A copper atom consists of copper nucleus surrounded by 29 electrons. The atomic weight of copper is 63.5gmol^-1. Let us now take two pieces of copper each weighing 10 g. Let us transfer one electron from one piece to another for every 1000 atoms in that piece The coulomb force between the two pieces after transfer of electrons if they are 1 cm apart is $p$$\times$10¹⁶ N. Find $p$. (Avogadro number N_A = 6 x 10²³ mole^-1, charge on an electron = -1.6 x 10^-19coulomb)
 

63.5 g copper contains N_A =6 X10²³ copper atoms. Therefore number of copper atoms in 10 g copper is $\mathrm{N}=\frac{6\times {10}^{23}}{63.5}\times 10$
Since only one electron is transferred for every 1000 atoms, therefore the number of electrons transferred, $\mathrm{n}=\frac{6\times {10}^{23}\times 10}{63.5\times 1000}$
Magnitude of charge q is given by $\mathrm{q}=\mathrm{ne}=\frac{6\times {10}^{23}\times 10}{63.5\times 1000}\times 1.6\times {10}^{-19}\mathrm{C}$
$\Rightarrow \mathrm{q}=\frac{1920}{127}\mathrm{C}$
Separation between pieces is $\mathrm{r}=1\mathrm{cm}={10}^{-2}\mathrm{m}$
One piece of copper has positive charge and the other negative charge, so force of attraction between the pieces is $\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$
$$\begin{gathered} \Rightarrow \mathrm{F}=\frac{9\times {10}^9\left(\frac{1920}{127}\right)\times \left(\frac{1920}{127}\right)}{{\left({10}^{-2}\right)}^2}\mathrm{N} \\ \Rightarrow \mathrm{F}=2.057\times {10}^{16}\mathrm{N} \end{gathered}$$
So, x = 2.06.