Re: Energy Attenuation, Stopping Power and Range

I try giving an answer starting from 2):
I presume that O-18 Water means water where only 18-O isotope has been
selected.
Havar is a register name for an alloy made of:
Co42.5/Cr20/Ni13/Fe/W/Mo/Mn
In FLUKA (see MATERIAL and COMPOUND commands) you can build any compound
and mixture. If you study the manual you can see that you may select
a given isotope of a given element defined with the MATERIAL card.
This is the way in which you can build the 18-O water.
I would do, for instance, as follows (defining a OXYGEN18 and a WATER18
material):

MATERIAL 1.0 1.0 .0000899 3.0 0.0 0.0 HYDROGEN
MATERIAL 8.0 18.0 0.00143 8.0 0.0 1.0 OXYGEN18
MATERIAL 0.0 0.0 1.0 20.0 0.0 0.0 WATER18
COMPOUND 2.0 HYDROGEN 1.0 OXYGEN18 0.0 0.0 WATER18

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.

        Regards
                Giuseppe Battistoni
                
On Thu, 1 May 2008, [BIG5] 李辰衍 wrote:

> Dear fluka discuss:
>
> 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 = 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 = 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?
>
>
> 2 Lack of the material I'm interested in
>
> I could not find a material of Havar and O-18 water from website.
> So ,for stopping power of proton for Havar foil and range in O-18
> water I really have no idea.
>
> Please advise me
>
> My Best Regards
>
> Chenyen Lee
> Cyclotron
> Deptment of Nuclear Medicine
> Chang Gung Memorial Hospital
> Linko, Taiwan.
>
Received on Thu May 1 12:35:40 2008