Question

The density of copper is $8.95$ g cm${}^{-3}$. It has a face-centered cubic structure. What is the radius of copper atom?
(Atomic mass of Cu $=63.5$ g mol${}^{-1}$, $N_A=6.02\times {10}^{23}$ mol${}^{-1}$)

Answer 1

Mass of one atom of copper $=\frac{\text{Molar mass of copper}}{\text{Avogadro's number}}=\frac{63.5}{6.023\times {10}^{23}}=1.054\times {10}^{-22}$ g
Mass of the unit cell $=4\times 1.054\times {10}^{-22}$ g
Volume of the unit cell $=\frac{\text{Mass of copper}}{\text{Density}}=\frac{4\times 1.054\times {10}^{-22}}{8.95}=4.71\times {10}^{-23}$ cm${}^3$
Volume of unit cell $=8\sqrt{8}{r}^3$
Hence, $r^3=\frac{4.71\times {10}^{-23}}{8\sqrt{8}}=2.07\times {10}^{-24}$ cm${}^3$
Hence, the radius of copper atom is $1.28\times {10}^{-8}$ cm.