Question

Calculate the lattice constant of Fe. Density of iron is 7800 Kg/m3, Atomic weight is 55.85 and Avagadro’s number is 6.023 x 1026 atoms/mol.
4 points
2.863 x 10^-10
2.683 x 10^-12
1.868 x 10 ^-12
1.868 x 10 ^-10

Answer 1

Given :

Density (ρ) = 7800kg/m³

Atomic weight (w) = 55.85

Avogadro's number (Nₐ)= 6.023×10²⁶ atom/mol

To find :

Lattice constant (a)

Formula :

a³ = (n × w) / (ρ × Nₐ)

Solution :

No. of atoms (n) = 2

a³ = (n × w) / (ρ × Nₐ)

a³ = (2 × 55.85) / (7800 × 6.023 × 10²⁶)

a = 2.87 × 10⁻¹⁰

Therefore the lattice constant is 2.87 × 10⁻¹⁰.