How am I obtaining higher energy photons from a fluorescein solution than the ones I'm exciting it with?

Hi all,

I have a theory question that is probably obvious to some of you!

I’m confocal scanning a FITC solution as follows: Imgur: The magic of the Internet with excitation at 520 nm and emission around 500 nm. I’m using time gating to eliminate laser reflection.

I thought I knew that fluorescence goes from lower wavelength excitation to higher wavelength emission according to Stokes, because a fluorophore producing fluorescence absorbs the higher energy photon, does some work thus losing energy in the process, and then emites a photon that must have a lower energy (higher wavelength) than the incoming one.

The anti-Stokes shift description on the Wiki page mentions a special case where the fluorophore receiving energy from a surrounding crystal lattice, but I’m scanning freely diffusing FITC molecules in water…

Please help me understand how I am receiving higher energy photons from FITC than the photons I’m pumping into it!

Thanks in advance.

Hi @JelleP,

This answer will probably be superseded by someone else’s more knowledgeable than me, but I think it basically boils down to this:

When you excite a solution of some fluorescent compound, their molecules will be distributed in the several vibrational energy associated with the ground state following a Boltzmann distribution, with the lower energy vibrational states more populated then the higher ones.

Nonetheless, there will always be a small fraction of molecules in higher vibrational states, which can be promoted to the next electronic (plus some vibrational) level with a less energetic photon that what the majority of molecules will need. Then, these same molecules will undergo internal conversion in the excited state and will emit a photon going back to one of the vibrational levels of the ground state (with a probability similar to their original distribution [I think—please correct me if I’m wrong]). So, a small fraction of the molecules will in fact emit at a longer wavelength than then excitation one. [As I understand it, the “extra energy” is just stolen from the thermal distribution].

Here’s a Jablonski diagram illustrating the case:

[modified from here]

As I understand it, by only picking emission around 500 nm, you’re effectively rejecting most of the photons that are actually emitted in your sample.

I hope this helps.

Cheers!

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Hi @NicoDF , that does help, many thanks!

I had not really considered the population statistics / distribution of energy states that the fluorophores could exist in. This does help a lot, also with the figure. Thanks again!

I am also ready for the obscure details if they exist, should a photophysicist come along.

Once, I remember a thread where people discussed about photobleaching during imaging. That phenomenon turned out to be much, much more complicated than I had expected, with several components playing a role…

Hi JelleP
Well i could see two processes at play.

The first one would simply be that your wavelength detection band is wide enough to collect some >520nm wavelength photons, even with 1-5% transmission you could collect something. If you need time gating to remove laser then you also collect a bit around 520nm ? (or at least OD is not extremely high). A look at the data would be nice to see if the level of signal you receive is comparable to classical fluorescence signals.

NicoDF theory is interesting. I would also suggest that you could have a low probability of performing two photon absorption, effectively creating emitted photon with 2times (minus a loss) the energy that you excited with.

Well, putting a spectrometer on the detection path could clear some doubts about this :slight_smile:

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