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 Henry Moseley 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 Bohr’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. 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!


1 Answer 1


The screening idea can be understood by thinking about the total electric field in the atom, which is created by the nucleus and electrons together. The contribution from the nucleus is the familiar $Z e / 4 \pi \epsilon_0 r^2$ (radially outwards). The contribution from the electrons is more complicated, but it can be approximated as $$ \frac{q(r)}{4\pi\epsilon_0 r^2} $$ where $$ q(r) = -e \int_0^r \sum_i |\psi_i^2| 4 \pi r^2 dr $$ is the total charge density within $r$ from the electrons with their wavefunctions $\psi_i$. There are two reasons why the screening is by more than 1 unit of charge:

  1. all the wavefunctions, not just the innermost shell, extend to the origin, so all the electrons give a little bit of negative charge at low values of $r$. This is especially the case for electrons in the $l=0$ states (called $S$ states). The terminology of "shell" and "subshell" is useful but it does not mean a sequence of separate regions of charge as a function of $r$. It refers to a sequence of overlapping distributions of charge as $n$ and $l$ increase, with the distribution for lower values of $n$ on average closer to the origin than the one for higher values of $n$.
  2. The other contribution to screening comes from the charge outside the shell you are thinking about. This may seem surprising, but it is easy to understand. The screening idea is roughly accounting for the total work required to take an electron and remove it from the atom. If there is negative charge located at values of $r$ larger than the one you started with, this negative charge will lower the electric field at places on the journey as the charge is pulled off the atom, thus reducing the total work required.

The equation $$ \frac{1}{\sqrt{\lambda_K}} = \sqrt{R} (Z - \sigma_K), $$ considered as an exact statement, says that $1/\sqrt{\lambda_K}$ is some complicated function of $Z$ because $\sigma_K$ itself depends on $Z$ in a complicated way. However $\sigma_K$ stays of order $1$ and therefore it is making only a relatively small correction when $Z$ is large. Rather than ignoring it altogether, a good approximation is to say that $\sigma_K$ is a constant, i.e. independent of $Z$, and then $1/\sqrt{\lambda_K}$ is a linear function of $Z$ (and $1/\lambda_K$ is a quadratic function of $Z$).

  • $\begingroup$ Many thanks for answer (I upvoted but rep is too low for it to count yet, sorry). Could you please address the last question I asked, about "equation $(2)$ becoming linear"? As I do not understand what this means. Regards. $\endgroup$
    – Skynet
    Nov 21, 2021 at 18:08
  • $\begingroup$ @Skynet ok; done. $\endgroup$ Nov 21, 2021 at 20:38
  • $\begingroup$ Thanks for update. "$\sigma_K$ itself depends on $Z$ in a complicated way", and what is this functional form of $\sigma_K=\sigma_K(Z)$? $\endgroup$
    – Skynet
    Nov 23, 2021 at 11:41
  • $\begingroup$ @Skynet That functional form is the result of the integrals over the charge density, so it has no simple expression. $\endgroup$ Nov 23, 2021 at 11:43
  • $\begingroup$ Okay, do you have link to these integrals over charge density please? $\endgroup$
    – Skynet
    Nov 23, 2021 at 11:45

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