Question

In the closest packing of spheres of uniform size, the ratio of radii of octahedral and tetrahedral holes is :

• A. $\frac{0.732}{0.414}$
• B. $\frac{0.414}{0.155}$
• C. $\frac{0.732}{0.225}$
• D. $\frac{0.414}{0.225}$

Answer 1

The correct option is D

$\frac{0.414}{0.225}$

If you put a small ball there, it will be in contact with all four spheres. Thus,

bd = 2 (R + r). r = (2.45 R) / 2 - R
= 1.225 R - R
= 0.225 R

For tetrahedral void, $\frac{r_{void}}{r_{sphere}}\approx 0.225;$

The octahedral hole is located at the centre of any four spheres that form a square. If we represent the radius of a ball fitting in the octahedral holes by r, and the radius of the sphere as R, then we have the relationship:

For Octahedral void, $\frac{r_{void}}{r_{sphere}}=0.414$