Alternate Answer
$[\text{V}_{d} =\frac{-eE\tau}{\text{m}}$ where $\tau$is the relaxation time.]
We have $\text{E} = - \frac{\text{V}}{\ell'}$
where V is potential difference across the length $\ell$
of the conductor
$\text{V}_{d} = \frac{\text{eV}\tau}{\text{m}\ell}$ Current flowing I = neAvd
I = neA$\text{V}_{d}\frac{\text{eV}\tau}{\text{ml}} = \frac{ne^{2}\text{AV}\tau}{\text{ml}}$
$\frac{\text{I}}{\text{V}} = \frac{\text{ne}^{2}A\tau}{ml} = \frac{1}{\text{R}}$ . . . . . (i)
Also, $\text{R} = \rho\frac{\ell}{\text{A}}$ . . . . (ii)
Comparing (i) and (ii)
$\rho = \frac{m}{ne^{2}\tau}$
Resistivity of the material of a conductor depends on the relaxation time, i.e., temperature and the number density of electrons.
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