Raman Spectroscopy: How to calculate their constants?

Question

$^1{\ce{H2}}$ 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.

Answer 1

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.

Answer 2

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.