Chemistry - Solid State Question with Solution | TestHub

Atom 'X' occupies FCC lattice points as well as alternate tetrahedral voids of the same lattice $\Rightarrow \frac{1}{4}$th distance of body diagonal.

$=\frac{\sqrt{3}\mathrm{a}}{4}=2{\mathrm{r}}_{\mathrm{X}}$

$\Rightarrow \mathrm{a}=\frac{8{\mathrm{r}}_{\mathrm{X}}}{\sqrt{3}}$

Number of atoms of X per unit cell:

$$=4+4=8$$

(FCC lattice points)

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

$$=\frac{8\times \frac{4}{3}\pi (r_{\mathrm{X}}{)}^3}{{\mathrm{a}}^3}\times 100$$

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

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

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

$$=34\%$$

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