Chemistry - Solid State Question with Solution | TestHub

Atom 'X' occupies FCC lattice points as well as alternate tetrahedral voids of the same lattice.

This means the distance of the body diagonal is $\frac{1}{4}$th of the total length.

$$=\frac{\sqrt{3}a}{4}=2r_X$$

$$\Rightarrow a=\frac{8r_X}{\sqrt{3}}$$

The number of atoms of X per unit cell is $4+4=8$ (4 from FCC lattice points and 4 from alternate tetrahedral voids).

The percentage packing efficiency is given by:

$$Packingefficiency=\frac{VolumeoccupiedbyX}{Volumeofcubicunitcell}\times 100$$

$$=\frac{8\times \frac{4}{3}\pi {\left(r_X\right)}^3}{a^3}\times 100$$

$$=\frac{8\times \frac{4}{3}\pi {\left(r_X\right)}^3}{{\left(\frac{8r_X}{\sqrt{3}}\right)}^3}\times 100$$

$$=\left(8\times \frac{4}{3}\times \pi \times \frac{1}{8^3}\times 3\sqrt{3}\right)\times 100$$

$$=\frac{\sqrt{3}\pi }{16}\times 100$$

$=34\%$

Hence, option (B) is the most appropriate option.