Re: [fluka-discuss]: Absolute efficiency of HPGe detector

From: PABLO CESAR ORTIZ RAMIREZ <rapeitor_at_ug.uchile.cl>
Date: Thu, 20 Oct 2016 15:43:15 -0300

Dear Andrea and Peter:

Thanks for your answer. Andrea, I have done and I know all that you have
mentioned in your response. My question specifically is: How can I obtain
the photopeak absolute efficiency of the detector for some source, for
instance a point source of 137Cs, from the simulation? I am using DETECT
card to indicate the detector and when I compile I obtain a output file
with extension .lis (Attached in this e-mail. Photopeak is in the 661,65
KeV), where the two first columns are the energy range of spectrum (E+DE)
and the third column correspond to the Response function R(E):={Probability
that a detected photon is registered in the energy range "E+DE"}. I want to
know how I can obtain the photopeak absolute efficiency of the detector
from the generated data.

2016-10-20 10:28 GMT-03:00 Andrea Tsinganis <Andrea.Tsinganis_at_cern.ch>:

> Dear Pablo,
>
> Your question is not perfectly clear to me and you also do not mention
> what source you are simulating, which might have helped. I am therefore
> assuming that you wish to study the absolute efficiency of the detector,
> understood as the probability of detection of a γ-ray of a given energy
> that enters the crystal, and perhaps compare with experimental data.
>
> Experimentally, you would do this e.g. with a 152Eu source of known
> activity at a fixed distance and looking at the known γ-peaks in the
> obtained spectrum. You have probably done this already, or will do it soon.
> If your goal here is to find the absolute efficiency of your simulated
> detector, then you could adopt a similar approach: define an isotropic
> source with a known gamma spectrum and then compare it with what is
> detected in the crystal.
>
> This can be done at different levels of sophistication, e.g. with
> monoenergetic sources at selected energies, or simulating a 152Eu source
> (for comparison with an experimental spectrum) or with a source with a flat
> energy distribution up to a few MeV. In this last case, your obtained
> spectrum would by definition give you the (simulated) detector's efficiency
> curve.
>
> Note, however, that the absolute value for the efficiency curve at this
> stage depends on the distance from the crystal at which you define your
> isotropic source (as is the case experimentally), since the angular
> acceptance changes. For the absolute efficiency, just direct the gammas so
> that they all enter the crystal, or just generate them inside the crystal
> to begin with.
>
> Hoping I have answered the right question(s),
> Regards,
> Andrea
>
>
>
> On Thu, Oct 20, 2016 at 8:41 AM, Peter Rubovič <peter.rubovic_at_suro.cz>
> wrote:
>
>> Hi Pablo,
>>
>> my guess would be to sum the efficiencies for every channel.
>>
>> Best regards,
>> Peter.
>>
>> Dne 20.10.2016 v 3:00 PABLO CESAR ORTIZ RAMIREZ napsal(a):
>>
>> Hello fluka users:
>>>
>>> I have been simulating a HPGe detector and I have been using the Detect
>>> card to obtain the energy spectrum. I have noticed that the obtained
>>> output file (.lis) give me the response function, i.e. {the number of
>>> detected photons by chanel}/{the total number of detected photons}. My
>>> question is: how can I obtain the absolute efficiency using this card?
>>> or how can I know the number of total detected photons by the detector?
>>>
>>> Thanks in advance.
>>>
>>> --
>>> Pablo Ortiz Ramírez
>>> Master of Science in Physics
>>> Faculty of Science, University of Chile
>>>
>>> Linux User #528950
>>> Ubuntu User #32991
>>>
>>
>> --
>> Peter Rubovič
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>


-- 
Pablo Ortiz Ramírez
Master of Science in Physics
Faculty of Science, University of Chile
Linux User #528950
Ubuntu User #32991



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Received on Thu Oct 20 2016 - 22:22:30 CEST

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