PHITS
Multi-Purpose Particle and Heavy Ion Transport code System
Exercise（II）:
How to stop a, b, g-rays and neutrons?
Feb. 2014 revised
title
1
Purpose of This Exercise
It is generally said that …
1. a-ray can be stopped by a piece of paper
2. b-ray can be stopped by an aluminum board
3. g-ray can be stopped by an lead block
4. neutron can penetrate all of these materials
Let’s check whether they are correct or not,
using PHITS!
Contents
2
Select an appropriate input file
from recommendation settings
Calculation condition
1. transport various particles
2. visualize the particle trajectories
An appropriate input file is …
PhotonTherapy.inp,
which is used for visualizing the particle
trajectories for X-ray therapy
But in this tutorial, an original input file “range.inp” was prepared
Copy Input File
3
Range.inpの確認
Calculation Condition
Incident particle： Electron (20MeV, 0.01cm radius beam）
Geometry： Cylindrical Al shielding (t = 2cm, r = 5cm) & Void
Tally： [t-track] for visualizing particle trajectories
[t-cross] for calculating particle fluxes behind the
shielding
20MeV
Electron
Al Void
Geometry
track.eps
Check Input File
cross.eps
4
Procedure for this exercise
1. Change the source to β-rays
2. Change the thickness of the shielding
3. Change the tallied region
4. Change the source to α-rays
5. Change the target to a piece of paper
6. Change the source to γ-rays, and the target to a lead block
7. Find an appropriate thickness of the lead block
8. Reduce the statistical uncertainty
9. Change the source to neutrons
10. Find an appropriate shielding material for neutrons
The input files for each procedure were prepared as “range*.inp”
Procedure
5
Step 1：Change Source
Change the source from 20 MeV to 1 MeV
electron (a typical energy of b-ray)
[Source]
s-type = 1
proj = electron
e0 = 20.00
r0 = 0.0100
x0 = 0.0000
y0 = 0.0000
z0 = -20.000
z1 = -20.000
dir = 1.0000
Change e0
& execute!
Source energy is
defined in [source]
section
Electron fluence (track.eps)
Shielded!
Step 1
6
Step2: Change the Thickness
Real aluminum boards are rather thin (~ 1mm)
Change the thickness of target to 1mm
[Surface]
set: c1[2.0] $ Thickness of Target (cm)
1
pz
0.0
2
pz
c1
3
pz 50.0
11
cz
5.0
999
so 100.0
• Thickness of target is
defined as a parameter c1
Electron fluence (track.eps)
1mm is too thin to stop!
Let’s investigate the minimum thickness of Al to stop 1 MeV electron
Step 2
7
Step3：Change Tallied Region
Let’s see the fluence distribution inside the target in more detail!
1. Target thickness (c1) should be 0.2 cm
2. Tally from -c1 to +c1 cm for X&Y directions
3. Tally from 0 to c1*2 cm for Z direction
[T-Track]
title = Track in
mesh = xyz
x-type = 2
xmin = -1.5
xmax = 1.5
nx = 50
y-type = 2
ymin = -1.5
ymax = 1.5
ny = 1
z-type = 2
zmin = 0.0
zmax = 3.0
nz = 90
Electron fluence
Stopped close to the
distal edge
Step 3
Photon fluence
Penetrated
8
Check Energy Spectrum
Integrated
value?
You can find the integrated
fluence per source at ”# sum
over” at the 94 line of flux.out
…
# sum over
Particle fluence behind
the target（flux.eps）
y(electron)... y(photon)
0.0000E+00 7.5439E-02
0.075 photons/incident electron
are escaped from the aluminum
board
An aluminum board
can stop b-ray, but not
secondary photon!
Photons with energies from 10 keV
to 100 keV are escaped
Step 3
9
Step4: How about α-rays?
1. Change the source from β-ray to α-ray with an energy
of 6 MeV (＝1.5MeV/u）
[Source]
s-type = 1
proj = electron
e0 = 1.00
r0 = 0.0100
x0 = 0.0000
y0 = 0.0000
z0 = -20.000
z1 = -20.000
dir = 1.0000
Fluence of α-rays
(3rd page of track.eps)
Stopped at the surface
Step 4
10
Step5: Change Geometry
1. Change the target to a piece of paper (C6H10O5)n
2. Assume density = 0.82g/cm3 & thickness = 0.01cm
[Material]
MAT[ 1 ] # Aluminum
27Al
1.0
[Cell]
1
1 -2.7 1 -2 -11 $ Target
2
0
2 -3 -11 $ Void
98
0
#1 #2 -999 $ Void
99
-1
999 $ Outer region
[Surface]
set: c1[0.2] $ Thickness of Target (cm)
1
pz
0.0
2
pz
c1
3
pz 50.0
11
cz
5.0
999
so 100.0
Fluence of α particle
•Stop at 0.006 cm in paper
•No secondary particle is generated
Step 5
11
Step 6: How about γ-rays?
1. Change the source to g-rays with energy of 0.662MeV
2. Change the target to a 1 cm lead block (11.34g/cm3)
204Pb
206Pb
207Pb
208Pb
0.014
0.241
0.221
0.524
Energy spectra behind the target
Fluence of photon
Many photons penetrate the
Target thickness is not enough
target without any interaction
Step 6
12
Step 7: Find an appropriate thickness
1. Change the target thickness to decrease the direct
penetration rate of photons down to 1/100
2. Check 75th line in cross.out
Energy spectrum
Penetration rate = 0.010
Fluence of photon for
the 4.3 cm lead target case
Step 7
13
Step8: Reduce Statistical Uncertainty
Estimate the penetration rate with statistical uncertainty below
10% by changing maxcas, maxbch, batch.now, istdev etc.
maxcas = 1000, maxbch = 1
3.1232E-01
3.8073E-01
4.6413E-01
5.6580E-01
6.8973E-01
3.8073E-01
4.6413E-01
5.6580E-01
6.8973E-01
8.4081E-01
0.0000E+00
0.0000E+00
0.0000E+00
0.0000E+00
0.0000E+00
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000E+00 0.0000
3.1116E-03 0.7100
1.1609E-03 1.0000
1.0050E-02 0.3148
0.0000E+00 0.0000
You can check the values in 75th line of cross.out
maxcas = 1000, maxbch = 14
3.1232E-01
3.8073E-01
4.6413E-01
5.6580E-01
6.8973E-01
3.8073E-01
4.6413E-01
5.6580E-01
6.8973E-01
8.4081E-01
0.0000E+00
0.0000E+00
0.0000E+00
0.0000E+00
0.0000E+00
0.0000
0.0000
0.0000
0.0000
0.0000
1.5474E-04 1.0000
8.8026E-04 0.3625
1.5655E-03 0.2444
9.0937E-03 0.0891
0.0000E+00 0.0000
0.0091 +- 9%
→ The penetration rate is certainly below 1/100
Step 8
14
Step 9: How about neutrons?
1. Change the source to neutron with energy of 1.0 MeV
2. Set “maxbch = 5”
Fluence of neutrons
Penetrated!
Energy spectra
80% of neutrons penetrate the
target without any interaction
Step 9
15
Step 10: Shielding material for neutrons
1. Change the target material and thickness in order to
decrease the penetration rate of neutrons down to 1/100
2. Try various materials for the target, and find an
appropriate shielding material for neutrons
Al (2.7g/cm2)
ca. 47 cm
C (1.77g/cm3)
H2O (1.0g/cm3)
ca. 32 cm
ca. 17 cm
Lighter nuclei such as hydrogen are suit for neutron shielding
Step 10
16
Summary
• Commonly views on the shielding
profiles of a, b, g-rays and neutrons
were verified using PHITS
• PHITS is useful for comprehensive
analysis of radiation transport owing to
its applicability to various particles
Summary
17
Homework (Hard work!)
1. Let’s design a shielding for high-energy neutron (100 MeV)
2. Index for the shielding is not the fluence but the effective
doses
3. Find the thinnest shielding that can reduce the doses by 2
order of the magnitude
4. You can combine 2 materials for the shielding
Hints
•
•
•
•
Use [t-track] in “h10multiplier.inp” in the recommendation settings
See the histogram of the doses by changing the axis from “xz” to “z”
Change “nx” parameter to 1 for avoiding to create too much files
Low-energy neutrons are effectively shielded by lighter nuclei, while
high-energy neutrons are shielded by inter-mediate mass nuclei
Homework
18
Air
Concrete
Iron
Example of Answer（answer1.inp）
2-layer shielding that consists of 80 cm iron and 25 cm concrete
Let’s Think
• How much photon can contribute to the dose?
• Why 2-layer shielding is more effective in comparison to mono-layer one?
• What’s happened when the order of the 2 layers would be changed?
Homework
19