Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I was wondering if there's a simple way to compute the mean free path of UV photons in a optically thick medium with density n.

I've looked up at the literature and found out that the mean free path of Lyman-Werner band photon is ~1 Mpc on cosmological scales but 0.1 pc inside clouds. However, I would like to read an article about that but I can't find any.

share|improve this question
    
Might be a stupid question, but what is the background you ask this, do you know anything about the absorption process? I don't know but I'd guess it heavily depends on the particular cross section. –  NiftyKitty95 Jul 19 '12 at 12:18
    
It's an expression of the cross section that I would like to find out. The problem is computing the mean free path of UV photons in a molecular cloud where dust and H2 are responsible for absorption. You can also assume that the dust density is proportional to the metallicity of the gas. –  Matteo Jul 19 '12 at 12:20
    
Okay, I don't know but maybe it goes somewhere in this and that direction. –  NiftyKitty95 Jul 19 '12 at 12:56
    
Can you clarify what you mean by "density"? Atoms per volume, or optical density, or index of refraction or what? Also, are the photons scattered, absorbed, or both? –  Steve B Jul 19 '12 at 14:13
    
By density I mean atoms per volume and the photons are only absorbed by the medium –  Matteo Jul 19 '12 at 14:37
add comment

2 Answers

If you know specifically what medium, you can look it up in the NIST XCOM database:

http://www.nist.gov/pml/data/xcom/index.cfm

Click on the "Database Search Form" and enter the details you have. Read the introduction text to determine how to use the results, but you will likely only need to multiply what the chart says by the density of the material.

share|improve this answer
add comment

For light traveling through a medium in the direction x,

$$I(x) = I_0 e^{-\alpha x}$$

where $I(x)$ is the intensity of light at position $x$ and $I_0$ is the intensity of light at position $x=0$. $\alpha$ is called "absorption coefficient". The absorption coefficient is related to absorption cross-section by

$$\alpha = n\sigma$$

where $n$ is the number of absorbing "things" (I guess atoms or molecules) per volume and $\sigma$ is the absorption cross-section of a single "thing". The mean free path is--I assume--defined as the mean distance which a photon can travel before getting absorbed. That would be $1/\alpha$.

share|improve this answer
    
Thanks, but the problem here is where to find the absorption cross section of UV photons, better as a function of wavelength. –  Matteo Jul 23 '12 at 10:55
    
What exactly is the absorber? An H2 gas molecule? Something else? –  Steve B Jul 23 '12 at 12:20
    
The absorber is another H2 molecules and dust particles (you can assume that the dust density scales with the gas density and is proportional to the metallicity of the gas). –  Matteo Jul 23 '12 at 12:59
    
I'm not much help. Keep trying literature search. For example, for H2 I see nist.gov/data/PDFfiles/jpcrd339.pdf –  Steve B Jul 23 '12 at 13:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.