(alpha, n) yield in aluminum

Dear all,

We are trying to simulate the (alpha, n) yield in aluminum for 8 MeV energy of alpha particle using RESNUCLEI card (alpha + Al27 → P30 + n).

The volume of the target is 1cm^3 and we normalized with the density of the target (in this case, 2.7 g/cm^3).
In order to obtain the (alpha, n) yield in aluminum, we divided the values written in
*** Residual nuclei distribution *** *** (nuclei/cmc/pr) *** (which are also normalized) by the total response and we obtained:

1E+06 * 1.04E-05/0.370 = 28.1 (neutrons/10^6 alpha) which is way higher than the experimental value for Al reported in the attached article (10 neutrons/10^6 alpha).

What are we doing wrong?

Kind regards,

alpha_Al_8MeV.inp (2.0 KB)

Comparison of thick-target (alpha,n) yield calculation codes (1).pdf (1.6 MB)

Hallo Ioana,
I do not understand the meaning of your normalization, although by chance it turns out to be substantially correct, because the RESNUCLEi total response almost coincides with the number of incident alpha particles, which stop inside your target without any nuclear interaction (only 50 alpha particles out of one million undergo a nuclear reaction). Also, I cannot see the meaning of the normalization by the density. In fact, in order to calculate the number of 30P nuclei generated by 10^6 alphas, you should just multiply the 30P value in the RESNUCLEi output (not normalized by the Al density, i.e. 2.8E-5 = your 1.04E-5 * 2.7) by 1e6, thereby getting 28 30P nuclei, which is what you found in a more adventurous way.
Now, when comparing the FLUKA 30P production cross section of alpha particles on aluminum with available data, one gets the following picture (note that the horizontal red line is an experimental uncertainty):

which at 8 MeV might be consistent with an overestimation of a factor of few.

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