Question

Derive an expression for resistivity of a conductor in terms of the number density of charge carriers in the conductor and relaxation time.

Answer 1

The relationship between the relaxation time (τ) and drift velocity ${\displaystyle \left(v_d\right)}$ is given by:

${\displaystyle v_d=-e\left(\frac{E\tau }{m}\right)}$

∴ ${\displaystyle \tau =\frac{\left(v_{d}m\right)}{e}\times E}$

Let L = Length of the conductor

A = Area of the conductor

n = free electron density

e = charge of the electron

E = Electric field

m = mass of the electron

τ = Relaxation time

The current flowing through the conductor is

I = ${\displaystyle neA{v}_d}$

I = ${\displaystyle neA\left(\frac{eE}{m}\right)\tau }$

Also, field E can be expressed as

E = ${\displaystyle \frac{V}{L}}$

The current flowing through the conductor is:

I = ${\displaystyle \frac{n{e}^{2}VA\tau }{mL}}$

or ${\displaystyle \frac{V}{I}=\frac{mL}{n{e}^2\tau A}}$

or ${\displaystyle R=\frac{mL}{n{e}^2\tau A}}$ ...${\displaystyle (\text{from Ohm's law}\frac{V}{I}=R)}$

or ${\displaystyle R=\frac{m}{n{e}^2\tau }\left(\frac{L}{A}\right)}$

Electrical resistivity, ${\displaystyle \rho =\frac{m}{n{e}^2\tau }}$ ${\displaystyle ...[\because R=\rho \frac{L}{A}]}$