Find the relation between drift velocity and relaxation time of carriers in a conductor. A conductor of length L is connected to a d.c.. The source of emf ‘E’. If the length of the conductor is tripled by stretching it, keeping ‘E’ constant, explain how its drift velocity would be affected.

Hint: Average relaxation time is the time between two successive collisions of the electrons in a conductor and drift velocity is the average velocity with which the electrons get drifted towards the positive terminal of the conductor in an electric field.

Complete step-by-step solution:

At any instant of time, the velocity acquired by the electrons with thermal velocity \[{{u}_{1}}\]and acceleration \[a\] is:
\[{{v}_{1}}={{u}_{1}}+a{{\tau }_{1}}\]
Average velocity \[{{v}_{d}}\] of all the electrons in the conductor would be the sum of averages of each term. The sum of averages of the thermal velocities of \[n\] electrons in the conductor is 0.
Hence,
\[{{v}_{d}}=0+a{{\tau }_{{}}}\]
Here
\[\tau =\dfrac{{{\tau }_{1}}+{{\tau }_{2}}+.....+{{\tau }_{n}}}{n}\]
The acceleration of an electron placed in an electric field is,
\[a=\dfrac{F}{m}=-\dfrac{eE}{M}\]
Substituting this in the formula of \[{{v}_{d}}\], we get
\[{{v}_{d}}=-\dfrac{eE}{M}\tau \]

For second part of the question, we know that the drift velocity is inversely proportional to the distance of the conductor:
\[{{v}_{d}}\propto \dfrac{1}{l}\]
Therefore, the correct answer is when the length of the conductor is tripled the drift velocity becomes one third.

Note: Students must remember that the negative sign in drift velocity formula shows that the drift velocity is in the opposite direction of the applied electric field. Drift velocity of electrons is of the order of \[{{10}^{-4}}m{{s}^{-1}}\] and the order of average relaxation time is \[{{10}^{-14}}s\].