Re: [fluka-discuss]: Question about 1 Moliere radius calculation for low energy electron

From: Paola Sala <paola.sala_at_mi.infn.it>
Date: Mon, 24 Aug 2020 17:51:06 +0200

Hello
answers below
> Dear all,
> I intend to determine the radius of a graphite cylinder in the air
when
> the radical energy leakage is 10%(1 Moliere radius), and the electron
energy 150MeV hit the graphite target.
> Therefore, I have four questions to consult you:
> 1) I have no idea which card in FLUKA should I use to calculate the
energy
> leakage to determine the 1 Moliere radius for 150MeV electron.
> Maybe I should use USRBIN card with energy to calculate energy
deposition
> or use USRBDX card with energy to calculate Kinetic energy or check the
xxx.out file for energy density per one beam particle (GeV/cm3). [a] If
use USRBIN card, how to calculate the total Energy deposition integrated
over volume and each part of the energy deposition. I have attatched the
alcu.inp file in the attatchment, however, the results are different
from xxx.out file. I don't know where I am wrong.
I would use a USERBIN, cylindrical symmetry, with many radial intervals in
order to calculate energy depositions within different radial extents.
When plotting the USRBIN results, you can limit the radial (or z)
interval.

The USRBIN results are normalized per cubic centimetre, the xxx.out are
NOT normalized unless you input the region volumes.
Thus, the two results should be connected through the volume of your
region, or the volume of the radial interval concerned.




> [b] If use USRBDX card, I found the Kinetic energy from graphite to air
region (area is 2pi*1*500) is 1GeV when the radius of graphite is 1cm
and
> length is 500cm (I set to represent infinite length) . I have no idea why?
No idea on how do you calculate this, can you please explain?

> Increase the radius, the energy value from graphite to air is decreased,
divide the energy value by 150MeV and then check if the result is equal
to
> 0.1 to determine whether the radius is suitable. I have attatched the
graphite.inp file in the attatchment.I wonder if I am right.
see previous answers.

> [c] If check xxx.out file, the energy density values are with no
statistical errors, and the values in multiple nodes are a little
different if I run with multiple nodes together. I have no idea which
one
> to use?

The results in xxx.out fluctuate statistically, and are naturally
(slightly) different for each cycle. The USRBIN and USRBDX results come
with statistical erros because thay are averaged over many cycles. If you
want to use the xxx.out results, you should average and calculate rms by
hand.

> I have no idea which method should I choose is right.

USRBIN
> 2) I have to change the radius every time to determine which radius
meets
> the energy leakage 10%, is there a convenient method in FLUKA.

USRBIN covering a "large" target with many radial intervals
> 3) The same problem about the calculation of energy leakage is 1% for
length direction.

Same, with many longitudinal intervals
> I use the radius determined in 2), and change the length of graphite
tatget every time. Then determine the proper length of graphite target
to
> meet the energy leakage equals to 1%. Am I right? Is there a convenient
method in FLUKA.
> 4) By the way, I am confused that is there any analytical formula for
low
> energy electron to calculate 1 Moliere radius. The value of radiation
length X0 and 1 Moliere radius for graphite in text book is for high
energy electron (GeV) and is it suitable for 150MeV electron?

I would say that 150 MeV is too low for the analytical formula
> Is the radiation length listed in xxx.out file for 150MeV electron? If
the radiation length X0 and 1 Moliere radius are both for 150MeV
electron, can I say that 5 Moliere radius and 20*X0 reach 1% energy
leakage for 150MeV electron.

Radiation length is independent from energy.

> Any help will be appreciated and thanks in advance.
> Best regards,
> Li


Paola Sala
INFN Milano
tel. Milano +39-0250317374
tel. CERN +41-227679148




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Received on Mon Aug 24 2020 - 20:30:34 CEST

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