Consider the packing fraction in a body centered cubic lattice. In this lattice, if the radius of the spheres is r, then the body diagonal of the BCC unit cell is 4 r and therefore the edge length l is
( 4r )2 = l2 + l2 + l2 or l = ( 4 /
) r
= 2.309 r
=
(17.3)
In the case of an FCC lattice, the face diagonal = 4r (See Fig 17.5 (c)), and therefore, the edge length l = 4r / . The number of spheres in the FCC unit cell is 4 and hence the packing fraction in the FCC lattice (closest packed) is
(17.4)
The highest packing fraction possible is 74.04 % and this is for the FCC lattice. The same value of packing fraction is for the HCP structure as well, which only differs from FCC in that the location of the third layer is different (ababab), but the number of atoms in a given volume is identical in FCC and HCP.
) r 
=
. The number of spheres in the FCC unit cell is 4 and hence the packing fraction in the FCC lattice (closest packed) is 
