Si cube doped 1 As in 10⁶

Assume: Si at 300 K, $n_i = 1.45\times10^{10}$ cm$^{-3}$, $\mu_e = 1350$ and $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. A 1 cm$^3$ cube has $L = 1$ cm and $A = 1$ cm$^2$, so $R$ (in $\Omega$) $= \rho$ (in $\Omega$ cm).

Pure silicon

$$\begin{aligned} \sigma_i &= en_i(\mu_e+\mu_h) = 1.602\times10^{-19}\times1.45\times10^{10}\times(1350+450) \\ &= 4.18\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{\rho L}{A} = \frac{1}{\sigma_i}\times\frac{1\ \text{cm}}{1\ \text{cm}^2} = 2.39\times10^{5}\ \Omega \end{aligned}$$

Doped with 1 As in $10^6$ Si atoms

Arsenic is a donor:

$$\begin{aligned} N_d &= \frac{5\times10^{22}}{10^6} = 5\times10^{16}\ \text{cm}^{-3}, \quad n \approx N_d \\ p &= \frac{n_i^2}{N_d} = \frac{(1.45\times10^{10})^2}{5\times10^{16}} = 4.2\times10^{3}\ \text{cm}^{-3}\ (\text{negligible}) \\ \sigma &= eN_d\mu_e = 1.602\times10^{-19}\times5\times10^{16}\times1350 = 10.81\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{10.81} = 0.0925\ \Omega \end{aligned}$$

The resistance falls by a factor of about $2.6\times10^{6}$. (At this doping, impurity scattering lowers $\mu_e$ to roughly 1000 cm$^2$/Vs in practice, giving about 0.12 $\Omega$; the textbook answer uses the pure-Si mobility.)

Answer: Pure Si: $R = 2.39\times10^{5}$ $\Omega$. Doped: $R \approx 0.0925$ $\Omega$.