Question

Metallic gold crystallizes in the face centred cubic lattice. What is the approximate number of unit cells in 20 g of gold? Atomic mass of gold is 197.
(1) $3.06 \times {10^{11}}$
(2) $6.02 \times {10^{23}}$
(3) $1.53 \times {10^{22}}$
(4) $1.25 \times {10^{22}}$

Answer 1

Hint: There are a total 4 atoms in a face-centred cubic unit cell or fcc. Mass of one mole of any substance is called its atomic or molar mass and it contains Avogadro number of atoms. You need to calculate the number of atoms that constitute 20 g of gold and then do calculation to find the number of unit cells in that calculated number of atoms.

Complete answer:
Firstly, let us discuss face centred cubic lattice (fcc).
A face-centred unit cell contains atoms at all the corners and at the centre of all the faces of the cubic unit cell. There are a total 8 corners atoms and 6 face-centre atoms in the cubic unit cell.
Now, contribution of an atom located at the corner in a fcc unit cell = $\dfrac{1}{8}$ per unit cell.
Contribution of an atom located at the face-centre in a fcc unit cell = $\dfrac{1}{2}$ per unit cell.
Thus, total number of atoms in the fcc unit cell = ${\text{8 corner atoms }} \times \dfrac{1}{8}{\text{ atom per unit cell + 6 face - centred atoms }} \times {\text{ }}\dfrac{1}{2}{\text{ atom per unit cell }}$
$\therefore $ Total number of atoms in a unit cell of fcc lattice = $1 + 3 = 4{\text{ atoms}}$
Given that, atomic mass of gold = 197g.
Number of atoms in 1mole of gold i.e., 197 g of gold = ${N_A} = 6.022 \times {10^{23}}{\text{ atoms}}$
$\therefore $ Number of atoms in 20 g of gold = $\dfrac{{6.022 \times {{10}^{23}}}}{{197}} \times 20 = \dfrac{{120.44}}{{197}} \times {10^{23}}{\text{ atoms}}$
And, we calculated above that there are 4 atoms in 1 unit cell of fcc lattice.
$\therefore $ When number of atoms are $\dfrac{{120.44}}{{197}} \times {10^{23}}{\text{ }}$, number of unit cells are = $\dfrac{1}{4} \times \dfrac{{120.44}}{{197}} \times {10^{23}}{\text{ }} = \dfrac{{30.11}}{{197}} \times {10^{23}} = 1.53 \times {10^{22}}{\text{ unit cells}}{\text{.}}$

Hence, option (3) is correct.

Note:
A key point to note that each atom located at the face centre is shared between two adjacent unit cells and hence only $\dfrac{1}{2}$ part of each atom belongs to the unit cell. An atom located at the corner is shared between 8 adjacent unit cells, hence only $\dfrac{1}{8}$ part of an atom belongs to the unit cell.