What do these wavenumbers correspond to in Raman spectroscopy?

In summary, when the molecule absorbs a photon, it then emits light of a higher intensity. The "hills" in the spectrum correspond to a change in energy of the molecule when it starts vibrate more.
  • #1
Lotto
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TL;DR Summary
When we have a Raman spectrum of a molecule, on the horizontal axis, there are values of wavenumbers. What does this wavenumbers correspond to?
I don't know whether it is an energy of a photon emitted by a deexciting molecule, or if it is an energy of laser's photons. Here is an example of such spectrum:
1678916731663.png

For example, that value of wavenumber ##3000\, \mathrm {cm^{-1}}## is an energy of an emitted photon or a photon from laser? And that high instensity is there why? It means that when the molecule absorbs the photon, it then emits light of a higher intesnity?
 
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  • #2
It seems like you’re trying to apply classical reasoning to a quantum problem. Photons are indistinguishable: they don’t have labels. (Note that it’s kind of difficult to discuss Raman spectroscopy at B level)

Raman spectroscopy relies on inelastic scattering of photons off a material. A rough classical picture of the Raman effect is that an incoming photon “bounces off” a molecule, causing the molecule to be vibrationally excited. Since energy is conserved, whatever energy is not deposited in the molecular vibration exits the system as a lower energy photon.

However, you can’t say this is the “same” photon or “different,” because photons are indistinguishable (NB—this is true even when the photons are the same energy, as in reflection or Rayleigh scattering).

In terms of intensity, certain vibrational modes of a molecule have certain cross sections, which will tell you how efficient the Raman scattering process is for them. The light detected in a Raman spectrum is much less intense than the Rayleigh scattered light, which is itself much less intense than the incoming laser light.
 
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  • #3
Lotto said:
When we have a Raman spectrum of a molecule, on the horizontal axis, there are values of wavenumbers. What does this wavenumbers correspond to?
As the google lemma explains, the horizontal axis is the Raman shift, i.e. the energy difference between the incoming and the scattered light. Traditionally the units used are cm-1. (8066 cm-1 = 1eV).

The vertical scale is relative.

For more, google ! e.g: this

##\ ##
 
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  • #4
BvU said:
As the google lemma explains, the horizontal axis is the Raman shift, i.e. the energy difference between the incoming and the scattered light. Traditionally the units used are cm-1. (8066 cm-1 = 1eV).

The vertical scale is relative.

For more, google ! e.g: this

##\ ##
So it is the energy that the molecule gained. So what do these "hills" mean? Does it correspond to a change in energy of the molecule when it starts vibrate more? So when the change in energy is ##3,000\,\mathrm {cm}^{-1}##, it vibrates with frequency ##3,000\,\mathrm {cm}^{-1}##?
 
  • #5
Lotto said:
So what do these "hills" mean?
Lineshape analysis is not particularly straightforward:
https://en.wikipedia.org/wiki/Spectral_line_shape
It could be that there are a lot of vibrational modes with relatively low cross section in that region of the spectrum. Or it could be some sort of pressure broadening. Or a combination of multiple factors.
Lotto said:
So when the change in energy is 3,000cm−1, it vibrates with frequency 3,000cm−1?
You're measuring the photons coming from the system in Raman spectroscopy. Since you're exciting with a known frequency, if you measuring frequencies that are lower than that, then you can assume via energy conservation that the missing energy was deposited in the system. For the range normally examined in Raman spectroscopy, yes, these will generally be vibrational modes.
 
  • #6
To add to what has already been said, an important distinction to make is the x-axis in wavenumbers is typically labeled as Raman Shift. This shift is relative to the Rayleigh scattering line at 0 cm
-1 and corresponds to the process by which the scattered light being measured is at the same frequency as the incident light source.

When a peak appears in a Raman spectrum, what is being measured is the frequency of a molecular vibration that is being driven by the incident light. You can think of this process as excitation from the ground state to some excited (virtual) state followed by de-excitation to a low-energy vibrational state with a corresponding emission of a photon. This emitted photon will be slightly different in energy (or frequency since E = hf) and this difference in the frequency of the incident photon and the frequency of the outgoing photon is what produces Raman peaks. The molecule is left in an excited vibrational state and we can directly measure the frequency (energy) of this vibrational mode.

You can go a lot deeper with this type of measurement in conjunction with group theory analysis and/or DFT calculations, but for simpler molecules, a database can typically tell you what each peak represents. For example, peaks that appear around 3000 cm-1 are typically related to C-H vibrations, so you can look at the Raman spectra of a given molecule and start to piece together what each vibration physically looks like.

As for line-shape and broadening, what TeethWhitener said is correct - it is not always immediately obvious what causes broadening (or narrowing) of vibrational modes, but there are typically many things at play. For example, researchers typically can get better resolution in Raman spectra by having the sample to be measured in a fridge or cooled to near liquid nitrogen temperatures: typically, the cooler the sample, the narrower the peak. Also, it is worth noting that it is very likely that there are just a large number of vibrational modes that are closely spaced and that is why the peaks don't look like pure Lorentzians. As a rule of thumb, there will typically be 3N-6 vibrational modes for a given molecule where N is the number of atoms in the molecule. There is a lot more nuance to this as not all of those modes will be Raman active (based on selection rules), but you can already see that for a decent-sized molecule, there is potentially a very large number of vibrational modes that you can measure.
 

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