Question

Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Answer 1

Hint: FCC or face-centred cubic structure has atoms at each of the corners as well as at each of the faces of the cubical cell. These atoms that are arranged in the cell are known as lattice points.

Complete step by step solution:
An arrangement of atoms in crystals in which the atomic centres are disposed in space in such a way that one atom is located at each of the corners of the cube and one at the centre of each face is known as the face-centred unit cell. It is relatively "tightly" packed (atomic packing factor = 0.74).

Now, in the question it is given that the atomic radius, r of gold is 0.144 nm. And the type of cubic cell is FCC. Therefore, let the length of the side of the cell be 'a'.
Thus, the diagonal of the side is given by,
$Diagonal \,of\, the\, side = 4r = a\sqrt {2}$
$\implies\,side\,length\,of \,cell, a \dfrac {4r}{\sqrt {2}} = 2\sqrt {2}r$
$\implies a = 2 \times 1.414 \times 0.144 \times {10}^{-9}$
$\implies a = 0.4073 \times {10}^{-9}m = 0.4073 nm$
Therefore, the length of the side of the cell is 0.4073 nm.

Note: It is important to note the structure of the cubic lattice before calculating the radius and side length of the cell. Each lattice has a different sort of relation between the radius and the side length.