Chemistry - Solid State Question with Solution | TestHub

The height of the$\mathrm{H}\mathrm{C}\mathrm{P}$unit cell$=4\sqrt{\frac{2}{3}}.\mathrm{r}=5.715$

$\text{r}=\frac{5.715\times \sqrt{3}}{\sqrt{2}\times 4}=\mathrm{1.75 }\text{Å}$

The base area of the unit cell$=6\sqrt{3}{\mathrm{r}}^2$

The volume of the unit cell$=\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{ }\mathrm{ }\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{ }\times$$\text{height}=24\sqrt{2}{\text{r}}^3=181.8$${\text{Å}}^3$