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

From: lzfneu <lzfneu_at_live.com>
Date: Sat, 19 Nov 2011 10:53:13 +0000

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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-^(Ķē*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

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<body class=3D'hmmessage'><div dir=3D'ltr'>
<BR>Dear Dr. Alberto Fasso' and all=2C<BR>&nbsp=3B<BR>Thanks Dr. Alberto Fa=
sso' for the warmly concern and kindly help !<BR>I still have some question=
s to consult you:<BR>1) Following Dr. Alberto Fasso''s suggestion=2C I try =
to set up a <BR>"blackhole" collimator to eliminate the back scatter neutro=
n to the <BR>detector. As a practice=2C I defined a pencile-like mono-direc=
tion neutron <BR>source and scored the neutron fluence from the sample to b=
alckhole using <BR>"USRBDX" card to eliminate the scatter neutron=2C but st=
ill the calculated <BR>neutron fluence of 0.5cm polyethe for 2.45MeV neutro=
n is higher than the <BR>incident one=2C which means that the scatter neutr=
on does not eliminate. Is <BR>my understanding right or how to set up a per=
fect collimation by means of <BR>a "blackhole" collimator=2C thank you !&nb=
sp=3B<BR>I have attatched my input file in the attatchment.<BR>2) I want to=
  calculate a shielding material's thermal neutron macro cross <BR>section u=
sing FLUKA. If I define a mono-direction pencile-like source <BR>whose ener=
gy is 0.0253eV and using "USRBDX" card to do the calculation <BR>with black=
hole collimator=2C is it right or what should I do to calculate a <BR>mater=
ial's macro cross section for thermal neutron using FLUKA=2C thanks !<BR>Th=
anks in advance and any help will be appreciated !<BR>&nbsp=3B<BR>Best rega=
rds=2C<BR>&nbsp=3B<BR>Z.F. Lee<BR><BR>&nbsp=3B<BR><BR>
<DIV>
<DIV dir=3Dltr>
<DIV>&gt=3B Date: Wed=2C 16 Nov 2011 16:02:29 -0800<BR>&gt=3B From: fasso_at_S=
LAC.Stanford.EDU<BR>&gt=3B To: fluka-discuss_at_fluka.org<BR>&gt=3B CC: lzfneu=
@live.com=3B francesco.cerutti_at_cern.ch<BR>&gt=3B Subject: Re: Another quest=
ion about neutron counting<BR>&gt=3B <BR>&gt=3B Dear Lee=2C<BR>&gt=3B <BR>&=
gt=3B your questions are based on a wrong idea of neutron dosimetry and tra=
nsport.<BR>&gt=3B It is difficult to give you a lecture on neutron dosimetr=
y in a simple email.<BR>&gt=3B You should read some good textbook.<BR>&gt=
=3B I will summarize here the main flaws in your concepts.<BR>&gt=3B 1) 2.4=
5 MeV neutrons do not produce any signal in a He-3 detector. To do that=2C<=
BR>&gt=3B they must first be moderated and slowed down to thermal energies.=
  FLUKA's<BR>&gt=3B primary particles (BEAMPART) in your case are 2.45 MeV n=
eutrons.<BR>&gt=3B There probably remain very few of them (or none at all) =
after 3 cm of<BR>&gt=3B polyethylene: the others have been scattered and ha=
ve lost some of their<BR>&gt=3B energy and are not "BEAMPART" particles any=
more. However=2C probably 3 cm are<BR>&gt=3B not sufficient to fully therma=
lize 2.45 MeV neutrons: most of them will have<BR>&gt=3B intermediate energ=
ies=2C mainly too high to be detected by He-3.<BR>&gt=3B It will be differe=
nt for 0.5 cm polyethylene: you can certainly score<BR>&gt=3B several BEAMP=
ART neutrons=2C but they will have nothing to do with what you<BR>&gt=3B ca=
n measure experimentally.<BR>&gt=3B 2) It will be very different if you sco=
re NEUTRONs instead of BEAMPART. In this<BR>&gt=3B case you will score all =
the neutrons=2C independent of their energy: that will<BR>&gt=3B include al=
l the neutrons which have been scattered and moderated. Some of<BR>&gt=3B t=
hem can be detected by He-3=2C others will have still energies too high.<BR=
>&gt=3B Again=2C you cannot expect to be able to predict the response of yo=
ur detector=2C<BR>&gt=3B unless you take the energy distribution carefully =
into account.<BR>&gt=3B I remind you that in any case the response of a rea=
l detector is<BR>&gt=3B proportional to fluence (of the right energy!)=2C a=
nd not to current.<BR>&gt=3B 3) To measure=2C and to calculate=2C a macrosc=
opic cross section=2C you need to<BR>&gt=3B set up a "good geometry". That =
is a geometry where both the source and<BR>&gt=3B the detector are collimat=
ed=2C so that the detector does not "see" any<BR>&gt=3B scattered neutron. =
It does not seem to be your case. In a "bad" geometry=2C<BR>&gt=3B i.e. wit=
hout a double collimation=2C the equation I=3DI0*e-^(=A6=B2*d) does<BR>&gt=
=3B not apply=2C but instead of an attenuation you will have a build-up<BR>=
&gt=3B factor due to neutrons which are not initially directed on the<BR>&g=
t=3B source-to-detector direction=2C but are scattered back to the detector=
.<BR>&gt=3B This is probably the reason why you find a fluence larger than =
expected.<BR>&gt=3B In FLUKA=2C perfect collimation can be achieved by mean=
s of a "blackhole"<BR>&gt=3B collimator. In the actual world=2C only an app=
roximation is possible.<BR>&gt=3B 4) In FLUKA=2C as already pointed out by =
Francesco=2C for low-energy neutrons<BR>&gt=3B elastic and inelastic scatte=
ring cannot be disentangled. The program only<BR>&gt=3B "knows" the average=
  energy loss=2C which can be a mixture of elastic and<BR>&gt=3B inelastic i=
nteractions.<BR>&gt=3B 5) Thermal neutrons (those that are detected by a He=
-3 detector) and also<BR>&gt=3B epithermal neutrons (of energies slightly h=
igher) are not moving in a<BR>&gt=3B definite direction. They are scattered=
  back and forth=2C exactly like atoms in<BR>&gt=3B a gas (that is why they =
are called "thermal") and the same neutron can<BR>&gt=3B cross a given poin=
t more than once. Any attempt to measure or calculate<BR>&gt=3B their atten=
uation in a given direction is meaningless.<BR>&gt=3B <BR>&gt=3B Best regar=
ds=2C<BR>&gt=3B <BR>&gt=3B Alberto</DIV></DIV></DIV> </div></bod=
y>
</html>=

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--_662d5e82-95f7-439a-95ed-6c9d3e04614a_--
Received on Sun Nov 20 2011 - 12:55:15 CET

This archive was generated by hypermail 2.2.0 : Sun Nov 20 2011 - 12:55:15 CET