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I was conducting a rovibrational IR experiment and noticed that for HCl, the P branch is always lower in intensity than the R branch.

I already understand that the intensity of the spectral lines is dependent on which rotational states are occupied at room temperature, via the Boltzmann distribution and application of the Lambert-Beer law.

However, what I don’t quite get is why the P branch should always be smaller than the R branch; for example, the R(2<-1) transition should be at the same intensity as the P(0<-1), as the initial state is J=1! So they should be the same in intensity. Why is this not the case?

enter image description here

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    $\begingroup$ There are transition moments to consider in addition to population, in electronic spectroscopy called Franck-Condon factors, I recall, although you will have to check, that they are called Honl_London terms in rotational spectroscopy. The effect is that the integral of overlap of wavefunctions is different depending on each particular transition. You should also consider that the transition propability depends on population difference between the levels involved, i.e. you cannot assume the the upper level has zero population, in HCl this may be ok as frequency is large wrt thermal energy. $\endgroup$
    – porphyrin
    Commented May 19 at 19:08
  • $\begingroup$ Additional considerations that might affect the intensity of the lines are the pressure of the sample (some transitions can saturate and its lines widen), and the resolution at which the spectrum is recorded. In the spectrum attached to your question, you can not distinguish beteewn the $\ce{H^{35}Cl}$ lines and its $\ce{H^{37}Cl}$ satellites. If the lines overlap this will alter the intensities and shape of the spectral lines. This is nice $\endgroup$
    – PAEP
    Commented May 19 at 21:37
  • $\begingroup$ You can find a higher resolution spectra in figure 5.2.2.1 of Chemistry LibreText. Physical Chemistry Lab Manual We regularly obtain a similar resolution spectrum using a PE IR Spectrometer with a resolution of 0.5 $\mathrm{cm^{-1}}$ @ 4000 $\mathrm{cm^{-1}}$. $\endgroup$
    – PAEP
    Commented May 19 at 21:46
  • $\begingroup$ @user146447, I do not think it is a good idea to conflate the spectra of both isotopomers using an average mass for HCl. Their natural abundance is 0.76:0.24, and since $\mathrm{ \nu \propto \sqrt{1/\mu} }$, each isotopomer has it own spectrum, they are very close, but you can distinguish them with enough resolution and if the transitions are not saturated. I think in this conditions is save to assume that the ratio of the intensities of each band in the spectrum reflects the natural abundance of each chlorine isotope. $\endgroup$
    – PAEP
    Commented May 29 at 16:39

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