, $Z\cdot e$ , of an atom is partially screened from the electron ejected from the $K$-shell (through absorption of the x-ray photon) by the remaining electrons of the atomic shell. 617
Therefore, on average, only the charge $(Z-\sigma_K)\cdot e$ acts on the electron during ionization. For sufficiently large $Z$ ($\gt \sim 30$), $\sigma_K$ is approximately constant and equation $(2)$ becomes linear.
I have some conceptual questions regarding these notes/script displayed above.
The text above says "remaining electrons of the atomic shell", but this makes no sense to me. The $K$ shell consists of just 2 electrons and when one electron is ejected (via x-ray absorption) there can only be 1 electron left in that innermost $K$ shell, so why is Moseley's law not being written as $\sqrt{\frac{1}{\lambda_k}}=\sqrt{R}(Z-1)$? The reason I write this is because there is only one electron left in the $K$ shell to provide any shielding from the ejected electron (that absorbed the x-ray photon).
The text then goes on to say "For sufficiently large $Z$ $(\gt \sim 30)$, $\sigma_K$​ is approximately constant and equation $(2)$ becomes linear". I have some questions regarding that quote, firstly, why restrict to high $Z$ ($\gt 30$) to get constant $\sigma_K$? Secondly, as mentioned before, $\sigma_K=1$ always (at least by my logic it is) so how can it possibly be "approximately constant"? Lastly, what does it mean to say that "equation $(2)$ becomes linear"? Even if $\sigma_K$ is a fixed numerical constant (say $\sigma_K=1$, in the case of my argument above), that equation is anything but linear. Linear equations have the form $y= mx + c$, this $(2)$ here is quadratic in $Z$.
I think that overall, I am misinterpreting some of the information provided to me, if anyone could help dispel my confusion with hints or tips I would be most grateful, many thanks! 617