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I was taking a look at Jablonski schemes and had some doubts regarding it. Now assume a regular molecule which follows this Jablonski scheme:

enter image description here

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? Is there such a rule?

Edit: As I am currently working on lanthanides, investigating the antenna effect. I was wondering which states the excited electronic atomic states (see the diagram) decay to; the picture shows all kinds of terms can be emitted to, but I am not sure in what proportion. Also, one usually only detects one emitted wavelength in these kinds of complexes (especially of Eu3+, since I worked with it), but I don't understand why because there should be many emission lines:

enter image description here

Mäßige
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    The de-excitation is into the manifold of vibrational levels of the ground electronic state, just as the cartoon figure shows. You can see this by looking at the fluorescence emission spectrum of, e.g., anthracene. See this excellent answer by @porphyrin: https://chemistry.stackexchange.com/a/77449/79678. – Ed V Jul 24 '23 at 12:21
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    By the way, azulene is a rare exception to Kasha’s rule: https://en.wikipedia.org/wiki/Azulene. – Ed V Jul 24 '23 at 12:45
  • Thats a really good answer, thanks; on another note: since atoms themselves can't rotate or vibrate, Kasha's rule shouldn't apply to atoms right? For example, since atoms can't relax vibronically, we should, in theory, see all emission bands from other electronically excited states? So for example, S1-S0, but also S2-S1 and S2-S0 etc. – Mäßige Jul 24 '23 at 14:44
  • (execpt intersystem crossing, which should be possible) – Mäßige Jul 24 '23 at 14:45
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    Atoms in the gas phase (or in plasmas) are different: they have energy levels. See, for example, the sodium emission 2D spectrum here: https://chemistry.stackexchange.com/a/164168/79678. – Ed V Jul 24 '23 at 14:59
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    For atoms, we have Grotrian diagrams, as shown in this excellent answer: https://chemistry.stackexchange.com/a/154055/79678. Off topic, but I had the honor of meeting Kasha in 1986. Amazing scientist and guitar maker. – Ed V Jul 24 '23 at 15:21
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    In the condensed phase (your figure) emission is usually from the lowest vibrational level and can occur to any vibrational level in the ground state according the the FC principle. (no rule other than this) In the vapour phase emission may occur from the vibrational level initially excited to the ground state and is also controlled by FC factors, or, if collisions occur , emission can be from any number of other vibrational levels in the excited state. It just depends on rate of collisions vs fluorescence decay rate constant. – porphyrin Jul 24 '23 at 16:40
  • Okay Grotrian diagrams I knew, but never knew that they were called like this, very interesting. I was asking because I am currently working in lanthanide chemistry which (for the most part) behave like isolated 3+ ions, but I was wondering to which states the electronic transitions could be attributed to. Look at the edit if you have the time. – Mäßige Jul 24 '23 at 19:19
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    It gets complicated. For example, neodymium ion fluorescence is the basis of the various solid state neodymium lasers: https://chemistry.stackexchange.com/a/158765/79678. Even europium (III) oxide strongly fluoresces in the red. There are plenty of papers you can find, so I think you have good search options. – Ed V Jul 24 '23 at 19:51

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