Question

The area of cross section, length, and density of a piece of metal of atomic weight 60 are ${10}^{-6}{m}^2$, 1 m, and $5\times {10}^3$kgm${}^{-3}$, respectively. If every atom contributes one free electron, find the drift velocity (in mms${}^{-1}$) of electrons in the metal when a current of 16 A passes through it. Avogadro's number is $N_A=6\times {10}^{23}/$mol and charge on an electron is $e=1.6\times {10}^{-19}C$

Answer 1

Given : $M=0.060$ kg $A={10}^{-6}{m}^2$ $\rho =5000$ $kg/m^3$ $I=16$ A

Number of atoms per unit volume $\frac{n}{V}=\frac{m{N}_A}{VM}=\frac{\rho N_A}{M}$

$\therefore$ $\frac{n}{V}=\frac{5000\times 6\times {10}^{23}}{0.060}=5\times {10}^{23}$ atoms per unit volume

As one Cu atom contributes one free electron, thus electron density $n_e=n=5\times {10}^{28}$ electrons per unit volume

Current density $J=\frac{I}{A}=\frac{16}{{10}^{-6}}=1.6\times {10}^7$ $A/m^2$

Drift velocity $v_d=\frac{J}{e{n}_e}=\frac{1.6\times {10}^7}{1.6\times {10}^{-19}\times 5\times {10}^{28}}=0.002$ m/s $=2$ mm/s