QB_PH_SEC7_028: Band Theory of Solids & Semiconductors

In an intrinsic semiconductor with effective densities of states $N_c$ and $N_v$ and bandgap $E_g$, the intrinsic Fermi level $E_i$ lies exactly at the middle of the bandgap ($E_i = (E_c + E_v)/2$) if and only if

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Thermal Carrier Densities**: $n = N_c e^{-(E_c - E_F)/k_B T}, \quad p = N_v e^{-(E_F - E_v)/k_B T}$ 2. **Intrinsic Condition ($n = p$)**: $N_c e^{-(E_c - E_i)/k_B T} = N_v e^{-(E_i - E_v)/k_B T}$ $e^{2E_i / k_B T} = \frac{N_v}{N_c} e^{(E_c + E_v)/k_B T}$ $E_i = \frac{E_c + E_v}{2} + \frac{1}{2} k_B T \ln\left(\frac{N_v}{N_c}\right)$ 3. **Effective Density of States Proportionality**: $N_c \propto (m_e^*)^{3/2}$ and $N_v \propto (m_h^*)^{3/2}$: $E_i = \frac{E_c + E_v}{2} + \frac{3}{4} k_B T \ln\left(\frac{m_h^*}{m_e^*}\right)$ 4. **Midgap Condition**: $E_i = \frac{E_c + E_v}{2}$ holds generally if $\ln(m_h^*/m_e^*) = 0 \implies m_e^* = m_h^*$. 5. **Conclusion**: Option (A) is correct.