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Radioactive flux attenuation in air - Need Help

  1. Nov 5, 2009 #1
    Hi,

    If we calculate the flux at a certain distance from a source of radiation by first neglecting the effect of attenuation in air, how does one factor in the effect of air attenuation?

    I do not know how to apply that theory. I'm am trying to understand what the difference is at sea level or at high elevation.

    Helping understanding both situations and how to factor it into the calculations would be great.

    Thanks.
     
  2. jcsd
  3. Nov 5, 2009 #2

    Astronuc

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    Staff: Mentor

    From a point source, the flux falls off as roughly 1/r2, which is a geometric effect. Attentuation by scattering or absorption is related to the atomic density and the microscopic cross-section. The product of atomic density (N) and microscopic (σ) = macroscopic cross-section (Σ), which gives the probability per unit length of a reaction (attentuation by scattering or absorption).

    Attenuation over a distance,d, is given by exp(-Σd). Charged particles lose energy by collision with atoms/molecules and electrons, as well as ionization of atoms/molecules.

    If one is considering gamma radiation, then one must consider scattering, e.g. Compton scattering, whereby the scattered photon has less energy than the initial photon. From a single monoergetic photon source, one would see a cascade (or spectrum) of photon energies as a function of distance from the source.
     
  4. Nov 5, 2009 #3
    Ok so I know the decay energy of a radioactive source and the decay energy. I know the sources isoytopically emits into 4pi which means that it emits into a sphere. I want to know the flux at some radius r so is the energy per second divided by the total area of 4pir^2. That is assuming purely geometric conditions of the flux reduction.

    Is that microscopic cross section dependent on the type of air you are in, sea level vs. high altitude? So if I get that attenuation (exp(-Σd)) do I just multiply that by the flux I previously calculated?

    What if the decay type is not gamma but beta- (electron emission)? How do I factor in Compton scattering?

    Thanks for the help.
     
  5. Nov 5, 2009 #4
    Here is what I have for positron and electron range in air (from Evans "The Atomic Nucleus, pgs 622-625)
    Range R in milligrams per cm2 of air

    R = 412 En

    where E is energy in MeV

    and n = 1.265 - 0.0954 Ln(E)

    For a radioactive source, the electrons and positrons have a continuous energy spectrum up to Emax.

    Electrons do not Compton scatter. Only photons do.

    Bob S
     
  6. Nov 6, 2009 #5
  7. Nov 6, 2009 #6
    I don't have this book so I do not know how to apply this. How does it change if one is at sea level verses at high altitude where the air is thinner?

    Also, why don't electron's compton scatter? I read that link posted above.

    Thanks.
     
  8. Nov 6, 2009 #7
    "Electrons do not Compton scatter. Only photons do."
    Here are the first two sentences from the URL.

    "In physics, Compton scattering or the Compton effect is the decrease in energy (increase in wavelength) of an X-ray or gamma ray photon, when it interacts with matter. Because of the change in photon energy, it is an inelastic scattering process"

    Bob S
     
  9. Nov 6, 2009 #8
    Here is what I have for positron and electron range in air (from Evans "The Atomic Nucleus, pgs 622-625)
    Range R in milligrams per cm2 of air

    R = 412 En

    where E is energy in MeV

    and n = 1.265 - 0.0954 Ln(E)

    For a radioactive source, the electrons and positrons have a continuous energy spectrum up to Emax.

    Air at sea level has a density of about 1.2 milligrams per cm3. At 7000 ft elevation, the density is only about 80% of this or 0.96 milligrams per cm3, so the electron's range in cm is about 25% longer. So use the formula in my previous post to calculate R in milligrams per cm2, and divide by the air density in milligrams per cm3 to get the range in cm.

    Bob S
     
    Last edited: Nov 6, 2009
  10. Nov 8, 2009 #9
    Hi Bob,

    Thanks a lot.

    So if an radioactive isotope has a decay energy of 'a' and a life time in years 'b' does the flux equal the (decay energy * activity)/area it emits into. For some reason I am thinking that the decay energy is the total amount of energy it can give off. Am I right or is the decay energy the constant energy it gives off?

    - - - - - - - - - - - - - - - - - - - -

    Can I also switch the conversation a little to the attenuation of a gamma ray through a solid of thickness x.

    I was using this equation:

    I(x) = Io * exp(-upx)

    where:
    u: mass attenuation coefficient
    p: density of material
    x:thickness of material

    So I have no problem calculating the % attenuation but how do I correlate this to the probability that it is stopped in the material.

    Does the material absorb the energy that was attenuated? So if it was a detector and the absorbed energy reaches the detection energy then the ray is detected right? Or am I completely wrong in my assumption?
     
    Last edited: Nov 8, 2009
  11. Nov 8, 2009 #10
    Hi infinite qbps-

    The " energy" refers to the energy in electron volts that is released during a single decay. It has no direct relation to the average lifetime of the radiosctive decay.

    In alpha decays, the range of the alpha particle is a constant, and the alpha goes nearly exactly the same distance every time a nucleus decays before it stops. In beta decays, this is not true for two reasons.

    1) In beta decay, the electron (or positron) has to shere energy with a neutriono in order to conserve lepton number. So the electron (or positron) energy spectrum is continuous from near zero to a fixed maximum, depending on the free energy available.

    2) The electron or positron range, for any fixed energy, straggles because of their light mass. So a definite range cannot be assigneed to any particular electron or positron energy.Range for alpha particles is based on the Bethe-Bloch dE/dx equation, which is based on many many Coulomb collisions with atomic electrons. This doesn't work too well for electrons and positrons because of their mass. Photon penetation , however, is well-suited to an exponential exp(-kx) range form, because a single photon absorption event (e.g., Compton scattering) removes the incident photon from the photon flux.
    Bob S
     
    Last edited: Nov 8, 2009
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