Question

The volume of the HCP unit cell is

• A. $24\sqrt{2}{r}^3$
• B. $\frac{16}{\sqrt{2}}{r}^3$
• C. $\frac{12}{\sqrt{2}}{r}^3$
• D. $\frac{64}{3\sqrt{3}}{r}^3$

Answer 1

The correct option is

A

$24\sqrt{2}{r}^3$

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

Therefore area of base is given as = $\frac{\sqrt{3}}{4}{a}^2\times 6$

Base area of regular hexagon = Area of 6 equilateral triangles each with side ‘a’.
$=\frac{6\sqrt{3}}{4}{a}^2$
Height of the unit cell(h) = 2 $\times$ Distance between adjacent layers $=2\times \sqrt{\frac{2}{3}}a$
$\therefore$ Volume of the unit cell = Base Area $\times$ height
$=\left(\frac{6\sqrt{3}}{4}{a}^2\right)\left(2\sqrt{\frac{2}{3}}a\right)=3\sqrt{2}{a}^3$
Putting a = 2r, volume = $24\sqrt{2}{r}^3$