Physics - Current Electricity Question with Solution | TestHub

The density of copper,

$\rho =9\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}=9\times {10}^6 \mathrm{g} {\mathrm{m}}^{-3}$

Avogadro number,$N_A=6.023\times {10}^{23}$

Mass of 1 mole of the copper atom,$M=63.5 \mathrm{g}$

Thus, the number of free electrons per volume is

$$\begin{gathered} n=\frac{N_A}{M}\rho =\frac{6.023\times {10}^{23}}{63.5}\times 9\times {10}^6 \\ =8.5\times {10}^{28} {\mathrm{m}}^{-3} \end{gathered}$$