DETECT card with an He-3 detector and Cadmium

Dear Fluka experts,

I simulated an isotropic AmBe source and an assembly of polyethylene which termalizes the neutrons emitted by the source; at around 30 cm from the source I placed a bare He3 detector. The assembly consists of a cylinder around the source and a kind of box of PE around the detector.
I want to predict the counts of the He3, so I assigned the card DETECT to the sensitive volume of the He3. I know that the proton and tritium produced by the (n,p) reaction are not trasported, but they deposit their enetgy locally, so I get a spectrum which is essentially a peak in correspondance of the Q-value. Anyway, I’m not interested in the spectrum so I calculated my counts just summing the event per pimary in the main peak.
First question: why the peak is not at the bin where764 keV belongs but at the bin right below? Is just a numerical reason?

I repeated the simulation surrounding the He-3 with a sphere of Cadmium (0.5 mm of thickness, because it resproduces a real one). With cadmium I expect to have the contribution of the gamma produced by the neutron capture and in fact I have counts below 100 keV that I neglect because they are below the threshold of my actual detector. However, I also got non negligible counts right below the Q-value, are these counts still gamma?
If yes, these counts represent the energy deposited by the gamma in the HE3, so I have to count them to reproduce a real detector? Or is there the possibility that these gammas release all their energy in the point where they are created (so I maybe have to discard them, because in the actual detector they release an energy which is under threshold?)

I attach the spectrum of the card DETECT without cadmium, with cadmium and the spectrum of neutrons that I got with the USRTRACK card (inside the He3) with the cadmium.

I read from the manual that the gamma are produce according to the group algorith but they are fully transported as all the other particle. However I read a very old comment (on the previous forum) which says that if there are no information about gamma production the Q value goes nto Kerma and stop. I don’t think i fully understood this sentence, do the card detect just count for the entire enrgy? This can happen with excited state of Cadmium produced by the neutron capture? I checked the table in the manual and both the RN and the GAM column are Y.
Sorry for the long post, I just want to be sure to count the right events coherently with the counts

Thank you,
Francesca

DETECT with cadmium DETECT without cadmium USRTRACK Neutron fluence with cadmium

Hi Francesca @fferrull .
The n + 3He → p + 3H Q-value actually sits between 763 and 764 keV, which appears in exact agreement with your plot.
The counts right below seem to be there also without cadmium, simply they are less visible due to the much more pronounced Q-value peak (40 times higher). To investigate their origin, you can add a few EVENTBIN per region (select a formatted unit) and associate each of them to a different particle type (NEUTRON, PHOTON, ELECTRON) by means of AUXSCORE: this way you will see the amount of energy deposited in the detector region on a event-by-event basis by the different radiation components (add also an EVENTBIN not linked to any AUXSCORE to get the total).
Counts due to gammas (and their secondary electrons) take into account their transport and possible escape (depending on the transport threshold you use), meaning that they do not necessarily deposit all their energy in the detector (what is scored by DETECT is what they actually deposit there, again depending on the adopted transport threshold).
As for the neutron spectrum, do not plot DPhi/DE x DE, rather DPhi/DE x E, i.e. DPhi/DlogE.
Best

Dear @fferrull

As a reasonable alternative (or additional test) you could also consider folding the estimated neutron spectrum in the active volume with the 3He(n,p) cross section to obtain the corresponding reaction rate and compare it to your experimental value.

Best regards

Dear all,

thanks for your suggestions i think I will follow both of them and i’ll come back if any problems.

Regards,
Francesca