A resistor is cut from n-type silicon in which ne = 1.0 × 10²¹ m⁻³…

Step 1. The field. $E = \dfrac{V}{L} = \dfrac{5.0\ \text{V}}{0.010\ \text{m}} = 500\ \text{V/m}$.

Step 2. The drift speed. $v_d = \mu_e E = 0.135\ \text{m}^2/\text{V s} \times 500\ \text{V/m} = 67.5\ \text{m/s}$.

Step 3. The resistance. The conductivity is $\sigma = e\, n_e \mu_e = 1.602 \times 10^{-19}\ \text{C} \times 1.0 \times 10^{21}\ \text{m}^{-3} \times 0.135\ \text{m}^2/\text{V s} = 21.6\ \text{S/m}$. So $\rho = 0.0462\ \Omega\ \text{m}$ and $R = \dfrac{\rho L}{A} = \dfrac{0.0462\ \Omega\ \text{m} \times 0.010\ \text{m}}{2.0 \times 10^{-6}\ \text{m}^2} = 231\ \Omega$.

Step 4. The current. $I = \dfrac{V}{R} = \dfrac{5.0\ \text{V}}{231\ \Omega} = 21.6$ mA.