In a solid, a Raman line observed at 300 cm-1 has intensity of Stokes line four
times that of the anti-Stokes line. The temperature of the sample is _______K.
(Round off to the nearest integer) (1 cm-1 ≡ 1.44 K)
Official Answer Key:
311 to 312
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.32)**
**1. Molecular Physics & Raman Spectroscopy**:
- In Raman scattering, the intensity ratio of the Stokes line ($I_S$) to the anti-Stokes line ($I_{aS}$) is governed by Boltzmann population statistics of vibrational/phonon levels:
$\frac{I_S}{I_{aS}} = \left(\frac{\omega_0 - \omega_{\text{vib}}}{\omega_0 + \omega_{\text{vib}}}\right)^4 \exp\left(\frac{\hbar\omega_{\text{vib}}}{k_B T}\right) \approx \exp\left(\frac{h c \tilde{\nu}}{k_B T}\right)$
**2. Detailed Step-by-Step Derivation**:
- Given:
- $\frac{I_S}{I_{aS}} = 4$
- Raman shift $\tilde{\nu} = 300 \text{ cm}^{-1}$
- Conversion factor: $1 \text{ cm}^{-1} \equiv 1.44$ K $\implies \frac{h c \tilde{\nu}}{k_B} = 300 \times 1.44 = 432$ K.
- Setting up the equation:
$\exp\left(\frac{432}{T}\right) = 4$
- Taking natural logarithm on both sides:
$\frac{432}{T} = \ln(4) \approx 1.38629$
$T = \frac{432}{1.38629} \approx 311.62 \text{ K}$
- Rounding to the nearest integer yields **312 K** (or within 311–312 K).
**3. Official Verification**:
The verified official IIT Master Answer range is **311 to 312**.
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