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In gas discharge lamps, for example, a current composed of ionized atoms excites the electrons of the atoms of the gas, and when they fall back to their ground state photons are emitted. Why is the reverse process (energy transfer of an excited electron to a free electron) never mentioned? Is it not possible or is the characteristic time involved too low for the electron to "wait" for an electron to pass?

I have read about losing excitation energy as heat in chemistry oriented articles, but they don't go too deep. Here is an example.

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  • $\begingroup$ not sure what you mean by reverse process - can you clarify - one reveerse process would be photon absorption and excitation of an electron. Am I correct in that you are thinking of a 'superelastic' collision where an excited atom loses energy in collision with a free electron and the free electron gains kinetic energy $\endgroup$ – tom Nov 6 '14 at 0:13
  • $\begingroup$ Yes, you are correct (about the last part). $\endgroup$ – Ant Nov 6 '14 at 0:15
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In principle there is nothing preventing that from happening, but ...

  • the population in a gas discharge tube is dominated by ground state atoms (contrast this with a laser which must obtain a "population inversion" to work), so there are more targets for the excitation than for the de-excitation.
  • as you suggest the relaxation time for most excited states is quite short so the odds of another free-electron coming along with just the right kinematics in the available time is pretty low

Asking students in a upper division quantum class to estimate the rate would be a neat exercise though. Must think about this.

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    $\begingroup$ Somehow I feel there ought to be an evil-cackle sound track as I read that last paragraph! $\endgroup$ – Floris Nov 5 '14 at 23:37
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    $\begingroup$ Oh, if only I could do the resonant bass tones at the end. $\endgroup$ – dmckee Nov 5 '14 at 23:39
  • $\begingroup$ Practice - it's all about practice. $\endgroup$ – Floris Nov 5 '14 at 23:40
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    $\begingroup$ "non-radiative recombination" is an important process in condensed matter, where the atoms are very close together and can exchange energy by collisions (phonons). $\endgroup$ – garyp Nov 6 '14 at 0:28
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Two thoughts for you in addition to dmckee's answer.

1) Andrew Murray's research group in Manchester has done some experiments on superelastic collisions of electrons with excited atoms, see for example here. In these collisions free electrons hit excited atoms and gain energy from the excitation energy of the atom. This is a difficult experiment - you need a nice laser system to make a large enough population of excited atoms to make it possible.

2) You mention losing energy as 'heat'. Energy loss to heat is particularly relevant to systems where the excited atoms (or molecules) are in liquid or solid. Effectively the excitation energy is converted by collisions with neighbouring atoms (or molecules) into vibrational / rotational motion, which goes into heat. This is relevant to Kasha's rule in classical photochemistry. Kasha's rule applies to molecules in liquids (or solids).

So in the case of the discharge lamp you are thinking of the excited atoms are in a gas phase environment and the likelihood of losing excitation energy to heat will depend on the rate of collisions and, hence, the gas pressure.

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You are wanting to know about the Auger process. The answer is yes. An excited electron can emit a photon when it falls to a lower orbital, or it can give its energy to another electron in the atom and eject it instead. The chances of this occurring are dependent on which atom, which energy transition, and to a very small extent the chemical environment.

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  • $\begingroup$ I think the question is about a free electron and an excited atom - Auger is about a highly excited atom where one electron falls down into a hole and another electron is ejected from the atom $\endgroup$ – tom Nov 6 '14 at 2:38

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