PHITSHow to stop , , -rays and neutrons?

Multi-Purpose Particle and Heavy Ion Transport code System

title 1

May 2015 revised

1. -ray can be stopped by a piece of paper

2. -ray can be stopped by an aluminum board

3. -ray can be stopped by an lead block

4. neutron can penetrate all of these materials

Contents 2

Purpose of This ExerciseIt is generally said that …

Let’s check whether they are correct or not, using PHITS!

Select an appropriate input file from recommendation settings

3Copy Input File

1. transport various particles2. visualize the particle trajectories

Calculation condition

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

Range.inp の確認

4Check Input File

Calculation Condition

20MeVElectron

Al

track.eps

Electron (20MeV, 0.01cm radius beam ）Cylindrical Al shielding (t = 2cm, r = 5cm) & Void

[t-track] for visualizing particle trajectories

[t-cross] for calculating particle fluxes behind the shielding

cross.eps

Void

Geometry

Incident particle ：

Geometry ：Tally ：

5Procedure

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 for this exercise

Step 1 ： Change Source

6Step 1

Source energy is defined in [source]

section

[ S o u r c e ] 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!

Change the source from 20 MeV to 1 MeV electron (a typical energy of -ray)

Electron fluence (track.eps)Shielded!

Step2: Change the Thickness

7Step 2

• Thickness of target is defined as a parameter c1

Real aluminum boards are rather thin (~ 1mm)Change the thickness of target to 1mm

Electron fluence (track.eps)1mm is too thin to stop!

[ S u r f a c e ]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

Let’s investigate the minimum thickness of Al to stop 1 MeV electron

Step3 ： Change Tallied Region

8Step 3

1. Target thickness (c1) should be 0.2 cm2. Tally from -c1 to +c1 cm for X&Y directions3. 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

Let’s see the fluence distribution inside the target in more detail!

Electron fluenceStopped close to the

distal edge

Photon fluencePenetrated

9Step 3

Particle fluence behind the target （ flux.eps ）

Photons with energies from 10 keV to 100 keV are escaped

Integrated value?

You can find the integrated fluence per source at ”# sum over” at the 94 line of flux.out

0.075 photons/incident electron are escaped from the aluminum board

An aluminum board can stop -ray, but not secondary photon!

Check Energy Spectrum

y(electron)... y(photon)…# sum over 0.0000E+00 7.5439E-02

Step4: How about α-rays?

10Step 4

Fluence of α-rays (3rd page of track.eps)Stopped at the surface

1. Change the source from β-ray to α-ray with an energy of 6 MeV ( ＝ 1.5MeV/u ）

[ S o u r c e ] 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

Step5: Change Geometry

11Step 5

Fluence of α particle•Stop at 0.006 cm in paper•No secondary particle is generated

[ M a t e r i a l ]MAT[ 1 ] # Aluminum 27Al 1.0

[ C e l l ] 1 1 -2.7 1 -2 -11 $ Target 2 0 2 -3 -11 $ Void98 0 #1 #2 -999 $ Void99 -1 999 $ Outer region

[ S u r f a c e ]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

1. Change the target to a piece of paper (C6H10O5)n

2. Assume density = 0.82g/cm3 & thickness = 0.01cm

Step 6: How about γ-rays?

12Step 6

1. Change the source to -rays with energy of 0.662MeV2. Change the target to a 1 cm lead block (11.34g/cm3)

204Pb 0.014206Pb 0.241207Pb 0.221208Pb 0.524

Fluence of photonTarget thickness is not enough

Energy spectra behind the targetMany photons penetrate the target without any interaction

13Step 7

Fluence of photon for the 4.3 cm lead target case

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 spectrumPenetration rate = 0.010

Step 7: Find an appropriate thickness

Step8: Reduce Statistical Uncertainty

14Step 8

Estimate the penetration rate with statistical uncertainty below 10% by changing maxcas, maxbch, batch.now, istdev etc.

3.1232E-01 3.8073E-01 0.0000E+00 0.0000 0.0000E+00 0.0000 3.8073E-01 4.6413E-01 0.0000E+00 0.0000 3.1116E-03 0.7100 4.6413E-01 5.6580E-01 0.0000E+00 0.0000 1.1609E-03 1.0000 5.6580E-01 6.8973E-01 0.0000E+00 0.0000 1.0050E-02 0.3148 6.8973E-01 8.4081E-01 0.0000E+00 0.0000 0.0000E+00 0.0000

You can check the values in 75th line of cross.out

3.1232E-01 3.8073E-01 0.0000E+00 0.0000 1.5474E-04 1.0000 3.8073E-01 4.6413E-01 0.0000E+00 0.0000 8.8026E-04 0.3625 4.6413E-01 5.6580E-01 0.0000E+00 0.0000 1.5655E-03 0.2444 5.6580E-01 6.8973E-01 0.0000E+00 0.0000 9.0937E-03 0.0891 6.8973E-01 8.4081E-01 0.0000E+00 0.0000 0.0000E+00 0.0000

0.0091 +- 9% → The penetration rate is certainly below 1/100

maxcas = 1000, maxbch = 1

maxcas = 1000, maxbch = 14

15Step 9

Step 9: How about neutrons?1. Change the source to neutron with energy of 1.0 MeV2. Set “maxbch = 5”

Fluence of neutronsPenetrated!

Energy spectra80% of neutrons penetrate the target without any interaction

16Step 10

Step 10: Shielding material for neutrons

C (1.77g/cm3)ca. 32 cm

1. Change the target material and thickness in order to decrease the penetration rate of neutrons down to 1/100 (see ”# sum over”)

2. Try various materials for the target, and find an appropriate shielding material for neutrons

Al (2.7g/cm2)ca. 47 cm

H2O (1.0g/cm3)ca. 17 cm

Lighter nuclei such as hydrogen are suit for neutron shielding

17

• Commonly views on the shielding profiles of , , -rays and neutrons were verified using PHITS

• PHITS is useful for comprehensive analysis of radiation transport owing to its applicability to various particles

Summary

Summary

Homework 18

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

Homework (Hard work!)

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 19

Example of Answer （ answer1.inp ）

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?

2-layer shielding that consists of 80 cm iron and 25 cm concrete

Iro

n

Co

ncr

ete

Air