Question

The percentage packing efficiency of the two dimensional hcp arrangement of sphere for unit cell ABCDEF shown below is:

• A. 90.64%
• B. 74.05%
• C. 68.02%
• D. 78.54%

Answer 1

The correct option is A 90.64%

Given,

$\text{Percentage packing efficiency}=\frac{\text{Area covered by atom in one unit cell}}{\text{Total area of the unit cell}}\times 100$
From the figure, the number of atoms in the unit cell is 3.
$\text{Area of unit cell}=\text{Area of hexagon}=\text{Area of ABCDEF}$
$\text{Area of unit cell}=6\times \text{Area of equilateral}△OED$
$\text{Area of}△OED=\frac{\sqrt{3}}{4}{a}^2$
where, a = length of the sides
$\text{Area of}△OED=\frac{\sqrt{3}}{4}\left(2r{)}^2\right[\because a=2r]$
$=\sqrt{3}{r}^2$
$\text{Area of unit cell}=6\times \text{Area of}△OED=6\times \sqrt{3}{r}^2$
$\text{Percentage packing fraction}=\frac{Z\times \text{area of atom}}{\text{Area of unit cell}}\times 100$
$=\frac{3\times \pi r^2}{6\times \sqrt{3}{r}^2}\times 100=90.64\%[\because Z=3]$
$\therefore$ Percentage packing efficiency is 90.64%