Answer:
To find the drift velocity, we can use the equation:I = nAvqWhere I is the current, n is the number of charge carriers per unit volume, A is the cross-sectional area of the wire, v is the drift velocity of the charge carriers, and q is the charge of an electron.We are given the electric field E as 4.5×10^−4 V/m. The current I is 2.8×10^17 electrons/s. The diameter of the wire is 1.9 mm, so the radius r is 0.95 mm or 0.00095 m. The cross-sectional area A is then πr^2 = 2.83×10^-6 m^2.The charge of an electron is q = 1.6×10^-19 C.We can rearrange the equation to solve for v:v = I/(nAq)To find n, we can use the density of aluminum and its atomic weight to calculate the number of atoms per unit volume, and then multiply by the number of electrons per atom:n = (density of aluminum * Avogadro's number) / (atomic weight of aluminum * volume per atom) * number of electrons per atomPlugging in the values, we get:n = (2.7 g/cm^3 * 6.02×10^23) / (26.98 g/mol * 4.05×10^-23 m^3/mol) * 3 = 1.85×10^29 electrons/m^3Now we can calculate the drift velocity:v = I/(nAq) = (2.8×10^17 electrons/s) / (1.85×10^29 electrons/m^3 * 2.83×10^-6 m^2 * 1.6×10^-19 C/electron) = 0.0025 m/sSo the drift velocity is 0.0025 m/s.To find the mean time between collisions, we can use the equation:τ = mv/eEwhere τ is the mean time between collisions, m is the mass of an electron, v is the drift velocity, e is the charge of an electron, and E is the electric field.The mass of an electron is 9.11×10^-31 kg. Plugging in the values, we get:τ = (9.11×10^-31 kg) * (0.0025 m/s) / (1.6×10^-19 C) / (4.5×10^-4 V/m) = 3.54×10^-14 sSo the mean time between collisions is 3.54×10^-14 s.