[miskovsky]

Could someone explain to me what exactly is a convergence limit in the vibronic spectrum of iodine? I know that when the v' increases, the vibrational energy spacing decreases. This is from a source I found but I don't quite understand:

"At a point called the convergence limit, the spacing between bands decreases to zero. Beyond this convergence limit, the spectrum is continuous because the excited state of the I2 molecule is not bound."

Why the excited state isn't "bound" anymore? And does the quantization of energy affect the fact that the spectrum becomes continuous at 500 nm?

Thank you!

[miskovsky]

Also, is the decreasing of the vibrational energy spacing the reason why the spacing of the fine structure and the intensity of the peaks decreases?

[Corribus]

Consider something like the Morse potential, which would be a decent approximation for the potential energy surface of a diatomic molecule like iodine. As you should hopefully know, the vibronic energy levels get closer together at higher energies because there is a dissociation point (dissociation energy), i.e., an energy above which the two iodine nuclei are no longer bonded. Above the dissociation energy, the molecule is no longer in a bound state. That is, the system is no longer confined by a potential well, particle energies/observables are no longer effectively quantized, etc. Like the states of a particle-in-a-box as the box dimension approaching infinity: the states become infinitely close together.

Intensity of peaks decreasing - no, this is not directly related to the energy level spacing. Rather, it is related to the Franck-Condon overlap integral between the ground and excited vibronic state wavefunctions.

(As a final note, the various factors that influence peak intensities and etc change if one is talking about vibronic spectra or pure vibrational spectra.)

[miskovsky]

Thank you very much! This answer was excellent. I didn't know that the Frank-Condon principle affects even the peaks.