# 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

From right angled triangle ABC, face diagonal

AB = $\sqrt{{\mathrm{AC}}^{2}{+ BC}^{2}}$ = $\sqrt{{a}^{2}{+ a}^{2}}$ = $\sqrt{2}$a

As spheres A and B are actually touching each other, face diagonal AB = 2R
therefore,  2R = $\sqrt{2}$a
R = 1/√2 a      ....(1)

Again from the right angled triangle ABD

AD= $\sqrt{{\mathrm{AB}}^{2}{+ BD}^{2}}$= $\sqrt{\sqrt{2{a}^{2}}{+ a}^{2}}$ = $\sqrt{3}$a

But as small sphere that is void touches other spheres, evidently body diagonal AD = 2(R + r)
therefore, 2(R + r) =
$\sqrt{3}$a
or R + r = $\sqrt{3}$/2 a ....(2)

Dividing equation (2) by equation (1)
(R + r)/R = $\frac{\sqrt{3}}{2}$a / $\frac{1}{\sqrt{2}}$a = $\sqrt{\frac{3}{2}}$
1 + r/R =
$\sqrt{\frac{3}{2}}$ = 1.225
r/R = 1.225 – 1 = 0.225

r = 0.225 R

• 48

root R=r

• -29
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