A metal crystallises in a face-centred cubic | vedclass

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(B) For an $FCC$ unit cell,the relationship between the edge length $a$ and the metallic radius $r$ is given by $4r = \sqrt{2}a$.Given $r = \sqrt{2} \

Vedclass · Accepted Answer

$6.4 	imes 10^{-29}$

Vedclass · Answer

(B) For an $FCC$ unit cell,the relationship between the edge length $a$ and the metallic radius $r$ is given by $4r = \sqrt{2}a$.Given $r = \sqrt{2} \ \mathring{A}$.Substituting the value of $r$: $a = \frac{4r}{\sqrt{2}} = \frac{4 	imes \sqrt{2}}{\sqrt{2}} = 4 \ \mathring{A}$.The volume of the unit cell is $V = a^{3} = (4 \ \mathring{A})^{3} = 64 \ \mathring{A}^{3}$.Since $1 \ \mathring{A} = 10^{-10} \ m$,then $1 \ \mathring{A}^{3} = 10^{-30} \ m^{3}$.Therefore,$V = 64 	imes 10^{-30} \ m^{3} = 6.4 	imes 10^{-29} \ m^{3}$.

$A$ metal crystallises in a face-centred cubic $(FCC)$ structure with a metallic radius of $\sqrt{2} \ \mathring{A}$. The volume of the unit cell (in $m^{3}$) is:

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