Normalization of results / Source routine with radioactive isotopes

Thank you so much @horvathd, for the detailed explanation. Just two more quires.

  1. The output of detect card is counts per primary weight, right ? So if the source strength is 1 micro Ci, then I have to multiply it with 1 micro Ci only, right ? There is no need to divide DETECT output by 2 in case of Co - 60 source, right ?

  2. If I use ISOTOPE option along with source.f, then I do not need to specify energy spectrum of source particle in source.f since it will be handled by ISOTOPE card. Then what should I keep as TKEFLK (NPFLKA) in source.f ?

Thanks and regards,
Riya

Dear Riya,

  1. Yes, the results are normalized per decay, so you only need to multiply the results with the activity of your source.

  2. If you look more closely at the source routine, you will see that the isotope, the heavy ion, and normal hadron primaries are treated separately when the energy / momentum assigned.

    In case of the radioactive isotopes the TKEFLK (NPFLKA) and PMOFLK (NPFLKA) and some other variables are not set, and you don’t have to add anything in this regard.

Cheers,
David

Thank you @horvathd, For my hpge, I gave two runs , one with Co 60 in isotope mode and other one using source.f with probability defined as

        EE = FLRNDM(XDUMMY)

        IF (EE .LE. 0.5D0) THEN 
          ENERGY = 1.17D-3
        ELSEIF (EE .GT. 0.5D0) THEN 
          ENERGY = 1.33D-3
        ENDIF

 		TKEFLK (NPFLKA) = ENERGY

Since final output of detect card is normalized to per primary, I expected that in both cases, the counts will be nearly same.

But it is not the case.
The results with isotope card is almost 1.65 times higher than without isotope card. Am I missing something ? Also with isotope card, there is a third high peak in higher energy which was not there when I was not using isotope card.

fluka_plot_with_isotope.pdf (91.5 KB) fluka_plot_without_isotope.pdf (39.7 KB) HPGe_17_tab_with_isotope.lis (720.0 KB) HPGe_17_tab_without_isotope.lis (720.0 KB)

regrads,
Riya

Dear Riya,

it is important to keep in mind, that in case of radioactive isotope sources FLUKA treats the decay as a primary “event”.

Which for Co60 means, one primary “event” consists (99.88% of the time) of the two gammas plus an electron from the beta- decay.

If you are simulating photos with the two energies, you are skipping half of the decay per primary “event”. So you need to multiply the results by 2 as well to get the same as with the isotope source. Which corresponds the 1.65 factor you were seeing. (The Gaussian broadening maybe plays a role why you don’t get approximately 2.)

The third peak can be easily understand: It is the peak, when both of the gammas stop in the detector with the total energy of ~2.5 MeV.

Cheers,
David

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Thank you so much @horvathd for explaining everything. It is clear to me now. So what I feel is if we want to compare gaussian broadened spectrum from our simulation with experimental results, it is better to use Isotope card instead of writing spectrum on our own. Am I right ?
And the peak at 2.5 MeV refers when the two photons completely transferring their energy, does it refer to photo electric effect ? Or does it refer the condition when these two photons lose their total energy inside the active detector region, no matter how many type of interactions they have undergone ?

Regards,
Riya

Dear Riya,

In case of Co 60, because it decays to a stable element, I would use the isotope source for its simplicity. For other isotopes, where the daughter is also radioactive, FLUKA also include those into the source spectrum. For these maybe a source routine would be preferred.

The peak (whether it is the 1.332 / 1.173 MeV one or both together) only tells us the amount of energy deposited in the detector region. It doesn’t have any indication how many or what kind of interaction took place.

Cheers,
David

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Thank you so much @horvathd, the concept is completely clear to me.

Regards,
Riya