What is the physical explanation for the red color in chlorophyll fluorescence?

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SUMMARY

The red color observed in chlorophyll fluorescence, when extracted using acetone and exposed to UV light, is primarily due to energy loss through vibrational and electronic state transitions. Upon absorption of a UV photon, chlorophyll undergoes excitation, and the vibrational energy dissipates before the emission of a red photon. This process aligns with the principles outlined in the Franck-Condon diagram, which illustrates the potential-energy surfaces for electronic states and their corresponding vibrational levels.

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Homework Statement



I am doing some research for a lesson pertaining to the fluorescence of chlorophyll.

When chlorophyll has been extracted from plant leaves using acetone, the solution will glow a bright red when exposed to UV light.

My question is: what is the physical cause for the red color? I have thought of two possible explanations: either (a) the emission of a red photon is preceded by the emission of a UV photon which is slightly lower in energy than the absorbed UV photon, with the difference in energy corresponding to the energy of the red photon, or (b) energy is lost through the usual processes of heat, vibrations, etc.

But I can't find documentation for either.

Can anybody out there point me to some reference material? Thanks
 
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(b) is correct. Fluorescence typically involves an excitation of both vibrational and electronic states when the photon is absorbed. The vibrational energy is lost before the photon is re-emitted.

See e.g. http://en.wikipedia.org/wiki/File:Franck-Condon-diagram.png" . The two big curves are the potential-energy surfaces for the two electronic states. The sub-levels are the vibrational levels.
 
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