2
$\begingroup$

In this site, https://www.princetoninstruments.com/userfiles/files/assetLibrary/Technical%20posters/Poster-Aberration-Free-Raman-Spectroscopy-with-the-IsoPlane.pdf the raman spectrum of cyclohexane shows higher peak at the right.

enter image description here

While in this site, http://group.nanovietech.com/products/advantage-raman.html, the raman is showing higher peak at the left:

enter image description here

Both are raman spectra of cyclohexane. My questions relate to the 2853 cm-1 peak and 802 cm-1 peak. Based on structure of cyclohexane. Which of them should be higher? Ignore the other peaks which I wasn't concerned.

I found this too:

enter image description here

Should the 802 cm-1 peak be higher than the 2853 cm-1 peak based on the molecular structure of cyclohexane?

$\endgroup$
  • 2
    $\begingroup$ Its not duplicate bec it doesnt show the exact spectra of cyclohexane. I was asking which of the above 2 is correct. Both are raman of cyclohexane. $\endgroup$ – Jtl Oct 15 '19 at 10:42
  • 1
    $\begingroup$ I've already removed the comment. I have been too quick and reading left right I thought you mean the axis reversal, sorry. For your actual Q, this should be related to the isoplane spectrometer having different performance. As they say reduced aberration, we can't expect to be frequency independent. Look at their example on peak shape and resolution . Also the detector can have a different response. This is absolute photons counting, by the way. $\endgroup$ – Alchimista Oct 15 '19 at 10:48
  • 1
    $\begingroup$ I an not aware of the calibration of shapes. You probably mean some internal calibration for quantitative purposes or otherwise Raman shift calibration. I don't even see the ground for a calibration in your actual sense. Can't judge FWHM from pictures, tough it seems that top spectrum has indeed high resolution. $\endgroup$ – Alchimista Oct 15 '19 at 11:08
  • 1
    $\begingroup$ If you have to choose a spectrometer is another story. But to be clear, it is like looking at two photos. There is no the right, or at least it is not straightforward. There is a better photo with high resolution, clarity, etc. Choosing the camera would depend on what and when you shoot. There is no Raman here. Not of the kind you mean. Both spectra are ok. They might even have the same peaks intensity. $\endgroup$ – Alchimista Oct 15 '19 at 11:21
  • 1
    $\begingroup$ That is exactly the point. Besides that I am not use to quantitative analysis by Raman, but it exists, it could be that the photons count is even the same internally. One company claims to get less aberration and high resolution with the isoplane line of spectrometer and apparently they are right. The intensity of a peak is area, and perhaps the ratio between the peaks in each spectrum is the same. You got focused on peak height rather but this could beneficial in terms of resolution, not itself. At least with no dramatic differences- that would be indeed point to something wrong. $\endgroup$ – Alchimista Oct 15 '19 at 11:42
2
$\begingroup$

First of all, you are extremely lucky in asking this question for cyclohexane: cyclohexane serves as reference material for checking your intensity calibration, so it is one of the few substances for which integrated band intensity ratios are available in the literature.

Integral intensity 2567 - 3068 cm$^{-1}$ is higher than integral intensity 700 - 900 (i.e. 801 cm$^{-1}$ band). At 532 nm excitation, the ratio is roughly 10.3 : 1.

  • You always have the fundamental $\nu^4$-dependence of Raman scattered intensity. If you look closely, that's $\nu_0 (\nu_0 - \nu_j)^3$ for the band with Raman shift $\nu_j$ for photon count and $(\nu_0 - \nu_j)^4$ for power. Thus, the relative band area differs depending on excitation wavelength.

  • For dispersive Raman instruments, the sensitivity is highly wavelength-dependent and this varies from instrument to instrument and for instruments with moving grating also depending on the grating position.

    • Thus, for intensity comparisons across different instruments, you'll need to calibrate the intensity axis for the spectrometers (and possibly fold the higher-resolved spectrum with the difference in line shape if that isn't negligible due to integration of broad spectral ranges)

    • However, when training chemometric models (calibration, quantitation, classification, ...) for one instrument (or maybe very similar instruments: say, same manufacturer & model), as long as that instrument is stable intensity calibration is not necessary as it will automatically become part of the model.
      Which is why many Raman spectra are shown as raw (uncalibrated) intensity.

  • This doesn't have much to do with "highly sophisticated" Raman spectrometer rather than doing sophisticated calibration or not. And among the highly sophisticated Raman spectrometers, you have tradeoffs that can make such a calibration more or less stable. In fact, different instruments are highly sophisticated in different aspects (e.g. spatial resolution, spectral resolution, signal stability/drift, versatility/possibility to adapt to different experimental setups, ...) some of these influence how easy intensity calibration is and/or how often it needs to be re-done.

Literature:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.