[ironnica]

Hi, I am in a module called analytical spectroscopy and I have some questions relating to the Jablonski diagram, I cannot get it.
I cannot find the particular image I'm viewing in my notes on the internet and dont know how to upload images. Basically I do not understand the basics unfortunately.

In the S1 energy level of fluorescence on the emission, it shows the small arrows for vibrational relaxation. I presume that the sub-levels are called vibrational levels. How does this energy come out from a physical point of view?

Im guessing that the drop from s1 to s0 is the emission of light

Thanks

[ironnica]

Ok, Jablonski diagram snip attached.

[ironnica]

A consequence of vibrational relation is that the fluorescence
bands are at longer wavelength than the aborption bands
(Stokes shift)


It may appear as though I'm not trying but I am. In this case, is it just that the wave is taken in at a higher energy than it emits, thus the shift in wavelength to a higher lambda? Is that the only resulting property of the vibrational relaxation, or is there heat given out or anything?

[DevaDevil]

You are right in asserting the energy-emissions will be in the form of photons.

(better to say this than to say "light" as only a limited wavelength of photons is visible to us.)

[ironnica]

sorry, that is correct. I need to perfect terminology. I assume from the diagram that the photon can only fall from the lowest vibrational level on s1 to any vibrational level on s0 for fluorescence. Is that a correct assumption?

Also, the vibrational energy that is lost through "vibrational relaxation". What is this in the form of? I'm sorry if theres an obvious resource with this but in a lot of searching i haven't found it yet.

[DevaDevil]

your assumption of s1 -> s0 transition being only from the lowest s1 excited state to the ground state is true.

The process in which the electron is excited by a photon of higher energy than (E(s₁)-E(s₀))  is called fluorescence. The energy difference between exciting photon and emitting photon is usually absorbed by the sample, but there can be internal conversion through coupling with a vibrational state in a lower electronic state, or intersystem crossing in which the singlet excited state transitions to a triplet excited state (phosphorescence).

[ironnica]

absorbed by the matrix of the sample or the fluorescent molecule itself? relating to the "energy difference between exciting photon and emitting photon is usually absorbed by the sample"

could this be a result of accidental quenching?

[DevaDevil]

absorbed by the matrix; a single molecule cannot absorb that energy.

and I do not know what you mean by quenching? The energy being absorbed is the result of irradiation of the sample with photons of an energy greater than that of the vibrational transition

[ironnica]

I understand that. Thank you very much for your assistance Deva