Calculate the packing efficiency of a metal crystal for a body-centred cubic $(BCC)$ structure.

(N/A) In a body-centred cubic $(BCC)$ unit cell,the atom at the centre is in contact with the other two atoms diagonally arranged along the body diagonal.
From the geometry of the cube,let the edge length be $a$ and the radius of the atom be $r$.
The face diagonal $b$ is given by:
$b^{2} = a^{2} + a^{2} = 2a^{2}$
$b = \sqrt{2}a$
The body diagonal $c$ is given by:
$c^{2} = a^{2} + b^{2} = a^{2} + 2a^{2} = 3a^{2}$
$c = \sqrt{3}a$
Since the atoms along the body diagonal touch each other,the length of the body diagonal is $c = 4r$.
Therefore,$\sqrt{3}a = 4r$,which gives $a = \frac{4r}{\sqrt{3}}$.
The volume of the unit cell is $a^{3} = \left(\frac{4r}{\sqrt{3}}\right)^{3} = \frac{64r^{3}}{3\sqrt{3}}$.
$A$ $BCC$ unit cell contains $2$ atoms. The volume occupied by these $2$ atoms is:
$V_{occupied} = 2 \times \frac{4}{3} \pi r^{3} = \frac{8}{3} \pi r^{3}$.
Packing efficiency is calculated as:
$\text{Packing Efficiency} = \frac{\text{Volume of } 2 \text{ atoms}}{\text{Total volume of unit cell}} \times 100\%$
$= \frac{\frac{8}{3} \pi r^{3}}{\frac{64r^{3}}{3\sqrt{3}}} \times 100\%$
$= \frac{8\pi}{3} \times \frac{3\sqrt{3}}{64} \times 100\%$
$= \frac{\sqrt{3}\pi}{8} \times 100\% \approx 68\%$.