-- IsabellaTepp - 14 Nov 2023

In ggland the Si-detector has been simulated as a cylinder (test-01) with the following bash-script:

--test-01=dz=36cm,z0=$((DIST-18))cm,ir=0.8cm,r=1.2cm,type=Si,enter,tree=1 \
--test-13=dz=0.01cm,z0=${DIST}cm,r=1.2cm,type=blackhole,enter,tree=1 \
--gun=d,dummy,T=0MeV,setboost,feed=target \
--gun=from=target,Ac229,dummy,T/u=8MeV/u,addboost,feed=cm \
-- gun=from=cm,p,Ac230:dummy:feed:setboost,Eexc=${Ex}MeV,recoil,phasespace,isotropic,feed=compound \
--gun-file=maxparticles=80,gun_230Ac_${Ex}MeV.txt,from=compound,boost \
--world=type=vacuum \
--fieldbox=d=200cm,Bz=${B}T \
--events=100000 \
--tree=helioscylinder${DIST}cm_${Ex}Mev.root

Where the test-13-object is a ''blackhole''-screen to absorb all the hits detected at the start of the detector. DIST in this case is the distance from the detector to the target and thiis parameter has been varied between 1.5, 2.0 and 2.5 cm. The excitation energy varied between 6, 7, 8, 9 and 10 MeV.

The ratio of detected protons to emitted ones have been calculated by:

if(gunpdg[i] == 2212){
total++
}

if(TST01inpdg[i] == 2212){
events++
}

ratio = events/total;

DIST [cm]	Ratio [%]
1.5	13.6
2.0	18.6
2.5	24.6
Ex [MeV]	Ratio [%]
6	27
7	23
8	19
9	13
10	5.1

Below follows some figures with the energy of the protons plotted against angles (both θ and the azimuthal angle φ).

The total kinetic energy of the protons plotted against θ for simulations with different excitation energy for the fissioning nucleus.

The total kinetic energy of the protons plotted against θ for simulations with different distance from target to the Si-detector.

The total kinetic energy of the protons plotted against φ for simulations with different excitation energy for the fissioning nucleus.

The total kinetic energy of the protons plotted against φ for simulations with different distance from target to the Si-detector.