Question

The maximum radius of sphere that can be fitted in the tetrahedral hole of cubical closed packing of spheres of radius $r$ :

• A. $0.732r$
• B. $0.414r$
• C. $0.225r$
• D. $0.155r$

Answer 1

The correct option is C $0.225r$

Let the face length of cube $=a$

And radii $=r$

And also

Let the radii of the sphere $=R$

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

$\therefore 2R=\sqrt{2a}...........\left(1\right)$

Along the body diagonal, two shapre of radius $R$ tetrahedral void of radius $r$

$\therefore 2R+2r=AD=\sqrt{3a}.........\left(2\right)$

Put the value of $R$ from equation (1) in equation (2)

$2r=\sqrt{3a}-\sqrt{2a}$

$\therefore \frac{r_{void}}{R_{sphere}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}$

$=\frac{\sqrt{3}}{\sqrt{2}}-1=\frac{1.732}{1.414}-1=0.225$

Thus for a tetrahedral void, $\frac{r_{void}}{R_{sphere}}=0.225$