Xenon crystallizes in FCC lattice and the edge of the unit cell is 620 pm, then the radius of xenon — Solid State Chemistry Question

💡 Solution & Explanation

For an FCC lattice, the relationship between edge length $a$ and radius $r$ is $4r = \sqrt{2}a$. Thus, $r = \frac{\sqrt{2} \times 620}{4} = \frac{1.414 \times 620}{4} = 219.20$ pm. Therefore, correct answer is A.