CsCl has a bcc arrangement. Its unit cell edg | vedclass

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(D) In a $bcc$ crystal structure like $CsCl$, the ions touch along the body diagonal.The body diagonal length is given by $a \sqrt{3}$, where $a$ is t

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} 	imes 400 \, pm$

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(D) In a $bcc$ crystal structure like $CsCl$, the ions touch along the body diagonal.The body diagonal length is given by $a \sqrt{3}$, where $a$ is the edge length.The body diagonal is also equal to $2(r_{+} + r_{-})$, where $(r_{+} + r_{-})$ is the inter-ionic distance.Therefore, $2(r_{+} + r_{-}) = a \sqrt{3}$.$(r_{+} + r_{-}) = \frac{a \sqrt{3}}{2}$.Given $a = 400 \, pm$, the inter-ionic distance is $\frac{400 \sqrt{3}}{2} \, pm$.

$CsCl$ has a $bcc$ arrangement. Its unit cell edge length is $400 \, pm$. What is its inter-ionic distance?

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