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

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

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

Table of contents Source with energy distribution Continuous energy distribution Discrete energy distribution Setup of multiple sources RI source Summary PHITS講習会　入門実習 Table of Contents 4

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

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

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 continuous distribution (e.g. e-type=1), specify the number of energy groups (ne), bin energy (e(i)), and intensity (w(i)). For discrete distribution (e.g. e-type=8), specify the number of energy peaks (ne), peak 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

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=1 format [ 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

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

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

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

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

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

Table of contents Source with energy distribution Continuous energy distribution Discrete energy distribution Setup of multiple sources RI source Summary PHITS講習会　入門実習 Table of Contents 14

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

Exercise 4 Simulate 60Co source. Energy distribution 16 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=8 format e-type = 8 ne = n e(1) w(1) ・ ・ ・ ・ ・ ・ e(n) w(n) Energy distribution 16

Answer 4 Simulate 60Co source. Energy distribution 17 sourceA.inp 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 [ 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 cross_eng.eps 60Co source track_xz.eps Energy distribution 17

Table of contents Source with energy distribution Continuous energy distribution Discrete energy distribution Setup of multiple sources RI source Summary PHITS講習会　入門実習 Table of Contents 18

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

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 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

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 60Co source 60Co source z=-10cm z=40cm Multi-source 21

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 z0 = 40. z1 = 40. dir = all e-type = 8 ne = 2 1.173 1 1.333 1 track_xz.eps Two 60Co sources. (With intensity ratio of right:left = 2:1) Multi-source 22

Table of contents Source having energy distribution Continuous energy distribution Discrete energy distribution Setup of multiple sources RI source Summary PHITS講習会　入門実習 Table of Contents 23

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 normalization 0: (/sec), 1: (/source) RI source 24

Exercise 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 = 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. 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. RI source 25

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

Decay time & Daughter nuclide γ-rays are not emitted from 137Cs without considering its daughter nuclide 137Cs （T1/2=30.04y） 137mBa（T1/2=2.552m） b- g (0.6617MeV) 85.1% 137Ba 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 ago Now 30 years later 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: 200 Bq 137mBa: 0 Bq 137Cs: 100 Bq 137mBa: 0 Bq 137Cs: 50 Bq 137mBa: 50 Bq Cs-137 100.0 dtime = -1.0 Cs-137 100.0 dtime = 0 Cs-137 100.0 dtime = 30.04*365.25*24*3600 137Cs: 100 Bq 137mBa: 100 Bq are considered as source RIs RI source 27

Most RIs reach equilibrium after 10 half-lives 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 = 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

Answer 7 Change 200Bq 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 cross_eng.eps 0.6617 MeV gamma-rays are emitted from 137mBa by defining 137Cs source. RI source 29

Let’s define 137Cs β-ray source 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

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

Table of contents Source having energy distribution Continuous energy distribution Discrete energy distribution Setup of multiple source RI source Summary PHITS講習会　入門実習 Table of Contents 32

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. 《休憩はさむ》 まとめ Refer to “setting of various source B” (phits-lec-sourceB-jp.ppt) for the setup of source using “Dump data”. Summary 33