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I was revisiting my notes for spectroscopy, and I came across the Franck-Condon-principle. I understand it well, but want to clear some doubts regarding this.

I know that the Franck-Condon-principle states that during an electronic transition from the ground state vibrational state (according to Boltzmann, the most populated), a transition happens, but not to the ground vibrational state of the electronically higher state, but to a vibrationally excited state of the electronically higher state, as seen here:

enter image description here

So, as seen by the excitation (highlighted in blue), the excited state of the electronically higher state decays through collision with other molecules to the ground vibrational state and fluorescences to a vibrationally excited state of the electronic ground state. That is the reason why the Emission spectrum is lower in energy than the absorption spectrum (Stokes shift): enter image description here

For me there remains a doubt: So basically for the transition to happen, the electronic energy difference needs to be overcome plus the vibrational difference, all by one photon that matches the energy difference as seen in picture 1. However, in statistical thermodynamics it is always assumed that electronic and vibrational transitions are uncoupled, so one photon should only ever excite an electronic transition, or a vibrational transition, but one photon can’t change both modes (electronic + vibrational) at the same time. Where did I think wrong?

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    $\begingroup$ No problem ! One photon can easily change both modes (electronic + vibration) at the same time. This is even what happens in nearly all photon absorptions in the visible or ultra-violet region. $\endgroup$
    – Maurice
    Commented Jun 16, 2023 at 11:56
  • $\begingroup$ Okay that makes sense then, but is the premise of statistical thermodynamics then wrong? $\endgroup$
    – Mäßige
    Commented Jun 16, 2023 at 12:26
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    $\begingroup$ Wrong ? What sort of "premise of statistical thermodynamics" are you speaking of ? $\endgroup$
    – Maurice
    Commented Jun 16, 2023 at 16:02
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    $\begingroup$ The geometry of the ground state and excited state are different so that transitions from any ground state vibrational level to any excited state one are allowed (supposing also spin allowed). The intensity of these vibronic transitions are determined by the wavefunction overlap which directly produces each FC factor. btw in the gas phase at low pressures there can be no collisions (during several excited state lifetimes) so that emission occurs from the initially excited level. $\endgroup$
    – porphyrin
    Commented Jun 16, 2023 at 19:36
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    $\begingroup$ As @Maurice says, you are messing up your Statistical Thermodynamics. When describing the population of the molecule in the ground state, it is usually assumed that the different molecular degrees of freedom are uncoupled so the electronic, vibrational, rotational and translational molecular degrees of freedom can be treated separately. Normally, the energy of the excited electronic excited electronic states is much higher than that of the ground state, in a good approximation the upper state is empty, and this state can only be accessed through photon absorption. $\endgroup$
    – PAEP
    Commented Jun 16, 2023 at 22:20

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