Question

The percentage of void space of a metallic element crystallising in a ABCABC..... type lattice pattern is:

Answer 1

The ABCABC… type of arrangement results into cubic close packing which resembles of face centered cubic structure in general.

In FCC, we have atoms on the corners of the cube as well as on the face centers.

So, total number of atoms in a cubic unit cell,

$n=\theta \times \frac{1}{\theta }+6\times \frac{1}{2}$

where $\left(\theta \right)$ is the number of corners

$\theta$ is each atom on corner is shared by $\theta$ unit cells

$6$ is number of faces

$\frac{1}{2}$ is each atom on face is shared by $2$ unit cells.

$\Rightarrow n=1+3=4$

Now, in FCC, atoms on face diagonals will be touching each other.

So, $4r=\sqrt{2}a$

$\Rightarrow r=\frac{a}{2\sqrt{2}}$ ($r$ is radius of each atom)

Now, volume of cubic unit cell of edge length $a=a^3$

Volume occupied $=4\times \frac{4}{3}\pi (r{)}^3$

$=4\times \frac{4}{3}\times \pi {\left(\frac{a}{2\sqrt{2}}\right)}^3$

$=\frac{16}{3}\times \pi \times a^{3}16\sqrt{2}$

$=\frac{\pi a^3}{3\sqrt{2}}$

% Volume occupied $=\frac{\frac{\pi a^3}{3\sqrt{2}}}{a^3}\times 100=74$%

% Void $=(100-74)$%$=26$%