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(C) For an $fcc$ unit cell, the relation between edge length $(a)$ and atomic radius $(r)$ is $a = 2\sqrt{2}r$. Given $r = 141.4 \ pm = 1.414 	imes 1

Vedclass · Accepted Answer

$6.4 	imes 10^{-23} \ cm^3$

Vedclass · Answer

(C) For an $fcc$ unit cell, the relation between edge length $(a)$ and atomic radius $(r)$ is $a = 2\sqrt{2}r$. Given $r = 141.4 \ pm = 1.414 	imes 10^{-8} \ cm$. Substituting the value of $r$: $a = 2 	imes 1.414 	imes 1.414 	imes 10^{-8} \ cm = 4.0 	imes 10^{-8} \ cm$. The volume of the unit cell is $V = a^3 = (4.0 	imes 10^{-8} \ cm)^3 = 64 	imes 10^{-24} \ cm^3 = 6.4 	imes 10^{-23} \ cm^3$.

Calculate the volume of a unit cell having an atomic radius of $141.4 \ pm$ forming an $fcc$ unit cell.

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