A certain element crystallises in a bcc latti | vedclass

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(A) For $bcc$ lattice,the relation between edge length $a_1$ and atomic radius $r$ is $\sqrt{3} a_1 = 4 r$,so $r = \frac{\sqrt{3}}{4} a_1$.Given $a_1

Vedclass · Accepted Answer

$33$

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(A) For $bcc$ lattice,the relation between edge length $a_1$ and atomic radius $r$ is $\sqrt{3} a_1 = 4 r$,so $r = \frac{\sqrt{3}}{4} a_1$.Given $a_1 = 27 \mathring{A}$,$r = \frac{\sqrt{3}}{4} 	imes 27$.For $fcc$ lattice,the relation between edge length $a_2$ and atomic radius $r$ is $\sqrt{2} a_2 = 4 r$,so $a_2 = \frac{4 r}{\sqrt{2}} = 2 \sqrt{2} r$.Substituting the value of $r$: $a_2 = 2 \sqrt{2} 	imes \frac{\sqrt{3}}{4} 	imes 27 = \frac{\sqrt{6}}{2} 	imes 27$.Using $\sqrt{6} = \sqrt{2} 	imes \sqrt{3} = 1.41 	imes 1.73 = 2.4393$.$a_2 = \frac{2.4393}{2} 	imes 27 = 1.21965 	imes 27 = 32.93$.Rounding off to the nearest integer,we get $33 \mathring{A}$.

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