What is the packing efficiency of fcc crystal | vedclass

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Question

(A) In an $fcc$ (face-centered cubic) unit cell,the atoms are present at the corners and the center of each face.The relation between the edge length

Vedclass · Accepted Answer

$74.0$

Vedclass · Answer

(A) In an $fcc$ (face-centered cubic) unit cell,the atoms are present at the corners and the center of each face.The relation between the edge length '$a$' and the radius of the atom '$r$' is $a = 2\sqrt{2}r$.The number of atoms per unit cell $(Z)$ for $fcc$ is $4$.The volume of $4$ spheres is $4 	imes (\frac{4}{3} \pi r^3) = \frac{16}{3} \pi r^3$.The volume of the unit cell is $a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3$.Packing efficiency = $\frac{	ext{Volume of 4 spheres}}{	ext{Volume of unit cell}} 	imes 100 = \frac{\frac{16}{3} \pi r^3}{16\sqrt{2}r^3} 	imes 100 = \frac{\pi}{3\sqrt{2}} 	imes 100 \approx 74.0 \%$.

What is the packing efficiency of $fcc$ crystal structure (in $\%$)?

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