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Question Detector Efficiency 

Forum: Event and Track Management
Date: 07 Nov, 2013
From: Ibrahim <Ibrahim>

Daer Geant4 users,

I calculated the full energy peak efficiency (Eff) of the detector.

The full energy peak efficiency (Eff) is defined as the number of counts in the photopeak, Np, divided by the number of the gamma rays of the photopeak energy emitted by the source N0: Eff= Np/N0

The simulated Eff is about twice larger than the experimental one. Do you know why? I highly appreciate your help.

Kind regards

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1 Idea: Re: Detector Efficiency   (Vladimir Ivanchenko - 18 Nov, 2013)
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