Gold crystallises in fcc lattice. The edge le | vedclass

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(B) For an $fcc$ lattice,the edge length $a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$.$1$. The closest distance between atoms $(x)$ in an $fcc$ latt

Vedclass · Accepted Answer

$2\sqrt{2}, 20.52$

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(B) For an $fcc$ lattice,the edge length $a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$.$1$. The closest distance between atoms $(x)$ in an $fcc$ lattice is given by $x = \frac{a}{\sqrt{2}}$.$x = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ \mathring{A} \approx 2.828 \ \mathring{A}$.$2$. The density $(y)$ is given by the formula $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For $fcc$,$Z = 4$. Given $M = 197 \ g \ mol^{-1}$,$N_A = 6 	imes 10^{23} \ mol^{-1}$,and $a = 4 	imes 10^{-8} \ cm$.$y = \frac{4 	imes 197}{6 	imes 10^{23} 	imes (4 	imes 10^{-8})^3} = \frac{788}{6 	imes 10^{23} 	imes 64 	imes 10^{-24}} = \frac{788}{38.4} \approx 20.52 \ g \ cm^{-3}$.Thus,$x = 2\sqrt{2}$ and $y = 20.52$.

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