Xenon crystallizes in fcc lattice and the edge length of unit cell is $620 \mathrm{pm}$. What is the radius…
Xenon crystallizes in fcc lattice and the edge length of unit cell is $620 \mathrm{pm}$. What is the radius of Xe atom?
- 219.2pm
- 438.5pm
- 265.5pm
- 536.9pm
Solution
$a=620 \mathrm{pm}$
For fcc unit cell, $r=\frac{a}{2 \sqrt{2}}$
$\therefore r=\frac{620}{2 \times 1.414}=219.2 \mathrm{pm}$
Asked in: MHT CET 2020 (16 Oct Shift 1)
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