Re: [fluka-discuss]: energy deposition loss

From: <>
Date: Fri, 23 Jul 2021 08:22:08 +0200

Dear Lisa

as soon as protons can make nuclear interactions, some energy is "lost"
because of binding energy losses.

Let me make an example: for a reaction like 12C(p,pn)11C, the reaction Q
is about 19 MeV, hence every time you have that reaction, you "lose" 19
MeV *). Obviously there are a variety of possible reactions, each one with
its own Q.

Of course this energy is not really lost, it goes into mass energy of the
residual products following the usual E=mc^2 formula. That is, if you
summed up also the masses of all involved nuclei before and after together
with the deposited energy you would get a perfect energy balance.

Obviously from a practical point of view, only the deposited (kinetic)
energy contributes to heating etc.

Please note that for 0.7 MeV beam the result should be 0% difference, the
fact that you see a 10^-5 difference is due to rounding since the
post-processing is done in single precision.


*) In reality if your cylinder is big enough to contain all neutrons until
they are captured, for the 12C(p,pn)11C reaction part of the "lost" energy
would be recovered by capture photons when the neutron will be eventually
captured on 12C (4.9 MeV) or 13C (8.2 MeV). Also, part of the "lost"
energy will reappear as decay heat, eg for 11C through the positron
emitted during the beta+ decay (and the neutrino which obviously will
escape), roughly 2 MeV including the neutrino and already accounting for
the 1.022 MeV annihilation photons; this energy will not appear in the
Fluka budget unless you ask for decays (semi-analogue mode) as well. Still
you can see from those numbers that in my example there is anyway a
significant energy "loss" to mass. Other more complex reactions can result
in even larger energy losses.

> Dear Fluka experts,
> I wanted to calculate the energy deposition in a graphite cylindrical
> target due to an axial proton beam. I've used the USRBIN card, asking
> for ENERGY and selecting the cylinder region. The beam is originated in
> the core of the target and it is completly stopped by the graphite:
> there is no particle loss outside the cylinder. However, I have noticed
> a sort of 'energy loss' in the results, and this loss increases as the
> beam energy increases:
> * Proton beam energy: 0.7 MeV --> Energy deposition: 0.699W --> %
> difference: 0.001%
> * Proton beam energy: 40 MeV --> Energy deposition: 39.65W --> %
> difference: 0.8%
> * Proton beam energy: 70 MeV --> Energy deposition: 68.62W --> %
> difference: 2%
> * Proton beam energy: 150 MeV --> Energy deposition: 142.2W --> %
> difference: 5.5%
> How can this be explained? Am I missing something?
> Thanks for your help!
> Best regards,
> Lisa

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Received on Fri Jul 23 2021 - 10:21:30 CEST

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