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

Ayame17
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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!
 
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