1. How to stop a, b, g-rays and neutrons?
PHITS
Multi-Purpose Particle and Heavy Ion Transport code System
How to stop a, b, g-rays and neutrons?
PHITS講習会　入門実習
Jul revised
title 1

2. Purpose of This Exercise
It is generally said that …
a-ray can be stopped by a piece of paper
b-ray can be stopped by an aluminum board
g-ray can be stopped by an lead block
neutron can penetrate all of these materials
Let’s check whether they are correct or not, using PHITS!
「実習」　前の基本的な話から
Contents 2

3. Check Range.inp Calculation Condition Geometry track.eps cross.eps
Incident particle：
Geometry：
Tally：
Photon with energy distribution (r = 0.01cm)
Cylindrical Al shielding (t = 0.1cm, r = 5cm) & Void
[t-track] for visualizing particle trajectories
[t-cross] for calculating particle fluxes behind the shielding
Photon Al
Void
Geometry
track.eps
cross.eps
Check Input File 3

4. Source with Energy Distribution
Energy distribution is defined by e-type subsection.
e-type = 21: number of energy bin (ne)
　　　　　　　　Lower boundary of each bin (MeV)
　　　　　　　　Relative flux of each bin (/MeV)
[ S o u r c e ]
s-type = 1
proj = photon
r0 =
x0 =
y0 =
z0 =
z1 =
dir =
e-type = 21
ne = 3
$ E_low Flux(/MeV)
3.0
1 : : 1
Relative flux of each bin (/MeV)
Lower boundary of each energy bin (ne+1)
See manual or lecture\advanced\sourceA in more detail 4

5. Procedure for this exercise
Change the source to β-rays
Change the thickness of the shielding
Change the tallied region
Change the source to α-rays
Change the target to a piece of paper
Change the source to γ-rays, and the target to a lead block
Find an appropriate thickness of the lead block
Reduce the statistical uncertainty
Change the source to neutrons
Find an appropriate shielding material for neutrons
The input files for each procedure were prepared as “range*.inp”
Procedure 5

6. Table 1 Relative flux of β-rays
Step 1：Change Source
Change energy distribution referring to Table 1 → Check cross.eps
Change particle to electron (proj) electron → Check track.eps
Table 1　Relative flux of β-rays
[ S o u r c e ]
s-type = 1
proj = photon
r0 =
x0 =
y0 =
z0 =
z1 =
dir =
e-type = 21
ne = 3
$ E_low Flux(/MeV)
3.0
Energy range
Relative Flux (/MeV)
0 ～ 0.5 MeV
2.0
0.5 ～ 1.0 MeV
1.0
1.0 ～ 1.5 MeV
0.5
Change red parameters
cross.eps after step1
track.eps after step2
(2nd page)
Step 1 6

7. Step2: Change the Thickness
Find an appropriate thickness of Al boards to completely stop electrons (check 1mm step)
[ S u r f a c e ]
set: c1[0.1] $ Thickness of Target (cm) pz
pz c1 pz cz so
Thickness of target is defined as a parameter c1
Electron fluence (track.eps, 2nd page)
3mm is enough thick to stop!
Step 2 7

8. Step3：Change Tallied Region
Let’s see the fluence distribution inside the target in more detail!
Target thickness (c1) should be 0.3 cm
Tally from -c1 to +c1 cm for X&Y directions
Tally from 0 to c1*2 cm for Z direction
[ T - T r a c k ]
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
Photon fluence
Penetrated
Step 3 8

9. Check Energy Spectrum Integrated value?
You can find the integrated fluence per source at ”# sum over” at the 70 line of cross.out
photon electron …
# sum over E E+00
0.032 photons/incident electron are escaped from the aluminum board
Particle fluence behind the target（cross.eps）
An aluminum board can stop b-ray, but not secondary photon!
Photons with energies from 10 keV to 500 keV are escaped
Step 3 9

10. Fluence of α-rays (3rd page of track.eps)
Step4: How about α-rays?
Change the source to α-ray with an energy of 6 MeV (＝1.5MeV/u）
Mono-energy should be defined by e0. (e-type is not necessary)
Particle type should be defined by “proj” (α-ray is “alpha”)
[ S o u r c e ]
s-type = 1
proj = electron
r0 =
x0 =
y0 =
z0 =
z1 =
dir =
e-type = 21
ne = 3
$ E_low Flux(/MeV)
1.5
After commenting out the e-type subsection, add e0=*** to define the mono-energy.
Fluence of α-rays (3rd page of track.eps)
Stopped at the surface
Step 4 10

11. Step5: Change Geometry
Change the target to a piece of paper (C6H10O5)n
Assume density = 0.82g/cm3 & thickness = 0.01cm
[ M a t e r i a l ]
MAT[ 1 ] # Aluminum
27Al
[ C e l l ]
$ Target
$ Void
#1 # $ Void
$ Outer region
[ S u r f a c e ]
set: c1[0.3] $ Thickness of Target (cm) pz
pz c1 pz cz so
Fluence of α particle
Stop at cm in paper
No secondary particle is generated
Step 5 11

12. Step 6: How about γ-rays?
Change the source to g-rays with energy of 0.662MeV
Change the target to a 1 cm lead (Pb) block (11.34g/cm3)
Energy spectra behind the target
Many photons penetrate the
target without any interaction
Fluence of photon
Target thickness is not enough
Step 6 12

13. Step 7: Find an appropriate thickness
Change the target thickness to decrease the direct penetration rate of photons down to 0.01
Check 45th line (0.6 MeV < E < 0.7 MeV) in cross.out
Fluence of photon for
the 4.1 cm lead target case
Energy spectrum
Penetration rate = 0.009
Step 7 13

14. Step8: Consider Statistical Uncertainty
Estimate the shielding thickness to give penetration rate below 0.01 with statistical significance, by changing maxcas, maxbch, batch.out, istdev etc.
Thickness = 4.1cm，maxcas = 2000, maxbch = 1
Elow(MeV) Ehigh(MeV) Photon r.err
2.0000E E E
3.0000E E E
4.0000E E E
5.0000E E E
6.0000E E E
See 45th line of cross.out
Statistical uncertainty is too large (~25%)
Thickness = 4.3cm，maxcas = 2000, maxbch = 12
2.0000E E E
3.0000E E E
4.0000E E E
5.0000E E E
6.0000E E E
% → The penetration rate is below 0.01 with statistically significant
Step 8 14

15. Step 9: How about neutrons?
Change the source to neutron with energy of 1.0 MeV
Set “maxbch = 3”
Fluence of neutrons
Penetrated!
Energy spectra
80% of neutrons penetrate the
target without any interaction
Step 9 15

16. Step 10: Shielding material for neutrons
Change the target material and thickness in order to decrease the penetration rate of neutrons down to 0.01 (see ”# sum line)
Try various materials for the target, and find an appropriate shielding material for neutrons
Al (2.7g/cm3)
ca. 40 cm
C (1.77g/cm3)
ca. 27 cm
H2O (1.0g/cm3)
ca. 18 cm
Lighter nuclei such as hydrogen are suit for neutron shielding
Step 10 16

17. Summary
Common sense on the shielding profiles of a, b, g-rays and neutrons were confirmed using PHITS
PHITS is useful for particle transport simulation in mixed radiation fields owing to its applicability to various radiation types
Summary 17

18. Homework (Difficult!)
Let’s design a shielding for high-energy neutron (100 MeV)
Index for the shielding is not the fluence but the effective doses
Find the thinnest shielding that can reduce the doses by 2 order of the magnitude
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

19. Example of Answer（answer1.inp）
Iron
Concrete
Air
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