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  4. Xenon crystallises in the fcc lattice and the edge of the unit cell is 620 pm. The nearest neighbour distance of Xe is

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Q. Xenon crystallises in the $fcc$ lattice and the edge of the unit cell is $620\, pm$. The nearest neighbour distance of $Xe$ is

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Solution:

As xenon $(Xe)$ crystallises in $fcc$ lattice, nearest neighbouring distance
$(d)=\frac{a}{\sqrt{2}}=\frac{620}{\sqrt{2}}$
$=\frac{620}{1.414}=438.5\, pm$