Given: mass number of gold - 197, density of gold - 19.7 g cm^-3, Avogadro's number - 6 x 10²³. The radius of the gold atom is approximately:

Volume occupied by one mole of gold$=\frac{197\mathrm{g}}{19.7\mathrm{g}{\mathrm{cm}}^{-3}}=10{\mathrm{cm}}^3$

Volume of one atom$=\frac{10}{6\times {10}^{23}}=\frac{5}{3}\times {10}^{-23}{\mathrm{cm}}^3=\frac{5}{3}\times {10}^{-29}{m}^3$

Let r be the radius of the atom. Therefore,

$$\begin{gathered} \frac{4}{3}\pi r^3=\frac{5}{3}\times {10}^{-29} \\ \Rightarrow r\cong 1.5\times {10}^{-10}\mathrm{m} \end{gathered}$$