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
$=\frac{\sqrt{3}a}{4}=2r_x$
$\Rightarrow a=\frac{8r_x}{\sqrt{3}}$
Number of atoms of $X$ per cell
$\underset{\text{(FCC lattice points)}}{=4}+\underset{\text{(Alternate tetrahedral voids)}}{4}=8$
$\%$ packing efficiency $=\frac{\text{ Volume occupied by }X}{\text{ Volume of cubic unit cell }}\times 100$
$=\frac{8\times \frac{4}{3}\pi {\left(r_x\right)}^3}{a^3}\times 100$
$=\frac{8\times \frac{4}{3}\pi {\left(r_x\right)}^3}{\left(\frac{{8r_x)}^3}{\sqrt{3}}\right)}\times 100$
$=\left(8\times \frac{4}{3}\times \pi \times \frac{1}{8^3}\times 3\sqrt{3}\right)\times 100$
$=\frac{\sqrt{3}\pi }{16}\times 100$
$=34\%$