Calculate the percentage of space occupied by the voids in body centred cubic unit cell.

The correct option is B 32%
For body centred cubic arrangement, $Z_{eff}=2$

$4r=\sqrt{3}a$
where,
r is radius of atom
a is edge length of cube
Thus, $a=\frac{4r}{\sqrt{3}}$
$\text{Volume of cell}=a^3$
$\text{Volume of cell}=(\frac{4r}{\sqrt{3}}{)}^3$
$\text{Packing efficiency}=\frac{\text{Total volume occupied by sphere}}{\text{Volume of unit cell}}\times 100$
$\text{Packing efficiency}=\frac{Z\times \frac{4}{3}\pi r^3}{a^3}\times 100$
$=\frac{2\times \frac{4}{3}4r^3}{{\left(\frac{4r}{\sqrt{2}}\right)}^3}\times 100=\frac{\sqrt{3}\pi \times 100}{8}=68\%$
$\text{Percentage of empty space}=100-68=32\%$