Question

Αt 20°C, elemental iron is bcc, a = 2.866 Å. At 950 °C, Fe is ccp, a = 3.43 Å. At each temperature, calculate: a) The density of iron, b) The metallic radius of iron atoms. For Fe: A = 55.85 g/mol

Answer 1

For elemental Fe

at 20 oC

a = 2.866 Angstrom = 2.866 x 10^-8 cm

Volume of unit cell = a^3 = (2.866 x 10^-8)^3 = 2.35 x 10^-23 cm^3

Mass of unit cell = 2 x 55.85/6.023 x 10^-23 = 1.85 x 10^-22 g

a) density of unit cell = 1.85 x 10^-22/2.35 x 10^-23 = 7.88 g/cm^3

b) radius = sq.rt.(3a/4) = sq.rt.(3 x 2.866 x 10^-8/4) = 1.241 x 10^-8 cm = 1.241 Angstrom

at 950 oC

a = 3.43 Angstrom = 3.43 x 10^-8 cm

Volume of unit cell = a^3 = (3.43 x 10^-8)^3 = 4.03 x 10^-23 cm^3

Mass of unit cell = 4 x 55.85/6.023 x 10^-23 = 3.71 x 10^-22 g

a) density of unit cell = 3.71 x 10^-22/4.03 x 10^-23 = 9.19 g/cm^3

b) radius = sq.rt.(2) a/4 = sq.rt.(2) x 2.866 x 10^-8/4 = 1.213 x 10^-8 cm = 1.213 Angstrom