C-atoms per cell of fcc lattice =4
No .of tetrahedal voids = 4
As C -atoms are present in alternate tetrahedral voids.
no. of C-atoms in tetrahedrl voids = 4
Total no. of C -atoms in the unit cell of diamed = 4+ 4 =8 ,i.e., Z=8
packing fraction =` ("Volume occupied by spheres")/("Volume of the unit cell") = (8 xx 4/3 pi r^(3))/a^(3) = 8 xx 4/3 xx (pir^(3))/a^(3)`
As alternate tetrahedral voids are also occupied by C-atoms , it can by seen that ` sqrt3 a = 8r or a = (8r)/sqrt3`
packing effeiciency = `(8 xx 4/3 pi r^(3))/ ((8r)/sqrt3)^(3) = 32/3 xx 22/7 xx (3sqrt3)/( 8 xx8xx8) = 0.43`
density ` p = (ZM)/(a^(3)N_(0)) = ( 8 xx 12)/((3.57xx10^(-8))^(3) (6.023 xx 10^(23)) = 3.5"g/cm"^(3)`