# Why can free lithium atoms not take part in an Auger process?

Shouldn't it be possible for an incoming photon to excite one of the 1s electrons to a 2p state (or one of even higher energy) and then for the excited electron to drop back to 1s and kick out the 2s electron?

Is there some fundamental reason that prohibits this process, or is it just very unlikely, or is it not called "Auger process" because the first electron is excited instead of kicked out?

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Is three enough? 2 in 1s and 1 in 2s, so if you knock out the 1s, the 2s falls in, and it just radiates. There's nothing else in 2s or 2p for it to knock off. (The 1s cannot be bumped off, because the energy is only enough to get it to 2s.) Or am I misunderstanding your question? –  Willie Wong Jun 28 '11 at 2:08
Isn't it possible for the 1s electron to be excited to 2s instead of being knocked out? In that case there are two electrons in 2s. –  Friedrich Jun 29 '11 at 12:11
Uh, how do you end up with 2 in 2s? The second photon in the Auger process comes from an outer-shell electron transitioning to a lower energy state. So at least one of the electrons will end up in 1s. Also, what are you thinking of as the Auger process? My understanding of Auger is "1 photon in, 2 electrons out", with the atom picking up +2 charge. –  Willie Wong Jun 29 '11 at 13:57
I was under the impression that "1 photon in, 1 electron out" would also be a possible Auger process, but maybe I am wrong. In that case, the incoming photon could excite one electron from the 1s shell. Then there could be two electrons in the 2s shell and one in the 1s shell. One 2s electron could fall back to 1s and kick the other 2s electron out. Doesn't this qualify as an Auger process? –  Friedrich Jun 29 '11 at 23:45
I thought "needs a virtual photon" was the same thing as "forbidden by the dipole selection rules". –  Dan Jul 1 '11 at 22:15

I am no expert on the Auger effect but the wikipedia page on Auger electron spectroscopy seems to indicate that Auger effects can be observed in lithium. As @WillieWong points out in the comments, the usual Auger effect does not work: if the laser simply removes one 1s electron, you are left with a 1s$^1$ 2s$^1$ configuration, which cannot ionize to Li$^{++}$ as the excited electron can only bump the ground one up to 2s (and even that transition is out since the final state is identical to the initial one!).
It is in principle possible, however, for the laser to simultaneously - or sequentially - ionize a lithium atom and excite the ion, leaving it in a configuration 2s$^1$ 2p$^1$ or higher. (As said in the comments, 2s$^2$ is out by dipole selection rules.) Such a configuration (or a similar one) would have enough energy to expel one of the electrons in an Auger transition.
The catch is that while a single laser pulse can do both things, a single laser photon can't. Leaving a doubly-excited ion is, for weak pulses, a two-photon process, since you need to do two transitions, each of which annihilates a photon. As such, it is twice as unlikely as the 1s$^1$ 2s$^1$ transition, or to put it another way it scales quadratically with the laser intensity. Fortunately, it is possible to produce pulses strong enough to use this to their advantage (or indeed to take the situation far out of the perturbation regime where photons are meaningful concepts) and use strong intensities to leave all sorts of interestingly excited ion electronic configurations. As long as some of these result in Auger electrons, these will be relatively easy to detect as they have quite characteristic signatures.
I believe when people state that an Auger effect can be observed in Lithium (as on Wikipedia), they usually mean bound Lithium, since it has more electrons available. You are saying that the laser would need to simultaneously ionize and excite the atom. But what about not ionizing it? Can't the laser just excite the atom to a 1s$^1$ 2s$^1$ 2p$^1$ state? –  Friedrich Jan 26 '13 at 22:00