Copper drift mobility from scattering cross-section

In the free electron (Drude) model, an electron moving at mean speed $u$ is scattered when it comes within the cross-sectional area $S$ of a scattering centre. With $N_s$ scatterers per unit volume:

$$\lambda = \frac{1}{S N_s}, \qquad \tau = \frac{\lambda}{u} = \frac{1}{S N_s u}, \qquad \mu_d = \frac{e\tau}{m_e}, \qquad \sigma = e n \mu_d$$

Each copper atom (ion) is taken as a scatterer, and each gives one conduction electron, so $N_s = n$.

Given: $u = 1.5 \times 10^{6}\ \text{m/s}$, $S = 3.9 \times 10^{-22}\ \text{m}^2$, $\rho = 8960\ \text{kg/m}^3$, $M = 63.56\ \text{g/mol}$.

Atomic (and electron) concentration

$$n = N_s = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.06356} = 8.49 \times 10^{28}\ \text{m}^{-3}$$

Mean free path and mean free time

$$\begin{aligned} \lambda &= \frac{1}{S N_s} = \frac{1}{3.9 \times 10^{-22} \times 8.49 \times 10^{28}} = 3.02 \times 10^{-8}\ \text{m} \\ \tau &= \frac{\lambda}{u} = \frac{3.02 \times 10^{-8}}{1.5 \times 10^{6}} = 2.01 \times 10^{-14}\ \text{s} \end{aligned}$$

Drift mobility

$$\mu_d = \frac{e\tau}{m_e} = \frac{1.602 \times 10^{-19} \times 2.01 \times 10^{-14}}{9.109 \times 10^{-31}} = 3.54 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 35.4\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$$

Conductivity

$$\sigma = e n \mu_d = 1.602 \times 10^{-19} \times 8.49 \times 10^{28} \times 3.54 \times 10^{-3} = 4.82 \times 10^{7}\ \text{S/m}$$

This is close to the measured value for copper ($5.9 \times 10^{7}$ S/m), showing that the simple model gives the right order of magnitude.

Answer: $\mu_d \approx 35.4\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $\sigma \approx 4.82 \times 10^{7}\ \text{S/m}$ (with $\lambda = 30.2$ nm, $\tau = 2.01 \times 10^{-14}$ s).