Re: Optical photons

From: Joseph Comfort <Joseph.Comfort_at_asu.edu>
Date: Thu, 31 May 2012 12:32:56 +0200 (CEST)

Hi Paola,

Thank you for checking the code and providing an answer. It makes sense
that the ceren process is much weaker than the scint process.

I am still learning how to work with optical photons. As a follow-up
question, where do I put in information regarding decay times (fast,
slow, very slow) for the radiation? The options in the manual do not
seem to apply (detector cooling times, etc.), but maybe I am not reading
them correctly.

Thank you,
Joe

On 05/25/2012 03:23 AM, Paola Sala wrote:
> Hello Joe
> a late answer..
> no, there is no competition. I just checked, enabling both cere and scint
> gives the sum of the optical photons generated in the two processes.
> However, the yield of cherenkov photons can b very small with respect to
> scintillation ones, to the point of being undetectable.
> Ciao
> Paola
>> I am studying the propagation of optical photons through a CsI(pure)
>> detector crystal, for both high-energy photons and other particles (e.g.,
>> neutrons), to look at timing distributions as they go out the end.
>>
>> A question. Is there a priority between Cherenkov production and
>> scintillation? In a test case, each process alone produces different
>> numbers of exiting optical photons. But if both processes are 'on,' the
>> results are the same as for scintillation alone. Both processes are
>> focussed on the 315-nm line, the maximum of the distribution.
>>
>> Joe Comfort
>>
>
>
> Paola Sala
> INFN Milano
> tel. Milano +39-0250317374
> tel. CERN +41-227679148
>
>
Received on Thu May 31 2012 - 13:24:30 CEST

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