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Raman effect

  1. Aug 4, 2010 #1
    I need certain clarifications in Raman effect.

    First of all, it is said that when a photon is incident on a liquid molecule, it gives part of its energy to the molecule, exciting it to the virtual level (higher one) and that the molecule when it returns not to the ground state from where it was originally excited but to a slightly higher state emits radiation of low frequency lines named as stokes' lines. Similarly opposite thing happenes in the case of anti stokes' lines.

    In this regard, I want to place the following questions or doubts.

    1. The incident photon loses some energy to the liquid molecule and gets scattered as a low frequency photon. Similarly, the excited molecule while returning to a level slightly higher than the ground state should also emit a photon of lesser frequency. Does it mean the stokes' lines consist of both these photons ie. the one that was scattered and the other that was emitted by the excited molecule dureing its return journey?

    2.Or, can we say that the incident photon was completely absorbed by the substance first,then the molecule uses part of that energy for exciting itself and gives back the remaining energy during its return as stokes' lines?

    3. If the second case is true, then the scattering will become another case of absorption and re-eimisstion of energy just as fluoresence?

    4. What about anti stokes' lines then? There no energy is taken by the liquid molecule but on the contrary it gives to the incident photon which is scattered as high frequency photon giving anti stokes lines.At the same time, the liquid molecule which is already in the excited state goes to virtual level and then returns, this time, to the ground state giving rise to anti stokes lines!
    Here. again I am confused. Does it mean that the photon emitted by the molecule, in the present case is also added to the scattered photon to contribute to the group of antistokes' lines?

    5. Then why is the intensity of stokes lines is greater than that of antistoks lines?

    6. In reality, is scattering a case of change in direction of the incident photon by the scattering medium or an absorption and re-emission process?

    I will be happy if some one comes with an elaborate and simple explanation.
     
  2. jcsd
  3. Aug 4, 2010 #2

    alxm

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    Science Advisor

    No, because you're typically seeing a transition between vibrational states, so the relaxation of the excited molecule occurs non-radiatively. And it's also in a different part of the spectrum if it does occur radiatively.

    You can look at it that way as well. But the excitation is then to a 'virtual' level, which doesn't actually exist. (This is one situation where it's valid to use the time-energy uncertainty relationship)

    The difference is that fluorescence is excitation to a real state. It can be observed in that excited state. You cannot observe the molecule excited to a virtual state.

    No, there's only one photon being emitted. You can view it either as the original photon being absorbed and re-emitted with higher energy, or you can view it as if the original photon was scattered off the molecule like a particle, gaining momentum in the process.

    Molecules are more often in their ground state than an excited state.
     
  4. Aug 6, 2010 #3
     
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