An element can crystallise in both face-centred cubic (fcc) and body-centred cubic (bcc) lattices. Assuming the atomic radius of the element remains the same in both lattices, let D be the length of the face diagonal of the fcc unit cell and a be the edge length of the bcc unit cell. The value of ( D a )^2 is _________.

Step-by-step solution

For an fcc lattice, the atoms touch along the face diagonal. Thus, the length of the face diagonal D is equal to 4r , where r is the atomic radius. For a bcc lattice, the atoms touch along the body diagonal. The relationship between the edge length a and the atomic radius r is 3 a = 4r , which gives a = 4r 3 . The ratio of the face diagonal of the fcc unit cell to the edge length of the bcc unit cell is: D a = 4r 4r 3 = 3 Squaring this ratio gives: ( D a )^2 = ( 3 )^2 = 3 Answer: 3