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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. Setting of various sources A. Aug. 2018 revised. Title. 1. Goal of this lecture. Transport simulation with various kinds of sources. Simulation with two 60 Co source.

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P HI T S

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  1. PHITS Multi-Purpose Particle and Heavy Ion Transport code System Setting of various sources A Aug. 2018 revised Title 1

  2. Goal of this lecture Transport simulation with various kinds of sources Simulation with two 60Co source. Source with continuous energy distribution. Purpose 2

  3. sourceA.inp Basic setup 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 Projectile: Geometry: Tally: Water 150MeV Proton Geometry track_xz.eps cross_eng.eps Check Input File 3

  4. Table of contents • Source with energy distribution • Continuous energy distribution • Discrete energy distribution • Setup of multiple sources • RI source • Summary Table of Contents 4

  5. Sources with energy distribution Source energy can be defined either as mono-energetic or distributed in PHITS Proton beam having energy distribution Energy distribution 5

  6. How to set 1 At [source]section, set e-type subsection (Unit of energy is MeV or angstrom). [ 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 [ 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 = 1 ne = 2 0.0 4 50.0 1 100.0 Energy distribution 6

  7. How to set ２ • 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. * To change weight or give energy with angstrom, use other e-type . （See Sec. 4.3.17 of the manual） For discrete distribution (e.g. e-type=8), specify the number of energy peaks (ne), peak energy (e(i)), and intensity (w(i)). For continuous distribution (e.g. e-type=1), specify the number of energy groups (ne), bin energy (e(i)), and intensity (w(i)). e-type = 1 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) e(n+1) e-type = 8 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) Number of e(i) is n+1 in total. Number of w(i) is n. (When “ne” is negative, energy distribution in each bin is uniform in [/Lethergy].) Numbers of e(i) and w(i) are n. Energy distribution 7

  8. Exercise 1 Set proton beam with energy distribution Bin : [0,50], [50,100], [100,150] in MeV Intensity : 1:3:2 in ratio. (See right figure) • Add e-type subsection and set energy distribution. • Comment out the line “e0=150”. sourceA.inp e-type=1format [ 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 = e-type = 1 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) e(n+1) Energy distribution 8

  9. Answer 1 Set proton beam with energy distribution Bin : [0,50], [50,100], [100,150] in MeV Intensity : 1:3:2 in ratio. sourceA.inp [ 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 = 1 ne = 3 0.0 1 50.0 3 100.0 2 150.0 cross_eng.eps The ratio of intensity is 1:3:2. (The ratio is given in integral value for e-type=1.) Energy distribution 9

  10. Exercise 2 Change the energy bin from [100:150] to [100:200]. • Change the energy range of the 3rd bin. • Check the energy distribution in the new energy bin setup. sourceA.inp [ S o u r c e ] ・ ・ ・ ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 150.0 e-type=1 : Source intensity is given by integral value. Energy distribution 10

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

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

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

  14. Table of contents • Source with energy distribution • Continuous energy distribution • Discrete energy distribution • Setup of multiple sources • RI source • Summary Table of Contents 14

  15. Source having discrete energy Sources having more than one energy peaks such as 60Co and 134Cs can be defined in PHITS. 60Cosource b- 60Co 60Co emits gamma-rays at two energies (1.173 and 1.333 MeV) after beta decay. g (1.173MeV) 100% g (1.333MeV) 100% 60Ni Energy distribution 15

  16. Exercise 4 Simulate 60Co source. sourceA.inp • Change the source particle from proton to photon. • Define an isotropic point source. (Change the source radius (r0) and direction (dir).) • Use e-type=8 and set the photon energies (1.173MeV and 1.333MeV) with intensity 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] [ S o u r c e ] totfact = 1.0 s-type = 1 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 e-type=8format e-type = 8 ne = n e(1) w(1) ・ ・ ・ ・ ・ ・ e(n) w(n) Energy distribution 16

  17. Answer 4 Simulate 60Co source. sourceA.inp [ 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 [ S o u r c e ] totfact = 1.0 s-type = 1 proj = photon $ e0 = 150. r0 = 0.0 z0 = -10. z1 = -10. dir = all e-type = 8 ne = 2 1.173 1 1.333 1 cross_eng.eps 60Cosource track_xz.eps Energy distribution 17

  18. Table of contents • Source with energy distribution • Continuous energy distribution • Discrete energy distribution • Setup of multiple sources • RI source • Summary Table of Contents 18

  19. Setup of multiple source Multiple source with different radiation types, positions, or energy distribution can be defined in PHITS. 60Cosource 60Cosource 60Co sources placed at right and left of target with the intensity ratio of 2:1. Multi-source 19

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

  21. Exercise 5 Set 60Co sources at the left and right of the cylindrical water (z=-10, 40cm) with intensity ratio of 2:1. • Add <source> lines to define two sources • Put two point sources at the position z=-10 and 40cm(Change z0 and z1 to define point sources) • Define the relative intensity of the left (z=-10cm) and the right (z=40cm) source to be 2:1. z axis 60Cosource 60Cosource z=-10cm z=40cm Multi-source 21

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

  23. Table of contents • Source having energy distribution • Continuous energy distribution • Discrete energy distribution • Setup of multiple sources • RI source • Summary Table of Contents 23

  24. RI (Radioactive Isotope) source α, β and γ radioisotope sources can be defined by simply specifying name of RI and its activity. (From PHITS2.86) How to set. • Set e-type=28 or 29. (29 for changing weight of source.) e-type=28 format e-type = 28 ni = n RI(1) A(1) ・ ・ ・ ・ ・ ・ RI(n) A(n) norm = *** Number of RIs. Name of the RIs and their activity (Bq). Option for normalization0: (/sec), 1: (/source) RI source 24

  25. Exercise 6 Set 60Co sources of 200 and 100Bq in the left- and right-sides, respectively, by using e-type=28. • Change e-type. • Specify 60Co of 200 and 100 Bq at z=-10 and 40 cm, respectively. (Name format is Co-60 or 60Co.) • Normalize tally results to the unit of (/sec). • In case of e-type=28, 29, set <source> to be 1.0 and totfact to be the number of <source> subsections (totfact=2.0 in this case) because the absolute activity of RIs is directly defined in Bq. sourceA.inp [ S o u r c e ] totfact = 1.0 <source> = 2.0 ・ ・ ・ ・ ・ ・ z1 = -10. dir = all e-type = 8 ne = 2 1.173 1 1.333 1 <source> = 1.0 ・ ・ ・ ・ ・ ・ z1 = 40. dir = all e-type = 8 ne = 2 1.173 1 1.333 1 RI source 25

  26. Answer 6 Set 60Co sources of 200 and 100Bq in the left- and right-sides, respectively, by using e-type=28. sourceA.inp [ S o u r c e ] totfact = 2.0 <source> = 1.0 ・ ・ ・ ・ ・ ・ z1 = -10. dir = all e-type = 28 ni = 1 Co-60 200.0 norm = 0 (continued) <source> = 1.0 ・ ・ ・ ・ ・ ・ z1 = 40. dir = all e-type = 28 ni = 1 Co-60 100.0 norm = 0 cross_eng.eps Gamma spectrum of 60Co sources (1.173 & 1.333 MeV) is realized. Note that the unit of these data is [1/cm2/sec], though plot axis label is [1/cm2/source]. RI source 26

  27. Decay time & Daughter nuclide γ-rays are not emitted from 137Cs without considering its daughter nuclide γ-rays of 0.6617 MeV are emitted from an isomer of Ba (137mBa) after the beta decay of 137Cs Specify decay time parameter “dtime” 30 years later 30 years ago Now For dtime < 0, activity at half-life x dtime prior are calculated, and then, current activities including daughter nuclides are considered For dtime > 0, activities at dtime (sec) later including daughter nuclides are considered 137Cs （T1/2=30.04y） 137Cs: 100 Bq 137mBa: 0 Bq 137Cs: 50 Bq 137mBa: 50 Bq 137Cs: 200 Bq 137mBa: 0 Bq 137mBa（T1/2=2.552m） b- g (0.6617MeV) 85.1% Cs-137 100.0 dtime = -1.0 Cs-137 100.0 dtime = 0 Cs-137 100.0 dtime = 30.04*365.25*24*3600 137Ba 137Cs: 100 Bq 137mBa: 100 Bq are considered as source RIs RI source 27

  28. Exercise 7 Change 200Bq 60Co to 200Bq 137Cs, and consider the radioactive equilibrium. • Change Co-60 to Cs-137 in the first <source> subsection. • Add dtime=-10 to the subsection. (-10 is default value) sourceA.inp [ S o u r c e ] totfact = 2.0 <source> = 1.0 ・ ・ ・ ・ ・ ・ z1 = -10. dir = all e-type = 28 ni = 1 Co-60 200.0 norm = 0 dtime = <source> = 1.0 ・ ・ ・ ・ ・ ・ Most RIs reach equilibrium after 10 half-lives dtime = -10 is convenient to reproduce most of RI sources in equilibrium. Notes • 10 x half-life is not enough to reach equilibrium for some RIs whose half-life is much shorter than that of its daughter nuclide; e.g. 105Ru（T1/2 = 4.44h）→ 105Rh（T1/2 = 35.36h） • Too large dtime (e.g. dtime = -1000.0) may cause an error RI source 28

  29. Answer 7 Change200Bq 60Co to 200Bq 137Cs, and consider the radioactive equilibrium. sourceA.inp [ S o u r c e ] totfact = 2.0 <source> = 1.0 ・ ・ ・ ・ ・ ・ z1 = -10. dir = all e-type = 28 ni = 1 Cs-137 200.0 norm = 0 dtime = -10 <source> = 1.0 ・ ・ ・ ・ ・ ・ cross_eng.eps 0.6617 MeV gamma-rays are emitted from 137mBa by defining 137Cs source. RI source 29

  30. Exercise8 Let’s define 137Cs β-ray source • Copy & paste <source> subsection for 137Cs • Change ‘proj’ in the new <source> subsection to electron • Set totfact = 3.0 • Set “part = photon electron” in [t-track] & [t-cross] to see the trajectories and energy spectra of electrons as well as photons (totfact should be always equal to the number of <source> subsections for RI source) RI source 30

  31. Answer8 sourceA.inp track_xz.eps (2nd page) [ S o u r c e ] totfact = 3.0 <source> = 1.0 proj = photon ・ ・ ・ ・ ・ ・ ni = 1 Cs-137 200.0 norm = 0 dtime = -10 <source> = 1.0 proj = electron ・ ・ ・ ・ ・ ・ ni = 1 Cs-137 200.0 norm = 0 dtime = -10 ・ ・ ・ ・ ・ ・ cross_eng.eps Continuum spectrum of β-rays RI source 31

  32. Table of contents • Source having energy distribution • Continuous energy distribution • Discrete energy distribution • Setup of multiple source • RI source • Summary Table of Contents 32

  33. Summary Source energy distribution (continuous and discrete) can be defined by specifying e-type in the [source] section. Multiple sources can be defined by setting <source> subsections. α, β, γ decay sources can be defined by directly specifying the name and activity of RIs. Refer to “setting of various source B” (phits-lec-sourceB-jp.ppt) for the setup of source using “Dump data”. Summary 33

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