QB_PH_SEC6_011: Atomic Spectra & Multi-Electron Atoms

In the normal Zeeman effect for a spectral line corresponding to the transition $^1D_2 \to ^1P_1$, the number of observed spectral components when viewed perpendicular to the applied magnetic field is

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Zeeman Splitting with Zero Spin ($S = 0$)**: The interaction Hamiltonian with a static magnetic field $\vec{B} = B\hat{z}$ is: $H' = \frac{e}{2m} (\vec{L} + 2\vec{S}) \cdot \vec{B} = \frac{e\hbar}{2m} L_z B = \mu_B B M_L$ 2. **Energy Levels**: - Upper state ($^1D_2$): $L = 2, M_{L1} \in \{-2, -1, 0, 1, 2\} \implies E_1(M_{L1}) = E_1^{(0)} + M_{L1} \mu_B B$ - Lower state ($^1P_1$): $L = 1, M_{L2} \in \{-1, 0, 1\} \implies E_2(M_{L2}) = E_2^{(0)} + M_{L2} \mu_B B$ 3. **Transition Frequency**: $\Delta E = E_1 - E_2 = (E_1^{(0)} - E_2^{(0)}) + (M_{L1} - M_{L2})\mu_B B = h\nu_0 + \Delta M_L \mu_B B$ 4. **Selection Rules**: Electric dipole transitions enforce $\Delta M_L = 0, \pm 1$: - $\Delta M_L = 0 \implies h\nu = h\nu_0$ ($\pi$-component, linearly polarized parallel to $\vec{B}$) - $\Delta M_L = +1 \implies h\nu = h\nu_0 + \mu_B B$ ($\sigma$-component) - $\Delta M_L = -1 \implies h\nu = h\nu_0 - \mu_B B$ ($\sigma$-component) 5. **Conclusion**: Exactly 3 spectral components are observed. Option (A) is correct.