Why does an electron in the ground state $n = 1$ appear to be screened by one electron?

Question

I have been trying to understand the reason why the $Z_{eff}$ value for an electron in the ground state equals $(P-1)$, where $P$ is number of protons.

My understanding of screening is that each electron in a higher subshell is screened by the electrons that are "under" it. so lets say there is an element with 11 electrons (sodium). My understanding is that the last electron in the 3p subshell should be screened by the 10 electrons that are "under" it which gives it a $Z_{eff}$ of $(11-10) = 1$ . Which is correct according to a book am reading. But the book also states that the $Z_{eff}$ of an electron located in 1s is $(11-1)$ even though there are no electrons below the ground state.

Can someone explain this behavior?

Answer 1

1. For the inner K-shell (also known as $1s$ or $n=1$ orbital), the formula $$Z_{\rm eff}~\approx~Z-1 \tag{1}$$ is essentially Moseley's experimental law from 1913, which predates QM. The main effect is because of electron-electron repulsion with the other $1s$ electron, i.e. a screening effect of the nucleus. A qualitative argument is that the $1s$ electron spends half the time further away from the nucleus than the other $1s$ electron, and therefore experience a screening effect, cf. OP's question.

2. A back-of-the-envelope calculation shows that repulsions from all the other $n>1$ orbitals are an order of magnitude too small to explain the observation (1).

3. Therefore we may for simplicity consider a 2-electron atom $$ H ~=~\frac{{\bf p}_1^2}{2m} + \frac{{\bf p}_2^2}{2m} + k_ee^2\left(-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{|{\bf r}_1-{\bf r}_2|} \right). \tag{2}$$ Here $k_e$ is Coulomb's constant and $a_0$ is the Bohr radius.

4. The ground state energy of the 2-electron atom (2) is then by definition written as $$ E_0 ~=~-\underbrace{\frac{k_ee^2}{2a_0}}_{\approx 13.6 eV} \left(Z_{\rm eff}^2 +Z^2\right). \tag{3}$$ The interpretation of eq. (3) is that the first ionization is screened, while the second ionization sees the bare nucleus.

5. A crude order-of-magnitude estimate of eq. (1) for $Z\gg 1$ can be argued as follows:

   • Use the virial theorem to argue that the kinetic terms are minus half the potential terms, cf. e.g. this Phys.SE post.
   • Assume that$^1$ $$\langle\frac{1}{r_1}\rangle~\sim~\frac{1}{a}~\equiv~\frac{Z}{a_0}~\sim~ \langle\frac{1}{r_2}\rangle \qquad\text{and}\qquad \langle\frac{1}{|{\bf r}_1-{\bf r}_2|}\rangle~\sim~ \frac{1}{2a},\tag{4}$$ which seem reasonable in the light of formulas for the hydrogenic atom.
   • This then leads to $$ E_0 ~\sim~-\underbrace{\frac{k_ee^2}{2a_0}}_{\approx 13.6 eV} \left(2Z^2-Z\right), \tag{5}$$ suggesting an intercept of a $\frac{1}{2}$ rather than $1$ in eq. (1). Anyway, it's in the right ballpark.

6. Let us for fun mention that the Helium atom $\text{He}$ and that the Lithium ion $\text{Li}^+$ have an intercept of $.66$ and $.64$, respectively.

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$^1$ Consider e.g. the wavefunction $$\psi(r_1,r_2)~=~\psi_{100}(r_1)~\psi_{100}(r_2), \qquad \psi_{100}(r_1)~:=~\frac{e^{-r/a}}{\sqrt{\pi a^3}}. \tag{6} $$ Then $$\langle\psi \mid \frac{1}{r_1}\mid \psi\rangle~=~\frac{1}{a}~=~ \langle\psi \mid\frac{1}{r_2}\mid \psi\rangle, \tag{7} $$ and $$\langle\psi \mid\frac{1}{|{\bf r}_1-{\bf r}_2|}\mid \psi\rangle~=~ \frac{8}{a^6} \int_{\mathbb{R}_+}\! dr_1~r_1^2 e^{-2r_1/a}\int_{\mathbb{R}_+}\! dr_2~r_2^2 e^{-2r_2/a}\int_{[-1,1]}\! d\chi~ \left(r_1^2 + r_2^2 -2r_1r_2\chi \right)^{-\frac{1}{2}}$$ $$~=~\ldots~=~\frac{16}{a^6} \int_{\mathbb{R}_+}\! dr_1~r_1^2 e^{-2r_1/a}\int_{\mathbb{R}_+}\! dr_2~r_2^2 e^{-2r_2/a}\frac{1}{\max(r_1,r_2)}~=~\ldots~=~\frac{5}{8a}.\tag{8}$$

Answer 2

Consider the two electrons in a $1s$ orbital, and we'll take the simplest example of a helium atom.

The existence of atomic orbitals is based on the assumption that the individual electron-electron repulsions average out on the timescale of our meaurements, so each of the two electrons in the $1s$ orbital sees the other as an averaged out distribution of negative charge centred on the nucleus.

So the electrons move in an averaged out potential consisting of a point positive charge at the nucleus and a smeared out negative charge centred on the nucleus. At distances very close to the nucleus the potential is dominated by the nucleus and we have $Z_{eff}\approx2$. At distances far from the nucleus the positive and negative charges both contribute and we end up with $Z_{eff}\approx 1$. At intermediate distances we end up some intermediate value of $Z$.

So we expect that the effective charge experienced by each electron in a helium atom is somewhere between $Z=1$ and $2$. I can't find a detailed calculation, but some Googling suggests $Z_{eff}\approx 1.7$. The effective charge is reduced (a bit) for the electrons in the lowest atomic orbital because even for electrons in the lowest orbital they are affected by the averaged out negative charge of the other electrons.

In a large atom, like the example of sodium ($Z=11$) that you cite, the 1s electrons move in a potential that is the sum of the nuclear charge and the averaged out charge of the other $10$ electrons. So as for helium, even the lowest energy $1s$ electrons will see a $Z_{eff} \lt 11$.