Question

An atom Y crystallizes in fcc unit cell having an edge length $a=200\sqrt{2}pm$ . What would be the radius of octahedral void present in the unit cell at the body centre?

• A. $100pm$
• B. $141.4pm$
• C. $73.2pm$
• D. $22.5pm$

Answer 1

The correct option is C $73.2pm$

For fcc:
$\sqrt{2}a=4\times R$
Here R is the radius of the atom Y
$\Rightarrow \sqrt{2}\times 200\sqrt{2}=4RR=100pm$

For a octahedral void :
$r=0.414R$
Here r is the radius of tetrahedral void and R is the radius of the atom Y
$r=(0.414\times 100)pmr=41.4pm$


In a fcc/ccp unit cell :
Octahedral Voids location :
1 at body centre and 12 voids at edge centres .
$\text{Contribution for voids at edges is}\frac{1}{12}$
$\text{Contribution for body centre void is}1$

$\text{Total octahedral voids}:1+4\times \frac{1}{12}=4$
There are 4 octahedral voids per unit cell