Raman Spectroscopy: How to calculate their constants?

$^1{\ce{H2}}$ is interrogated with a laser of wavelength lambda = 532 nm and vibrational Stokes transitions are observed at S(0) = 14302.206 cm$^{-1}$, Q(0) = 14638.442 and O(2) = 14992.823 cm$^{-1}$.


1. Estimate ωe (cm$^{-1}$)
2. Determine B1 (cm$^{-1}$)


I know the band origin = Q(0). I also know from rovibrational spectra that


$ ω\_0 = ω\_e -2ω\_eX\_e$.


* Could someone please help me answer this question and explain?


If there are any additional resources I could read on these topics that would be great too.




In Raman spectroscopy the difference in frequency between the exciting line and the measured signal frequencies contains the information you want. These values, made positive if necessary, give the ro-vibrational transition frequencies.


You are assuming that the molecule is vibrating with an anharmonic potential, such as a Morse potential. The energy levels are given by $E\_n= \omega\_e(n+1/2) - x\_e\omega\_e(n+1/2)^2$ (in wavenumber units) with quantum number $ n = 0, 1, ...~$, and $\omega\_e$ the frequency and $x\_e$ is the dimensionless anharmonicity. 


In addition to the vibrational energy there is rotational energy whose levels are $E\_J =B\_eJ(J+1)$ for rotational constant $B\_e$ (in wavenumbers) and quantum number $J=0, 1, 2,...$ and ignoring centrifugal distortion. You will also have to use different $B\_e$ values for the $n=0$ and $n=1$ vibrational levels since the bond length is slightly different between (anharmonic) vibrational levels.


You know that the rotational-vibrational levels appear in groups O, P, Q, R, S in increasing frequency and that Raman has O, Q & S bands only. In the Q band $\Delta J = 0$ between vibrational levels *n* to $n+1$ and is $-2$ and $+2$ for O & S branches respectively. The Raman selection rules are $\Delta n = \pm 1$ and $\Delta J = 0, \pm 2 $. 


If you look at a rotation-vibration energy level diagram for IR transitions you should by adapting this be able to work out your values for your Raman transitions.


There is a figure showing energy levels at
 <https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy>


The Q band transition ($\Delta n =0, \Delta J =0$) in your notation is $\omega\_0 = \omega\_e - 2x\_e \omega \_e $ and the other branches calculated similarly with $\Delta J = \pm 2$ as appropriate.


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Good explanation is given by @porphyrin's answer and I am going to get into details now.


The laser is at 532 nm hence in absolute wavenumbers this comes to $18789.9\hspace{1ex} cm^{-1}$.


$S(0)$ appears at $14302.2\hspace{1ex}cm^{-1} \rightarrow 4487.6 \hspace{1ex}cm^{-1} (Ramanshift)$ 


$Q(0)$ appears at $14638.4\hspace{1ex}cm^{-1} \rightarrow 4151.4 \hspace{1ex}cm^{-1} (Ramanshift)$ 


$O(0)$ appears at $14992.8\hspace{1ex}cm^{-1} \rightarrow 3797.1 \hspace{1ex}cm^{-1} (Ramanshift)$


\*Just the difference from the incident laser in wavenumbers.


Next, rotational constant is easy to estimate roughly,
$E\_{J}=6\times B\_{0}$ for $J=0$ to $J=2$.


$354.3=6\times B\_{0}$


Rotational constant, $B\_{0}=59.05 \hspace{1ex}cm^{-1} $




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In this equation, $E\_n=ω\_e(n+1/2)−x\_eω\_e(n+1/2)^2$ , we already know $E\_n$ and $n$. ($n=1$ and $E\_n$=energy of the Q branch in wavenumbers, relative to laser). then $\omega\_e$ calculation is straightforward.


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