Metals — Question 5

Question 5

Fermi motion determines the mean free path

A metal has $v_F=1.50\times10^6\,\mathrm{m/s}$ and $\tau=2.00\times10^{-14}\,\mathrm s$. In a weak field its drift speed is only $v_d=2.00\times10^{-4}\,\mathrm{m/s}$.

1. Find the mean free path and compare it with the drift displacement during one relaxation time.

2. For lattice spacing $a=.300\,\mathrm{nm}$, find the number of spacings in the mean free path and explain the distinct velocity scales.

Question 5 - Solution

Fermi motion determines the mean free path

(a) Method and calculation

$$\boxed{\ell=v_F\tau=3.00\times10^{-8}\,\mathrm m=30.0\,\mathrm{nm}}.$$$$\boxed{\Delta x_d=v_d\tau=4.00\times10^{-18}\,\mathrm m,\quad\ell/\Delta x_d=7.50\times10^9}.$$

(b) Interpretation and further calculation

$$\boxed{\ell/a=30.0/.300=100}.$$$$\langle\mathbf v\rangle=\mathbf v_d,\qquad |\mathbf v|\sim v_F\gg|\langle\mathbf v\rangle|.$$

Physical check

The large microscopic velocities nearly cancel in equilibrium. Current measures the small directional imbalance, whereas scattering paths involve the microscopic speed.