12 octahedral voids positioned at each edge and shared between four unit cells = 12 x 1 / 4 = 3
Total number of octahedral voids = 4
In ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to this number.
Packing Efficiency in hcp and ccp Structures let the unit cell edge length be‘ a’ and face diagonal AC = b.
Let r be the radius of the sphere. So we get
Each unit cell in ccp structure has effectively 4 spheres.
Hence total volume of four spheres is equal to 4 X( 4 / 3) Π r 3 and volume of the cube is a 3 or( 2√2r) 3.