# Does the emission spectrum for a dye depend on the frequency of light used to excite the dye?

Take a look at the emission spectrum for an arbitrary fluorescent dye: http://en.wikipedia.org/wiki/File:Fluorescein_spectra.jpg

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$? Consider that there is going to be a spread of eigenstates around a particular excited state (http://www.olympusmicro.com/primer/java/jablonski/jabintro/jablonskijavafigure1.jpg), corresponding to particular molecular vibration or rotational modes.

The thought is that, if we excite a dye at a wavelength slightly red-shifted with respect to $\lambda_{max}$, or slightly blue-shifted with respect to $\lambda_{max}$, we'd start out at a different eigenstate clustered around the same excited state (say $S_1$). I imagine there's a probability of decaying from each of these eigenstates in $S_1$, which gives us the spread for the emission spectrum, so if we bias the initial eigenstate, we'd achieve a different emission spectrum (presumably "statistically" red-shifted with respect to the excitation frequency). Or is the fluorophore taking a rapid random walk through all accessible eigenstates belonging to a particular excited state?

I suppose we could also ask a similar question with respect to the photoelectric effect in terms of the energy provided to the excited electron (conditioned on having a threshold frequency to span a band gap).

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?

No, it is independent of that. The excitation $S_0 \rightarrow S_1$ is vertical, bond length are not changed in the excitation process. Compare the mass of the nuclei with that of the electrons and you know why. You'll just end up in a different vibronic level of the electronically excited state.

However, before the emission takes place, this initial Franck_condon states relaxes to the vibrational ground state of $S_1$ with an equilibrated geometry and emission occurs from that (Kasha's rule).

• Thank you very much for your answer, and for pointing out Kasha's rule. I wasn't sure what you meant by "...bond length are not changed in the excitation process...", I don't think I was suggesting this? Mar 4, 2014 at 23:23
• From Kasha's rule I can extrapolate that Stokes shifts are more likely than Anti-Stokes shifts. And can I also understand that the spread of the emission spectrum is due to IC to different vibrational eigenstates in the ground state? Mar 4, 2014 at 23:25
• In organic molecular photochemistry in solution, I've personally never observed any Anti-Stokes shifts for the fluorescence. But i've seen strong and solvent-dependent Stokes shifts. Mar 4, 2014 at 23:35
• The literature and references for Kasha's rule is exactly what I was looking for. Mar 5, 2014 at 0:17

The Jablonski diagram usually shown (as in your web page reference) is that for a molecules in solution and, as has been explained in other, answers the vibrational deactivation is generally fast so that it cannot compete with fluorescence so generally fluorescence occurs from the lowest excited vibrational level of the excited state.
In Azulene, a naphthalene isomer with a 5 and 7 member rings, fluorescence comes mainly from the S$_2$ excited state to the ground state as internal conversion S$_2$-S$_1$ is slow compared to fluorescence. This occurs because of the very large S$_2$ to S$_1$ energy gap. Some benzanthracenes also show emission from S$_2$ . The vibrational transitions in fluorescence spectra in solution are broadened due to collisions and solvent perturbation effects.

In the vapour phase when molecules are at low pressure and there are no collisions during the excited state lifetime, fluorescence arises from the level excited and to various vibrational levels in the ground state. In this case the emission spectrum does depend on the level excited, and the spectrum is now a series of sharp lines, rather than broad vibrational bands observed in solution.