QB_PH_SEC6_019: Molecular Spectroscopy

A diatomic molecule has fundamental vibrational frequency $\omega_e = 2000\text{ cm}^{-1}$ and anharmonicity constant $x_e = 0.01$. What is the wavenumber (in $\text{cm}^{-1}$) of the first overtone ($v = 0 \to v = 2$) transition? (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Anharmonic Vibrational Energy Expression**: $G(v) = \omega_e\left(v + \frac{1}{2}\right) - \omega_e x_e \left(v + \frac{1}{2}\right)^2$ 2. **Energy for Specific Levels**: - Ground state ($v = 0$): $G(0) = \frac{1}{2}\omega_e - \frac{1}{4}\omega_e x_e$ - Second excited state ($v = 2$): $G(2) = \frac{5}{2}\omega_e - \frac{25}{4}\omega_e x_e$ 3. **Transition Wavenumber for First Overtone** ($v = 0 \to 2$): $\nu_{0\to 2} = G(2) - G(0) = 2\omega_e - 6\omega_e x_e = 2\omega_e(1 - 3x_e)$ Wait, let's verify: $25/4 - 1/4 = 24/4 = 6$. Yes, $2\omega_e - 6\omega_e x_e = 2\omega_e(1 - 3x_e)$! $2\omega_e - 6\omega_e x_e = 2(2000) - 6(2000)(0.01) = 4000 - 120 = 3880\text{ cm}^{-1}$ 4. **Re-Verification**: $G(2) - G(0) = (5/2 - 1/2)\omega_e - (25/4 - 1/4)\omega_e x_e = 2\omega_e - 6\omega_e x_e = 4000 - 120 = 3880\text{ cm}^{-1}$. 5. **Official Exact Value**: 3880.