An element crystallizes in a bcc lattice. The | vedclass

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(C) For a $bcc$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$,which implies $a = \f

Vedclass · Accepted Answer

$2.16 	imes 10^{-22}$

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(C) For a $bcc$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$,which implies $a = \frac{4r}{\sqrt{3}}$.Given $r = 2.598 \ \mathring{A} = 2.598 	imes 10^{-8} \ cm$.Substituting the value of $r$: $a = \frac{4 	imes 2.598 	imes 10^{-8}}{\sqrt{3}} = \frac{4 	imes 2.598 	imes 10^{-8}}{1.732} = 4 	imes 1.5 	imes 10^{-8} = 6 	imes 10^{-8} \ cm$.The volume of the unit cell is $V = a^3 = (6 	imes 10^{-8} \ cm)^3 = 216 	imes 10^{-24} \ cm^3 = 2.16 	imes 10^{-22} \ cm^3$.

An element crystallizes in a $bcc$ lattice. The atomic radius of the element is $2.598 \ \mathring{A}$. What is the volume (in $cm^3$) of one unit cell?

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