Subtopic: Rigid Body Dynamics & Moment of Inertia Tensor
Question Statement:
In a diatomic molecule of mass M, electronic, rotational and vibrational energy
scales are of magnitude E, E and EV, respectively. The spring constant for the
vibrational energy is determined by E. If the electron mass is m then
Options:
(A) $E_R \sim \frac{m}{M} E_e$
(B) $E_R \sim \sqrt{\frac{m}{M}} E_e$
(C) $E_V \sim \sqrt{\frac{m}{M}} E_e$
(D) $E_V \sim \left(\frac{m}{M}\right)^{1/4} E_e$
Official Answer Key:
A,C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.50)**
**1. Molecular Physics & Born-Oppenheimer Scaling Laws**:
- In a diatomic molecule, electron mass is $m$ and reduced nuclear mass is $M \gg m$.
- Electronic energy scale: $E_e \sim \frac{\hbar^2}{m a_0^2}$.
**2. Detailed Step-by-Step Scaling Derivation**:
- **Vibrational Energy Scale ($E_v$)**:
- Effective nuclear spring constant $k \sim \frac{d^2 E_e}{dr^2} \sim \frac{E_e}{a_0^2} \sim \frac{m}{\hbar^2} E_e^2$.
- Vibrational frequency $\omega_v = \sqrt{\frac{k}{M}} \sim \sqrt{\frac{m}{M}} \frac{E_e}{\hbar}$.
- $E_v = \hbar \omega_v \sim \sqrt{\frac{m}{M}} E_e$. **Statement (C) is TRUE**.
- **Rotational Energy Scale ($E_R$)**:
- Moment of inertia $I \sim M a_0^2$.
- Rotational energy $E_R = \frac{\hbar^2}{2I} \sim \frac{\hbar^2}{M a_0^2} = \left(\frac{m}{M}\right) \frac{\hbar^2}{m a_0^2} \sim \left(\frac{m}{M}\right) E_e$.
- **Statement (A) is TRUE**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, C)**.
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