QJEEAdv08Chc132: JEE Advanced 2008 Chemistry - Class 12 JEEAdv08Chc132 Solution: The volume of this HCP unit cell is :

Volume of unit cell = Area of base $$\times$$ Height of the unit cell

The (HCP) is a hexagonal prism structure. The base of (HCP) is a hexagon. We know that the hexagon is made of the 6 equilateral triangles.

Area of equilateral triangle $$=\frac{\sqrt{3}}{4}{a}^2$$

Therefore, the area of the base is given as,

Area of base $$=6\times \frac{\sqrt{3}}{4}{a}^2$$

Let's the 'h' be the height of the hexagonal unit

Therefore, we have

$$\frac{h}{a}=\sqrt{\frac{8}{3}}$$

Now, substitute these values in the volumes of (HCP) we have,

Volume of HCP = Area of base $$\times$$ height of the hexagon

$$\therefore$$ Volume of HCP $$=6\times \frac{\sqrt{3}}{4}{a}^2\times \sqrt{\frac{8}{3}}a$$

Since, spheres touch each other, the edge length 'a' is equal to twice the radius of the atom. That is a = 2r

Substitute these values in the volume we get,

Volume of HCP $$=6\times \frac{\sqrt{3}}{4}{a}^2\times \sqrt{\frac{8}{3}}a$$

$$=6\times \frac{\sqrt{3}}{4}(2r{)}^2\times \frac{\sqrt{8}}{\sqrt{2}}2r=24\sqrt{2}{r}^3$$