$I = nqAv_{d}$

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.