0

I'm wondering where the energy difference between the two electron states $2 \:^1 S$ and $2\: ^1 P$ comes from in a helium atom. I thought it is the fine structure, but the fact that $S = 0$ is confusing me because there shouldn't be any fine structure influence for the two states.

Qmechanic
  • 201,751
Nigel N.
  • 103

1 Answers1

1

I imagine that you're referring to the second and fourth excited states, $^1S$ with configuration $1s2s$ and $^1P$ with configuration $1s2p$.

Frankly, I'm surprised that the difference between those two states isn't larger: the $^1S$ state sits $20.62 \:\rm eV$ above the ground state, and the $^1P$ state is $21.22 \:\rm eV$ above the ground state, and the relative difference at $0.6\:\rm eV$ is frankly pretty minor.

This difference can mostly be attributed to differences in electrostatic shielding produced by the $1s$ electron. It's a relatively standard fact that $s$ electrons wander further inwards towards the core than $p$ electrons on the same shell, because of the absence of the centrifugal barrier. This means that the $2s$ electron spends more 'time' at low-$r$ regions where it experiences a higher effective nuclear charge, and that brings its energy down.

In addition to that, there probably are indeed fine-structure effects that meddle with both of those energies at a level that's not too far below the $0.6\:\rm eV$ energy difference, including both a spin-orbit coupling on the $^1P$ state as well as kinetic-energy and Darwin terms on both states. But you don't need to bring those out to explain the difference.

Simply put, multi-electron atoms are messy places. The nice symmetries of the hydrogen atom, where e.g. equal-$\ell$ subshells were magically degenerate (so long as you're happy to ignore all the ways in which they're not degenerate) fly straight out of the window starting with helium upwards.

probably_someone
  • 35,459
Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
  • 1
    Indeed. In the semiclassical Sommerfeld picture, the orbitals look like this: https://commons.wikimedia.org/wiki/File:Sommerfeld_ellipses.svg –  Oct 09 '18 at 13:55
  • And the reason why the effect is not that large in helium is probably that a single $1s$ electron has a lot of wiggle room for correlations. The difference in neutral lithium is larger: 1.8 eV. –  Oct 09 '18 at 14:12
  • @Pieter That's a good point. The NIST ASD entry for helium doesn't report how much the states veer off from the Hartree-Fock picture, but I suspect that there's some substantial correlation and that labels like $1s2p$ might not be very accurate. (Or probably: that to the extent that it can be defined, the $1s$ orbital in the $1s2s$ and $1s2p$ states is actually rather different.) – Emilio Pisanty Oct 09 '18 at 14:35