1.

give a relationship between nearest neighbour distance(d),radius of atom(r), edge  of unit cell(a), for fcc and BCC crystal

Answer»

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.



Discussion

No Comment Found

Related InterviewSolutions