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AlKindi
Oct8-11, 09:33 AM
If the fluorescence is the re-emitting of a photon with a larger wave lenght due to the transition from a higher energy state to a lower energy state in the case of resonance Raman (where there aren't any virtual states) seems be equal to the fluorescence. Which differences are there?

Claude Bile
Oct13-11, 09:27 AM
Fluorescence occurs due to real electronic transitions, while Raman scattering occurs as a result of virtual electronic-vibrational transitions.

That is, fluorescence is an absorption/re-emission process while Raman scattering is an inelastic scattering process.

Claude.

fsonnichsen
Oct14-11, 07:09 AM
Claude's answer brings up an interesting point about this problem: when does a photon "lose its identity" and become classified as a "re-emitted" photon?

AlKindi
Oct15-11, 04:32 AM
Fluorescence occurs due to real electronic transitions, while Raman scattering occurs as a result of virtual electronic-vibrational transitions.

That is, fluorescence is an absorption/re-emission process while Raman scattering is an inelastic scattering process.

Claude.

Thanks Claude, but what's about Resonace Raman scattering are there any virtual electronic states involve? I don't think so... [Modern Raman Spectroscopy– A Practical Approach p. 94]

AlKindi
Oct15-11, 04:45 AM
Claude's answer brings up an interesting point about this problem: when does a photon "lose its identity" and become classified as a "re-emitted" photon?

I think when a photon (and the wave associeted) interact costructively with a system there is a time in which the photon "lose its identity" (e.g. the 10-5 seconds for the fluorescence) the energy is non more in the photon but in the system e.g. molecule. So we can say that it is re-emitted because of this transition I think we can also say the system (the molecule, the atoms etc. etc.) is re-formed.