# Mean free path of electron in air?

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

$$$$\lambda \approx \frac{1}{n\sigma}$$$$

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

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

The collision cross-section is

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

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?

• 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. Commented Feb 18, 2021 at 11:15

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.