Given data:
Density of Iridium (Ir) = 22.4 g/cm3
Atomic mass of Ir = 192.2 g/mol
Therefore, volume of Ir atom/mol = 192.2 g.mol-1/22.4 g.cm-3 = 8.58 cm3/mol
1 mole of Ir has 6.023 * 10^23 Ir atoms
Hence, Volume/atom = 8.58 / 6.023 * 10^23 = 1.425*10^-23 cm3
Now, a Face centered cubic cell has 4 atoms/cell
Volume of a single unit cell = 4 * 1.425*10^-23 cm3 = 5.7 * 10^-23 cm3
Volume (V) of a cube is related to the length of its side (a) as:
V = a^3
a = (V)^1/3 = (5.7 *10^-23)^1/3 = 3.91*10^-8 cm
For an FCC cell, the length of the side (a) is related to the atomic radius (R) as:
a = 2.83 R
R = a/2.83 = 3.91*10^-8/2.83 = 1.38 *10^-8 cm
