Practice question

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 2.0 × 10²⁵) × 3.14 × 6.0 × 10⁻⁷ = (1)/(5.33 × 10⁻¹⁹) ≈ 1.88 × 10⁻²⁰.d = √(1.88 × 10⁻²⁰) ≈ 1.37 × 10⁻¹⁰ m. Substituting values gives 1.37 ×