Packing efficiency of metal crystal in face-centred cubic lattice (or ccp or hcp lattice)

The packing efficiency of a metal crystal in the face-centred cubic lattice (which equals that of the ccp and hcp lattices) is obtained by the same four steps.

Step 1 : Radius of particle/sphere : In an fcc unit cell, a corner particle touches the particle at the centre of the adjacent face along the face diagonal (Fig. 1.11). For the face $ABCD$, apply the Pythagoras theorem to the right-angled triangle $ABC$, with $\angle ABC = 90^\circ$:

The face diagonal spans a corner sphere's radius, the full diameter of the face-centre sphere, and the opposite corner sphere's radius, so $AC = 4r$. Hence $\sqrt{2}\,a = 4r$, that is

Step 2 : Volume of sphere : $\text{volume of one particle} = \dfrac{4}{3}\pi\left(\dfrac{\sqrt{2}}{4}a\right)^{3} = \dfrac{\pi a^3}{12\sqrt{2}}$

Step 3 : Total volume of particles : An fcc unit cell contains 4 particles, so the total occupied volume is $4\times\dfrac{\pi a^3}{12\sqrt{2}} = \dfrac{\pi a^3}{3\sqrt{2}}$

Step 4 : Packing efficiency :

Thus 74 % of the volume of the fcc (equally, ccp or hcp) unit cell is occupied by particles, and only 26 % is void space -- the most efficient packing of the three cubic lattices.

Table 1.3 collects the relation between $r$ and $a$ and the occupied volumes for all three cubic lattices, and Table 1.4 summarises their coordination numbers alongside their packing efficiencies.

Table 1.3 : Edge length and particle parameters in cubic system