Question 1

Find electron density ($n$): Number of atoms per unit volume $n = \frac{\rho N_A}{M}$. Given $\rho = 6.35 \times 10^3\text{ kg/m}^3$, $M = 63.5\text{ g/mol} = 63.5 \times 10^{-3}\text{ kg/mol}$, $N_A = 6 \times 10^{23}$. $n = \frac{6.35 \times 10^3 \times 6 \times 10^{23}}{63.5 \times 10^{-3}} = 6 \times 10^{28}\text{ m}^{-3}$ Since there is 1 conduction electron per atom, $n = 6 \times 10^{28}\text{ electrons/m}^3$.

Find resistance of the wire ($R_w$): $R_w = \frac{l}{\sigma A} = \frac{100}{2 \times 10^8 \times 0.5 \times 10^{-6}} = \frac{100}{100} = 1\text{ }\Omega$.

Find current ($I$): Total resistance $R_{eq} = R_w + r = 1 + 1 = 2\text{ }\Omega$. $I = \frac{E}{R_{eq}} = \frac{2}{2} = 1\text{ A}$.

Find drift velocity ($v_d$): $I = nAev_d \implies v_d = \frac{I}{nAe}$ $v_d = \frac{1}{(6 \times 10^{28}) \times (0.5 \times 10^{-6}) \times (1.6 \times 10^{-19})} = \frac{1}{4.8 \times 10^3}\text{ m/s}$ $v_d \approx 0.208 \times 10^{-3}\text{ m/s} = 0.208\text{ mm/s}$.