Calculate packing efficiency in body-centred cubic lattice. - Vidyadip

Question

Calculate packing efficiency in body-centred cubic lattice.

✓

Answer

Step 1 : Radius of sphere :
In the unit cell of body-centred cubic lattice, there are 8 atoms at 8 corners and one atom at the centre of the cube.

The atoms are in contact along the body diagonal BF. Let a be the edge length and $r$ the radius of an atom.
Consider a triangle $BCE$.
$B E^2=B C^2+C E^2=a^2+a^2=2 a^2$
Consider triangle $BEF$,
$\begin{aligned}
& B F^2=B E^2+E F^2=2 a^2+a^2=3 a^2 \\
& B F=\sqrt{3} a .
\end{aligned}$
From figure, $B F=4 r$
$\begin{aligned}
& \therefore 4 r=\sqrt{3} a \\
& \therefore r=\frac{\sqrt{3}}{4} a
\end{aligned}$

Step 2 : Volume of sphere :
Volume of one particle
$\begin{aligned}
e & =\frac{4}{3} \pi r^3 \\
& =\frac{4}{3} \pi\left(\frac{\sqrt{3}}{4} a\right)^3 \\
& =\frac{4}{3} \pi \times \frac{(\sqrt{3})^3}{64} \times a^3
\end{aligned}$
$=\frac{\sqrt{3} \pi a^3}{16}$

Step 3 : Total volume of particles :
The unit cell of bcc structure contains 2 particles.
$\begin{aligned}
& \therefore \text { Volume occupied by } 2 \text { particles }=\frac{2 \times \sqrt{3} \pi a^3}{16} \\
&=\frac{\sqrt{3} \pi a^3}{8}
\end{aligned}$

Step 4 : Packing efficiency
Packing efficiency
$\begin{aligned}
& =\frac{\text { Volume occupied by particles in unit cell }}{\text { Volume of unit cell }} \times 100 \\
& =\frac{\sqrt{3} \pi a^3 / 8}{a^3} \times 100 \quad \\
& =\frac{\sqrt{3} \times 3.142}{8} \times 100 \\
& =68 \%
\end{aligned}$
$\therefore$ Packing efficiency $=68 \%$
$\therefore$ Percentage of void space $=100-68$
$=32 \%$

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