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

Αt 20°C

Begin by finding the mass of the unit cell. Obtain the mass of an Fe atom from its molar mass. Since the BCC unit cell contains 2 atoms per unit cell, multiply the mass of Fe by 2 to get the mass of a unit cell.

m (Cr atom) = 55.85 x 1 / 6.023 x 10²³ = 9.27 x 10^-23

m (unit cell) = 9.27 x 10^-23 x 2 atoms = 18.54 x 10^-23

Compute the volume of the unit cell (in cm) by converting the edge length to cm and cubing the edge length. (We use centimeters because we want to report the density in units of g/cm3.)

V = (a)³

    = (2.866 x 10-8)³

    = 2.354 x 10^-23

Density = m(unit cell) / V = 18.54 x 10^-23  /2.354 x 10^-23   = 7.8759 g/cm3

Αt 950°C

Since the BCC unit cell contains 4 atoms per unit cell, multiply the mass of Fe by 4 to get the mass of a unit cell.

m (Cr atom) = 55.85 x 1 / 6.023 x 10²³ = 9.27 x 10^-23

m (unit cell) = 9.27 x 10^-23 x 2 atoms = 37.08 x 10^-23

Density = m(unit cell) / V = 37.08 x 10^-23  / 2.354 x 10^-23   = 6.69155 g/cm3