Chemistry - Solid State Question with Solution | TestHub

Packing fraction in hcp$=\frac{6\times \frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}$

$=\frac{6\times 4\mathrm{\pi }{\mathrm{r}}^3}{3\times \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\times \mathrm{H}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}$

Area (of base) of unit cell in hcp$=6\sqrt{3}{\mathrm{r}}^2$(follow text)

height of unit cell in hcp$=4\mathrm{r}\times \sqrt{\frac{2}{3}}$(follow text)

So packing fraction$=\frac{6\times 4\mathrm{\pi }{\mathrm{r}}^3}{3\times 6\sqrt{3}{\mathrm{r}}^2\times 4\mathrm{r}\times \sqrt{\frac{2}{3}}}$

$=\frac{\mathrm{\pi }}{3\sqrt{2}}=0.74$

or = 74%

Thus, empty space in hcp = 1 - 0.74 = 0.26

or = 26%.