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Emission probability calculation (quick and easy question)

  1. Apr 2, 2010 #1
    Hi!

    This is a very basic question. How do I calculate the total probability for a certain gamma of a decay? For example, I know from empiric method that the 2 gammas from Co60 have a probability of 50%. But for Eu152 I have lots of gammas and I need to compute them all in a simulation program called PENCYL (from the PENELOPE package), so, I have this data:

    http://ie.lbl.gov/toi/nuclide.asp?iZA=630152

    But the probabilities next to the energies are not the total probabilities... How do I calculate them?

    Thanks :)
    tirwit

    PS: WE'RE ARE ALL IN A BLACK HOLE!!! We just don't know it yet =P
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 3, 2010 #2

    phyzguy

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    I don't think you can calculate them. They are empirically measured. Why do you think you can calculate them?
     
  4. Apr 3, 2010 #3
    The table for nuclide search for Co60 at http://ie.lbl.gov/toi/nuclide.asp?iZA=270060, give intensity for it's 2 main gammas. Each gamma has a intensity of approximately 99,9%, but these 2 if sum, give more than 100%. But, for Co60 (a simple case just to demonstrate), we know that each of those gammas is seen 50% of the times for each of the gammas. So, my question is, how do we pass of a intensity of 99,9% to 50%? If you can show me the way or tell me where can I get info about that, I'll calculate the same thing for Eu152. :)
     
    Last edited by a moderator: Apr 25, 2017
  5. Apr 3, 2010 #4

    phyzguy

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    You're misunderstanding the decay chain. Try going to this link http://en.wikipedia.org/wiki/Decay_scheme , where there is a diagram of the Co-60 energy levels. Almost all of the time, both the 1.17 MeV and the 1.33 MeV gammas are produced by a single atom as the nucleus decays from Co-60 to Ni-60. That's why the probabilities sum to >100%, because the number of gammas produced per decay is >1. So we never have to pass from 99.9% to 50%.
     
  6. Apr 3, 2010 #5
    So when I'm writing the code and the instructions for the source definition say:

    C SPECTR : For a source with continuous (stepwise constant) spectrum,
    C each 'SPECTR' line gives the lower end-point of an energy
    C bin of the source spectrum (Ei) and the associated relative
    C probability (Pi), integrated over the bin. Up to NSEM=200
    C lines, in arbitrary order. The upper end of the spectrum is
    C defined by entering a line with Ei equal to the upper energy
    C value and with Pi set to a negative value.
    C DEFAULT: none

    how do I calculate the associated relative probability? The code example which had the Co60 defined, was like this:

    SPECTR 1173.237e3 0.5e0
    SPECTR 1173.273e3
    SPECTR 1332.501e3 0.5e0
    SPECTR 1332.501E3 -1.0e-1
     
  7. Apr 3, 2010 #6

    phyzguy

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    I'm not sure, but it looks to me like whoever wrote the Co-60 code example also misunderstood and that the code example is wrong.
     
  8. Apr 3, 2010 #7
    Nah, the code example is right ;). It comes in the PENELOPE package. It's a Monte Carlo Simulation for energy loss of particles in materials. It's used, among other things, to calibrate the efficiency of radiation detectors. It's driving me crazy lol :eek:

    My supposition is that for a spectrum of Co60, the 2 gammas represent 0,5 (each of them), of the gammas observed, where the whole spectrum of Co60 would correspond to 100%.

    Thanks for the help anyway =)
     
  9. Apr 3, 2010 #8

    phyzguy

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    I see. I think the difference is that your code is calculating the percentage of total gammas which are of each energy, while the nuclide tables are giving the number of gammas per decay which are of each energy. So I think the answer to your original question is that you just need to add up all of the gammas in the nuclide tables and divide by this number so that the sum of the total percentages are normalized to 1.
     
  10. Apr 3, 2010 #9
    Go to

    http://ie.lbl.gov/toi/nuclide.asp?iZA=630152

    Look in Most Recent ENSDF Data (12/2002) and click on the gammas for both the electron capture and the beta decay modes; they list the gamma probabilities in the Igamma column. [STRIKE]The sum for each decay mode is 100%.[/STRIKE] To get absolute intensity, multiply by the EC or beta (-) branching ratios listed at bottom of page. Eu [STRIKE]252[/STRIKE] 152 (odd-odd) decays to both Z=64 and Z=62 nuclei.

    Bob S
     
    Last edited by a moderator: Apr 25, 2017
  11. Apr 4, 2010 #10
    Following your method I also checked the datasheet for 60Co. The listed branching ratio in the bottom of the page was 1. Which would give the same 0,99 intensity for the two 60Co main gammas, when it should give 0,5 for each gamma.

    Am I confusing the concepts?
     
  12. Apr 4, 2010 #11
    In cobalt 60, the probability of each gamma being emitted in a single beta decay is 99.98%. Look at the Java level scheme. The 1.1 and 1.3 MeV gammas are in series. The sum of the two probabilities is meaningless.

    Bob S
     
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