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packing fraction of fcc lattice?

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Packing Fraction of Face-Centred Cubic  Lattice


Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube

As there are 4 sphere in fcc unit cell

∴ Volume of four spheres = 4 (4/3 πr3

In fcc, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere

AC= 4r

But from the right angled triangle ACD 

AC = √AD2 + DC2 = √a2 + a2= √2a 

4r = √2a

or a = 4/√2 r

∴ volume of cube = (2/√2 r)3 


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