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?

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

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