Step-by-step Derivation of the Volume of an HCP Unit Cell

1. Understanding the Geometry of the HCP Unit Cell The hexagonal close-packed (HCP) unit cell is a hexagonal prism. It consists of two hexagonal layers (top and bottom) and a middle layer of three atoms sandwiched between them. The atoms are considered to be spheres of radius 'r'.

2. Determining the Relationship between Unit Cell Edge Length 'a' and Atomic Radius 'r' In the hexagonal layers, the atoms are close-packed. Each atom at a corner touches its neighbors. The distance between the centers of two adjacent atoms is equal to the sum of their radii, which is r + r = 2r. This distance corresponds to the edge length 'a' of the hexagonal base. $a = 2r$

3. Calculating the Area of the Hexagonal Base The base of the unit cell is a regular hexagon with side length 'a'. A regular hexagon can be divided into six identical equilateral triangles, each with a side length of 'a'. The area of one equilateral triangle is given by the formula: $A_{triangle} = \frac{\sqrt{3}}{4} a^2$ The total area of the hexagonal base is six times the area of one triangle: $A_{base} = 6 \times A_{triangle} = 6 \times \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} a^2$ Substituting a = 2r into the equation for the base area: $A_{base} = \frac{3\sqrt{3}}{2} (2r)^2 = \frac{3\sqrt{3}}{2} (4r^2) = 6\sqrt{3} r^2$

4. Calculating the Height 'c' of the Unit Cell The height 'c' of the unit cell is the perpendicular distance between the top and bottom hexagonal layers. The three atoms in the middle layer are situated in the depressions of the bottom layer. Consider an atom from the middle layer resting on three atoms of the bottom layer. The centers of these four touching spheres form a regular tetrahedron with an edge length equal to the distance between the centers of any two touching spheres, which is a = 2r. Let's find the height of this tetrahedron. Consider the base of the tetrahedron, which is an equilateral triangle with side 'a'. The centroid of this base is directly below the apex atom (the one from the middle layer). The distance from any vertex of this base triangle to its centroid is $d = \frac{a}{\sqrt{3}}$. Now, we have a right-angled triangle formed by the apex of the tetrahedron, the centroid of the base, and one vertex of the base. The hypotenuse is the edge of the tetrahedron ('a'), one side is the distance from the vertex to the centroid ('d'), and the other side is half the height of the unit cell (c/2). Using the Pythagorean theorem: $a^2 = d^2 + (\frac{c}{2})^2$ $a^2 = (\frac{a}{\sqrt{3}})^2 + (\frac{c}{2})^2$ $a^2 = \frac{a^2}{3} + \frac{c^2}{4}$ $\frac{c^2}{4} = a^2 - \frac{a^2}{3} = \frac{2a^2}{3}$ $c^2 = \frac{8a^2}{3}$ $c = \sqrt{\frac{8a^2}{3}} = a \sqrt{\frac{8}{3}} = a \frac{2\sqrt{2}}{\sqrt{3}}$ Now, substitute a = 2r: $c = (2r) \frac{2\sqrt{2}}{\sqrt{3}} = \frac{4\sqrt{2}}{\sqrt{3}} r$

5. Calculating the Volume of the HCP Unit Cell The volume of the unit cell is the product of its base area and its height. $V = A_{base} \times c$ $V = (6\sqrt{3} r^2) \times (\frac{4\sqrt{2}}{\sqrt{3}} r)$ The $\sqrt{3}$ terms cancel out: $V = (6 r^2) \times (4\sqrt{2} r)$ $V = 24\sqrt{2} r^3$

6. Conclusion The calculated volume of the HCP unit cell is $24\sqrt{2} r^3$. This matches option A.

Final Answer is option A.