Gamma photon spread over 20 keV USRBDX

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Dear FLUKA experts,

I have simulated thermal neutrons on boron and I am monitoring the photons that are leaving the material using USRBDX. We are expecting photons of 478 keV due to Li-7 de-excitation which is produced in B10(n,alfa)Li7. If I define the energy bin of 1 keV the plot obtained with USRBDX around the 478 keV energy is the one in the attached figure. The question is why does the peak spread over this large range of energy? Shouldn’t it be found in a single 1 keV bin? (for other energies, the peaks are located in a single bin)

478 keV
478 keV692×537 46.8 KB

This is the input file: Forum.inp (1.2 KB)

Kind regards,
Denis Barbu

3

Dear Denis,

In a nutshell: the expectation of a sharp gamma energy holds only in the centre-of-mass frame.

If you boost to the lab, the energy of the photon is necessarily Doppler broadened (the intermediate ^7\textrm{Li}_\textrm{1st} is not at rest).

The kinematics indeed checks out. If you picture the process in two steps, in the first you’d have

\textrm{n}+^{10}\textrm{B}\rightarrow\alpha+^7\textrm{Li}_\textrm{1st}.

Assuming a 1 meV neutron and the ^{10}\textrm{B} at rest, the ^7\textrm{Li}_\textrm{1st} takes between 839.7222 and 839.7502 keV (triple check in case of arithmetic hiccups on my side in this quick check). Not a huge spread, but still that’s considerable kinetic energy in this scale. Proceeding analogously for the second step,

^7\textrm{Li}_\textrm{1st}\rightarrow\gamma+^7\textrm{Li}_\textrm{gs} ,

you see that the minimum possible energy of the photon is 469.999 keV and the maximum is 485.313 keV, which pretty much matches what you report.

So nothing mysterious, just good old Doppler broadening.

Cheers,

Cesc

PS: see also Prompt gamma spectrum from a proton beam - #4 by cesc

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4

Dear Cesc,

Thank you very much for the clarification! Now it all makes sense!

Kind regards,
Denis

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