I know this sounds dumb, but it puzzles me when I search for ArI levels catalogue in NIST. It shows argon's first ionisation level as $15.7596119 \; \text{eV}$ in row 427, with more rows after with energy above that. This is puzzling. How can atomic energy levels go beyond ionisation level? Shouldn't the atom be stripped off an electron and becoming an ion on those levels? NIST documentation put some words on this, but it doesn't explain the question I want to know. Chances are that this is an idiotic question out of poor knowledge, please forgive my ignorance if that's true.
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$\begingroup$ I don't know enough details to fully understand that table, but I think it means that it takes 15.75...eV to remove the outer electron. But instead of removing one electron entirely, the energy can be used to excite more than one electron to a higher orbital, and that can take more energy than only removing one electron. There is an upper limit to the energy states, but that is when all electrons are removed and the atom is fully ionised. (Unless you start including energy states of the nucleus.) $\endgroup$– JanKanisCommented Jun 7, 2021 at 22:28
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2 Answers
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When you ionise an atom typically we assume the electron is removed and the ion is left behind in its ground state. Then the ionisation energy is just the energy required to do this. But you could remove the electron and then leave the ion in an excited state, not the ground state. This would take more energy because you need to supply energy not only to remove the electron, but also to promote the remaining ion to the excited state.
So in this sense there isn't a single ionisation energy, but a whole range of ionisation energies corresponding to the different excited states of the ion left behind after the electron is removed. When listing ionisation energies we don't normally worry about this. We just assume the ion will be left in the ground state and list that energy as the ionisation energy. However in spectra you may see lines corresponding to transitions that leave the ion in an excited state, and it's these transition energies that the NIST article is listing.
Note that for the energies below $15.7596119$ eV the final state of the Argon atom is given as ${}^2P^\circ{}_{3/2}$. This is a term symbol and describes the overall state of the atom. The exact details of what the term symbol means need not worry us. The key point is that if you look at the energies above $15.7596119$ eV you'll see that that the final state of the Argon atom is ${}^2P^\circ{}_{1/2}$ i.e. it is in a different final state.
If you look at the entries in the NIST document around the ionisation energy:
then $15.7596119$ eV is the energy needed to ionise the atom and leave the ion in the ${}^2P^\circ{}_{3/2}$ state. The higher energy $15.7622$ eV is the energy needed to excite an electron to the 9d level and also excite the whole atom to the ${}^2P^\circ{}_{1/2}$ state.
answered Jun 7, 2021 at 7:05
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$\begingroup$ "also excite the whole atom to the ${}^2P^\circ{}_{1/2}$ state" — doesn't this term symbol define the state of the remaining ion (i.e. without the $9d$ electron), rather than the whole atom? $\endgroup$– RuslanCommented Jun 7, 2021 at 16:21
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The atom is a whole quantum mechanical entity, the electrons occupying orbitals at definite energy levels. Scattering photons of the appropriae frequency an electron has a probability to be removed, it does not need to be in the shallowest energy level.
If one could radiates the atom with the frequency/energy of a lower level , there will be a quantum mechanical probability that the electron from that level would be extracted, and the atom would be ionized because it would have excess positive charge corresponding to an inner electron leaving. See this enter link description here
