Re: Score neutron from 16O(p,xn) reaction

From: Francesco Cerutti <>
Date: Mon, 1 Dec 2008 18:23:15 +0100

Dear Lee,

> Without normalization, could I interpret the result like single proton
> contribute around 0.0008 neutron production within bin 0-18deg,
> 0-1.5MeV.

not at all. What you plot is dN/[dOmega dE] i.e. neutrons/[sr GeV] per
primary proton. In other words, the number of neutrons per primary proton
is divided by the solid angle corresponding to each angular bin (by the
way, you have no 0-18deg bin, but 0-10, 10-20, 20-30..., as you can easily
realize translating into degrees the bin limits in the _tab.lis file,
which are printed in radians) and by the kinetic energy interval width
(in GeV).

> The population of neutron in energy bin 0-3.5MeV is less than bin
> 0-1.5MeV ,the angular distribution deviates compared with ICRU.

Of course the population in the 0-3.5MeV interval is NOT less than that in
the 0-1.5MeV one. As pointed out above, the two curves are divided by the
the kinetic energy interval width, which is different in the two cases.
Note that also the total response printed in the _sum.lis file - for
USRYIELD only! - is integrated over the first scoring quantity but not the
second (you have to multiply it by the width of the second quantity only
bin in order to get neutrons per primary proton).

For 0-1.5MeV neutrons, you asked for neutrons as EMERGING from
interactions, whereas for the other energy intervals, you score neutrons
ESCAPING your target.

Since the 20MeV proton range in water (around 4mm) is less than your
target length, the latter is indeed not relevant to the choice of the
biasing factor for the inelastic scattering length.



Francesco Cerutti
CH-1211 Geneva 23
tel. ++41 22 7678962
fax ++41 22 7668854
Received on Tue Dec 02 2008 - 15:43:46 CET

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