The mass number of atom of gold is 197 and the density is 19.7 gm per cm³ . The radius of the gold is approximately (given that N = 6´10²³ )

 

Atomic wt of ‘Au’ = 197

Gram atomic weight of ‘Au’ = 197g 6.023´10²³ atoms of 'Au' ® 197 g 1atom of Au ® ?

$=\frac{197}{6.023\mathrm{X}{10}^{23}}=\frac{197}{6.023}\mathrm{X}{10}^{-23}\mathrm{g}$

$=32.70795\mathrm{X}{10}^{-23}\mathrm{g}$

$\mathrm{density}=\frac{\mathrm{mass}}{\mathrm{volume}}\Rightarrow \mathrm{volume}=\frac{\mathrm{mass}}{\mathrm{density}}$

$\mathrm{density} \left(\mathrm{d}\right)=19.7 \mathrm{g}/{\mathrm{cm}}^3$

$\mathrm{m}=32.70795\mathrm{X}{10}^{-23}\mathrm{g}$

$\therefore \mathrm{v}=\frac{\mathrm{m}}{\mathrm{d}}=\frac{32.70795\mathrm{X}{10}^{-23}}{19.7}$

$=1.66\mathrm{X}{10}^{-23}{\mathrm{cm}}^3$

$\mathrm{v}=\frac{4}{3}{\mathrm{\pi r}}^3$

${\mathrm{r}}^3=\frac{3}{4}\mathrm{X}\frac{1.66\mathrm{X}{10}^{-23}}{3.14}$

${\mathrm{r}}^3=\frac{4.98}{12.56}\mathrm{X}{10}^{-23}$

${\mathrm{r}}^3=0.3964\mathrm{X}{10}^{-23} =3.964\mathrm{X}{10}^{-24}$

$\mathrm{r}=\sqrt[3]{3.964\mathrm{X}{10}^{-24}}=1.58\mathrm{x}{10}^{-8}\mathrm{cm}$

$\therefore \mathrm{r}=1.58\mathrm{X}{10}^{-10}\mathrm{m}$