CsBr has a bcc structure. If the edge length | vedclass

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(B) In a $bcc$ structure,the distance between the body-centered ion and the corner ions is given by the formula: $d = \frac{\sqrt{3}a}{2}$.Given the e

Vedclass · Accepted Answer

$3.72$

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(B) In a $bcc$ structure,the distance between the body-centered ion and the corner ions is given by the formula: $d = \frac{\sqrt{3}a}{2}$.Given the edge length $a = 4.3 \ \overset{\circ}{A}$.Substituting the values: $d = \frac{1.732 	imes 4.3 \ \overset{\circ}{A}}{2}$.$d = \frac{7.4476}{2} \ \overset{\circ}{A} = 3.7238 \ \overset{\circ}{A}$.Rounding to two decimal places,the distance is $3.72 \ \overset{\circ}{A}$.

$CsBr$ has a $bcc$ structure. If the edge length is $4.3 \ \overset{\circ}{A}$,what is the minimum distance between $Cs^+$ and $Br^-$ ions in $\overset{\circ}{A}$?

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