3
$\begingroup$

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

$\endgroup$
9
  • 3
    $\begingroup$ 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: chemistry.stackexchange.com/a/77449/79678. $\endgroup$
    – Ed V
    Jul 24 at 12:21
  • 3
    $\begingroup$ By the way, azulene is a rare exception to Kasha’s rule: en.wikipedia.org/wiki/Azulene. $\endgroup$
    – Ed V
    Jul 24 at 12:45
  • 2
    $\begingroup$ Atoms in the gas phase (or in plasmas) are different: they have energy levels. See, for example, the sodium emission 2D spectrum here: chemistry.stackexchange.com/a/164168/79678. $\endgroup$
    – Ed V
    Jul 24 at 14:59
  • 2
    $\begingroup$ For atoms, we have Grotrian diagrams, as shown in this excellent answer: chemistry.stackexchange.com/a/154055/79678. Off topic, but I had the honor of meeting Kasha in 1986. Amazing scientist and guitar maker. $\endgroup$
    – Ed V
    Jul 24 at 15:21
  • 3
    $\begingroup$ It gets complicated. For example, neodymium ion fluorescence is the basis of the various solid state neodymium lasers: 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. $\endgroup$
    – Ed V
    Jul 24 at 19:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.