give a relationship between nearest neighbour distance(d),radius of atom(r), edge of unit cell(a), for fcc and BCC crystal
give a relationship between nearest neighbour distance(d),radius of atom(r), edge of unit cell(a), for fcc and BCC crystal
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Simple cubic -
Nearest dist - a, radius a/2 , z(no of atoms per unit cel) = 1
Fcc -
Nd = a/(2)1/2 , radius = a/2 root2 z=4
Bcc-
Nd- a root 3 /2 ; radius = a root3 /4 z= 2
For the derivation and detailed answer must refer to master NCERT book.
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