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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 NA = 6.022 x 1023.
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Solution
Edge length of unit cell (a) = 288pm
volume of unit cell = (a)3
=(288)3 = 2.389 × 10-23 cm3
density of iron = ρ = 7.86g.cm3
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 1023
= 2.0191
≈ 2 atoms per unit cell
In a body centered cubic structure (bcc), the total number of atoms equals 2.
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