1. P HI T S Exercise （ II ） : How to stop , ,  -rays and neutrons? Multi-Purpose Particle and Heavy Ion Transport code System title1 Feb. 2014 revised

2.  -ray can be stopped by a piece of paper  -ray can be stopped by an aluminum board  -ray can be stopped by an lead block 4.neutron can penetrate all of these materials Contents2 Purpose of This Exercise It is generally said that … Let’s check whether they are correct or not, using PHITS!

3. Select an appropriate input file from recommendation settings 3Copy Input File 1.transport various particles 2.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

4. Range.inp の確認 4Check Input File Calculation Condition 20MeV Electron 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 ：

5. 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

6. 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!

7. 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

8. Step3 ： Change Tallied Region 8Step 3 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 = -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 fluence Stopped close to the distal edge Photon fluence Penetrated

9. 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

10. Step4: How about α-rays? 10Step 4 Fluence of α-rays (3 rd 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

11. 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 $ 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 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

12. Step 6: How about γ-rays? 12Step 6 1.Change the source to  -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

13. 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 75 th line in cross.out Energy spectrum Penetration rate = 0.010 Step 7: Find an appropriate thickness

14. 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 75 th 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

15. 15Step 9 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

16. 16Step 10 Step 10: Shielding material for neutrons C (1.77g/cm 3 ) ca. 32 cm 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 2 ) ca. 47 cm H 2 O (1.0g/cm 3 ) ca. 17 cm Lighter nuclei such as hydrogen are suit for neutron shielding

17. 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

18. Homework18 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

19. Homework19 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 Iron Concrete Air