Problem 5.176 — Scattering Cross Section

Problem Statement

The Rayleigh scattering cross section of an air molecule at $\lambda = 500$ nm is approximately $\sigma = 5\times10^{-31}$ m². The number density of air at STP is $n = 2.69\times10^{25}$ m$^{-3}$. Find the scattering mean free path.

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

• See the step-by-step solution for the specific equations applied.
• All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: $$l_{mfp} = \frac{1}{n\sigma} = \frac{1}{2.69\times10^{25}\times5\times10^{-31}} = \frac{1}{1.345\times10^{-5}} \approx \boxed{74\text{ km}}$$

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$l_{mfp} = \frac{1}{n\sigma} = \frac{1}{2.69\times10^{25}\times5\times10^{-31}} = \frac{1}{1.345\times10^{-5}} \approx \boxed{74\text{ km}}$$

$$\text{Numerical result} = \text{given expression substituted with values}$$

$$\boxed{l_{mfp} = \frac{1}{n\sigma} = \frac{1}{2.69\times10^{25}\times5\times10^{-31}} = \frac{1}{1.345\times10^{-5}} \approx \boxed{74\text{ km}}}$$

$$l_{mfp} = \frac{1}{n\sigma} = \frac{1}{2.69\times10^{25}\times5\times10^{-31}} = \frac{1}{1.345\times10^{-5}} \approx \boxed{74\text{ km}}$$

Answer

$$l_{mfp} = \frac{1}{n\sigma} = \frac{1}{2.69\times10^{25}\times5\times10^{-31}} = \frac{1}{1.345\times10^{-5}} \approx \boxed{74\text{ km}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.