Question

Xenon crystallises in face centered cubic lattice and the edge of the unit cell is $620\text{pm}$, then the radius of xenon-atom is:

• A. $219.20pm$
• B. $536.94pm$
• C. $265.5pm$
• D. $438.5pm$

Answer 1

The correct option is A $219.20\text{pm}$

For fcc lattice:
$4r=\sqrt{2}a\text{where}a=620\text{pm}\Rightarrow r=\frac{1}{2\sqrt{2}}\times a\Rightarrow \frac{1}{2\sqrt{2}}\times 620\text{pm}=219.20\text{pm}$.