Question

Silver has a cubic unit cell with a cell edge of $408\text{pm}$. Its density is $10{\text{g cm}}^{-3}$. How many atoms of silver are there in the unit cell?
Give the nearest integer value in this case.
(Take $N_A=6\times {10}^{23}$)

Answer 1

Length of the edge of unit cell $=408\text{pm}$
Volume of unit cell $=(408\text{pm}{)}^3=67.92\times {10}^{-24}{\text{cm}}^3$
Mass of unit cell = Density $\times$ Volume $=10{\text{g cm}}^{-3}\times 67.92\times {10}^{-24}{\text{cm}}^3=679.2\times {10}^{-24}\text{g}$
$\text{Mass of unit cell = No. of atoms in unit cell}\times \text{Mass of each atom}$
Now, mass of each atom = $\frac{\text{Atomic mass}}{\text{Avogadro number}}=\frac{108}{6\times {10}^{23}}=18\times {10}^{-23}\text{g}$
Let the unit cell contains ${}^{\prime }{n}^{\prime }$ atoms.
So the mass of unit cell $=n\times 18\times {10}^{-23}=679.2\times {10}^{-24}$
$\Rightarrow n=\frac{679.2\times {10}^{-24}}{18\times {10}^{-23}}=3.77$
$\therefore$ Number of atoms present in a unit cell $=4$
Since, the unit cell contains $4$ atoms per unit cell, silver has face centred cubic lattice.