QB_PH_SEC6_005: Lasers and Quantum Optics

The ratio of the Einstein spontaneous emission coefficient $A_{21}$ to the stimulated emission coefficient $B_{21}$ for transition frequency $\nu$ is proportional to

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Thermal Equilibrium Balance**: In thermal cavity with radiation density $\rho(\nu)$: $N_1 B_{12} \rho(\nu) = N_2 [A_{21} + B_{21} \rho(\nu)]$ 2. **Boltzmann Ratio**: $N_1/N_2 = e^{h\nu / k_B T}$ with $g_1 = g_2 \implies B_{12} = B_{21}$: $\rho(\nu) = \frac{A_{21}/B_{21}}{e^{h\nu / k_B T} - 1}$ 3. **Comparison with Planck's Law**: $\rho(\nu) = \frac{8\pi h \nu^3}{c^3} \frac{1}{e^{h\nu / k_B T} - 1}$. 4. **Ratio**: $\frac{A_{21}}{B_{21}} = \frac{8\pi h \nu^3}{c^3} \propto \nu^3$. Option (C) is correct.