2
$\begingroup$

The Beer-Lambert law is meant to describe the interaction between a solute and incident light.

But, does the frequency of the light here change due to the interaction with molecules?

I think there is a change in the intensity of light, not in the frequency. But this seems counter-intuitive.

Is there no change, or is it so small that it's just neglected?

$\endgroup$
  • 2
    $\begingroup$ Mostly, it doesn't change at all. Some tiny portion does (see Raman scattering). $\endgroup$ – Ivan Neretin Feb 15 '17 at 9:05
  • 2
    $\begingroup$ In absorption the photon is destroyed and it energy taken up by the molecule, so thats that. Any photons not absorbed thus lead to a reduction in the intensity of light transmitted. As @Ivan Neretin writes, in Raman Scattering there is a change in the photon's frequency, but in this case no photon is absorbed just scattered, i.e. an excited state is not produced. $\endgroup$ – porphyrin Feb 15 '17 at 17:48
  • $\begingroup$ But what happens to the absorbed energy of the molecules? $\endgroup$ – Mockingbird Feb 15 '17 at 22:49
  • 1
    $\begingroup$ All the absorbed energy is returned. Some fraction of this energy is re-emitted as fluorescence and/or phosphorescence. Another fraction may cause a bond to be broken, but if not any remaining ends up as heat. In femtosecond time scale experiments it is possible to see the solvent warm up and then cool as the energy diffuses away from the molecule. $\endgroup$ – porphyrin Feb 17 '17 at 17:39
  • 1
    $\begingroup$ The Beer-Lambert law was never intended to apply to complicated processes like inelastic scattering. The law is best used when you can just treat the problem as a decrease in the number of photons, and don't have to start thinking about altering or re-radiating those photons. $\endgroup$ – uhoh May 23 '17 at 3:41

Your Answer

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

Browse other questions tagged or ask your own question.