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Q: Calculate average length of photon path: emit->absorb

  1. Nov 24, 2015 #1
    Assume a very large volume of a gas of a specific molecule. This molecule can absorb a photon and spontaneously quickly re-emit a photon of the same energy before the molecule can physically interact with another molecule. Therefore the energy of these photons never participate in the equilibrium temperature of the gas. Given the specific molecule, its numeric density (number of molecules per m3), and the energy of the photon, how does one calculate the average distance such a photon will travel between emission and absorption by another molecule?

    I am not sure if this is the right forum to pursue this question. It seems like physical chemistry, but I may be mistaken. I have been unable to find any reference to explain this calculation. I have found references to absorption coefficient, cross section, and mean free path, but nothing that puts all the pieces together.

    I would appreciate any help.
     
  2. jcsd
  3. Nov 24, 2015 #2
  4. Nov 24, 2015 #3
    Hi Scott:

    Thanks for your post.

    I have read the Wikipedia article on cross section. It doesn't explain how to get the mean path length from the cross section. Also
    it says:
    σ = μ/n​
    where
    • σ is the cross section of this event (SI units: m2),
    • μ is the attenuation coefficient due to the occurrence of this event (SI units: m−1), and
    • n is the number density of the target particles (SI units: m-3).
    In another article I am instructed to use the absorption coefficient, rather than the attenuation coefficient, since I am interested in the absorption cross section. I know the value for n, but I cannot find a clear explanation about the value of the absorption coefficient. I am guessing It comes from the molecule's absorption spectrum for the particular wave length of the photon, but exactly how is unclear. If that is right, I can get the μ value, which on absorption spectrum charts, but it is shown as a percent. How do I get from a percent to the m-1 unit? For example, in some charts for some wavelengths the absorption is 100%. The m-1 unit might mean I am to assume that the photon doesn't have a specific wavelength but a range, and the absorption coefficient is the inverse of this range. Although unit-wise this seems plausible, intuitively it seems very strange.

    So, if I find a way to do calculate μ, then I have n and μ and can calculate σ. How do I get from their to the average path length.

    Also, I think the probability distribution of the path length is exponential, but it would be useful to have that verified also.

    Regards,
    Buzz
     
    Last edited: Nov 24, 2015
  5. Nov 24, 2015 #4
  6. Nov 25, 2015 #5
    Hi Scott:

    I appreciate your trying to help.

    I have also read the Wikipedia article on Mean Free Path (MFP). It gives as a reference
    This discussion of MFP seems to be about molecules hitting molecules, rather than the path length of a photon.

    Regards,
    Buzz
     
  7. Nov 25, 2015 #6
    There is cross section of molecule-molecule collisions and another one for photon absorption. If you use the last one, you get the mean free path of photons.
    The formula is the same.
    λ=1/(n σ) where n is the concentration of absorption centers (here molecules).
     
  8. Nov 25, 2015 #7

    jtbell

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    Staff: Mentor

    The m-1 unit on μ has nothing to do with wavelength. (Although the value of μ does generally depend on the wavelength.) It has to do with the distance traveled. Generally you have an "exponential absorption law" $$N = N_0 e^{-\mu x}$$ where N0 is the initial number of particles (photons) and N is the number that survive after a path length x. If x is in m, then μ must be in m-1 in order to make the exponent dimensionless. Going further, μ depends on both the concentration of "targets" (n) and the "cross section" per target (σ): μ = nσ so $$N = N_0 e^{-n \sigma x}$$ Finally, it can be shown with some calculus, that for exponential absorption the mean path λ equals 1/μ = 1/(nσ) as nasu noted. So yet another way to write the exponential absorption law is $$N = N_0 e^{-x/\lambda}$$
     
    Last edited: Nov 25, 2015
  9. Nov 25, 2015 #8
    Hi nasu:

    Thanks for your post.

    I am not sure which article you are referring to:
    (1) The Wikipedia article on cross section
    (2) The Wikipedia article on MFP
    (3) The "The Mean Free Path in Air: article form the Journal of Aerosol Physics

    (1) seems to only about scattering
    (2) has a section "Mean free path in radiography" which discusses photons, but not in a useful way. At the beginning of this article the following MFP value is given for air at sea level: 68 nm. If I were interested in molecules hitting each other, that would be useful. The later discussion about radiography gives
    Radiography eq.png
    which is not helpful since there is no discussion of how to calculate μ. A link to
    https://en.wikipedia.org/wiki/Attenuation_coefficient
    is also not helpful.
    (3) I cannot get access to reading the article.

    I would much appreciate it if you could post something specific about calculating μ from some available information about a specified molecule. The molecule I am particularly interest in is CO2.

    Regards,
    Buzz
     
  10. Nov 25, 2015 #9

    Ygggdrasil

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    Science Advisor

    A good starting point for calculating the absorption cross section of a molecule is to start with the Beer-Lambert law, which states that the absorbance of a solution (at a particular wavelength) is given by the formula A = εlc where l is the pathlength of your spetrophotometer, c is the concentration of your molecule of interest, and ε is the molar extinction coefficient. ε is typically given in units of L mol-1 cm-1, that is inverse molar concentration times inverse length to give a unitless absorbance when you multiply εlc together.

    However, if you perform dimensional analysis on the molar extinction coefficient, you will see that it has units of area per mole. Thus, if you divide the molar extinction coefficient by avogadro's number, you recover the absorption cross section of your molecule of interest.

    Alternatively, you can use the A = εlc formula to determine the mean free path of a photon directly by converting absorbance to a probability of absorption.
     
  11. Nov 25, 2015 #10
  12. Nov 25, 2015 #11
    Hi nasu:

    I apologize for misunderstand your post #6. Thanks you for clarifying in your post #10, and for the citation.

    I now get that λ=1/(n σ) is the means free path that length I want. I still don't know how to get the value of σ.

    You seem to know about much more this topic than I do, so I am hopeful you might be able to teach me something about μ I don't understand. Perhaps you can show me how to get a value for μ for a specific example. How about CO2 and a photon wave length of 6 μm.

    BTW, I noticed that in the cited paper there were many typographical errors substituting "-" for an intended "=".

    Regards,
    Buzz
     
    Last edited: Nov 25, 2015
  13. Nov 25, 2015 #12
    Hi jtbell:

    Thank you for your post.

    If I understand the equation for the absorption law correctly, it implies the following as the probability distribution function p(x) for the path length x:
    p(x) = λ ex/λ .​
    Is this correct?

    Regards,
    Buzz
     
  14. Nov 25, 2015 #13
    Hi Ygggdrasil:

    Thank you for your post.

    I looked at the Beer-Lambert law article, and I confess I found it to be unintelligible.

    As I asked nasu in my post #12, perhaps you can help me by explaining specifically how to calculate the value of μ for CO2 and a photon wavelength of 6 μm.

    Regards,
    Buzz
     
  15. Nov 25, 2015 #14

    Ygggdrasil

    User Avatar
    Science Advisor

    I took a look at the wiki article for the Beer-Lambert law and agree that it is very poorly written. Here is a site with perhaps a more clear explanation: http://chemwiki.ucdavis.edu/Physica...onic_Spectroscopy_Basics/The_Beer-Lambert_Law

    In general, the attenuation coefficient (µ) is something that is measured experimentally. This is likely something you would have to look up or extract from experimental data, and the information in your opening post does not provide enough information to calculate such a figure. Looking at the absoption spectrum of CO2, however, shows very little absorbance at 6 µm (1666 cm-1). http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Type=IR-SPEC&Index=1#IR-SPEC
     
  16. Nov 25, 2015 #15
    Hi Ygggdrasil:

    Thank you for your post and the links.

    It only shows the extent of my ignorance that when I looked at a spectrum for CO2, I interpreted the chart values near 1.0 as meaning near 100% absorption rather than transparency.

    OK, I now choose a wavenumber of 2350 cm-1. This appears to have a transmittance value of about 2.5%. Can σ be calculated from that? If not, can you suggest a reference with absorption values for σ for CO2 for photons with IR wavelengths, frequencies, or wavenumbers?

    Regards,
    Buzz
     
  17. Nov 25, 2015 #16

    jtbell

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    Staff: Mentor

    Using the notation of the equations that I posted, the transmittance is N/N0. In order to calculate σ using my second equation in post #7, you need to know n and x. Clicking the "Help / Software credits" tab on the graph linked in Yggdrasil's post indicates that the path length (x) is 10 cm. However, directly underneath the graph is the notice "Concentration information is not available for this spectrum and, therefore, molar absorptivity values cannot be derived." This seems to indicate that n (molecules/cm3) is unknown. Therefore you can't calculate σ from this graph. If you can find similar data or a graph that includes the density or concentration of CO2 used, it should be possible to do the calculation.
     
  18. Nov 25, 2015 #17
    Hi jtbell:

    Thanks for your post.

    The link in Yggdrasil's post is
    I see that the "Path length" is 10 cm.

    I gather than that the 10 cm value relates to some part of the experimental setup. Can you explain what part of the apparatus is 10 cm?

    There is also an entry:
    Name(s) dioxomethane .​
    What does this substance have to do with CO2?

    Another entry:
    State GAS (200 mmHg DILUTED TO A TOTAL PRESSURE OF 600 mmHg WITH N2) .​
    This seems to say that the gas of interest (CO2 or dioxomethane) has a partial pressure of 200 mmHg. Unfortunately we don't have a temperature. With a temperature and pressure we could calculate a volume and a density.

    Regards,
    Buzz
     
  19. Nov 25, 2015 #18

    jtbell

    User Avatar

    Staff: Mentor

    I interpret that to mean that there is a cell containing the gas under study, 10 cm long, and that the light passes through it.

    Disclaimer: This is educated guessing on my part. I am not a spectroscopist.
     
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