Packing fraction of a layer in 2D hexagonal c | vedclass

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Question

(C) In a $2D$ hexagonal close packing,each sphere is in contact with $6$ other spheres. Consider a unit cell area formed by the centers of the spheres

Vedclass · Accepted Answer

$\frac{\pi}{2\sqrt{3}}$

Vedclass · Answer

(C) In a $2D$ hexagonal close packing,each sphere is in contact with $6$ other spheres. Consider a unit cell area formed by the centers of the spheres. The area of the hexagon formed by the centers of $6$ spheres around a central sphere is $6 	imes (\frac{\sqrt{3}}{4} 	imes (2r)^2) = 6\sqrt{3}r^2$. However,the effective area occupied by the spheres within this unit is $3 	imes (\pi r^2)$. The packing fraction is calculated as: $	ext{Packing Fraction} = \frac{	ext{Area occupied by spheres}}{	ext{Total area of the unit cell}}$. For $2D$ hexagonal packing,the packing fraction is $\frac{\pi}{2\sqrt{3}} \approx 0.906$.

Packing fraction of a layer in $2D$ hexagonal close packing arrangement is:

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