Enthalpy,
There are a number of reasons there aren't many strong UV-emitting (organic) fluorophores. I could elaborate at length, but generally the fluorescence efficiency is related to the ratio of the radiative rate of excited state deactivation to the sum of the radiative and nonradiative rates,
[tex]\phi_f = \frac {k_r}{k_r + k_{nr}}[/tex]
Not surprisingly, fluorohpores with high efficiency are those in which the radiative rate far exceeds the nonradiative rate. As the emission wavelength shifts to the red, k
nr gets very large due to efficient coupling of vibronic states along the ground and excited electronic surfaces, hence why there are very few good NIR-emitting fluorophores. (See:
http://www.chemicalforums.com/index.php?topic=81175.msg296146#msg296146) k
r is a bit more difficult to model, but in the atomic limit, the (spontaneous) radiative rate is essentially related to the Einstein A coefficient, which has a linear dependency on the Einstein B coefficient, which is basically an absorption rate. Molecules absorbing in the UV tend to have lower absorption cross-sections than those in the visible, for a variety of reasons, such that the radiative rate tends to be relatively small at very high absorption energies. (The Einstein coefficients are applicable to very simple atomic systems. Strickler and Berg published a landmark modification applicable to molecules in the 1960s, a good paper for you to read if you're interested in this kind of thing. But, the basic principle of the Einstein coefficient holds.) The end result is that for many related systems, fluorescence yields tend to be highest in the visible range, and lower especially as you get into the UV and NIR regions. For instance, if you plot out the fluorescence yields of many conjugated oligomers as a function of their emission wavelength, you get an inverted U shape. An example can be found in one of my own papers, figure attached below. This is for oxidized MEH-PPV polymers. As you move toward monomeric systems, the triplet yield - essentially a form of nonradiative deactivation - also often becomes competitive with fluorescence.
So, basically - things like benzene, or even ethene or acetylene - while having absorption and fluorescence in the UV region, have fairly low fluorescence yields. Benzene is only a few percent, and ethene and acetylene are practically nonfluorescent. As you get to smaller molecule, the high symmetry also causes some problems for you because of the presence of low-lying electron states with extremely low absorptivity. My old book of laser dyes from Kodak lists only one example with an emission wavelength below 400 nm (terphenyl, emission peak at 354 nm in cyclohexane) with a QY of 17% (according to another source). The smaller analogue, biphenyl, has an emission wavelength of 315 nm and a QY of 7%. The extinction coefficient is reasonable at around 18000 M-1 cm-1. But it's just going to get worse as you move deeper into UV region.
There is also the matter of absorption intensity that you'd need to take into account for practical usage. UV-emitting atoms have a very high fluorescence yield (there are few routes for non-radiative relaxation). But excitation efficiency is also very low, so it's difficult to prepare enough excited-state species to make them a practical source of UV fluorescence. (Consider, even if you're fluorescence yield is 100%, if your emitters don't absorb any photons, they can't emit any photons.) Mercury vapor is very bright when a large portion of atoms are excited, but it's not easy or efficient to prepare this body of excited species optically. This is why an electric arc is used (as in a mercury lamp) - much more efficient. Based on your stated application of converting a higher energy fluorescence to a slightly lower one (like a phosphor coating on a UV lamp tube), you'll need to take this into consideration, because your emitter first has to absorb energy to then emit it.
In the end you may be better off with inorganic emitters (semiconductor QDs), which play by different rules, but of course they have their own problems as you move toward the UV end of the spectrum - although I understand UV LEDs are now available even deep into the UV region.
Not an easy problem to solve. If I could, I'd probably be rich. Strong Far-UV-emitting OLED with tunable emission wavelength would be huge, lot of potential applications. Corribus the billionaire.
ChemBuddy | 
Chem Reddit | 
Topic: Ultraviolet Fluorescence? (Read 3944 times)