Single correct · +4 / −1
The formula for the mean free path λ in an ideal gas (number density n, molecular diameter d) is:
- (a) λ = 1/(√2 nπ d²)
- (b) λ = nπ d²
- (c) λ = RT/(nπ d²)
- (d) λ = d/n
Answer and solution
Answer (a) λ = 1/(√2 nπ d²)
Solution
Physics sir · 4 steps
Picture one molecule sweeping out a cylinder of radius d; anything inside it gets hit.
- Collision cylinder. Two molecules touch when their centres come within one diameter d. So a moving molecule hits every centre lying inside a cylinder of cross-section π d² around its path.
- Count collisions. In time t the cylinder has volume π d² v t and holds nπ d² v t molecules, so that many collisions happen. The distance travelled is vt, so the path per collision is 1/(nπ d²).
- Others move too. The other molecules are not at rest. Collisions depend on the average relative speed, which is √2 times the average speed. That raises the collision count by √2.
λ = 1/(√2 nπ d²)
- So. The answer is (a). Units check: nπ d² has units m⁻³ × m² = m⁻¹, so its reciprocal is a length, as a mean free path must be.
The idea to keep. Mean free path goes down when molecules are more crowded (n) or bigger (d²). The √2 comes from relative motion. Always check that your answer comes out as a length.
Answer from the NTA final answer key. Solution by PrepWiser.