Derive `sigma = (n e^(2) tau)/(m)`
where the symbols have their usual meaning.

Explanation

It is the time interval between two successive collisions of electron with the vibrating atoms of a current carrying conductor.
We know that the relation between electric current (I) and drift velocity `(V_d)`.
`I=n eAV_d`….(1)
Expression for drift velocity in terms of electric field and relaxation time is given by
`V_d=(eEtau)/m`....(2)
Substituting (2) in (1) `rArr I=(n eAeEtau)/m`
`rArr I=(nAe^2 Etau)/m`
But `E=V/I` ,
Where `V to` Potential across the conductor
l `to` length of the conductor
`I=(nAe^2 Vtau)/(ml)`
`I/V=(nAe^2 tau)/(ml)`
`V/I=(ml)/(nAe^2tau) rArr R=(m/(nAe^2tau))l`
`R=(m/(n e^2tau))l/A`
`R=(rhol)/A`
Here `rho=m/(n e^2tau)`
The conductivity of the material
`sigma=1/rho=1/(m/(n e^2tau))`
`sigma=(n e^2tau)/m`
Mobility : It is drift velocity acquired by free electrons per unit electric field .
i.e., Mobility , `mu=V_d/E`