Question

The unit cell of metallic gold is face-centred cubic.
(a) How many atoms occupy the gold unit cell?
(b) What is the mass of a gold unit cell?

Answer 1

Hint: Firstly, we all know that the unit cell is the smallest unit of any crystal lattice. In case of a simple cubic cell containing there are eight atoms, ions or molecules present at the corners of a cube. In a body-centered cubic (bcc) cell there is an additional component present at the center of the cube. So, there are 2 atoms per unit call having coordination number 8.
Also, we know that in a face-centered cubic (fcc) there are 4 atoms per unit cell and the coordination number is 12. The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell.

Complete step by step answer:
For a face centered cubic cell, we know that there are six different faces. So, there are atoms per unit cell. On each face of the cube, there are 8 corner atoms and one atom of each face. So, the face atoms share atoms with their adjacent cells. Thus, each cell will contain 1/2 of a face atom.
So, we can easily calculate the total number of atoms in a unit cell by the 4 lattice points per unit cell.
$
  {Z_{effective}} = 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} \\
  {Z_{effective}} = 4atoms \\
 $
Thus, in each gold unit cell there are 4 gold atoms.
So, the mass of gold unit cell containing 4 gold atoms is

Mass of gold unit cell = $4 \times {M_{gold}}$
As we know, mass of gold = \[196.96{\text{ }}u = 197u\]
Thus, mass of gold unit cell = $4 \times 197 = 788ammu$

Note:
FCC is stronger than BCC as the atoms are properly arranged in FCC crystal. There are seven different crystal systems: monoclinic triclinic, trigonal, orthorhombic, cubic, tetragonal, and hexagonal. We know that the body-centered cubic (bcc) has a coordination number 8 and it contains 2 atoms per unit cell.