Question

What is $\%$ packing fraction of a diamond type crystal? (given $\sqrt{3}\pi =5.44$)
Given,($\pi =3.2,\sqrt{3}=1.7$

Answer 1

In a diamond type crystal, the main lattice is FCC and alternate tetrahedral voids are occupied by the same atoms as the main lattice.
So, the effective number of atoms in one unit cell = $\underset{fccunitcell}{\frac{1}{8}\times 8+\frac{1}{2}\times 6}+\underset{tetrahedralvoid}{1\times 4}=8$

If a cube of edge length 'a' is intersected in such a way that it divides the cube into 8 small cubes then this tedrahedral void will lie in the body center of this cube.

So, the distance between the nearest two atoms i.e. corner atom to the atom in tetrahedral void $=\frac{\sqrt{3}a}{4}=2r$ where r = radius of atom.
So, packing fraction $=\frac{8\times \frac{4}{3}\pi \left(r^3\right)}{a^3}\times 100=\frac{8\times 4\times \pi \times (\sqrt{3}{)}^3\times a^3}{3\times 8^3\times a^3}\times 100=\frac{\sqrt{3}\pi }{16}\times 100=34\%$