If the radius of the tetrahedral void is r and radius of the atoms in close packing is R, derive the relation between r and R?
A tetrahedral void may be represented in a cube. In which there spheres form the triangular base, the fourth lies at the top and the sphere occupies the tetrahedral void.
Let the length of the side of the cube = a
Radius of sphere = R
Radius of void = r
From right angled triangle ABC, face diagonal
AB = = = a
As spheres A and B are actually touching each other, face diagonal AB = 2R
therefore, 2R = a
R = 1/√2 a ....(1)
Again from the right angled triangle ABD
AD= = = a
But as small sphere that is void touches other spheres, evidently body diagonal AD = 2(R + r)
therefore, 2(R + r) = a
or R + r = /2 a ....(2)
Dividing equation (2) by equation (1)
(R + r)/R = a / a =
1 + r/R = = 1.225
r/R = 1.225 – 1 = 0.225
r = 0.225 R
Let the length of the side of the cube = a
Radius of sphere = R
Radius of void = r
From right angled triangle ABC, face diagonal
AB = = = a
As spheres A and B are actually touching each other, face diagonal AB = 2R
therefore, 2R = a
R = 1/√2 a ....(1)
Again from the right angled triangle ABD
AD= = = a
But as small sphere that is void touches other spheres, evidently body diagonal AD = 2(R + r)
therefore, 2(R + r) = a
or R + r = /2 a ....(2)
Dividing equation (2) by equation (1)
(R + r)/R = a / a =
1 + r/R = = 1.225
r/R = 1.225 – 1 = 0.225
r = 0.225 R