That's more of a spectroscopy than a chromatography questions, and I'm certainly no spectroscopist. There is a good general description of the fluorescence process on Wikipedia:
http://en.wikipedia.org/wiki/Fluorescence_spectroscopy
In practice, almost any molecule that has a double bond will have an energy level that allows absorption of a photon somewhere in the UV (the more conjugated the double-bond system, the lower the energy gap, and the longer the wavelength). Most of these compounds decay back to their ground state via a series of excited vibrational states. In essence, the absorbed energy is dissipated as heat.
If a molecule has large energy gaps between vibrational states, however, the most probable transition may be to decay directly back to the ground electronic state by emitting another photon. Since some of the energy has already been dissipated, that photon will have lower energy (i.e. a longer wavelength) than the one originally absorbed.
In general, fluorescent molecules have fairly rigid, multi-ring structures that lack closely spaced vibrational modes. The excitation wavelength will generally correspond to a UV absorbance maximum; the emission wavelength will be longer.
So: look at the UV spectrum. Set the excitation wavelength of the spectrofluorometer at the absorbance maximum, and scan the emission side out to longer wavelength to see if there is any fluorescence.
