Excess electrons are placed on a small lead sphere withmass8.00g so that its net charge is$\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{C}$. (a) Find thenumber of excess electrons on the sphere. (b) How many excesselectrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol.

Electric charge can exist only as an integral multiple of charge on an electron.The formula of the charge is given by,

$\mathit{Q}\mathbf{=}\mathbf{\pm }\mathit{N}\mathit{e}$

Here, Nis an integer, and eis the charge of the electron.

(a)

The charge is anumerical number of electron charge.So,

$$\begin{gathered} Q=-Ne \\ N=-\frac{Q}{e} \end{gathered}$$

Substitute the charge,$Q=3.2\times {10}^{-9}\mathrm{C}$ and$e=-1.6\times {10}^{-19}\mathrm{C}/\mathrm{electron}$ in the equation to get the number of excess electrons.

$$\begin{gathered} \mathrm{N}=\frac{-3.2\times {10}^{-9}\mathrm{C}}{-1.6\times {10}^{-19}\mathrm{C}/\mathrm{electron}} \\ =2\times {10}^{10}\mathrm{electron} \end{gathered}$$

Hence, the number of excess electrons on the sphere are $2\times {10}^{10}$electron.

(b)

First, calculate the number of moles. The atomic mass of lead is$m_A=207\times {10}^{-3}\mathrm{kg}/\mathrm{mole}$ and the number of moles is, $m_b=8\times {10}^{-3}\mathrm{kg}$.

$$\begin{gathered} k=\frac{m_b}{m_A} \\ =\frac{8\times {10}^{-3}\mathrm{kg}}{207\times {10}^{-3}\mathrm{kg}/\mathrm{mole}} \\ =0.0386\mathrm{mole} \end{gathered}$$

Then, the number of lead atoms is given by $N_l=k{N}_A$.

$$\begin{gathered} N_l=0.0386\mathrm{mole}\times 6.023\times {10}^{23}\mathrm{atom}/\mathrm{mole} \\ =2.32\times {10}^{22}\mathrm{atom} \end{gathered}$$

Substitute the known values in the ratio$N_{\epsilon /a}=\frac{N}{N_l}$ to get the number of exceed electron per lead atom.

$$\begin{gathered} N_{\epsilon /a}=\frac{2\times {10}^{10}\mathrm{electron}}{2.23\times {10}^{22}\mathrm{atom}} \\ =8.6\times {10}^{-13}\mathrm{electron}/\mathrm{atom} \end{gathered}$$

Hence, there are$8.6\times {10}^{-13}\mathrm{electron}/\mathrm{atom}$ excess electrons per lead atom.