Intrinsic Si from effective densities of states

For an intrinsic semiconductor

$$n_i = \sqrt{N_cN_v}\,\exp\left(-\frac{E_g}{2kT}\right),\qquad \rho_i = \frac{1}{en_i(\mu_e+\mu_h)}$$

Given: $N_c = 2.9\times10^{19}$, $N_v = 1.1\times10^{19}\ \text{cm}^{-3}$, T = 300 K. Data used (Si): $E_g = 1.1$ eV, $kT = 0.02585$ eV, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}$ C.

Intrinsic concentration

$$\begin{aligned} \sqrt{N_cN_v} &= \sqrt{2.9\times10^{19}\times1.1\times10^{19}} = 1.786\times10^{19}\ \text{cm}^{-3}\\ \frac{E_g}{2kT} &= \frac{1.1}{2\times0.02585} = 21.28\\ e^{-21.28} &= 5.750\times10^{-10}\\ n_i &= 1.786\times10^{19}\times5.750\times10^{-10}\\ &= 1.03\times10^{10}\ \text{cm}^{-3} \end{aligned}$$

Intrinsic conductivity and resistivity

$$\begin{aligned} \sigma_i &= en_i(\mu_e+\mu_h)\\ &= 1.6\times10^{-19}\times1.027\times10^{10}\times1800\\ &= 2.96\times10^{-6}\ \text{S cm}^{-1}\\ \rho_i &= \frac{1}{\sigma_i} = 3.38\times10^{5}\ \Omega\,\text{cm} \end{aligned}$$

Answer: $n_i \approx 1.03\times10^{10}\ \text{cm}^{-3}$; $\rho_i \approx 3.38\times10^{5}\ \Omega$ cm ($3.38\times10^{3}\ \Omega$ m).

Note: $n_i$ is very sensitive to $E_g$ and T. Because of the factor $e^{-E_g/2kT}$, using $E_g = 1.12$ eV instead of 1.1 eV would reduce $n_i$ by a factor of about 1.5.