CBSE Class 12-science Answered
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 | 24 Jan, 2019, 19:08: PM
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](https://images.topperlearning.com/topper/tinymce/cache/31f294ffda022b0e4f9a432d490b2e33.png)
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](https://images.topperlearning.com/topper/tinymce/cache/4b54cb7d32f67a05cd624185dd88394e.png)
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](https://images.topperlearning.com/topper/tinymce/cache/546ea764e51f4a821cfa3f3979327b20.png)
Answered by Varsha | 25 Jan, 2019, 11:43: AM
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