Question

I was taking a look at Jablonski schemes and had some doubts regarding it. Now assume a regular molecule which follows this Jablonski scheme: We can see that when molecules absorb certain quantized wavelengths, they get promoted to vibrationally excited, excited electronic states. But when they emit the photon again, they do that from the vibrationally lowest, lowest excited electronic state (vibrational ground state of S1 (or T1 for phosphorescence)), this is a result from Kasha's rule. However, when they emit photons, does it happen to any of the vibrationally excited states of the electronic ground state (although the probability into which vibrational mode of S0 is given by the Franck-Condon-principle) or into a specific one?

Answer 1

During fluorescence, molecules typically emit photons from the vibrationally lowest, lowest excited electronic state, while the specific vibrational state of the ground state depends on the Franck-Condon principle.

Step-by-step explanation:

When molecules emit photons during fluorescence, they generally do so from the vibrationally lowest, lowest excited electronic state. This is known as Kasha's rule. However, the specific vibrational state of the electronic ground state from which the photons are emitted depends on the Franck-Condon principle, which determines the probability of the molecule being in a given vibrational mode of the ground state.

In other words, the emission can happen to any of the vibrationally excited states of the electronic ground state, but the probability is determined by the Franck-Condon principle.