Question

Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

• A. 25
• B. 35
• C. 55
• D. 75

Answer 1

Answer: (B)

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} a}{4}=2 r_x $
$\Rightarrow a =\frac{8 r_x}{\sqrt{3}}$
Number of atoms of $X$ per cell
$ \underset{\text{(FCC lattice points)}}{=4} + \underset{\text{(Alternate tetrahedral voids)}}{4 } = 8$
$\% $ packing efficiency $ =\frac{\text { Volume occupied by } X }{\text { Volume of cubic unit cell }} \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{\left.8 r _{ x }\right)^3}{\sqrt{3}}\right)} \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 \%$