Questions on Raman Effect and its Clarifications

[ananthu]

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.

 

[alxm]

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?

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.

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?

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)

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

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.

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?

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.

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

Molecules are more often in their ground state than an excited state.

 

[ananthu]

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.

Thank you for your enlightening explanation.

Now, when the excitement of the liquid molecule does not in any way interfere with the scattered photons, why should we bring it to the picture at all? Why not we simply say that a photon which loses part of its energy to the molecule during collision gets scattered as a low frequency photon giving rise to stokes lines and so on..? In what way the phenomenon of the liquid molecule rising to the virtual level and falling to different lower levels contribute to the explanation of scattering of photons by the liquid molecule?

Also let us take the analogy of a ball thrown upwards from ground floor of a multi-storied building, reaching its maximum height and falls back under gravity to the same point of projection ie. the ground floor itself. In this case there is no reason why the ball should
land on the first floor instead of reaching the ground floor as the gravity itself will bring it back to the starting point. In the case of the excited electron of the molecule. of course. it is the electrostatic force.

But the incident photon being absorbed by the excited molecule and re-emitted as a photon of same or higher or lower frequency, is logically satisfying.

But this logic contradicts with the very definition of scattering.

I remember that some where I have studied that no absorption takes place in the case of scattering especially Raman scattering.