Relaxation time and mean free path

The drift mobility and the mean free (relaxation) time $\tau$ are related by

$$\mu_d = \frac{e\tau}{m_e} \quad\Rightarrow\quad \tau = \frac{\mu_d m_e}{e}, \qquad \lambda = u\tau$$

where $u$ is the mean speed and $\lambda$ the mean free path.

Given: $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 4.3 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $u = 2 \times 10^{7}\ \text{m/s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$, $e = 1.602 \times 10^{-19}\ \text{C}$.

Relaxation time

$$\tau = \frac{4.3 \times 10^{-3} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.44 \times 10^{-14}\ \text{s}$$

Mean free path

$$\lambda = u\tau = 2 \times 10^{7} \times 2.44 \times 10^{-14} = 4.89 \times 10^{-7}\ \text{m} \approx 489\ \text{nm}$$

(Using the speed as given. The usual Fermi speed in copper is about $1.6 \times 10^{6}$ m/s, which would give $\lambda \approx 40$ nm; the mean free path scales directly with the speed used.)

Answer: $\tau = 2.44 \times 10^{-14}\ \text{s}$, $\lambda = 4.89 \times 10^{-7}\ \text{m}$.