How can I estimate the $E_{0-0}$ value of an organic dye? I have obtained the UV-vis absorption and fluorescence spectra of this dye in solution. This article talks about (in Table 2 notes) the intercept between the normalized absorption and emission spectra, but I don't quite understand what they mean.

Does someone here know? Attached is the (non-normalized) spectra.

enter image description here


You're looking for the point where the normalized absorption and emission spectra cross each other.

The UV/Vis absorption spectra is almost always higher in energy (shorter wavelengths) than the emission spectra, because the molecule goes from the electronic and vibrational ground state to the electronic excited state. Usually this also means some vibrational modes are also excited. Since the molecule is typically in either the vibrational ground state or a low-energy vibrational state, the absorption process can go up several vibrational levels (i.e, the energy absorbed is higher than just the basic electronic excitation).

For emission, the molecule often loses some vibrational energy, so the emission comes from the ground vibrational state of that electronic excited state. The molecule ends up in the ground electronic state, but a vibrationally excited state (i.e., it loses a bit less energy than the $E_{0-0}$ energy difference).

So if you normalize the absorption and emission spectra, the crossing point should roughly correspond to the $E_{0-0}$ difference.

  • $\begingroup$ When I have a chance, I'll add some graphics to make these points clearer. $\endgroup$ – Geoff Hutchison Feb 15 '16 at 19:59
  • $\begingroup$ Area normalized to 1, or max intensity normalized to 1? $\endgroup$ – Yoda Feb 15 '16 at 20:48
  • $\begingroup$ I would normalize the peak intensity. $\endgroup$ – Geoff Hutchison Feb 15 '16 at 20:57

An alternative and probably more accurate way to estimate the $E_{0{-}0}$ for a molecule in solution is to take half the difference between absorption and fluorescence peaks. To do this first convert these wavelengths to wavenumbers, then add half the difference to the fluorescence peak (or subtract from absorption) and convert back to wavelengths.
(As an experimental observation you should also try, if possible, to get better quality absorption and emission spectra as negative values are meaningless and this suggest that your background subtraction is not quite correct).

The figure below shows the relationship between absorption and fluorescence spectra when the absorption is the mirror image of the fluorescence. (Note that the spectra on the right of the figure are not on the same scale with the energy levels on the left.) A mirror image does not quite occur with the spectra in your question, but the principle is the same. The reason that the fluorescence comes from the lowest vibrational level in the excited state is that the rate of vibrational relaxation (thin wavy line), caused by energy transfer to surroundings (usually solvent), is almost always far faster than emission. If the potential energy surfaces are not displaced horizontally, indicating different bond lengths in ground and excited state, then by the Franck-Condon Principle almost no transitions will be observed.

mirror image


Taking the crossing point between normalized and reduced absorption and emission spectra will give you the $E_{0-0}$.

Take care that reduced here means that the absorption spectrum $A(E)$ needs to be divided by the photon energy $E$, and the emission spectrum $N(E)$ needs to be divided by $E^3$. By normalizing the reduced spectra to the maximum of the corresponding peak, the two spectra intersect exaclty at $E=E_{0-0}$.

In the case of mirror-image spectra, $E_{0-0}$ is the midpoint between the absorption and emission maxima, separated by 2$λ$.

You can find more details at:

Vandewal et al., How to determine optical gaps and voltage losses in organic photovoltaic materials. Sustainable Energy & Fuels (2018)


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