Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.

#### Solution

If there are *N* electrons and the velocity of the *i*^{th} electron at a given time is *v*_{i} where, *i* = (1, 2, 3, …N), then

`1/N sum_(i-1)^N V_1 = 0` (If there is no external field)

When an external electric field is present, the electrons will be accelerated due to this field by

`veca = (-evecE)/m`

Where,

− *e* = Negative charge of the electron

*E* = External field

*m* = Mass of an electron

Let *v*_{i} be the velocity immediately after the last collision after which external field was experienced by the electron.

If *v*_{i} is the velocity at any time *t*, then from the equation *V* = *u* + *at*, we obtain

`vecV_i = vecv_i - (evecE)/m t ........ (1)`

For all the electrons in the conductor, average value of *v*_{i} is zero.

The average of *v*_{i} is *v*_{d} or drift velocity.

This is the average velocity experienced by an electron in an external electric field.

There is no fixed time after which each collision occurs. Therefore, we take the average time after which one collision takes place by an electron.

Let this time, also known as relaxation time, be*τ**.* Substituting this in equation (i)

`vecv_i - o`

`t = tau`

`vecV_i = vecv_d`

Then,

`vecv_d = (-evecE)/m`

Negative sign shows that electrons drift opposite to the applied field.