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Q. Xenon crystallises in face-centered cubic lattice, and the edge of the unit cell is $620 \,pm$. The radius of a xenon atom is

The Solid State

Solution:

For an fcc lattice, $4r = \sqrt{2}a$, where $ a = 620\,pm$
Or $r = \frac{\sqrt{2}}{4} . a = \frac{1}{2\sqrt{2}}$.
$a = \frac{1}{2\sqrt{2}} . 620 = 219.20\,pm$