How to set up a blackhole collimator and calculate thermal neutron cross section

Dear Dr. Alberto Fasso' and all
Thanks Dr. Alberto Fasso' for the warmly concern and kindly help !
I still have some questions to consult you:
1) Following Dr. Alberto Fasso''s suggestion I try to set up a
"blackhole" collimator to eliminate the back scatter neutron to the detector.
As a practice I defined a pencile-like mono-direction neutron
source and scored the neutron fluence from the sample to balckhole using
"USRBDX" card to eliminate the scatter neutron=2C but still the calculated
neutron fluence of 0.5cm polyethe for 2.45MeV neutron is higher than the
incident one=2C which means that the scatter neutron does not eliminate. Is
my understanding right or how to set up a perfect collimation by means of
a "blackhole" collimator=2C thank you !
I have attatched my input file in the attatchment.
2) I want to calculate a shielding material's thermal neutron macro cross
section using FLUKA. If I define a mono-direction pencile-like source
whose energy is 0.0253eV and using "USRBDX" card to do the calculation
with blackhole collimator=2C is it right or what should I do to calculate a
material's macro cross section for thermal neutron using FLUKA
thanks !
Thanks in advance and any help will be appreciated !
Best regards
Z.F. Lee

> Date: Wed=2C 16 Nov 2011 16:02:29 -0800
> From: fasso_at_SLAC.Stanford.EDU
> To: fluka-discuss_at_fluka.org
> CC: lzfneu_at_live.com=3B francesco.cerutti_at_cern.ch
> Subject: Re: Another question about neutron counting
>
> Dear Lee
>
> your questions are based on a wrong idea of neutron dosimetry and transport.
> It is difficult to give you a lecture on neutron dosimetry in a simple email.
> You should read some good textbook.
> I will summarize here the main flaws in your concepts.
> 1) 2.45 MeV neutrons do not produce any signal in a He-3 detector. To dothat
> they must first be moderated and slowed down to thermal energies. FLUKA's
> primary particles (BEAMPART) in your case are 2.45 MeV neutrons.
> There probably remain very few of them (or none at all) after 3 cm of
> polyethylene: the others have been scattered and have lost some of their
> energy and are not "BEAMPART" particles anymore. However probably 3 cmare
> not sufficient to fully thermalize 2.45 MeV neutrons: most of them will have
> intermediate energies mainly too high to be detected by He-3.
> It will be different for 0.5 cm polyethylene: you can certainly score
> several BEAMPART neutrons but they will have nothing to do with what you
> can measure experimentally.
> 2) It will be very different if you score NEUTRONs instead of BEAMPART. In this
> case you will score all the neutrons independent of their energy: thatwill
> include all the neutrons which have been scattered and moderated. Some of
> them can be detected by He-3 others will have still energies too high.
> Again you cannot expect to be able to predict the response of your detector
> unless you take the energy distribution carefully into account.
> I remind you that in any case the response of a real detector is
> proportional to fluence (of the right energy!) and not to current.
> 3) To measure and to calculate a macroscopic cross section you need to
> set up a "good geometry". That is a geometry where both the source and
> the detector are collimated so that the detector does not "see" any
> scattered neutron. It does not seem to be your case. In a "bad" geometry
> i.e. without a double collimation the equation I=I0*e-^(Š²*d) does
> not apply but instead of an attenuation you will have a build-up
> factor due to neutrons which are not initially directed on the
> source-to-detector direction but are scattered back to the detector.
> This is probably the reason why you find a fluence larger than expected.
> In FLUKA perfect collimation can be achieved by means of a "blackhole"
> collimator. In the actual world only an approximation is possible.
> 4) In FLUKA as already pointed out by Francesco for low-energy neutrons
> elastic and inelastic scattering cannot be disentangled. The program only
> "knows" the average energy loss which can be a mixture of elastic and
> inelastic interactions.
> 5) Thermal neutrons (those that are detected by a He-3 detector) and also
> epithermal neutrons (of energies slightly higher) are not moving in a
> definite direction. They are scattered back and forth exactly like atoms in
> a gas (that is why they are called "thermal") and the same neutron can
> cross a given point more than once. Any attempt to measure or calculate
> their attenuation in a given direction is meaningless.
>
> Best regards
>
> Alberto