Question

Gold crystallizes in the face centred cubic lattice. Calculate the approximate number of unit cells in $2mg$ of gold. (atomic mass of gold $=197u)$.

Answer 1

Contribution of an atom present at the corner of the fcc unit cell = $\frac{1}{8}$ per unit cell

Contribution of an atom present at the face-centre of the fcc unit cell = $\frac{1}{2}$ per unit cell

Therefore, the total number of atoms in a fcc unit cell =

$8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$

Mass of gold (Given) = 197g

No. of atoms in i mole of gold, $N_A$ = $6.022\times {10}^{23}$ atoms

Therefore, number of atoms in 20g of gold = $\frac{6.022\times {10}^{23}}{197}\times 20=\text{x (Let's say this value is equal to x})$

As there are 4 atoms in 1 unit cell of fcc lattice

Hence, number of unit cells = $\frac{1}{4}\times x=1.53\times {10}^{22}$unit cells