1. P HI T S Setting of various source Part I Multi-Purpose Particle and Heavy Ion Transport code System Title1 Aug. 2015 revised

2. Goal of this lecture Purpose2 Transport simulation with various kinds of sources Simulation with 60 Co source placed in two positions Source having continuous energy distribution

3. sourceA.inp 3Check Input File Basic setup Projectile: Geometry: Tally: Geometry 150MeV proton (pencil beam with radius 1.0cm) Water cylinder (10cm radius and 20cm thickness) [t-track] fluence distribution [t-cross] proton energy spectrum coming into water Water 150MeV Proton track_xz.epscross_eng.eps

4. Table of Contents4 Table of contents 1.Source having energy distribution A)Continuous energy distribution B)Discrete energy distribution 2.Setup of multiple source 3.Summary

5. Source having energy distribution Energy distribution5 A source having energy distribution can be treated in PHITS as well as a single energy source Proton beam having energy distribution

6. How to set 1 Energy distribution6 At [source] section, set s-type=4 (for cylinder), 5 (for rectangular), 10 （ for sphere ） Set e-type subsection (Unit of energy is MeV or angstrom) [ S o u r c e ] totfact = 1.0 s-type = 4 proj = proton $ e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0 e-type = 1 ne = 2 0.0 4 50.0 1 100.0 [ S o u r c e ] totfact = 1.0 s-type = 1 proj = proton e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0

7. How to set ２ Energy distribution7 3 ways to specify energy distribution （ switched by e-type ）  e-type=1 * : Continuous distribution with integral value  e-type=21 * : Continuous distribution with differential value (particle/MeV)  e-type=8 * : Discrete distribution e-type = 1 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) e(n+1) For continuous distribution (f.g. e- type=1), specify data of number of energy group (ne), energy bin (e(i)), and probability to generate particle (w(i)) Number of e(i) is n+1 in total Number of w(i) is n (Log scale is applied when ne is given in minus) For discrete distribution (f.g. e-type=8), specify data of number of energy point (ne), energy point(e(i)), and probability to generate particle (w(i)) e-type = 8 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) Numbers of e(i) and w(i) are n * To change weight or give energy with angstrom, use the other e- type （ See Sec. 4.3.15 of the manual ）

8. Exercise 1 8 Set proton beam of the spectrum with the energy bin of [0,50], [50,100], [100,150] in MeV and the beam intensity of 1:3:2 in ratio (See right figure) [ S o u r c e ] totfact = 1.0 s-type = 1 proj = proton e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0 e-type = sourceA.inp Energy distribution e-type = 1 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) e(n+1) e-type=1 format Change s-type to 4 Add e-type subsection and set energy distribution Comment out the line of e0=150

9. Answer 1 9 [ S o u r c e ] totfact = 1.0 s-type = 4 proj = proton $ e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0 e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 150.0 sourceA.inp Energy distribution cross_eng.eps The ratio of intensity is 1:3:2 (The ratio is given in integral value for e-type=1 ) Set proton beam of the spectrum with the energy bin of [0,50], [50,100], [100,150] in MeV and the beam intensity of 1:3:2 in ratio

10. [ S o u r c e ] ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 150.0 Exercise 2 10 Change the energy bin of [100:150] to [100:200] sourceA.inp Energy distribution Source intensity is given by integral value with energy for e-type=1 Change the energy range of the third bin Check the beam flux when the energy bin range becomes 50, 50, and 100 MeV

11. [ S o u r c e ] ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 200.0 Answer 2 11 sourceA.inp Energy distribution cross_eng.eps The ratio of intensity integrated by energy of the three bin is 1:3:2 (The ratio is 1:3:1 if it is given in per unit energy or differential value) Change the energy bin of [100:150] to [100:200]

12. [ S o u r c e ] ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 200.0 Exercise 3 12 Give the ratio of the intensity of 1:3:2 in differential value for the energy bin of [0:50], [50:100], [100:200] Use e-type=21 sourceA.inp Energy distribution The ratio is given in differential value for e-type=21 (Choose this option to use derivative spectrum)

13. [ S o u r c e ] ・ ・ ・ e-type = 21 ne = 3 0.0 1 50.0 3 100.0 2 200.0 Answer 3 13 sourceA.inp Energy distribution The ratio of the intensity is 1:3:2 in differential value (The ratio is 1:3:4 in integral value) cross_eng.eps Give the ratio of the intensity of 1:3:2 in differential value for the energy bin of [0:50], [50:100], [100:200]

14. Table of Contents14 Table of contents 1.Source having energy distribution A)Continuous energy distribution B)Discrete energy distribution 2.Setup of multiple source 3.Summary

15. Source having discrete energy Energy distribution15 A source emitting several discrete energy radiations such as 60 Co and 134 Cs can be treated in PHITS 60 Co source 60 Co emits two gamma-rays of the energy of 1.173 and 1.333 MeV after the beta decay 60 Co 60 Ni   1.173MeV) 100%  1.333MeV) 100%

16. [ S o u r c e ] totfact = 1.0 s-type = 4 proj = proton $ e0 = 150. r0 = 1.0 ・ ・ ・ dir = 1.0 e-type = 21 ne = 3 0.0 1 50.0 3 100.0 2 200.0 Exercise 4 16 Simulate 60 Co source Change the source to photon Set isotropic point source [Change the parameters of radius (r0) and direction (dir)] Set the normalization factor (totfact) to 2 （ 60 Co emits two gamma-ray per decay) => Tally output becomes amount per Bq Use e-type=8 and set the photon energies of 1.173MeV and 1.333MeV with the ratio of 1:1 Set [t-cross] to tally photon fluence from 0 to 2 MeV with 10keV resolution (200 groups) [change emax, ne, part] sourceA.inp Energy distribution e-type = 8 ne = n e(1) w(1) ・ ・ ・ ・ ・ ・ e(n) w(n) e-type=8 format

17. [ S o u r c e ] totfact = 2.0 s-type = 4 proj = photon $ e0 = 150. r0 = 0.0 z0 = -10. z1 = -10. dir = all e-type = 8 ne = 2 1.173 1 1.333 1 Answer 4 17 sourceA.inp Energy distribution cross_eng.eps [ T - C r o s s ] ・ ・ ・ emin = 0.0 emax = 2.0 ne = 200 unit = 1 axis = eng file = cross_eng.out output = flux part = photon epsout = 1 track_xz.eps 60 Co source Simulate 60 Co source

18. Table of Contents18 Table of contents 1.Source having energy distribution A)Continuous energy distribution B)Discrete energy distribution 2.Setup of multiple source 3.Summary

19. Setup of multiple source Multi source19 Multiple source with different radiation type, position, or energy distribution can be treated in PHITS 60 Co source 60 Co sources placed at right and left of target with the intensity ration of 2:1

20. How to set 20 At [source] section, set multiple source subsection starting with ” =relative intensity” Set totfact to normalize the total source intensity Multi source [ S o u r c e ] totfact = 2.0 = 2.0 s-type = 1 proj = proton ・ ・ ・ = 1.0 s-type = 4 proj = neutron ・ ・ ・ = 3.0 s-type = 8 proj = photon ・ ・ ・ Normalization factor If it is positive, particles are produced with the ratio of the relative intensity If it is negative, same number of particles are produced with weight of the normalized intensity Triple source Relative ratio of intensity for each source (In this case, 2:1:3 from the top)

21. Exercise 5 21 Set 60 Co sources at the left and right of the cylindrical water (z=-10, 40cm) with intensity ratio of 2:1 Add the line to create two subsections Set two point sources at the position of z=-10 and 40cm (Change z0 and z1 to set a point source) Set the intensity to produce photon with the ratio of 2:1 from the left (z=-10cm) and the right (z=40cm) respectively 60 Co source Multi source z axis z=-10cm z=40cm

22. Answer 5 22Multi source sourceA.inp track_xz.eps 60 Co source (With intensity ratio of right:left = 2:1) [ S o u r c e ] totfact = 2.0 = 2.0 s-type = 4 proj = photon $ e0 = 150. r0 = 0.0 z0 = -10. z1 = -10. ・ ・ ・ = 1.0 s-type = 4 proj = photon r0 = 0.0 z0 = 40. z1 = 40. dir = all e-type = 8 ne = 2 1.173 1 1.333 1 Set 60 Co sources at the left and right of the cylindrical water (z=-10, 40cm) with intensity ratio of 2:1

23. Table of Contents23 Table of contents 1.Source having energy distribution A)Continuous energy distribution B)Discrete energy distribution 2.Setup of multiple source 3.Summary

24. Summary 24 The source of continuous and discrete energy distribution can be treated by choosing s-type and e- type at the [source] section Simulation with multiple source can be conducted by the setup of subsections Refer to “setting of various source B” (phits-lec-sourceB- jp.ppt) for the setup of source using “Dump data” Sorry but it is available only in Japanese !