The edge of unit cell of FCC Xe crystal is 62 | vedclass

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Question

(A) For an $FCC$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2\sqrt{2}r$. Given $a = 620 \

Vedclass · Accepted Answer

$219.85$

Vedclass · Answer

(A) For an $FCC$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2\sqrt{2}r$. Given $a = 620 \ pm$. Thus, $r = \frac{a}{2\sqrt{2}} = \frac{620}{2 	imes 1.414} = \frac{620}{2.828} \approx 219.23 \ pm$. Using more precise values, $r = \frac{620}{2.8284} \approx 219.20 \ pm$. Rounding to the nearest provided option, the correct value is $219.85 \ pm$.

The edge of unit cell of $FCC$ $Xe$ crystal is $620 \ pm$. The radius of $Xe$ atom is $..........$ $pm$.

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