Volume of Hexagonal Close Packed Unit Cell

To find the volume of the hexagonal close-packed (HCP) unit cell in terms of the radius $r$ of the spheres:

1. Relation between edge length ($a$) and radius ($r$):
   In a hexagonal close-packed lattice, the spheres on the base touch each other along the edge of the regular hexagon. Therefore, the side length $a$ of the hexagonal base is:
   $a = 2r$

2. Area of the base:
   The base of the HCP unit cell is a regular hexagon with edge length $a$, which is composed of six equilateral triangles of side $a$. The area of the base is:
   $\text{Area of base} = 6 \times \left(\frac{\sqrt{3}}{4} a^2\right) = \frac{3\sqrt{3}}{2} a^2$
   Substituting $a = 2r$:
   $\text{Area of base} = \frac{3\sqrt{3}}{2} (2r)^2 = \frac{3\sqrt{3}}{2} \times 4r^2 = 6\sqrt{3}r^2$

3. Height of the unit cell ($c$):
   The middle layer of three spheres rests in the depressions formed by three touching spheres in the bottom layer. The centers of these four spheres form a regular tetrahedron with edge length $a$.

The distance from a vertex to the base of this regular tetrahedron gives the vertical distance between the bottom layer and the middle layer:
$h = \sqrt{a^2 - \left(\frac{a}{\sqrt{3}}\right)^2} = \sqrt{a^2 - \frac{a^2}{3}} = \sqrt{\frac{2}{3}}\,a$

Since the unit cell spans two such layers (from bottom layer to middle layer, and middle layer to top layer), the total height $c$ of the unit cell is:
$c = 2h = 2\sqrt{\frac{2}{3}}\,a$
Substituting $a = 2r$:
$c = 2\sqrt{\frac{2}{3}}\,(2r) = 4\sqrt{\frac{2}{3}}\,r$

1. Volume of the unit cell ($V$):
   The volume of the prism is given by the base area multiplied by the height:
   $V = \text{Area of base} \times c$
   $V = \left(6\sqrt{3}r^2\right) \times \left(4\sqrt{\frac{2}{3}}\,r\right)$
   $V = 24\sqrt{3} \times \frac{\sqrt{2}}{\sqrt{3}}\,r^3 = 24\sqrt{2}\,r^3$

Thus, the correct option is (A).