D2O photodisintegration simulation

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File:D2O photodis.pdf

File:Dustin anal.pdf

The result of GEANT4 simulation of neutron photoproduction on deuterium and oxygen-16 is presented below:

X section compare.gif


Neutron yield ([math]4\pi[/math]) from [math]D_2O[/math] (90%[math]D_2[/math] concentration) in reaction [math]O16(\gamma,n)O15[/math] is presented below. Initially [math]5\cdot 10^7[/math] brem photons with 25 MeV end-point energy were thrown on the target. 410 neutrons initially produced got absorbed in the target (E=0MeV) and 2509 neutrons will likely escape the target. Figure in blue on the left is the total energy spectrum of neutrons created in reaction [math]O16(\gamma,n)O15[/math]. Figure in blue on the right is the neutron energy spectrum with energy cut applied.

O16 neutron yield.gif O16 neutron yield cut.gif

Neutron yield ([math]4\pi[/math]) from [math]D_2O[/math] (90%[math]D_2[/math] concentration) in reaction [math]D_2(\gamma,n)p[/math] is presented below. Initially [math]5\cdot 10^7[/math] brem photons with 25 MeV end-point energy were thrown on the target. 857 neutrons initially produced got absorbed in the target (E=0MeV) and 16799 neutrons will likely escape the target. Figure in black on the left is the total energy spectrum of neutrons created in reaction [math]D_2(\gamma,n)p[/math]. Figure in blue on the right is the neutron energy spectrum with energy cut applied.

D2 neutron yield2.gif D2 neutron yield cut.gif

So, if we apply the energy cut in neutron spectrum which is a sum of neutrons coming from [math]O16(\gamma,n)O15[/math] and [math]D_2(\gamma,n)p[/math] reactions at (0MeV)<E<=5MeV, then the ratio of unpolarized photons from [math]O16(\gamma,n)O15[/math] and polarized photons from [math]D_2(\gamma,n)p[/math] will be [math]855/15128=0.056[/math]. In the same time we will loose [math](16799-15128)/15128=11%[/math] of the "good" high energy neutrons produced in [math]D_2(\gamma,n)p[/math] reaction. Binning is the same for all the plots.

I will verify if those spectra and all the ratios are the same in the case where we use the geometrical constrains applied by the detectors placement.