Resistance of arsenic-doped germanium wire

Arsenic is a donor; each As atom gives one free electron. Find the donor concentration from the masses, then the conductivity and resistance.

Given: mass of Ge = 50 g, density 5.46 g cm⁻³; mass of As = 5 µg = $5\times10^{-6}$ g, M = 74.92 g mol⁻¹; $\mu_e = 3600\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$; L = 1 cm; $A = 2.25\times10^{-4}\ \text{cm}^2$; $N_A = 6.022\times10^{23}$.

Step 1: Volume of germanium

$$V = \frac{m}{\rho} = \frac{50}{5.46} = 9.158\ \text{cm}^3$$

(The tiny mass of As does not change the volume.)

Step 2: Number of As atoms

$$N_{As} = \frac{5\times10^{-6}}{74.92}\times6.022\times10^{23} = 4.019\times10^{16}$$

Step 3: Donor (electron) concentration

$$N_d = \frac{4.019\times10^{16}}{9.158} = 4.389\times10^{15}\ \text{cm}^{-3}$$

All donors are ionised at room temperature and $N_d \gg n_i(\text{Ge}) \approx 2.4\times10^{13}$, so $n \approx N_d$ and the hole contribution is negligible.

Step 4: Conductivity and resistivity

$$\begin{aligned} \sigma &= eN_d\mu_e = 1.6\times10^{-19}\times4.389\times10^{15}\times3600\\ &= 2.528\ \text{S cm}^{-1}\\ \rho &= 1/\sigma = 0.3956\ \Omega\,\text{cm} \end{aligned}$$

Step 5: Resistance of the wire

$$\begin{aligned} R &= \frac{\rho L}{A} = \frac{0.3956\times1}{2.25\times10^{-4}}\\ &= 1758\ \Omega \end{aligned}$$

Answer: $N_d \approx 4.39\times10^{15}\ \text{cm}^{-3}$, $\rho \approx 0.396\ \Omega$ cm, and the wire resistance $R \approx 1.76\ \text{k}\Omega$.