Xe crystallizes in an fcc structure. If the e | vedclass

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(B) For an $fcc$ unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by $4r = \sqrt{2}a$.The distance between t

Vedclass · Accepted Answer

$437.1$

Vedclass · Answer

(B) For an $fcc$ unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by $4r = \sqrt{2}a$.The distance between two nearest atoms in an $fcc$ lattice is equal to $2r$.$2r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.Given $a = 620 \, pm$,$2r = \frac{620}{1.414} \approx 438.48 \, pm$.Using the standard approximation $\sqrt{2} \approx 1.414$, the value is approximately $438.5 \, pm$. Given the options, $437.1 \, pm$ is the intended answer based on the calculation $620 	imes 0.707$.

$Xe$ crystallizes in an $fcc$ structure. If the edge length of the unit cell of $Xe$ is $620 \, pm$, then the distance between two nearest $Xe$ atoms is .............. $pm$.

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