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

From: PABLO CESAR ORTIZ RAMIREZ <rapeitor_at_ug.uchile.cl>
Date: Thu, 20 Oct 2016 19:05:43 -0300

Dear Andrea and Peter:

Thanks again, I already have resolved my doubts. I was confused due to that
I was summing the numbers in the third column (absolute efficiency) over
every channel and the result was 1. From here, I thought that the third
column represented to the response function and not to the absolute
efficiency, but this occured because in the first channel of the spectrum
are registered all photons that are not detected, i.e the photons whose
deposited energy is zero. Then, {detected photons} + {undetected photons} =
{emitted photons}, and, {emitted photon}/{emitted photon}=1.

Thank you very much.

Best regards.

2016-10-20 15:43 GMT-03:00 PABLO CESAR ORTIZ RAMIREZ <rapeitor_at_ug.uchile.cl>
:

> 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
>



-- 
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 Fri Oct 21 2016 - 01:36:19 CEST

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