In diamond lattice, all lattice points and alternate tetrahedral voids are occupied by carbon atoms. If diamond crystallizes in fcc form with edge length 'a' , find out (a)number of next nearest neighbours in diamond lattice (b)distance between the next nearest neighbours.

In diamond lattice, all lattice points and alternate tetrahedral voids

1.	In diamond lattice, all lattice points and alternate tetrahedral voids are occupied by carbon atoms. If diamond crystallizes in fcc form with edge length 'a' , find out (a)number of next nearest neighbours in diamond lattice (b)distance between the next nearest neighbours.
Answer» Solution :(a)Nearest neighbours are atoms at the LATTICE POINS and tetrahedral voids. Hence, NEXT nearest neighbours are atoms at the lattice POINTS. `therefore` Number of next nearest neighbours =12 (being fcc lattice ) (B) For fcc, nearest neighbour distance (d)=`a/sqrt2`=0.707 a