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Does molecular vibrational transition and consequent emission of infrared radiation involve electrons changing energy level? In wikipedia, about vibronic transitions it says "Most processes leading to the absorption and emission of visible light, are due to vibronic transitions. This may be contrasted to pure electronic transitions which occur in atoms and lead to sharp monochromatic lines (e.g. in a sodium vapor lamp) or pure vibrational transitions which only absorb or emit infrared light.". Does this mean infrared radiation is emitted without electrons playing a direct part? What about changing in molecular rotational energies? Also, I understand reflection is not a radiation absorption-emission phenomena, so can you explain what actually happens to the photons? Same thing for radiation scattering.

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3 Answers 3

In the case of the IR photons the system (i.e. molecule) does change it's state (by changing its vibrational or rotational mode).

You can also imagine such a thing happening in a purely atomic context. This would be a direct analogy of the Mössbauer Effect, only happening in an atomic context rather than nuclear one. By the totalitarian rule I have to imagine that it is possible, but it must be strongly suppressed.

I've never heard of a measurement making use of it in a purely atomic sense.

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Thx for your answer! You mean the electrons don't change energy levels, just the molecule as a whole, and that causes the emission of a photon? And what about the question about reflection and scattering? The reflected/refracted radiation consists of the same photon that was incident? How do you conceive the process? – João Feb 7 '12 at 4:27
You should also include in the discussion bremstrahlung, which is photons emitted in a continuous spectrum when a charged particle accelerates/decelerates. – anna v Feb 8 '12 at 12:26
Anna, again the bremstrahlung effect is related to charged particles, particularly electrons, my question is what about changing in vibrational and rotational molecular energies... – João Feb 8 '12 at 18:19
@João Well, the molecule is also a charged particle, as are the atoms and electrons etc. it's composed of. – Luaan Aug 6 '14 at 10:53

Regarding the general question posed in the title, yes, it is possible in general to have photon emission (or absorption) without electrons changing energy levels. For example, in nuclear magnetic resonance (NMR), EM radiation (photons) are absorbed and re-emitted with changes in nuclear spin state. NMR spectroscopy relies on splitting between nuclear spin states due to a large applied magnetic field. This splitting causes absorption and stimulated emission of EM radiation with energy equal to the splitting. So NMR is one example of photon emission without changes in electronic energy levels playing a direct role.

However, with regards to your specific examples, electrons are directly involved. In the vibrational modes, the resonant frequencies are determined by the stiffnesses of the molecular bonds. Those stiffnesses in turn depend on the electronic structure of the molecule. The question is, how much does the electronic energy change when the bond lengths and angles are changed by small amounts? This can be calculated using electronic structure packages like Gaussian and ABINIT. The nuclei are also involved; the masses of the nuclei also factor into the resonant frequencies. The picture is similar for (dihedral) rotation of chemical bonds.

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See also my closely related question:… – Douglas B. Staple Apr 1 '13 at 19:56
I just noticed that the question was posted in Feb. 2012, not Feb. 2013... meaning that it's over a year old... Oh well. – Douglas B. Staple Apr 1 '13 at 19:56

Infrared radiation is electromagnetic radiation, a quantum photonic energy phenomenon. Photons are generated for any form of matter above absolute zero. ANY motion of ANY subatomic particle generates photons. Infrared photons are ALWAYS being generated, all around, from every object not at absolute zero. Is this not true?

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For the record, you should try to avoid asking a question in your answer. Just a tip. This is more or less fine though – Jim Jun 19 '13 at 14:14

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