Question

In a ccp type structure, if half of the face-centred atoms are removed, then percentage void in unit cell is approximately:

• A. $54\%$
• B. $46.25\%$
• C. $63\%$
• D. $37\%$

Answer 1

The correct option is A $54\%$

Half of the face centered atoms are removed.

New $Z_{effective}=8\times \frac{1}{8}+3\times \frac{1}{2}=2.5$

Volume of atoms $=Z\times$ Volume of single atom $=2.5\times \frac{4}{3}\pi r^3$

Volume of unit cell $=a^3$

In FCC $4r=a\sqrt{2}$ (along the face diagonal)

$⟹r=\frac{a}{2\sqrt{2}}$

Packing fraction= $\frac{4/3\pi r^3}{a^3}$

Packing fraction $=\frac{2.5\times 4}{3}\pi \frac{a^3}{16\sqrt{2}}\times \frac{1}{a^3}=0.46$

Packing percentage= 46%

Percentage void $=100-46=54\%$