Respuesta :
The electron relaxation time is [tex]8.0605 \times 10^{-7} s[/tex]
The mean free path is [tex]9.67260 \times 10^{-2} m[/tex]
Explanation:
Electron density, [tex]n=18 \times 10^{22} \mathrm{cm}^{-3}=18 \times 10^{20} \mathrm{m}^{-3}[/tex]
Aluminium resistivity, [tex]\rho=2.45 \mu \Omega . c m[/tex]
[tex]\rho=2.45 \times 10^{-6} \times 10^{-2}=2.45 \times 10^{-8} \Omega . m[/tex]
From the Drude's model we have:
[tex]\tau=\frac{m}{n \times q^{2} \times \rho}[/tex]
Where:
τ= Electron relaxation time
m = mass of a charge
q = magnitude of a charge
We know, electron mass = [tex]9.1 \times 10^{-31} \mathrm{kg}[/tex]
Charge of electron = [tex]1.6 \times 10^{-19} \mathrm{C}[/tex]
By substituting all given values for electron, we get
[tex]\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}}[/tex]
[tex]\tau=\frac{9.1 \times 10^{-31}}{112.896 \times 10^{20-38-8}}[/tex]
[tex]\tau=\left(\frac{9.1}{112.896}\right) \times 10^{-31+26}=0.080605 \times 10^{-5}[/tex]
When multiply by 100 and divide by 100, we get
[tex]\tau=8.0605 \times 10^{-7} \mathrm{s}[/tex]
Mean free path is given as:
[tex]l=v_{a} \times \tau[/tex]
where:
l = Mean free path
[tex]v_{a}[/tex] = Average velocity of electrons
We know the general value for average velocity of electrons at room temperature:
[tex]v_{a}=120000 \mathrm{m} / \mathrm{s}[/tex]
Therefore,
[tex]l=120000 \times 8.0605 \times 10^{-7}=967260 \times 10^{-7} \mathrm{m}=9.6726 \times 10^{-2} \mathrm{m}[/tex]
