1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Radiation propagation

  1. Jun 24, 2009 #1
    I am trying to do a simple calculation on radiation propagation using Cs-137 as an example.

    Cs-137 has the following properties:

    • Activity = 3.400e+15 Bq/kg
    • Decay energy = 188.096e-12 J (1.174 MeV)

    This results in 639.526e+03 J/s of radiation, which is then inversly proportional to the distance from the source (or is it squared...I think that was for intensity...?). But...this is at distance 0, so it doesn't work. Could I perhaps somehow use the cross-sectional area of the nuclei?

    I have a PDF document with algorithms for more accurately calculating radiation propagation, with the specific distance propagated and taking into account absorption by a medium etc., but first I would like to understand the basics. Primarily the inversly proportional thingy, as the energies aren't reference distance at 1 m (or any other reference distance), but at s=0.

    From there on I should then be able to calculate the absorbed dose, and then the equivalent dose.
     
  2. jcsd
  3. Jun 24, 2009 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Assuming the source is emitting in all directions - it isn't in a sheilded housing with a window - then it must be the square of the distance.
    Think of concentric spheres around the source, the same number of particles must go through each sphere but the surface area is the square of the radius.
     
  4. Jun 24, 2009 #3
    Cs-137 has two types of radiation, a 661 KeV photon and a 512 KeV (max energy) beta- (95%); and a 1.174 MeV (max energy) beta w/o photon(5%). The betas and the photon obey different absorption characteristics in tissue.
    1 Rad = 100 ergs per gram of absorbed dose, and
    1 erg = 10-7 joules
     
  5. Jun 25, 2009 #4
    Does radiation behave similarly to a sound source? I.e. is the directivity coefficient the same?
     
  6. Jun 25, 2009 #5

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Any source that emits equally in all directions is going to have an inverse square law - just from the pure geometry example given above.
     
  7. Jun 25, 2009 #6
    I'm talking about shielding/reflection, being a radiant source, would a radioactive source placed in the corner of three surfaces be stronger than a source placed in an open field?

    Something similar to http://www.engineeringtoolbox.com/directivity-coefficient-sound-d_67.html

    Or does ionising radiation not reflect that well off common surfaces/materials?
     
    Last edited: Jun 25, 2009
  8. Jun 26, 2009 #7
    In general, radiation from radioactive sources (photons, betas, alphas, neutrons) does not reflect at all off of any surface, but there are exceptions. Neutron sources (PuBe or plutonium-beryllium) are reflected. Also, photon sources, like from Cs-137, do "reflect" some scattered Compton gamma rays, often referred to as backscattered x-rays. See
    http://www.larrylawson.net/compton.htm
    Another "backscatter" source could be positrons from beta+ decay, where the positrons annihilate with electrons and emit isotropic 511-KeV gammas.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook