Si with 10¹⁵ As: resistance and Fermi level

Assume: Si at 300 K ($27^\circ$C), $kT = 0.02586$ eV, $n_i = 1.45\times10^{10}$ cm$^{-3}$, $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. Sample size is not given, so take a 1 cm cube: $R$ ($\Omega$) $= \rho$ ($\Omega$ cm).

Resistance with $10^{15}$ As cm$^{-3}$

$$\begin{aligned} n &\approx N_d = 10^{15}\ \text{cm}^{-3}, \quad p = \frac{n_i^2}{n} = 2.1\times10^{5}\ \text{cm}^{-3}\ (\text{negligible}) \\ \sigma &= eN_d\mu_e = 1.602\times10^{-19}\times10^{15}\times1350 = 0.216\ \Omega^{-1}\text{cm}^{-1} \\ \rho &= 4.62\ \Omega\,\text{cm} \;\Rightarrow\; R = 4.62\ \Omega \end{aligned}$$

Fermi level

$$\begin{aligned} E_F - E_{Fi} &= kT\ln\frac{N_d}{n_i} = 0.02586\ln\frac{10^{15}}{1.45\times10^{10}} \\ &= 0.02586\times11.14 = 0.288\ \text{eV} \end{aligned}$$

$E_F$ is 0.288 eV above $E_{Fi}$.

Further doped with $10^{22}$ B cm$^{-3}$

$N_a \gg N_d$, so the sample becomes strongly p-type: $p = N_a - N_d \approx 10^{22}$ cm$^{-3}$.

$$\begin{aligned} \sigma &= ep\mu_h = 1.602\times10^{-19}\times10^{22}\times450 = 7.21\times10^{5}\ \Omega^{-1}\text{cm}^{-1} \\ \rho &= 1.39\times10^{-6}\ \Omega\,\text{cm} \;\Rightarrow\; R = 1.39\times10^{-6}\ \Omega \end{aligned}$$

The resistance falls by a factor of $4.62/1.39\times10^{-6} \approx 3.3\times10^{6}$.

Note: $10^{22}$ cm$^{-3}$ is 20% of all Si atoms, so the material would be heavily degenerate and the real hole mobility far below 450 cm$^2$/Vs; the figure above follows the simple model as asked. (If the intended value is $10^{17}$ cm$^{-3}$, then $p = 9.9\times10^{16}$ cm$^{-3}$ and $R = 0.140$ $\Omega$.)

Answer: $R = 4.62$ $\Omega$, $E_F - E_{Fi} = 0.288$ eV; after boron doping, p-type with $R \approx 1.39\times10^{-6}$ $\Omega$.