Question

Consider Al at room temperature. Its electron density is n = 18 x 1022 cm-3 and its electrical resistivity is rho = 2.45 μΩ-cm. Find its electron relaxation time τ and electron mean free path in the Drude model.

Answer 1

Answer:

$\tau=1.28968\times 10^{-8}\ s$ is the electron relaxation time

$l=9.6726\times 10^{-4}\ m$ is the mean free path

Explanation:

Given:

• electron density of aluminium at room temperature, $n=18\times 10^{22}\ cm^{-3}$

• resistivity of aluminium, $\rho=2.45\times 10^{-8}\ \Omega.m$

From the Drude's model we have:

$\tau=\frac{m}{\rho.n.q^2}$

where:

$\tau=$ electron relaxation time

$m=$ mass of a charge

$q=$ magnitude of a charge

Putting respective values for electron:

$\tau=\frac{9.1\times 10^{-31}}{2.45\times 10^{-8}\times 18\times 10^{22}\times (1.6\times 10^{-19})^2}$

$\tau=8.0605\times 10^{-9}\ s$

Mean free path is given as:

$l=v_a.\tau$

where:

$l=$ mean free path

$v_a=$ average velocity of electrons

• Now we have the the general value of average velocity of electrons at room temperature:

$v_a=120000\ m.s^{-1}$

So,

$l=120000\times 8.0605\times 10^{-9}$

$l=9.6726\times 10^{-4}\ m$