I'm interested in approximating the mean free path of an electron in air. I think I'm going to need to add something more into my approximation because currently I calculate $400m$ for the mean free path at atmospheric pressure. Say the mean radius of an air molecule (either $\text{O}_{2}$ or $\text{N}_{2}$) is about $R=0.15nm$, the approximate mean free path of the electron, $\lambda$ at atmospheric pressure and room temperature is

\begin{equation} \lambda \approx \frac{1}{n\sigma} \end{equation}

where $n$ is the number density and $\sigma$ is the collision cross section. The number density at atmospheric pressure ($1 \ \text{atm} = 1.01 \times 10^{5} \ \text{Pa}$), is

\begin{equation} n = \frac{N}{V} = \frac{P}{k_{b}T} = 2.45 \times 10^{25} \ \text{m}^{-3} \end{equation}

The collision cross-section is

\begin{equation} \sigma = \pi (2r)^{2} = 10 \times 10^{-29} \ \text{m}^{2}, \end{equation}

using the classical electron radius of $r= 2.8 \times 10^{-15} \ \text{m}$. The mean free path of the electron is then $400$ m. I recognise the assumption of the radius and nature of the collisions does not make sense for interactions of charged particles. However, working out proper collision cross-section is quite hard.

What even is a reasonable mean free path of an electron in air at atmospheric pressure, and is there any smart way to approximate it?

  • $\begingroup$ Do $r$ and $R$ mean the same thing here? Apart from that, when we talk about a charged particle, taking the density-based radii as the basis for the cross-section is bound to overestimate the mean free path. $\endgroup$
    – Roger V.
    Commented Feb 18, 2021 at 11:15

1 Answer 1


I am very surprised nobody has commented on this.

You are making the mistake of using the electron radius instead of the radius of the molecule. Since the molecule larger in size, using the hard sphere collision model, it will dominate the cross section.

For hard sphere collisions, the cross section is: $$ \sigma = \pi (r_\mathrm{el} + r_\mathrm{molecule})^2 $$

Let's take an estimate of the radius of the molecule. According to wiki the Van der Waals radius of Nitrogen is $\approx 10^{-10}~$m. Let's say, the molecule will have $\approx 2 \cdot 10^{-10}~$m.

Then using your equation for the mean free path: $$ \lambda = \frac{1}{2.45 \cdot 10^{25} \pi 2^2 10^{-20}} \approx 3 \cdot 10^{-7} \mathrm{m} $$

According to this article the mean free path of electron in air is $6.8 \cdot 10^{-7}~$ m, so this estimate is not so bad.


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