Question

Face centred cubic crystal lattice of copper has density of $8.996g.c{m}^{-3}$. Calculate the volume of the unit cell.
Given molar mass of copper is $63.5g.mo{l}^{-1}$ and Avogadro number $N_A$ is $6.022\times {10}^{23}mo{l}^{-1}$

Answer 1

Density of the unit cell $\left(d\right)=\frac{\text{Mass of the unit cell (M)}}{\text{Volume of the unit cell (V)}}....\left(i\right)$
Let number of atoms in the unit cell be $Z$ and mass of each atom $m$
then $M=Z\times m$
Mass of each atom $\left(m\right)=\frac{\text{Atomic mass}}{Avogadr{o}^{\prime }sno.}$
Thus from equation (i)
$d=\frac{Z\times Atomicmass}{Avogadr{o}^{\prime }sno.\times V}....\left(ii\right)$
For face centered cubic lattice (FCC), the total no. of atoms in the unit cell $\left(Z\right)=1/8\times 8+1/2\times 6=1+3=4$
So, $8.966gc{m}^{-3}=\frac{4\times 63.5gmo{l}^{-1}}{6.022\times {10}^{23}mo{l}^{-1}\times V}$
or, Volume of the unit cell $\left(V\right)=\frac{254}{53.99\times {10}^{23}}c{m}^3=4.7\times {10}^{-23}c{m}^3$
Hence, the volume of the unit cell is $4.7\times {10}^{-23}c{m}^3$