Efficiency of packing in body centered cubic structures is found to b…

Body centered cubic unit cell (BCC)

Atom at the centre will be in touch with the other two atoms diagonally arranged.

In ∆ EFD,

b² = a² + a² = 2a²

b = \(\sqrt{2}\)a

Now in ∆ AFD

c² = a² + b² = a² + 2a² = 3a²

c = \(\sqrt{3}\)a

The length of the body diagonal c is equal to 4r, where r is the radius of the sphere (atom), as all the three spheres along the diagonal touch each other.

Therefore, \(\sqrt{3}\)a = 4r

a = \(\frac{4r}{\sqrt{3}}\)

Also we can write, r = \(\frac{\sqrt{3}}{4}\)a

In this type of structure, total number of atoms is 2 and their volume is 2 x \(\frac{4}{3}\)πr³

Volume of the cube, a³ will be equal to (\(\frac{4}{\sqrt{3}}\)r)³ or a³ = (\(\frac{4}{\sqrt{3}}\)r)³

Packing efficiency = \(\frac{\text{Volume occupied by four spheres in the unit cell}}{\text{Total volume of the unit cell}}\) x 100%

P.E. = \(\frac{2 × \frac{4}{3}πr^3}{(\frac{4}{\sqrt{3}})^3}\) x 100 = 68%

% of free space in BCC unit cell i.e., Void efficiency = 100 - 68 = 32%