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(C) $1$. For a $BCC$ structure, the interatomic distance $(d)$ is the distance between the nearest neighbours, which is given by $d = \frac{\sqrt{3}}{

Vedclass · Accepted Answer

$346.4 \ pm, 600 \ pm$

Vedclass · Answer

(C) $1$. For a $BCC$ structure, the interatomic distance $(d)$ is the distance between the nearest neighbours, which is given by $d = \frac{\sqrt{3}}{2} a$. Given edge length $a = 400 \ pm$, $d = \frac{\sqrt{3}}{2} 	imes 400 = 1.732 	imes 200 = 346.4 \ pm$.$2$. For an $FCC$ structure, the distance between the nearest neighbours is equal to $2r$, where $r$ is the radius of the atom. Given $r = 300 \ pm$, the distance $d = 2 	imes 300 = 600 \ pm$.

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