Answer
The ion drift speed is $~~0.289~m/s$
Work Step by Step
We can find the magnitude of the current density:
$J = \sigma~E$
$J = (2.70\times 10^{-14}~(\Omega\cdot m)^{-1})(120~V/m)$
$J = 3.24\times 10^{-12}~A/m^2$
We can find the density of the net charge carriers:
$620~cm^{-3} - 550~cm^{-3} = 70~cm^{-3}$
We can find the ion drift speed:
$J = (ne)~v_d$
$v_d = \frac{J}{ne}$
$v_d = \frac{3.24\times 10^{-12}~A/m^2}{(70\times 10^6/m^3)(1.6\times 10^{-19}~C)}$
$v_d = 0.289~m/s$
The ion drift speed is $~~0.289~m/s$
