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  4. Packing efficiency by arrangement of atoms in two dimensional hexagonal close packing is

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Q. Packing efficiency by arrangement of atoms in two dimensional hexagonal close packing is

The Solid State

Solution:

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Let radius of the sphere $=r$
Area occupied by sphere in hexagonal close packing
$\pi r^2+6 \times\left(\frac{1}{6} \times \pi r^2\right)=2 \pi r^2$
Area of hexagonal $=6 \times\left[\frac{\sqrt{3}}{4} \times(2 r)^2\right]$
$ 6 \times \frac{\sqrt{3}}{4} \times 4 r^2=6 \sqrt{3} \times r^2 $
$\% $ occupied by $= \frac{2 \pi r^2}{6 \times \sqrt{3} \times r^2} \times 100 $
$ =\frac{2 \times 3.14}{6 \times \sqrt{3}} \times 100=60.43 \%$