Exercises (II): How to stop radiations?

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Transcript Exercises (II): How to stop radiations?

P HI T S Multi-Purpose P article and H eavy I on T ransport code S ystem

Exercise

（

II

）

: How to stop

,

-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 (

= 2cm,

= 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

Procedure for this exercise

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) [ 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!

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 [ 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

• 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 - T r a c k ] title = Track in mesh = xyz x-type = 2 xmin = ymin = -1.5 xmax = ymax = zmax = 1.5

nx = 50 y-type = 2 -1.5 1.5

ny = 1 z-type = 2 zmin = 0.0 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

y(electron)... y(photon) … # sum over 0.0000E+00 7.5439E-02 Particle fluence behind the target （ flux.eps

） Photons with energies from 10 keV to 100 keV are escaped Step 3 0.075 photons/incident electron are escaped from the aluminum board An aluminum board can stop b -ray, but not secondary photon!

Step4: How about α-rays?

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 Step 4 Fluence of α-rays (3 rd page of track.eps) Stopped at the surface 10

Step5: Change Geometry

1. Change the target to a piece of paper (C 6 H 10 O 5 ) n 2. Assume density = 0.82g/cm 3 & thickness = 0.01cm

[ 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 $ Void 98 0 #1 #2 -999 $ Void 99 -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

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/cm 3 ) 204Pb 0.014

206Pb 0.241

207Pb 0.221

208Pb 0.524

Fluence of photon Target thickness is not enough Energy spectra behind the target Many photons penetrate the 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 75 th line in cross.out

Fluence of photon for the 4.3 cm lead target case Step 7 Energy spectrum Penetration rate = 0.010

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 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 75 th line of cross.out

maxcas = 1000, maxbch = 14 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 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/cm ca. 47 cm 2 ) C (1.77g/cm ca. 32 cm 3 ) H 2 O (1.0g/cm 3 ) ca. 17 cm Lighter nuclei such as hydrogen are suit for neutron shielding Step 10 16

Summary • 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 17

Homework (Hard work!)

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

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