# Another Raman spectrum question!

1. Feb 4, 2006

### photon79

A laser in visible is used in Raman spectroscopy and the resultant spectrum is in IR and Microwave region(which means molecular vibrations and rotations are excited), what type of interaction is this that induces these levels? ( I cannot think more than..."To be Raman active molecules should posses anisotropic polarizability".)

2. Feb 4, 2006

### inha

What's meant by interaction here? Because obviously the photons scatter inelastically exciting the rotational and vibrational modes and in my books the inelastic scattering part is the interaction.

3. Feb 4, 2006

### photon79

Thank you inha!
My question is How this inelastic scattering of visible light excites rotations or vibrations ?

4. Feb 4, 2006

### Gokul43201

Staff Emeritus
Where did you get this from (I'm not saying it's wrong - I don't know enough to categorically say that) ? From my basic understanding of Raman, if the incident frequency is $\omega_0$, and you excite say, a vibrational mode with eigenfrequency $\omega_n$, the Raman spectrum you see will have peaks at $\omega_0 + \omega_n$ and $\omega_0 - \omega_n$ - the stokes and antistokes lines.

5. Feb 4, 2006

### photon79

Yes, but my question is how vibration can be escited during the scattering? (As the source used is visible, I think some basic understanding is missing from my side)

6. Feb 4, 2006

### Gokul43201

Staff Emeritus
There's a nice classical picture and a simplistic QM picture that can be provided that provide a useful intuition. I think I'm going to give a brief intro to both pictures in a thread on Spectroscopy in the Chemistry subforum (later tonight or tomorrow). Feel free to join this thread and ask questions. Hopefully, someone more knowledgeable than me can tackle the tough questions.

7. Feb 5, 2006

### Claude Bile

Keep in mind that the molecules are being excited via scattering events not absorption events. In a scattering event, the molecule is not excited to a higher electronic energy, because it is not being excited at an electronic resonance. This is why you only get excitation of Raman-active vibrational modes and not electronic modes.

Claude.

8. Feb 6, 2006

9. Feb 6, 2006

### Reshma

When a substance is irradiated with monochromatic radiation, it is found that part of the incident radiation is scattered, in general this scattering is due to the presence of dust particles and inhomogenities in the medium (also known as Tyndall scattering) and to some extent due to the polarizibility of the medium(known as Rayleigh scattering). In both the cases, the scattered radiation has the same frequency as that of the incident frequency.

But, in Raman effect the frequency of the scattered radiation is different from the incident radiation. The Raman scattering is very weak compared to the Tyndall and Rayleigh scattering. However, Raman scattering and Rayleigh scattering can be explained in terms of the quantum theory of radiation.

Rayleigh scattering is a case of elastic scattering, an alternate way to describe this is to think of it as simultaneous absorption and re-emission of the photon of the same frequency. Raman scattering is an example of inelastic scattering.

10. Feb 6, 2006

### Reshma

Umm...if my sources are correct, then according to Stokes hypothesis, the wavelength of the secondary radiation is always greater than that of the radiation incident on the substance. So, frequency of scattered radiation will be lower.So,
$\omega_0 - \omega_n$--->Stokes line &
$\omega_0 + \omega_n$--->Anti-Stokes line.

11. Feb 6, 2006

### Gokul43201

Staff Emeritus
If you are pointing out my carelessness with labeling the Stokes and Antistokes lines, thanks - I got them reversed.

If you are saying that the wavelength of Raman scattered radiation is "always greater than that of the [incident] radiation", I'll have to disagree.

To address the original question : the short answer is that IR emission happens due to subsequent transitions between vibrational states following the excitation to a virtual state and a relaxation to an excited vibrational state.

12. Feb 6, 2006

### photon79

Now its much clear to me.
My question was "How can we absrove rotations and/or vibrations in Raman spectra where we use visible source?"
Here what happens is molecule is excited to a virtual state (as Gokul said)say 20,000 Cm-1 which is in visible and scattered at lower vibrational say, 19,000 Cm-1 which is also in visible , what we measure in Raman spectra is difference in these levels 1,000 Cm-1 which is in IR. This is the basis and we get spectral lines in visible but we take the difference which will be in Microwave or IR.

13. Feb 6, 2006

### Claude Bile

Scattering occurs off resonance, that is, the incident photon does not correspond to any energy gaps that correspond to an allowed transition (some transitions are forbidden by selection rules).

Absorption occurs on resonance, that is, the incident photon does match an energy gap of an allowed transition and the atom or molecule undergoes a change in energy and reconfigures its orbitals appropriately.

To describe scattering, the term virtual state is sometimes used. The word virtual is used to indicate the fact the excited state is not a true excited state in that there are no changes in the orbitals of the atom/molecule.

Claude.

14. Feb 7, 2006

### Reshma

No, only according to Stokes hypothesis not according to Raman effect. That's way they are called anti-Stokes lines.

15. Oct 23, 2010

### mjkigat2

Hello!
I am using T64000 Raman and I get some additional spikes when I am trying to do a Raman spectra. I don't know why...

16. Oct 23, 2010

### mjkigat2

I'm using a Raman spectrometer to analyse PMMA samples and I have some additional spikes ...I don't know why

17. Oct 23, 2010

### Reshma

Have you checked the spectrum of a standard sample?

18. Oct 24, 2010

### Tzontonel

additional spikes - but say where? xx cm$$^{-1}$$

19. Nov 3, 2010

### Rajini

It could be noises. (From external light)..

20. Nov 5, 2010

### narra

This is starting to get a bit confusing because different people describe scattering differently. Claude, does an excitation of a vibrational state require an absorption of energy to do so?

Perhaps someone can provide a clear description of the differences between absorption and scattering processes, because some sources describe scattering as an absorption followed by an emission which suggests to me that scattering is includes an absorption.