So in my physical chemistry class we have been looking into radiation, selection rules, etc.

Taking into account the discrete levels of energy an electron can have, I cannot understand why fluorescence and absorption spectra of compounds have continuous lines instead of discrete values.

If electrons in a system can only have a select number of energies, why is it that these graphs show a continuum of absorption and emission?

I also have this problem with blackbody radiation, since it is a continuous line for any temperature.

  • $\begingroup$ Related: chemistry.stackexchange.com/q/35658/16683 This answer may help you. I distinctly remember we had a question on this, but maybe I'm just imagining it. Or maybe I'm thinking of the one I linked you. $\endgroup$
    – orthocresol
    Oct 8 '16 at 16:17
  • $\begingroup$ Related: 1, 2. $\endgroup$ Oct 8 '16 at 16:23

You are right that absorption and emission are quantized processes, but UV-visible peaks represent a broad range of wavelengths because there are multiple vibrational states of their representative ground and excited electronic states.

vibrational states

In the case on my diagram, you have six different quantized electronic transitions with slightly different ΔE values. In reality, there are more energy levels than that, plus the rotations and vibrations are constantly changing the orbital energy levels as a result of interactions with the matrix, so that helps fill in the peaks. The peak shape can actually be changed by altering the temperature. This gives rise to high-resolution UV-Visible spectrophotometric techniques, which have a few specific applications.


One reason, not mentioned in the previous comment, why this is the case is that in most cases, some of the photon energy can be accounted for by the translational energy levels, which are so close together, as to be functionally continuous, hence, various photon energies can cause the main transition plus or minus some translational energy

  • $\begingroup$ The difference in the energy scale between translational motion and most optical absorption/fluorescence spectroscopies means that you won't see this effect directly from translation. It would appear due to changing local environments from solute diffusion. $\endgroup$ Oct 8 '16 at 16:31

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.