Question

The resistivity of a pure semiconductor at 298 K is \(3000\,\Omega\text{m}\). Assume the number of electrons excited \((n_e)\) across the band gap is \[ n_e = N_A \exp\left(-\frac{E_g}{k_B T}\right) \] Given: \[ N_A = 6.02 \times 10^{23}\ \text{mol}^{-1},\quad k_B = 8.62 \times 10^{-5}\ \text{eV/K},\quad T = 298\ \text{K} \] Mobilities: \[ \mu_e = 0.14\ \text{m}^2/(\text{V·s}),\quad \mu_h = 0.06\ \text{m}^2/(\text{V·s}) \] Absolute electron charge: \[ q = 1.60 \times 10^{-19}\ \text{C} \]

Hint:

Intrinsic carrier concentration depends exponentially on band gap. Even small errors in \(n_e\) cause large changes in \(E_g\).

Answer 1

Correct Answer: 0.45

Resistivity relation for intrinsic semiconductor:
\[ \rho = \frac{1}{q\, n_e (\mu_e + \mu_h)} \] Thus:
\[ n_e = \frac{1}{\rho\, q\, (\mu_e + \mu_h)} \] Substitute values:
\[ n_e = \frac{1}{3000 \times (1.6 \times 10^{-19}) \times (0.14 + 0.06)} \] \[ n_e = \frac{1}{3000 \times 1.6 \times 10^{-19} \times 0.20} \] \[ n_e = \frac{1}{9.6 \times 10^{-17}} = 1.04 \times 10^{16} \] Using excitation relation:
\[ n_e = N_A \exp\left(-\frac{E_g}{k_BT}\right) \] \[ \exp\left(-\frac{E_g}{k_BT}\right) = \frac{n_e}{N_A} \] \[ \frac{n_e}{N_A} = \frac{1.04 \times 10^{16}}{6.02 \times 10^{23}} = 1.73 \times 10^{-8} \] Take natural logarithm:
\[ -\frac{E_g}{k_BT} = \ln(1.73 \times 10^{-8}) \] \[ \ln(1.73 \times 10^{-8}) = -17.87 \] Thus:
\[ E_g = 17.87 \times k_B T \] \[ E_g = 17.87 \times (8.62 \times 10^{-5}) \times 298 \] \[ E_g = 0.46\ \text{eV} \] Rounded to 2 decimals: \[ E_g = 0.46\ \text{eV} \]