Si cube with As and B per million

The cube is 1 cm³, so L = 1 cm, A = 1 cm² and $R = \rho L/A$ equals $\rho$ numerically.

Data used (the "list" in the paper): $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}$ C, Si atoms $5\times10^{22}\ \text{cm}^{-3}$.

Pure silicon

$$\begin{aligned} \sigma_i &= n_ie(\mu_e+\mu_h)\\ &= 1.45\times10^{10}\times1.6\times10^{-19}\times(1350+450)\\ &= 4.176\times10^{-6}\ \text{S cm}^{-1}\\ R_i &= \frac{1}{\sigma_i} = 2.39\times10^{5}\ \Omega \end{aligned}$$

Doped with As and B

$$N_d = \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3},\qquad N_a = \frac{5\times10^{22}}{10^6} = 5\times10^{16}\ \text{cm}^{-3}$$

$N_a \gg N_d$, so the cube is p-type (compensated):

$$p = N_a - N_d = 4.995\times10^{16}\ \text{cm}^{-3},\qquad n = \frac{n_i^2}{p} = 4.2\times10^{3}\ \text{cm}^{-3}$$ $$\begin{aligned} \sigma &= e(n\mu_e + p\mu_h) \approx ep\mu_h\\ &= 1.6\times10^{-19}\times4.995\times10^{16}\times450\\ &= 3.596\ \text{S cm}^{-1}\\ R &= \frac{1}{3.596} = 0.278\ \Omega \end{aligned}$$

Answer: pure cube $R \approx 2.39\times10^{5}\ \Omega$; doped cube $R \approx 0.278\ \Omega$ (about $8.6\times10^5$ times smaller). Boron dominates because it is 1000 times more concentrated than arsenic.