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If all the excited molecules at higher S1 vibrational states rapidly relax to the lowest S1 vibrational state, why do the fluorescence transitions from v' = 1, v' = 2, v' = 3, and other states to v" = 0 have different wavelengths? Why don't they all relax to v' = 0 first and then all transition to the same vibrational level of v", releasing the same wavelength?

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    $\begingroup$ Because there are vibrational levels in the ground state and the Franck Condon factors from the v=0 (ex state) to each of these lower levels has a finite value. Both electronic states have a finite width in terms of bond extension and so the overlap probability between them varies between vibrational levels. Sketch out two harmonic potentials displaced along x a bit and add vibrational levels and wavefunctions to each to see the effect. Wikipedia for 'Franck-Condon Principle' has a good picture. $\endgroup$ – porphyrin Feb 26 at 8:58

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