Chemistry - Solid State Question with Solution | TestHub

In a face-centered cubic (FCC) unit cell, atoms are present at the $8$ corners and the centers of the $6$ faces.

Number of atoms from corners $=8\times \frac{1}{8}=1$

Number of atoms from faces $=6\times \frac{1}{2}=3$

Total number of atoms per unit cell ($Z$) $=1+3=4$

Let $a$ be the edge length of the unit cell and $r$ be the radius of the atom. In an FCC lattice, atoms touch each other along the face diagonal.

Face diagonal $=\sqrt{2}a=4r\Rightarrow a=2\sqrt{2}r$

Volume of the unit cell $=a^3=(2\sqrt{2}r{)}^3=16\sqrt{2}{r}^3$

Volume occupied by $4$ atoms $=4\times \frac{4}{3}\pi r^3=\frac{16}{3}\pi r^3$

Packing efficiency $=\frac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}}\times 100$

Packing efficiency $=\frac{\frac{16}{3}\pi r^3}{16\sqrt{2}{r}^3}\times 100=\frac{\pi }{3\sqrt{2}}\times 100\approx 74\%$

Answer: Packing efficiency $=74\%$ and number of atoms per unit cell $=4$