Question

Ferrite (a - Fe), which has a BCC crystal structure, is the stable form of pure metallic iron from absolute zero to 900°c. Using the density of iron estimate it's atomic radius.

Answer 1

density of ferrite is 5 g/cm3

we have formula d = ( M x z) / ( N x a^3)

where M is atomic mass of Fe = 55.845 g/mol , N = avagadro number = 6.023 x 10^23

z = number of atoms per unit cell = 2 , a = edge length of unit cell

we get 5g/cm3 = ( 55.845 g/mol x 2 ) / ( 6.023 x 10^23 x a^3)

a = 3.33 x 10^-8 cm = 3.33 A

for BCC atoms connected along body diagonal

hence   4r = 1.732 a

r = ( 1.732 x 3.33 A) / ( 4) = 1.44 A     ( where A = angstrom units = 10^-8 cm)