Answer
(a) $v_d = 7.4\times 10^{-6}~m/s$
(b) $\tau = 2.1\times 10^{-14}~s$
Work Step by Step
(a) We can find the drift speed:
$v_d = \frac{i_e}{n_e~A}$
$v_d = \frac{i_e}{n_e~\pi~r^2}$
$v_d = \frac{3.5\times 10^{17}~electrons/s}{(6.0\times 10^{28}~electrons/m^3)~(\pi)~(0.5\times 10^{-3}~m)^2}$
$v_d = 7.4\times 10^{-6}~m/s$
(b) We can find the mean time between collisions:
$v_d = \frac{e~\tau~E}{m}$
$\tau = \frac{v_d~m}{e~E}$
$\tau = \frac{(7.4\times 10^{-6}~m/s)(9.1\times 10^{-31}~kg)}{(1.6\times 10^{-19}~C)(2.0\times 10^{-3}~V/m)}$
$\tau = 2.1\times 10^{-14}~s$
