From: G. Battistoni <giuseppe.battistoni_at_mi.infn.it>

Date: Sun, 02 Aug 2009 18:35:39 +0200

Date: Sun, 02 Aug 2009 18:35:39 +0200

The answer is not easy and not univoque.

1) First of all the minimum thickness in a transport code like FLUKA is

meaningful if there are enough atoms to make valid the use of multiple

coulomb scattering approach and energy loss

calculations. Therefore it depends on the medium, its atomic properties

and the density.

I am aware of successful FLUKA calculations with silicon detectors of

thickness of the order of few tens of microns. I doubt that a gaseous

medium of the same thickness (in space) would obtain the same success.

2) There is at least a second consideration: in a given geometry the

minimum thickness that can be appreciated depends on its ratio with

respect to the maximum size of the problem itself. In other words, the

scale of the minimum size has to be such to be within the resolution

given by the precision accuracy. Take into account that FLUKA makes use

of double precision. In order to exploit it, you have to write the

geometrical parameters using the maximum precision (see the manual)

Hope that these concept can guide you...

Giuseppe Battistoni

Mr. Bhushankumar Jagnnath Patil. wrote:

*> Dear Fluka experts,
*

*> I am studding the scattering of electron beam by using various
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*> materials as scatterer(cylindrical geometry). I have a question that
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*> how smallest thickness of the material I can give? Is FLUKA consider
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*> micron thickness or not?
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*> Thanking you.
*

*>
*

*> Mr. Bhushankumar Jagnnath Patil.
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*> Microtron Accelerator Lab.
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*> Department of Physics,
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*> University of Pune,
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*> Ganeshkhind, Pune 411007
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*> Ph. No. Off. 020-25692678 Ext. 421
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*> Mob. 9823968377
*

*> Webpage http://physics.unipune.ernet.in/~bjp
*

*> Alternate e-mail amolbhushan_at_gmail.com
*

*>
*

Received on Sun Aug 02 2009 - 19:11:02 CEST

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