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

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SUMMARY

The discussion focuses on calculating the mean free path (mfp) of photons in a neutral atomic hydrogen cloud with varying densities (n_H = 1, 10, and 100 cm^{-3}) and an absorption cross section of σ₀ = 6.3 × 10⁻¹⁸ cm². The relevant equation for optical depth is τₙ = σ₀ * (ν/ν₀)⁻³.⁵ * nₕ. The mean free path for photons with energy of 20 eV is derived using the formula l = 1/(nₕ * σ), where nₕ represents the hydrogen density. The comparison with the mean free path for photons at the ionization energy of hydrogen (13.6 eV) is also discussed.

PREREQUISITES
  • Understanding of mean free path calculations in physics
  • Familiarity with absorption cross sections
  • Knowledge of optical depth and its significance
  • Basic concepts of photon energy and ionization energy
NEXT STEPS
  • Calculate the mean free path for photons at different energies using the formula l = 1/(nₕ * σ)
  • Explore the implications of varying densities on photon behavior in neutral atomic hydrogen clouds
  • Investigate the relationship between optical depth and mean free path in astrophysical contexts
  • Study the effects of different absorption cross sections on photon interactions
USEFUL FOR

Students and researchers in astrophysics, particularly those studying photon interactions in neutral atomic hydrogen environments, as well as educators teaching concepts related to mean free path and optical depth.

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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Anyone got any idea?
 

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