Resistance of antimony-doped germanium bar

Antimony (group V) is a donor; each Sb atom donates one electron.

Given: $m_{Ge} = 100$ g, $\rho_{Ge} = 5.46\ \text{g cm}^{-3}$, $m_{Sb} = 4\ \mu\text{g} = 4\times10^{-6}$ g, $M_{Sb} = 121.76$; bar L = 2 cm, $A = 0.012\times0.012 = 1.44\times10^{-4}\ \text{cm}^2$. Assumed: electron mobility in Ge $\mu_e = 3900\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (standard value, not given), $N_A = 6.022\times10^{23}$.

Step 1: Volume of germanium

$$V = \frac{100}{5.46} = 18.32\ \text{cm}^3$$

Step 2: Number of Sb atoms

$$N_{Sb} = \frac{4\times10^{-6}}{121.76}\times6.022\times10^{23} = 1.978\times10^{16}$$

Step 3: Density of antimony atoms

$$N_d = \frac{1.978\times10^{16}}{18.32} = 1.08\times10^{15}\ \text{cm}^{-3}$$

Step 4: Density of donated electrons

At room temperature every Sb atom is ionised, and $N_d \gg n_i(\text{Ge}) = 2.4\times10^{13}$, so

$$n \approx N_d = 1.08\times10^{15}\ \text{cm}^{-3}$$

Step 5: Conductivity and resistance

$$\begin{aligned} \sigma &= eN_d\mu_e = 1.6\times10^{-19}\times1.08\times10^{15}\times3900\\ &= 0.674\ \text{S cm}^{-1}\\ \rho &= 1/\sigma = 1.484\ \Omega\,\text{cm}\\ R &= \frac{\rho L}{A} = \frac{1.484\times2}{1.44\times10^{-4}}\\ &= 2.06\times10^{4}\ \Omega \end{aligned}$$

Answer: $N_d = n \approx 1.08\times10^{15}\ \text{cm}^{-3}$; $R \approx 20.6\ \text{k}\Omega$ (with $\mu_e = 3800$ the resistance would be about 21.1 kΩ).