Question

The percentage of free space in a face centered cubic (fcc) is :

• A. 52%
• B. 48%
• C. 74%
• D. 26%

Answer 1

The correct option is D 26%


In FCC:

$\sqrt{2}\times a=4ra=2\sqrt{2}\times r$

$Z_{eff}\text{for FCC}=4$
So,

$\text{The total volume occupied by sphere}V=4\times \frac{4}{3}\times \pi r^3$
$\text{Volume of the cube}=a^3=(2\sqrt{2}r{)}^3$
$\text{Packing efficiency (P.F)}=\frac{4\times \frac{4}{3}\times \pi r^3}{(2\sqrt{2}r{)}^3}\times 100\%P.F=\frac{\pi \times 100}{3\sqrt{2}}\%\approx 74\%$
$\text{Free space}=26\%$