# Why does the excitation and emission spectrum of a fluorescent molecule have overlap?

Question is rather self-explanatory. Came up during a lecture without a concrete answer.

I understand that the differences in emission wavelengths is due to relaxation to the lowest energy level of S1, but why do fluorescent molecules necessarily overlap in their excitation and emission spectra?

• If you are asking why there is no overlapping of lines in emission or excitation spectra, the answer to this question also explains why Rutherford model of atom was rejected. If lines were overlapping, it mean that atoms emitted energy (in the form of radiation) in continuous manner, if this was the case, electrons would collapse into the nuclei giving Thomson model of atom. But this never happens. So, atoms can emit energy in discrete manner, which was explained by Bohr. Thus, there will be no overlap of lines in emission or excitation spectra. – Immortal Player Dec 7 '13 at 12:20

This is most easily seen in solution. The overlap in molecules occurs only when the electronically excited states of a molecule have rather similar geometries and so transitions between vibrational levels in the two states occur and the same wavelength. After excitation the excited state rapidly looses energy until only the v=0 level is populated. This takes about $10^{-13}$ sec, far faster than fluorescence.