Question

Does the spacing between the vibronic peaks in the fluorescence emission spectra have to be the same as those in absorption spectra? why?

Answer 1

First of all, many compounds do not emit light after absorption of radiation. so these molecules do not have an excitation spectrum. The reason is that the absorbed energy can dissipate in the molecule without emitting light/fotons.

Secondly, molecules absorb light to different excited states. Emission comes from the lowest of these states, called the S1 state. When a molecule is excited to a higher state it often ends up in this lowest excited state S1 and then emits radiation. In this case the excitation spectrum is the same as the absorption spectrum. However from a higher excited state a molecule does not have to go to the lowest excited state.

An absorption spectrum measures wavelengths at which a molecule absorbs light , while an excitation spectrum determines the wavelengths of light necessary to produce emission or fluorescence from the molecule, as monitored at a particular wavelength. One can think of an excitation spectrum as "fluorescence detected absorption'" . For many aromatic organic molecules,S1, S2, S3, etc., emission (fluorescence) normally occurs only from the lowest excited singlet, S1. When one excites the higher states S2, S3 etc., one typically has rapid radiationless relaxation down to S1 on the picosecond timescale, followed by emission to the ground singlet, S0. So in such molecules one will indeed observe peaks in the excitation spectrum (obtained while monitoring S1 fluorescence) corresponding to absorptions to S2, S3, etc., because such absorptions do ultimately lead to emission from S1. If, by contrast, absorption to the higher S2, S3 states gives rise only to radiationless deactivation to the ground state (the case described by Dr. ten Brink), then the excitation spectrum for S1 fluorescence would not show the S2, S3 absorptions.