# raman spectroscopy question

by n0_3sc
Tags: raman, spectroscopy
 P: 265 I read in a journal article that in raman spectroscopy "it is well known that signal yield increases with shorter wavelength". Can anyone please explain why? I would've thought longer wavelengths = shorter frequencies = higher susceptibilities?
 P: 265 Sorry to bring this post up again, but I found the answer to my question if anyone is interested: Turns out the Raman process does NOT depend much on wavelength. It is the detection of the CCD's that have the highest quantum efficiencies in the visible. Between 600nm and 1000nm there is a 60% difference in detection efficiency.
P: 37
 Quote by n0_3sc Sorry to bring this post up again, but I found the answer to my question if anyone is interested: Turns out the Raman process does NOT depend much on wavelength. It is the detection of the CCD's that have the highest quantum efficiencies in the visible. Between 600nm and 1000nm there is a 60% difference in detection efficiency.
well the Raman process DO depent much on wavelengt (to the fourth power). And it's well understood.
The fourth power can appear through a lengthy derivation using timedependent pertubation theory, second quantization of E-field and derivation of interaction between charges and E-field. You don't wanner mess with this.....

P: 265

## raman spectroscopy question

 Quote by evidenso well the Raman process DO depent much on wavelengt (to the fourth power). And it's well understood. The fourth power can appear through a lengthy derivation using timedependent pertubation theory, second quantization of E-field and derivation of interaction between charges and E-field. You don't wanner mess with this.....
Rayleigh Scattering depends on wavelength to the fourth power not Raman Scattering.
P: 37
 Quote by n0_3sc Rayleigh Scattering depends on wavelength to the fourth power not Raman Scattering.
Sorry to disapoint ya.
Raman intensity DO varies with the fourth power of the observed frequency for normal Raman scattering, which, in turn, depends on laser frequency.
It can be derived from the classical treatment of scattering from an oscillating induced dipole, with the intensity expressed in watts.

Therefore, in biological Raman scattering one has to find a compromise between reduced fluorescence and reduced Raman signal by choosing bigger wavelength to obtain signals.

In the full quantum picture it appears when you combine final density of states, laser intensity and scattered intensity with the Kramer Heisenberg equation

You should note that modern Raman spectrometers,
which usually measure photons/seconds rather than watts, are governed by a
slightly different frequency dependence.
 P: 265 I'm now confused :( Do you have any material I can refer too? In all the theory I've gone through I don't see any fourth power dependence. (haven't done QM though) Is it: Raman intensity varies with fourth power wavelength or, Raman intensity varies with fourth power frequency? Why are modern spectrometers different?
P: 37
 Quote by n0_3sc I'm now confused :( Do you have any material I can refer too? In all the theory I've gone through I don't see any fourth power dependence. (haven't done QM though)
This is a basic non QM practical guide
"Raman Spectroscopy in Chemical Analysis wiley 2000"

 Quote by n0_3sc Is it: Raman intensity varies with fourth power wavelength or, Raman intensity varies with fourth power frequency?
Raman intensity varies with fourth power of frequency
and you know that $$c=\lambda*frequency$$

 Quote by n0_3sc Why are modern spectrometers different?
This is explained in "Raman Spectroscopy in Chemical Analysis wiley 2000"
It's because there is a factor $$E=hv$$ in difference between watts and photon/sec
 P: 52 hi, can you point me to a reference for the quantized field treatment? googling didn't help much, seems this is really a topic that not many are willing to touch. thx in advance
 Sci Advisor P: 1,867 "Atom-photon interactions" by Cohen-Tannodji for instance?

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