Calculating Mean Free Path of Photons: Neutral Atomic Hydrogen Cloud Comparison

[Ayame17]

Homework Statement

Neutral atomic hydrogen cloud, density $$n_H$$. Absorption cross section $$\sigma_{0} = 6.3*10^{-18}cm^{2}$$. Determine the mean free path of photons with energy of 20eV, for densities $$n_H$$ = 1, 10 and 100 $$cm^{-3}$$. Compare this to mfp for photons with ionisation energy of hydrogen (13.6eV) at the same densities.

Homework Equations

The only possibly relevant equation given in our notes is $$\tau_{\nu}=\sigma_{0}*(\frac{\nu}{\nu_{0}})^{-3.5}*n_{H^{0}}$$

The Attempt at a Solution

I'll be able to try the second bit once I figure out the first bit! I looked up some stuff on the mean free path, and figured that it could've just been $$l=\frac{1}{n_{i}*\sigma}$$, but then is $$n_i$$ the same as $$n_H$$? And the equation given in the notes (above), all that is said about it is that the optical depth of photons above the Lyman limit $$\nu_0$$ can be derived from it. I simply can't see where to put in the amount of energy so that it will make a difference!

 

[Ayame17]

Anyone got any idea?