Question

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?

Answer 1

Answer:

Let the face length of the cube and radii of the void be a and r respectively. Let the radii of the sphere be R. Two spheres touch each other along the face diagonal AB.

Length of the face diagonal =  $\bold{\sqrt{2} a}$

$\therefore 2 R=\sqrt{2} a---(i)$

Along the body diagonal AD, two spheres radius R each touch  the tetrahedral  void of radius r.

$\therefore 2 R+2 r=A D=\sqrt{3} a----(i i)$

Putting the value of R from equations (i) and (ii)

$\therefore \frac{r_{v o i d}}{R_{s p h e r e}}$ $=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}-1=0.225$

Thus, for tetrahedral void = $\frac{r_{v o i d}}{R_{s p h e r e}}$ = 0.225