Hi Riya,
the short answer is you cannot.
The two things "total no of particles" vs "fluence of particles" are completely different.
Look the FLUKA beginners course the lecture on scoring. The first slides, describe
the quantities and their relation.
Fluence is defined/equivalent to "density of tracks" with units cm/cm^3 (not surface).
So if you multiply with your scoring volume you get the total length of all scored
particle tracks, and if you divide by the average track length of your particles,
you should get something that will be close to "total no of particles".
But don't do it.... :)
Vasilis
________________________________
From: owner-fluka-discuss_at_mi.infn.it [owner-fluka-discuss_at_mi.infn.it] on behalf of riya dey [riyadey_at_barc.gov.in]
Sent: Friday, April 12, 2019 08:18
To: fluka-discuss_at_fluka.org
Subject: [fluka-discuss]: How to get total no of particles using USRTRACK
Dear FLUKA Expert,
A RPP having dimension -1000 1000 -50 50 -50 50 (total volume 5E+6 cm^3) has been created as a target and at the same position a volume source (emitting positron with energy 633 keV) having the same dimension has been introduced using source.f card.
Now consider the source contains 1Ci/m^3 activity and all the region is filled with vacuum. This means total no of positron inside the source is 3.7E+10 (i.e Ci to Bq conversion) x 5E+6 particles per s.
Now if we use USRTRACK card inside the same geometry it should reproduce this number i.e 3.7E+10 x 5E+6 particles per s.
How can we convert the volume averaged fluence value to total no of particle inside the volume ? The USRTRACK output has a unit of particles cm^2 per primary. To get total no of particle inside the volume, which area (cm^2) do we have to multiply ? I have tried by multiplying the surface area of target ( here source and target dimension and location are same) , but I did not get the above mentioned number (3.7E+10 x 5E+6 particles per s). I got values higher than that.
Thanks and Regards
Riya Dey
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Received on Fri Apr 12 2019 - 15:38:45 CEST