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

Asked by ap996969 | 24th Jan, 2019, 07:08: PM

Expert Answer:

In the simple cubic lattice, the nearest neighbours touch along the edge. 
 
therefore space straight d space equals space straight a

space space space space space straight r space equals straight a over 2
 
In the body centred cubic lattice (bcc) the nearest neighbours touch along the body diagonal. 
therefore space straight d space equals space fraction numerator square root of 3 over denominator 2 end fraction straight a

space space space space straight d space equals space 0.866 straight a

straight r space equals space fraction numerator square root of 3 over denominator 4 end fraction straight a space

straight r space equals space 0.0433 straight a
 
 
 
In the face centred cubic lattice (fcc) the nearest neighbours touch along the face diagonal. 
 
therefore space straight d space equals fraction numerator straight a over denominator square root of 2 end fraction

space space space space straight d space equals 0.707 straight a

space space space straight r space equals space fraction numerator straight a over denominator 2 square root of 2 end fraction

space space space straight r space equals space 0.353 straight a
 

Answered by Varsha | 25th Jan, 2019, 11:43: AM