$I = nqAv_{d}$
Substituting $V_{d} = \frac{qE}{m} \left( \frac{\lambda}{ < v >} \right)$ into the above eqn,
$I = nqA \left( \frac{qE \lambda}{m < v >} \right)$
Substituting $E = \frac{V}{l}$ into the above eqn,
$I = nq^{2} A \left( \frac{V \lambda}{m l < v >} \right)$, where V is potential difference, l is distance
$I = \left( \frac{nq^{2} A \lambda}{m l < v >} \right) V$
Comparing $I = \left( \frac{1}{R} \right) V$ with the above equation,
$R = \frac{ m < v >}{nq^{2} \lambda} \left( \frac{l}{A} \right)$
Recall that $R = \rho \frac{l}{A}$, hence
$\rho = \frac{ m < v >}{nq^{2} \lambda}$
Drude’s classical theory of the electron gas correctly predicts that the current in a metal obeys Ohm’s law and provides an expression for resistivity that can be tested by experiment.
