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(B) Gold crystallizes in a face-centered cubic $(fcc)$ lattice,which is a type of cubic close-packed structure. For an $fcc$ unit cell,the number of a

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$4.07 	imes 10^{-8} \ cm$

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(B) Gold crystallizes in a face-centered cubic $(fcc)$ lattice,which is a type of cubic close-packed structure. For an $fcc$ unit cell,the number of atoms per unit cell,$Z = 4$. The density formula is given by $ho = \frac{Z 	imes M}{a^3 	imes N_A}$. Here,$ho = 19.3 \ g/cm^3$,$M = 197 \ g/mol$,and $N_A = 6.023 	imes 10^{23} \ mol^{-1}$. Substituting the values: $19.3 = \frac{4 	imes 197}{a^3 	imes 6.023 	imes 10^{23}}$. $a^3 = \frac{788}{19.3 	imes 6.023 	imes 10^{23}} \approx 6.78 	imes 10^{-23} \ cm^3$. Taking the cube root,$a \approx 4.07 	imes 10^{-8} \ cm$.

Gold has a cubic close-packed structure which can be viewed as spheres occupying $74\%$ of the total volume. What is the edge length of the unit cell if the density of gold is $19.3 \ g/cm^3$? $(Au = 197 \ amu)$

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