To solve this problem, we need to understand the selection rules for rotational Raman transitions and how these affect the spacing of lines in the rotational Raman spectrum.
In the given problem, we have the energy levels for the rotational motion of a molecule expressed as:
\(E_J = BJ(J+1) \text{ cm}^{-1}\) where \(J = 0, 1, 2, \ldots\).
The selection rule for pure rotational Raman transitions is \(\Delta J = 0, \pm 2\). For Raman lines, we focus on transitions where \(\Delta J = \pm 2\). The Stokes and anti-Stokes Raman lines stem from these types of transitions:
- Stokes lines: represent transitions \(J \rightarrow J+2\), resulting in the energy difference:
- Initial energy: \(BJ(J+1)\)
- Final energy: \(B(J+2)(J+3)\)
- Energy difference (Stokes shift): \(\Delta E_{\text{Stokes}} = B(J+2)(J+3) - BJ(J+1) = 4B(J+1) + 2B\)
- Anti-Stokes lines: represent transitions \(J+2 \rightarrow J\), with the energy change:
- Initial energy: \(B(J+2)(J+3)\)
- Final energy: \(BJ(J+1)\)
- Energy difference (Anti-Stokes shift): \(\Delta E_{\text{Anti-Stokes}} = BJ(J+1) - B(J+2)(J+3) = -4B(J+1) - 2B\)
For the separation between the closest Stokes and anti-Stokes lines, we calculate the absolute difference between these two energy transitions:
\(\Delta \nu = |\Delta E_{\text{Stokes}} - \Delta E_{\text{Anti-Stokes}}| = |(4B(J+1) + 2B) - (-(4B(J+1) + 2B))| = 2 \times (4B(J+1) + 2B) = 12B\)
Thus, the separation between the closest Stokes and anti-Stokes lines is \(12B\) cm-1.
Correct Answer:
\(12B\)
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