How does the drift velocity of electrons in a conductor depend on the temperature?

I have two contradicting views for this.

  • First, we can say that increasing the temperature of the conductor will increase the kinetic energy of the electrons. Hence, their drift velocity should increase with increase in temperature.
  • Or, from the relation $v_d = \frac{eE}{m}T$ ($T$ is the relaxation time) we can say that the drift velocity is directly proportional to the relaxation time. Increasing the temperature will obviously decrease the relaxation time - as collisions will become more frequent - and thus decrease the drift velocity. Hence, an increase in the temperature will cause a decrease in the drift velocity.

So which view is correct?


4 Answers 4


Drift Velocity - It is defined as the velocity gained by free electron of conductor in the opposite direction of applied electric field

We know that:-

1- E = F/q = F/e (here, the charge is electron)

Where, E=electric field, F=Force applied, e=charge on the electron

2-F= mass × acceleration = m × a


F= m × a

here, m= mass of electron

a = F/m

a= eE/m


V= u + at (kinematics equation)

V(drift)= 0 + (eE/m)t (initially, u=O)

So, Drift Velocity= (eE/m)t

where, t= relaxation time and it is denoted by τ (Pronounced as tau)

  • 3
    $\begingroup$ Hi Venomous Viper, welcome to physics.SE! Please note that this site supports Mathjax, which is an engine that allows you to typeset math using Latex-like commands. Please see here for a tutorial: math.meta.stackexchange.com/q/5020/289977 Let me encourage you to edit your answer using mathjax; it will look better and it'll be easier to read. Cheers! $\endgroup$ Jul 22, 2018 at 22:26

You can think of it in simple terms.
The average kinetic energy of the lattice ions increases as the temperature increases.
Between "collisions" with the metal lattice ions the free electrons are accelerated by the electric field and so increase their velocity along the electric field direction.
However because the lattice ions are vibrating more, that increase in velocity will not be as great, so the drift velocity is less.

In the macroscopic world that decrease in the drift velocity would manifest itself as an increase in the resistance of the (metal) conductor.

Note that the free electrons also increase their kinetic energy due to the increase in temperature but that is an increase in their root mean square velocity which a measure of their random movement throughout the conductor.


Both the cases lead to the same result: decrease in drift velocity!

In the first case the drift velocity is also going to decrease. Increase in KE will increase the speed of the electrons and not their drift velocity. An increase in speed will cause the electron to cover the same distance in a smaller interval of time, hence decreasing the mean relaxation time. This, in turn, decreases the drift velocity in accordance with the equation: $v_d=\frac{eE}{m}T$.


Let us consider the following equation:


where $i$ is the current passing through the material, $e$ the amount of charge on the charge carrier (here it's the charge on an electron), $v_d$ is the drift velocity, $A$ is the area of cross section of the material and $n$ is the number of charge carriers per unit volume of the material.

For the sake of simplicity, let's assume the material's dimension doesn't change on increasing the temperature. So $A$ is a constant. Also $e$ is invariant of temperature. In a conductor, we know that on increasing the temperature, the current passing through it decreases when the terminals are maintained at a constant potential difference. This is because its resistance increases.

Let's replace the constant terms in the above equation by $k$:


As $i$ decreases, the product $v_dn$ must also decrease to maintain the equality. In a conductor, the value of $n$ remains fairly constant and hence the decrease in current is due to decrease in the drift velocity. And hence the drift velocity decreases with increase in temperature. This is just a reverse reasoning based on our observations from experiments.

An interesting note on semiconductors: As we increase the temperature, the value of $n$ increases in a semiconductor. Even if $v_d$ decreases with increase in temperature, the increase in $n$ outshines it and hence the product $v_dn$ increases on the right hand side and finally the current increases with increase in temperature.

For further reading: Doris Jeanne Wagner and Rensselaer Polytechnic Institute


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.