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## Defining the integration time using a primary event number

Forum: Space Applications
Date: 14 Oct, 2010
From: Manuel Castro Avila <Manuel Castro Avila>

 Hi everyone, I like your help. I have to simulate the behavior of an instrument during a exposition time to the radiation, in this case, to gamma photons. The instrument has a detector with size of 169 cm^{2} and 50 cm away from the detector, there is an aperture of 1332.25 cm^{2}. So, I have the power law that represent the energy distribution of the photon spectrum that I go to use. The power law is the following: 5.16x10^{-2} E^{-1.81} ph cm^{-2} Sr^{-1} s^{-1} MeV^{-1} This power law is valid between 0.04 MeV to 10 Mev. I need to calculate the number of primary events that represent a exposition time, for instance, 4 hours. To calculate this number, I calculate the integral among the energy range and I get: 0.854 ph cm^{-2} Sr^{-1} s^{-1} MeV{-1} -The integration time is: 4h=14400 s -I want to integrate over 4*PI Sr (I considere a isotropic flux into my simulation using GPS from GEANT 4) -The detector area is of 169 cm^{2} Now, to get the total number photons, I multiply the number above (0.854 ph ...) by 14400*4*PI*169 and I get the total primary event number equals 26116617.68 photons. My macro file is: /gps/particle gamma /gps/pos/type Surface /gps/pos/shape Sphere /gps/pos/radius 60. cm /gps/pos/centre 0.0 0.0 0.0 cm /gps/ang/type cos /gps/ene/type Pow /gps/ene/min 1 MeV /gps/ene/max 10 MeV /gps/ene/alpha -1.8 /run/beamOn 26116617 My question is: Is this procedure above correct?? Am I using the correct way to calculate the integration time?? By the other hand, the power law represent the photon background measured in the atmosphere to a atmospheric depth of 3.5 g cm^{-2}, is there a way to adapt this power law to the case where the atmospheric depth is of 2.8 g cm^{-2}??? Thanks, Manuel

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