QB_PH_SEC7_022: Band Theory of Solids & Semiconductors

Near the top of the valence band in a semiconductor, the curvature of the energy-wavevector dispersion relation $d^2 E / dk^2$ is negative. This implies that the effective mass $m^*$ of an electron located near the top of the band is

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Dynamic Response in a Periodic Lattice**: Under an external force $F = -e\mathcal{E}$, acceleration is: $a = \frac{dv_g}{dt} = \frac{d}{dt}\left(\frac{1}{\hbar}\frac{dE}{dk}\right) = \frac{1}{\hbar}\frac{d^2 E}{dk^2}\frac{dk}{dt} = \frac{1}{\hbar^2}\frac{d^2 E}{dk^2} F$ 2. **Effective Mass Comparison with Newton's Second Law**: $a = \frac{F}{m^*} \implies \frac{1}{m^*} = \frac{1}{\hbar^2} \frac{d^2 E}{dk^2}$ 3. **Band Extremum Analysis**: - Bottom of conduction band: Local minimum $\implies \frac{d^2 E}{dk^2} > 0 \implies m^* > 0$ (positive). - Top of valence band: Local maximum $\implies \frac{d^2 E}{dk^2} < 0 \implies m^* < 0$ (negative). 4. **Conclusion**: Effective mass is negative. Option (A) is correct.