Question

Chromium metal crystallizes as a body-centered cubic lattice. If the atomic radius of Cr is 1.25 angstroms, what is the density of Cr metal in g/cm3?

Answer 1

Density = mass / volume

Mass of unit cell = No. of particles in unit cell X mass of one particle

No. of particles in BCC unit cell = 2

( atoms at corners contribute one atom to the unit cell. In addition there is one atom in the body of the unit cell. So total atoms = 2)

Mass of one atom = Atomic mass / Avagadroes No. = 52.0 / 6.022 X10²³ = 8.64 X 10^-23 g

Hence mass of unit cell = No. of particles in unit cell X mass of one particle

Hence mass of unit cell = 2 X 8.64 X 10^-23 g = 17.28 X 10^-23 g

Volume of unit cell = a³

a = length of the cube

It can be shown that a is related to radius of atom in BCC unit cell as

a = 4 r / (3)^1/2

r = radius of Cr = 1.25Ao = 1.25 X 10^-10m = 1.25 X 10^-10 X 10² = 1.25 X 10^-8cm

a = 4 X 1.25 X 10^-8 / (3)^1/2 = 2.89 X 10^-8 cm

Volume = a³ = (2.89 X 10^-8 )³ = 2.4 X 10^-23cm^-3

Density = mass / volume = 17.28 X 10^-23 g / 2.4 X 10^-23cm^-3 = 7.18g/cm^-3