I'm in the process of trying to understand X-Ray Absorption Fine Structure (XAFS) spectroscopy, but I think this question is more fundamental than merely as it pertains to XAFS. I think I don't entirely understand the interplay between electronic, vibrational, and rotational excitations in molecules. I understand that vibrational fine structure can be seen in electronic spectra. But how do we know which vibrational state a transition will end up in? Is it a Boltzmann distribution?

So to my main question. I realize that on the energy scale of XAFS, vibrational energy levels are not resolved; I just want to know if these transitions lead to the same type of vibrational state distribution that valence electronic transitions yield. Is that the case? Or if not, why not?


You have different questions in the title and in the body of your post and I will answer both.

Electronic excitation happens very fast compared to nuclear motion so the molecule remains "frozen" when it is brought from an equilibrium to an excited state. If the equilibrium geometry of the excited state is very similar to the ground state then the molecule will remain still, but if this is not the case, the molecule will start to oscillate around the equilibrium point. This is known as Frank-Condon principle and Wikipedia article has nice figures to illustrate this point.

Exciting different electrons leads to different electronic states that have different equilibrium geometries so you cannot expect same vibrational distributions.

In your title you also ask about relaxation of excited molecule. Since the energy of the X-Ray photon is much higher than the ionization energy, the excited molecule will be ionized before it can relax vibrationally.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.