A metal has a bcc structure. If the distance | vedclass

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(C) For a $bcc$ (body-centered cubic) structure, the distance between the nearest atoms $(d)$ is given by the formula $d = \frac{\sqrt{3}}{2} a$, wher

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$200$

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(C) For a $bcc$ (body-centered cubic) structure, the distance between the nearest atoms $(d)$ is given by the formula $d = \frac{\sqrt{3}}{2} a$, where $a$ is the edge length of the unit cell.Given that the distance between nearest atoms $d = 1.73 \ \mathring{A}$.Since $\sqrt{3} \approx 1.732$, we substitute the values into the formula: $1.73 = \frac{1.732}{2} a$.Solving for $a$: $a = \frac{1.73 	imes 2}{1.732} \approx \frac{3.46}{1.732} \approx 1.9976 \ \mathring{A} \approx 2 \ \mathring{A}$.To convert the edge length from $\mathring{A}$ to $pm$, we use the conversion factor $1 \ \mathring{A} = 100 \ pm$.Therefore, $a = 2 \ \mathring{A} 	imes 100 \ pm/\mathring{A} = 200 \ pm$.

$A$ metal has a $bcc$ structure. If the distance between two nearest atoms is $1.73 \ \mathring{A}$, what is the edge length of the unit cell in $pm$?

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