Practice question
Question
What is the collision frequency of a gas molecule with a mean free path of 3.0 × 10⁻⁷ m and average speed of 600 m/s?
Explanation
**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. Collision frequency = ()/(l).(600)/(3.0 × 10⁻⁷) = 2.0 × 10⁹ s⁻¹. Substituting values gives 2.0 × 10⁹ s⁻¹, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.
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