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Q. Xenon crystallizes 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, $4 r=\sqrt{2} a$, where $a=620\, pm$

$r=\frac{\sqrt{2}}{4} \cdot a=\frac{1}{2 \sqrt{2}} \cdot a=\frac{1}{2 \sqrt{2}} \cdot 620=219.20\, pm$