Resistance of 8 cm³ Si cube

The cube volume is 8 cm³, so each side is $L = 2$ cm and the face area is $A = 4\ \text{cm}^2$. For a cube

$$R = \frac{\rho L}{A} = \frac{\rho}{L} = \frac{\rho}{2\ \text{cm}}$$

Data used (Si, 300 K): $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT = 8.617\times10^{-5}\times300 = 0.02585\ \text{eV}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}\ \text{C}$.

Pure (intrinsic) silicon

$$\begin{aligned} \sigma_i &= n_i e(\mu_e+\mu_h) \\ &= 1.45\times10^{10}\times1.6\times10^{-19}\times1800 \\ &= 4.176\times10^{-6}\ \text{S cm}^{-1}\\ \rho_i &= 1/\sigma_i = 2.395\times10^{5}\ \Omega\,\text{cm}\\ R_i &= \frac{2.395\times10^{5}}{2} = 1.197\times10^{5}\ \Omega \end{aligned}$$

Doped with 1 As per $5\times10^9$ Si atoms

Arsenic is a donor:

$$N_d = \frac{5\times10^{22}}{5\times10^{9}} = 10^{13}\ \text{cm}^{-3}$$

Since $N_d \gg n_i$, $n \approx N_d = 10^{13}\ \text{cm}^{-3}$ and $p = n_i^2/N_d = 2.1\times10^{7}\ \text{cm}^{-3}$ (negligible).

$$\begin{aligned} \sigma &= e(n\mu_e + p\mu_h) \approx eN_d\mu_e \\ &= 1.6\times10^{-19}\times10^{13}\times1350 \\ &= 2.16\times10^{-3}\ \text{S cm}^{-1}\\ \rho &= 463\ \Omega\,\text{cm}\\ R &= \frac{463}{2} = 231.5\ \Omega \end{aligned}$$

Answer: pure cube $R \approx 1.20\times10^{5}\ \Omega$; doped cube $R \approx 231.5\ \Omega$ — about 517 times smaller, even though only one atom in five billion was replaced.