Answer to Question #268330 in Molecular Physics | Thermodynamics for Yusuf

Question

Gold has a face centered cubic unit cell and has the density 𝟏𝟗. 𝟑 𝒈 𝒄𝒎−𝟑. Calculate the

length along an edge. (𝐴𝑢 = 197 𝑔 𝑚𝑜𝑙−1)

Answer 1

FCC type gold contain atom

$\frac{1}{8}\times8+6\times\frac{1}{2}=4$

Mass of unit cell of FCC type

$4\times\frac{197}{6.023\times10^{23}}=1.30\times10^{-21}$ g

Density of gold =19.3g/cm³

Density=$\frac{mass of unit cell }{volume of unit cell}$

Volume of unit cell =$\frac{mass of unit cell }{density of gold}$

$$V=\frac{1.30\times10^{-21}}{19.3}=6.73\times10^{-23}cm^3$$

$V=a^3=6.73\times10^{-23}cm^3$

$a=4.06\times10^{-8}cm$

For FCC type unit cell

$a=\sqrt{8}r$

$$r=\frac{a}{\sqrt8}=\frac{4.06\times10^{-8}}{\sqrt8}=1.43\times10^{-8}cm$$