Question

In a Raman spectroscopy experiment done at 300 K, a Raman line is observed at $200 \, cm^{-1} \, (\sim 25 \, meV)$. The ratio of the intensity of the Stokes line to that of the Anti-Stokes line is ............ (rounded off to two decimal places). Given: Boltzmann constant, $k = 8.62 \times 10^{-5} \, eV K^{-1}$

Hint:

Stokes lines are stronger than Anti-Stokes at room temperature because population of excited vibrational states is much smaller. Use $\frac{I_S}{I_{AS}} = e^{\Delta E/kT}$ for quick calculation.

Answer 1

Correct Answer: 2.45

Step 1: Recall intensity ratio relation.
The ratio of Stokes to Anti-Stokes line intensities in Raman scattering is given by: \[ \frac{I_S}{I_{AS}} = \exp\left( \frac{\Delta E}{kT} \right) \] where $\Delta E$ is the energy difference between ground and excited vibrational states.
Step 2: Identify given vibrational energy.
We are told: \[ \Delta E \approx 25 \, meV = 0.025 \, eV \]
Step 3: Calculate $kT$ at 300 K.
\[ kT = (8.62 \times 10^{-5})(300) \, eV \] \[ kT = 0.02586 \, eV \]
Step 4: Compute ratio exponent.
\[ \frac{\Delta E}{kT} = \frac{0.025}{0.02586} \approx 0.967 \]

Step 5: Evaluate exponential.
\[ \frac{I_S}{I_{AS}} = \exp(0.967) \approx 2.63 \]
Step 6: Interpretation.
- Stokes lines are more intense than Anti-Stokes because most molecules are in the ground vibrational state at room temperature. - The ratio $\sim 2.6$ reflects this population imbalance. Final Answer: \[ \boxed{2.63} \]