How much energy does it take to singly ionize a neutral lithium atom by removing an electron from the $1s$ orbital?

  • $\begingroup$ Note that the $1s$ orbital is not the 'ground state'. The ground state is exclusively a multi-electron state, with the configuration $1s^2 \, 2s$, and its energy cannot be split up between the different electrons. The lowest-lying orbital does exist as a concept within Hartree-Fock theory, but it is not the ground state in any useful physical sense. $\endgroup$ – Emilio Pisanty Jun 7 '19 at 21:28
  • $\begingroup$ Thanks for your answer, I'm actually trying to simulate Lithium with hartree fock and want to compare my results to some real data but you are of course right that in reality, the electron states do not decouple. I should have stated that in my question.. $\endgroup$ – nerror Jun 7 '19 at 21:33
  • $\begingroup$ But still, particular orbitals have their energy, that is changing across the element table, due kernel charge, kernel shielding and electron repulsing effects. $\endgroup$ – Poutnik Jun 8 '19 at 4:07

You can obtain this directly from the energy-levels part of the NIST Atomic Spectra Database.

  • In Spectrum, put Li I for the energy levels of neutral lithium, or Li II for the energy levels of the first cation.
  • Choose whatever Level units you're most comfortable with.

Click Retrieve Data to get the results.

The ionization energy you want is not listed explicitly, and you need to get it as the sum of two factors:

  • The ionization energy from Li I to Li II, which is listed in the spectrum of Li I, at the entry Li II ($1s^2$ $^1S_0$) at the left-hand column.
  • The excitation energy from the $1s^2$ of the cation to the relevant $1s\,2s$ of the cation, which are listed in the Li II spectrum.

Note that there are two such states, with a small $\sim 1\:\rm eV$ energy difference between them, depending on how the two remaining spins are aligned. (Though, frankly, given how ridiculously hard it is to excite the cation to anything other than the ground state, this difference may well be negligible.)

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.