Question

Calculate the radius of iridium atom, given that Ir has an FFC crystal structure, a density of 22.4 g/cm3 and atomic weight of 192.2g/mol.

Answer 1

d = ZM/Na^3

in FCC no of atoms 4 in unit cell z = 4

M = 192.2g/mole

d = 22.4g/cm^3

N = 6.023*10^23

d = ZM/Na^3

22.4 = 4*192.2/6.023*10^23 * a^3

a^3   = 4*192.2/6.023*10^23*22.4

a^3   = 5.7*10^-23

a    = 3.8*10^-8 cm

for FCC

a = 22 r

3.8*10^-8 = 2*1.414*r

r              = 3.8*10^-8/2.828    = 1.34*10^-8cm

The radius of iridium atom = 1.34*10^-8cm = 1.34A⁰ >>>>answer