#### Question

A unit cell of iron crystal has edge length 288 pm and density 7.86 g.cm^{-3}. Find the number of atoms per unit cell and type of the crystal lattice.

Given : Molar mass of iron = 56 g.mol^{-1}; Avogadro's number N_{A} = 6.022 x 10^{23}.

#### Solution

Edge length of unit cell (a) = 288pm

volume of unit cell = (a)^{3}

=(288)^{3} = 2.389 × 10^{-23} cm^{3}

density of iron = ρ = 7.86g.cm^{3}

Mass of iron unit cell (M) = Density × Volume

= 7.86 × 2.389 × 10^{-23}

= 18.778 × 10^{-23} g

molar mass of iron = 56 g.mol^{-1}

moles of iron in a unit cell = mass of iron in unit cell / molar mass of iron

= 18.778 × 10^{-23} / 56

= 3.353 × 10^{-24} mol

number of atoms per unit cell = moles of iron in a unit cell × Avogadro's number

= 3.353 × 10^{-24} × 6.022 x 10^{23}

= 2.0191

≈ 2 atoms per unit cell

In a body centered cubic structure (bcc), the total number of atoms equals 2.