Intensity of stokes and anti-stokes lines?

[itari1985]

According to Raman effect, the intensity is directly proportional to the 4th power of the wavelength. Then how come stokes lines, which have higher wavelengths than anti-stokes lines, are more intense than the anti-stokes lines?

 

[inha]

uh? I think you answered your own question. If I is propotional to lambda to power 4 and the stokes line has higher wavelengths of course they're more intense then.

 

[MiGUi]

If I remember well, the anti-stokes line is near absorption bands. Its gain is too low there, so it is less intense than stokes line.

 

[itari1985]

Sorry intensity is inversely proportional to the 4th power of wavelength.

 

[Claude Bile]

Isn't the lambda^-4 dependence characteristic of Rayleigh scattering?

Claude.

 

[Barth]

To have a Raman anti-stoke diffusion, we need to have a transition of an atom initially in a excited vibrationnal level to the ground level (if we forget the intermediate virtual state). The stoke diffusion, instead, relies on a transition from the ground level to an excited vibrationnal level.

At thermal equilibrium, the populations of the different levels follows the Boltzmann distribution. So the population of the ground level is higher that the excited level. So the stokes transition is much more probable that the anti-stokes transition.

Does it answer to the question ?

Barth

 

[Gokul43201]

To add a touch to Claude's and Barth's posts : vibrational eigenmodes (or phonons in solids) are bosons. Their distribution among different levels is given by the Bose occupation number. At low temperatures, the different levels have significantly different populations.

 

[itari1985]

yes it does answer my question. Thanx