From: Alberto Fasso' <fasso@slac.stanford.edu>

Date: Thu May 01 2008 - 18:15:16 CEST

Date: Thu May 01 2008 - 18:15:16 CEST

Dear Chenyen Lee,

I would like to add some considerations to those of Giuseppe Battistoni

on the difficulty of calculating ranges by Monte Carlo.

The continuous slowing down approximation (CSDA) is, indeed, an

approximation (I was tempted to say: a fiction). Charged particles

are not slowed down continuously. The approximation is particularly bad

when the stopping power in question concerns a very thin layer, and

35 micrometer are indeed very thin.

FLUKA is designed to simulate the reality, not the fiction: and in a thin

layer of the real world there is a broad distribution of energy losses

(ionization fluctuations). I don't know which defaults you have chosen

(you have sent no input), but in the most common set of defaults

(NEW-DEFA) delta rays are produced with threshold of 1 MeV, and restricted=

=20

ionization fluctuations are simulated below that threshold.

If you want to obtain a result consistent with the CSDA, you must make sure

that no delta rays are produced, and that ionization fluctuations are=20

suppressed. To this purpose, use command DELTARAY to set a threshold

higher than that of your primary particles, and command IONFLUCT to switch

of ionization fluctuations (WHAT(1) =3D -1.0).

Best regards,

Alberto

On Thu, 1 May 2008, Giuseppe Battistoni wrote:

*> Now, coming to point 1):
*

*> of course the back-on-the-envelope calculation that you propose
*

*> for 9.5 MeV protons is too much rough and your suspicion is right.
*

*> By Monte Carlo you can probably calculate the average residual energy of
*

*> protons exiting the foil after injecting them at 9.5 MeV. The result will
*

*> allow you to evaluate the energy attenuation.
*

*> Range calculations by MonteCarlo are cumbersome: you can divide a
*

*> water volume in many tiny slices (USRBIN) and measure the average
*

*> number of crossed slices...
*

*> In general people refers to separated numerical programs to calculate the
*

*> energy loss and range.
*

*>
*

*> =09Regards
*

*> =09=09Giuseppe Battistoni
*

*>
*

*> On Thu, 1 May 2008, [BIG5] =A7=F5=A8=B0=ADl wrote:
*

*>
*

*>> I want to calculate the energy attenuation of proton passing through
*

*>> Havar foil and range of proton in O-18 water
*

*>> ,but I face problems as below.
*

*>>
*

*>> 1 continuous slowing down
*

*>> I check ICRU report 49 and NIST website for stopping power.
*

*>> http://physics.nist.gov/PhysRefData/Star/Text/PSTAR.html
*

*>> Material: Silver
*

*>> (MeV) Stopping Power (MeV cm2/g)
*

*>> 9.500E+00 2.388E+01
*

*>> Silver: mass density: 10.49g/cm^3
*

*>> mass density * stopping power =3D 250.5 MeV/cm
*

*>> So , if the proton of initial energy 9.5MeV went through 35 micro
*

*>> meter depth of silver.
*

*>> Energy loss would be 240.5MeV/cm * 35*10^-6 =3D 0.8767 MeV
*

*>> I think this neglects the fact that proton is continuous slowing down,
*

*>> so the result would be overestimate.
*

*>> How to get the accurate result?
*

*>>
*

*>> Chenyen Lee
*

*>> Cyclotron
*

*>> Deptment of Nuclear Medicine
*

*>> Chang Gung Memorial Hospital
*

*>> Linko, Taiwan.
*

*>>
*

*>
*

*>
*

--=20

Alberto Fass=F2

SLAC-RP, MS 48, 2575 Sand Hill Road, Menlo Park CA 94025

Phone: (1 650) 926 4762 Fax: (1 650) 926 3569

fasso@slac.stanford.edu

--1334196310-1328798022-1209658516=3D:27719--

Received on Fri May 2 00:20:56 2008

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