Mean Free Path — Question 1

Question 1

Collision diameter determines an excluded cross section

For identical dilute hard-sphere molecules of diameter $d$ and number density $\nu$, use $\lambda=1/(\sqrt2\pi d^2\nu)$. Take $d=.400$ nm and $\nu=1.00\times10^{25}\,\mathrm{m^{-3}}$. A collision occurs when the centers of two spheres approach within distance $d$.

1. Explain why the collision cross section is $\pi d^2$, and calculate $\lambda$.

2. If diameter were mistakenly treated as a radius in $4\pi r^2$, by what factor would the inferred mean free path change? Compare with the result of neglecting the $\sqrt2$ motion correction.

Question 1 - Solution

Collision diameter determines an excluded cross section

(a) Method and calculation

$$r_1+r_2=d/2+d/2=d\ \Rightarrow\ \sigma_c=\pi d^2=\pi(.400\times10^{-9})^2.$$$$\lambda=\frac1{\sqrt2\pi(.400\times10^{-9})^2(10^{25})}\approx1.40674\times10^{-7}\,\mathrm m.$$

(b) Interpretation and further calculation

$$\sigma_{\mathrm{wrong}}=4\pi d^2=4\sigma_c\ \Rightarrow\lambda_{\mathrm{wrong}}=\lambda/4.$$$$\lambda_{\mathrm{stationary}}=1/(\pi d^2\nu)=\sqrt2\lambda\approx1.98944\times10^{-7}\,\mathrm m.$$The radius mistake and the neglected relative-motion correction are separate errors.

Physical check

The collision area is four times the geometric cross-sectional area of one molecule.