Question

On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?

Answer 1

Free electrons are in continuous random motion. They undergo change in direction at each collision and the thermal velocities are randomly distributed in all directions.

∴ Average thermal velocity

`u=(u_1+u_2...u_n)/n = 0 `

The electric field E exerts an electrostatic force ‘−Ee’

The acceleration of each electron is

`veca=(-evecE)/m `

Here,

m → Mass of an electron

e → Charge on an electron

Drift velocity is given by

`vecv_d=(v_1+v_2+....+v_n)/n`

`vec(V_d)=((vecu_1+vecat_1)+(vecu_2+vecat_2)+...+(vecu_n_vecat_n))/n`

Here,

`vecu_1,vecu_2->` Thermal velocities of the electrons

`vecatau_1,vectau_2` Velocities acquired by electrons

τ₁, τ₂ → Time elapsed after the collision

`vecv_d=(vecu_1+vecu_2+...vecu_n)/n+(veca(t_1+t_2+...t_n))/n`

Since`(vecu_1+vecu_2+...+u_n)/n=0`

∴ vd = a τ .....(3)

Here `t=(t_1+t_2+t_3...+t_n)/n` is the average time elapsed.

Substituting for a from equation (2),

`vec(V_d)=(-evecE)/mt...(4)`

Because of the external electric field, electrons are accelerated. They move from one place to another and current is produced.
For small intervals dt, we have

Idt = −q

Here, q is the total c

(i) The average velocity of all the free electrons in the conductor with which they get drifted towards the positive end of the conductor under the influence of an external electric field is called the drift velocity.

 (ii) Free electrons are in continuous random motion. They undergo a change in direction at each collision and the thermal velocities are randomly distributed in all directions.

∴ Average thermal velocity

`u=(u_1+u_2...u_n)/n = 0 `

The electric field E exerts an electrostatic force ‘−Ee’

The acceleration of each electron is

`veca=(-evecE)/m `

Here,

m → Mass of an electron

e → Charge on an electron

Drift velocity is given by

`vecv_d=(v_1+v_2+....+v_n)/n`

`vec(V_d)=((vecu_1+vecat_1)+(vecu_2+vecat_2)+...+(vecu_n_vecat_n))/n`

Here,

`vecu_1,vecu_2->` Thermal velocities of the electrons

`vecatau_1,vecatau_2->` Velocities acquired by electrons

τ1, τ2 → Time elapsed after the collision

`vecv_d=(vecu_1+vecu_2+...vecu_n)/n+(veca(t_1+t_2+...t_n))/n`

Since`(vecu_1+vecu_2+...+u_n)/n=0`

∴ vd = a τ .....(3)

Here `t=(t_1+t_2+t_3...+t_n)/n`is the average time elapsed.

Substituting for a from equation (2),

`vec(V_d)=(-evecE)/mt...(4)`

Because of the external electric field, electrons are accelerated. They move from one place to another and current is produced.
For small intervals dt, we have

Idt = −q

Here, q is the total charge flowing.

Let n be the free electrons per unit area. Then, the total charge crossing area A in time dt is given by
Idt = neAvddt

Substituting the value of vd, we obtain

`Idt= n eA(-eE/m) dt`

 I/A = J

 Here, J is the current density.

`|J|=n e^2/m |E|T`

From Ohm’s law,

J = σE

Here, σ is the conductivity of the material through which the current is flowing.

Thus,

`sigma =(n e^2)/mT`

`sigma=1/rho`

or

`rho=1/sigma`

Substituting the value of conductivity, we obtain

`rho=m/(n e^2T)`

Here, τ is the relaxation time.