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

Explanation

Let .n. be the number density of electrons , .L. be the length of the conductora and .A. be the area of cross -section of the conductor .
Let `v_d` be the drift velocity of the electrons and Dx be a small length .
Electrons drift in a direction opposite to the electric field . Number of electrons in .Dx. and area of cross -section .A. is equal to (ADx) n.
Charge on the these electrons =(nADx)e.
If Dt is the time taken for effective displacement of electrons then rate of flow charge = `nAe ((Deltax)/(Deltat))` .
By definition, electric current I=rate of flow of charge
i.e., `I=nAev_d`...(1)
where , average velocity of electrons with which it drifts against the direction of electric field is known as drift velocity `(v_d)`.
Let .a. be the acceleration of electrons . Let .E. be the electric field intensity . Force on electrons ,
F=ma
i.e., eE=ma
But = `a=v_dt` where .t. is the relaxation time .
Relaxation time represents the average time taken for two successive collisions of electrons and ions in the lattice.

Hence , `eE=mv_dt` or `v_d=(eE)/(mtau)` ...(2)
We also know that electric potential difference between the ends of the conductor V=EL
`v_d =(eV)/(mtauL)` substituting this is the expression (1)
We write , `I=nAe ((eV)/(mtauL))`
i.e., `I=((nAe^2)/(mtauL))V`
`R=(mtau L)/(nAe^2)` is called the electrical resistance of a conductor . and `K=((nAe^2)/(mtauL))` is called the electric conductance of a conductor . the expression `sigma =("ne"^2)/(mtau)` is called the electrical conductivity and `1/sigma=(mtau)/("ne"^2)` is called electrical resistivity .