I am trying to find the mean free path of silver atoms travelling through air (I am given the temperature, pressure and that they are attenuated by a factor $2.72$ in a distance of $10^{-2}$m).
I know that $\lambda=\frac{k_B T}{\sqrt{2}p\sigma}$ where $\sigma$ is the collisional cross-section, however I don't know how to find this from the attenuation?
I suggest looking into the Beer-Lambert law for further details on what I'm about to explain. It is known that the beam of (in this case) silver atoms will be attenuated in the form of $e^{-n\sigma d}$ where $d$ is the distance, $n$ is the gas (i.e., air) density and $\sigma$ is the collisional cross section. It's interesting that the amount of attenuation is 2.72 since this is very close to Euler's number (i.e., Napier's constant) $e^{1}$. Note that the beam is being attenuated by this amount such that $-n\sigma d$ should be equal to -1. You can relate density to pressure through the ideal gas law $p=nk_{\rm B}T$ which in constant temperature amounts to $p=\gamma_{0}n$ where $\gamma_{0}=k_{\rm B}T$. Based on your expression for $\lambda$, I ended up with $\lambda=\frac{d}{\sqrt{2}}\simeq 0.7$ cm. Please let me know if anything is unclear or if you have other questions.
