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P HI T S

P HI T S. Multi-Purpose P article and H eavy I on T ransport code S ystem. Exercise （ II ） : How to stop a , b , g -rays and neutrons?. Feb. 2014 revised. title. 1. a -ray can be stopped by a piece of paper b -ray can be stopped by an aluminum board

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P HI T S

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  1. 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

  2. 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 Purpose of This Exercise It is generally said that … Let’s check whether they are correct or not, using PHITS! Contents 2

  3. Select an appropriate input file from recommendation settings Calculation condition • transport various particles • 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

  4. Range.inpの確認 Calculation Condition Incident particle： Geometry： Tally： 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 20MeV Electron Al Void Geometry cross.eps track.eps Check Input File 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. 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

  7. 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) 1mmis too thin to stop! Let’s investigate the minimum thickness of Al to stop 1 MeV electron 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 0.2 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 94 line of flux.out y(electron)... y(photon) … # sum over 0.0000E+00 7.5439E-02 0.075 photons/incident electron are escaped from the aluminum board Particle fluence behind the target（flux.eps） 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

  10. 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 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 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

  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 block (11.34g/cm3) 204Pb 0.014 206Pb 0.241 207Pb 0.221 208Pb 0.524 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 1/100 • 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

  14. 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 75th 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

  15. Step 9: How about neutrons? • Change the source to neutron with energy of 1.0 MeV • Set “maxbch = 5” Energy spectra 80% of neutrons penetrate the target without any interaction Fluence of neutrons Penetrated! 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 1/100 • 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) ca. 32 cm H2O (1.0g/cm3) ca. 17 cm Lighter nuclei such as hydrogen are suit for neutron shielding Step 10 16

  17. 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 Summary 17

  18. 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 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 18

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

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