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

Explanation

`Volume =A(Deltax)=A v_(d)(Deltat)`
Let .n, be the number of free electrons per unit volume (number density of free electrons ) in the material ,then are n `v_(d)(Deltat)` A electrons,A-area of cross section and `v_(d)`-drift velocity.
The amonunt of charge crossing the atea A to the left time `Deltat` is ,
I`(Deltat)=neAv_(d)(Deltat)`
Current :`I=neAv_(d)`
Magnitude of drift velocity of electrons is
`v_(d)=((etau)/(m))E` ...(1)
e-magnitude of electron charge `tau`-relaxation time
E-electric field m-mass of electron
Substituting `v_(d)` from eq. (1)
`I=neA((etau)/(m))E=((ne^(2)tauA)/(m))E`
Thus, current density ,j `(I)/(A)=(("ne"^(2)tau)/(m))E` ....(2)
But j=`sigmaE` ...(3)
Thus ,from eq. (2) and eq. (3),conductivity ,`sigma=("ne"^(2)tau)/(m)`