Question

(i) Derive an expression for drift velocity of free electrons.

(ii) How does the drift velocity of electrons in a metallic conductor vary with an increase in temperature ? Explain.

Answer 1

(i) Consider a conductor in which an electric field E is produced. Let a free electron experience a force (-eE) in this electric field. So, the acceleration of free electron is

$a=F/m=-eE/m....\left(i\right)$


Here, e= Charge on electron.
m= Mass of an electron.

So, the final velocity of the free electron in time interval $t_1$ is,
$v_1=u_1+a{t}_1$

For n free electrons, the final velocities be $v_2,v_3,...,v_n.$
So, the average velocity of the free electrons or the drift velocity
$v_d=\frac{(v_1+v_2+v_3+....+v_n)}{n}$

$or{v}_d=\frac{(u_1+a{t}_1+u_2+a{t}_2+....+u_n+a{t}_n)}{n}$

$or{v}_d=\frac{\left[\right(u_1+u_2++....+u_n)+(a{t}_1++a{t}_2+....+a{t}_n\left)\right]}{n}$

$or{v}_d=\frac{\left[\right(u_1+u_2++....+u_n)+a(t_1++t_2+....+t_n\left)\right]}{n}$

But,$\frac{(u_1+u_2++....+u_n)}{n}=$average initial velocity of free electorns=0.

and $\frac{(t_1+t_2++....+t_n)}{n}=$average time taken between two consecutive collision=$\tau$

Where $\tau$ is relaxation time.

So, $v_d=a\tau$

or, $v_d=\frac{-e\underset{E}{\to }}{m}\tau$ [From(i)]

This is the required relation.

(ii) The drift velocity of free electrons in a metallic conductor decreases with an increase in temperature.The reason being, if we increase the temperature of the metallic conductor, the collisions between the electrons and ions increase, which in turn leads to a decrease in the relaxation time. Hence, drift velocity decreases.