If we observe isolated chlorophyl solution with UV light, chlorophyl is seen as red. Explanation suggests that when a specific atom of chlorophyl absorbs UV light, the atom gets excited and in a short time electrons still goes from excited state to ground state. But why it is not emits same wavelength light, and instead emits lower energy light - red ?
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2$\begingroup$ Many materials do that. Search for fluorescence. $\endgroup$– PoutnikCommented Nov 26, 2021 at 20:28
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1$\begingroup$ Besides generally searching for “fluorescence”, see here and perhaps search in this stack exchange for other answers. $\endgroup$– Ed VCommented Nov 27, 2021 at 0:03
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2$\begingroup$ Very basic and quick - I think is what you are asking for. As molecules are constituted of different part linked and vibrating, the upwards path can be closed once absorption occurred. Part of the energy is dissipated in rearranging the molecule and by vibrations. Then from a certain already lowered down level, emission occurs. As a starting point for more look indeed at the thread linked by Ed V. Wikipedia on the subject is also nice. Note that "isolated"guarantee emission, it is not functional to the fact that UV in / red out. A molecule which is not isolated can have its emission quenched.. $\endgroup$– AlchimistaCommented Nov 27, 2021 at 12:14
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2$\begingroup$ ... because the surrounding can interact with it and remove the otherwise "to be emitted" energy. $\endgroup$– AlchimistaCommented Nov 27, 2021 at 12:17
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2$\begingroup$ @EdV indeed the "guarantee" I have used above is with the context given. OP: Isolated molecules can be not fluorescent, or doing it with a very small efficiency nevertheless. $\endgroup$– AlchimistaCommented Nov 27, 2021 at 15:52
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1 Answer
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There are three types of re-emission when photons interact with matter:
- Emitting the same wavelength, e.g., simple reflection, resonance fluorescence and stimulated emission.
- Emitting longer wavelength due to losses in absorption of initial photon, e.g., fluorescence and phosphorescence and Compton scattering.
- Emitting shorter wavelength than that incident. This occurs in two photon or multiphoton fluorescence. It requires the molecule stay in a metastable state long enough for a second photon to raise it to a higher yet state, requiring intense light. AFAIK, there is some little literature on UV-incident multiphoton fluorescence, but I don't know if chlorophyll has been investigated.
answered Nov 28, 2021 at 1:02
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$\begingroup$ I think it will be better to call "emit same wavelength" condition as resonance fluorescence rather than reflection. Atoms show this phenomenon in flames. If we excite Na atoms in flame by 589 nm, they will emit 589 nm. $\endgroup$– ACRCommented Nov 28, 2021 at 1:45
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$\begingroup$ Upvoted, but it is not clear to me what is puzzling the OP. Is it the matter of the “chlorophyl atom” not behaving like a hydrogen atom, i.e., it can absorb UV light, but fluoresces red? Or is the red fluorescence itself the issue? I think the former is what they are asking about, but unless they clarify or elaborate, not much more to say, I guess. $\endgroup$– Ed VCommented Nov 28, 2021 at 21:51