A metal wire of cross-sectional area 0.5 mm ^2 and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1 . The density, atomic mass and electrical conductivity of the metal are 6.35 10^3 kg m ⁻³ , 63.5 gm/mole and 2 10^8 mho m ⁻¹ , respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s ⁻¹ ) of the electrons in the wire is: [Take Avogadro's number

Step-by-step solution

The number density of conduction electrons n is given by the number of atoms per unit volume, since there is one conduction electron per atom: n = d N_A M Substituting the given values: n = 6.35 10^3 6 10²³ 63.5 10⁻³ = 6 10²⁸ m ⁻³ The resistance of the wire R is: R = L A R = 100 2 10^8 0.5 10⁻⁶ = 100 10^2 = 1 The total resistance of the circuit is R_ total = R + r = 1 + 1 = 2 . The current in the circuit is: I = E R_ total = 2 2 = 1 A The drift velocity v_d is given by the relation I = n e A v_d : v_d = I n e A v_d