The percentage of empty space in a body centred cubic arrangement is ________.

• In a body-centred cubic arrangement of lattices, the constituent atoms are located on the 8 corners of the cube with one atom at the centre of the cube.

Let the edge length of the cube be ‘a’ (ED or BC in the fig) and radius of each constituent spherical particle be ‘r’ and the arrangement is such as that atom at the centre is in touch with other two atoms diagonally.

• Hence the body diagonal AF will be equal to = r +2r + r = 4r (1)

In the right angle ∆ EFD,

FD² or b²⁼ FE² + ED² ( by Pythagoras’s theorem ) = a² + a² = 2a²

Or, FD or b = √2a

• In the right-angled triangle, ∆ AFD

AF² or c²= AD² + FD² = a² + b² = a² + 2a² (as b²= 2a² ) =3a²

Or, AF or c =√3a (2)

Hence, from equations (1) and (2) we can get ,

AF = 4r =√3a a

Or, r = √3a/4

And a = 4r/√3

• Then, the volume of 1 cubic unit cell is a³ = (4r/√3) = 64r³/ 3√3

• Since in body centred arrangement , there are 2 atoms per unit cell

Volume occupied by the atoms = 2 X 4/3 π r³ = 8/3 π r³

Since, Packing efficiency = Volume occupied by two spheres in the unit cell X 100 %

Total volume of the unit cell

= 8/3 π r³ X 100 % = 68.04 % = 68%_appx.

64r³/ 3√3

Therefore , Packing efficiency in a body centred cubic lattice is 68 %

So, empty space or void space will be = (100 -68 )% = 32 %