Question

An element crystallizes in 3-D hexagonal closed packed structure but 2 atoms from the corner of the hexagonal unit cell are absent.
The packing fraction of such lattice would be :

• A. $\frac{17\pi }{54\sqrt{2}}$
• B. $\frac{8\pi }{26\sqrt{2}}$
• C. $\frac{13\pi }{38\sqrt{3}}$
• D. $\frac{\pi }{2\sqrt{3}}$

Answer 1

The correct option is A $\frac{17\pi }{54\sqrt{2}}$

Elements that crystallize in hcp unit cell are assumed to be spherical .
$\text{Total no. of spheres in a hcp unit cell}=(10\times \frac{1}{6})+(2\times \frac{1}{2})+3=\frac{17}{3}$
where,
$(10\times \frac{1}{6})\text{is for corners atoms of two hexagonal layers}$
$(2\times \frac{1}{2})\text{is for the centre atoms of two hexagonal layers}$
$3\text{is for atoms in middle layer}$


$\text{Total volume occupied}=\frac{17}{3}\times \frac{4}{3}\times \pi \times r^3=\frac{68}{9}\pi r^3$
$\text{Volume of hcp unit cell (hexagon)}=\text{Base area}\times \text{Height}$


$\text{Base area of a regular hexagon}=6\times \frac{\sqrt{3}}{4}(2r{)}^2=6\times \sqrt{3}{r}^2$

$\text{Height of Unit Cell (h)}=4r\times \sqrt{\frac{2}{3}}$

$\text{Volume of hexagon (V)}=(6\times \sqrt{3}{r}^2)\times (4r\times \sqrt{\frac{2}{3}})V=24\sqrt{2}{r}^3$
$\text{Packing fraction}=\frac{\frac{68}{9}\pi r^3}{24\sqrt{2}{r}^3}=\frac{17\pi }{54\sqrt{2}}$