Question

Consider Aluminium at room temperature. Its electron density is n = 18 x 10^22 cm-3 and its electrical resistivity is rho = 2.45 μΩ * cm.
a. Find its electron relaxation time τ and electron mean free path in the Drude model.

Answer 1

The electron relaxation time is $8.0605 \times 10^{-7} s$

The mean free path is $9.67260 \times 10^{-2} m$

Explanation:

Electron density, $n=18 \times 10^{22} \mathrm{cm}^{-3}=18 \times 10^{20} \mathrm{m}^{-3}$

Aluminium resistivity,  $\rho=2.45 \mu \Omega . c m$

$\rho=2.45 \times 10^{-6} \times 10^{-2}=2.45 \times 10^{-8} \Omega . m$

From the Drude's model we have:

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

Where:

τ= Electron relaxation time

m = mass of a charge

q = magnitude of a charge

We know, electron mass = $9.1 \times 10^{-31} \mathrm{kg}$

Charge of electron = $1.6 \times 10^{-19} \mathrm{C}$

By substituting all given values for electron, we get

$\tau=\frac{9.1 \times 10^{-31}}{18 \times 10^{20} \times\left(1.6 \times 10^{-19}\right)^{2} \times 2.45 \times 10^{-8}}$

$\tau=\frac{9.1 \times 10^{-31}}{112.896 \times 10^{20-38-8}}$

$\tau=\left(\frac{9.1}{112.896}\right) \times 10^{-31+26}=0.080605 \times 10^{-5}$

When multiply by 100 and divide by 100, we get

$\tau=8.0605 \times 10^{-7} \mathrm{s}$

Mean free path is given as:

$l=v_{a} \times \tau$

where:

l = Mean free path

$v_{a}$ = Average velocity of electrons

We know the general value for average velocity of electrons at room temperature:

$v_{a}=120000 \mathrm{m} / \mathrm{s}$

Therefore,

$l=120000 \times 8.0605 \times 10^{-7}=967260 \times 10^{-7} \mathrm{m}=9.6726 \times 10^{-2} \mathrm{m}$