Mike's notes Flashcards

Question

why is TVL18 MV = TVL6 MV for lead?

Answer 1

high Z material pair production proprtional to Z ln (E) TVL1= TVLe = 5.7 cm as compton goes down as 1/ln(E), PP picks up

Answer 2

n = log10(1/B) B = I/Io t= TVL1 + (n-1)TVLe

Answer 3

t = TVL1 + (n-1)TVLe

Answer 4

10 cm at 6 MV, 11 cm for 18 MV

Answer 5

fraction of time linac is directed towards a primary barrier walls: U = 1/4 floor: U = 1 ceiling: U = 1/4

Answer 6

office, console = 1 adjacent tx room = 0.5 staff washroom = 1/5 vault door = 1/8 storage room = 1/20 outdoor areas = 1/40

Answer 7

linac weekly output at iso (1 m) including tx and QA ex. 35 patients per day, 2.5 y/fx, 5 days per week: W = 35 X 5 X 2.5 Gy = 440 Gy/week physics QA 160 Gy/week Total = 600 Gy/week

Answer 8

W is higher because of larger tx distance for the same pt dose W = D * d^2 d about 4 m typically ex. D = 12 Gy, W = 192 Gy/week for 1 TBI/week

Answer 9

IMRT delivers a radiation field in many segments More MUs per pt than conventional radiotherapy for same pt dose leakage works thus increases by factor of 2-15 (Ci = MU imrt/ MU conv) IMRT factor does not significantly affect the primary or scatter workload since both IMRT and conventional 3DRT deliver the same dose to the tumor. For VMAT, SBRT, SRS, the same concept applies, i.e. multiply by CI.

Answer 10

Bpri = P dpri^2/ (WUT) P = max permissible dose equivalent (0.1 mSv/wk for controlled areas and 0.02 mSv/wk for uncontrolled areas) dpri= distance from source to point protected W = workload, linac output at iso i.e. 1 m (ex 1000 Gy/wk) U= use factor T = occupancy factor B = I/Io patient attenuation is ignored to be conservative

Answer 11

6 TVL = 120 cm heavy concrete

Answer 12

4 TVL = 80 cm heavy concrete

Answer 13

B= (P/(alphaWT)*dsca^2*dsec^2*400/F P= max permissible dose equivalent alpha = scatter fraction at 1 m from patient for 400 cm^2 beam incident (tabulated) W = workload, linac output at iso U=1 T= occupancy factor dsca= distance from source to patient dsec= distance from patient to point protected F= field area at patient mid-depth at 1 m (40x40)

Answer 14

-scattered angle and energy -also material but this is always water for the patient

Answer 15

The value for α(θ) is given for field size 20x20=400 cm^2, thus need to scale up to the max field size F=40x40 cm^2.

Answer 16

Bl = P* dl^2/ (10^-3 * WT) P= max permissible dose equivalent dl= distance from source to point protected 10^ -3 is allowed head leakage (0.1%) W = workload, linac output at iso T= occupancy factor U=1

Answer 17

two source rule If a barrier must shield from 2 different sources and individual calculated barriers differ by more than 1 TVL, the thinner source barrier may be ignored. Otherwise use the thicker barrier + 1 HVL if tleak>tscat by 1 TVL ⇒use tleak otherwise ⇒add 1 HVL to tleak

Answer 18

leakage energy> scatter energy Scatter energy is always lower than the primary energy due to Compton interaction.

Answer 19

HVL=TVL x ln(2) / ln(10)

Answer 20

No, depending on the workload, the barrier thickness for 6MV could be larger (or smaller) than 18MV.

Answer 21

Think about this equation as how much of the dose (Gy) given to the patients (workload) will reach the maze door due to this specific mechanism (scattered primary off walls). Hs= WUg alphao Ao alphaz Az/ (dh dr dz)^2 W = primary workload Ug= use factor for wall g (first wall the primary scatters off of) alpha o = relfection coefficient at first scattering surface Ao= area of 1st scattering surface alpha z = reflection coeff for 2nd reflection from maze surface Az (assume 0.5 MeV as this is max compton scatter energy) Az= area of maze inner entry projected onto maze wakk from irradiated primary barrier Ao dh = distance from target to 1st reflection surface (add 1 m distance for SAD) dr = distance from first reflection, past maze edge, to be on maze midline (b) dz= centerline distance along maze from b to maze door

Answer 22

(1) incident angle, (2) reflection angle, (3) incident photon energy, and (4) wall material. They are tabulated in NCRP151

Answer 23

H = W Ug a(theta) (F/400) alpha1 Ai/(dsa dsec dzz)^2 a(theta) = scatter fraction for patient scattered radiation at angle theta W = primary workload Ug = use factor for wall G (where linac points) F = field area at patient mid depth at 1 m alpha1 = reflection coefficient for wall G for pt scattered radiation (assume E = 0.5 MeV) A1= area of wall G seen from maze door dsca= target to patient distance = 1m dsec= patient to maze centerline distance at wall z dzz= centerline distance along maze

Answer 24

H = L Wl Ug alpha1 A1/ (dsec dzz)^2 Lf= head leakage at 1 m from target (0.1 %) Wl = head leakage workload (may differ from primary W) Ug = use factor for wall G (where linac aims) alpha1= reflection coeff for leakage scatter from wall G A1= area of wll G seen from maze door dsec= target to maze centerline distance at wall G dzz= centerline distance along maze

Answer 25

due to IMRT factor

Answer 26

H = L Wl Ug B/ (dl^2) L= head leakage ration at 1 m from target = 0.1% Wl= leakage workload- may differ from primary W Ug = use factor for wall G where linac aims B= wall transmission through path dl = target to maze door center entrance (through wall)

Answer 27

Hg = fHs + Hps+ Hls + Hlt f= fraction of primary beam tansmitted through patient = 0.25 for 6-10 MV Hs = dose due to primary scattered from room surfaces Hps = dpse duer tp scattered primary from patient Hls= dose due to single scatter head leakage Hlt= dose due to head leakage transmitted through the wall Htot = 2.64 Hg = total photon dose at maze door for 4 cardinal angles (for a "typical" maze as detailed in NCRP) Bdoor = P/Htot P is 0.1 mSv/ wk (controlled area)

Answer 28

6 mm of Pb

Answer 29

stray photons are scattered by sky to ground outside the tx room H = (2.5*10^7)(Bxs)Do gamma^1.3/(di ds)^2 Bxs= roof shielding transmission factor for photons Do= x-ray output dose rate at 1 m from target gamma = max beam solid angle di= target to 2 m above roof distance 2.5*10^7 includes conversion of Gy to nSv as H is given in nSv/h Equation is only (an order of magnitude) estimate and it is just to indicate which room design parameters could have an impact due to skyshine.

Answer 30

laterally scattered x-ray from roof barriers to adjacent buildings Hss= Do F f(theta)/ (Xr^2 * 10^(1+(t-TVL1)/TVLe) Hss = side scattered dose equivalent rate (Sv/h) Do= x-ray output dose rate of iso F= area of square field at 1 m from target f(theta) = angular distribution of roof-scattered photons 9tabulated) Xr= distance from beam center roof-top to point of interest t= roof thickness TVL1 and TVLe= first and equilbirum TVLs of roof shielding material It ignores oblique photonincidence & photoneutrons It dominate overboth leakageradiation & skyshine 10^etc is empirical equation for roof transmission

Answer 31

Compton effect

Answer 32

3O2 interacts with electron beam to produce 2O3 lethal gas e- beams produce O3 more than photon beams O3 concentration should be < 0.1 ppm ventilation: 3 room changes/hr is adequate for health protection 0.1 ppm ozone in air has odour

Answer 33

before linac operational to ensure meeting design goals first linac beam on - preliminary survey after inital linac calib- energy check- complete survey

Answer 34

locate hot spot via head-wrapped film, and quantify dose rate with ion chamber

Answer 35

measure dose equivalent/MU and dose rate at the hottest spot beyond each barrier (30 cm beyond) search for voids; cracks, or other defects in shielding using a sensitive photon rate meter with a fast response time (eg Geiger Mueller or scintillation detector)

Answer 36

no phantom, max dose rate and field size, 4 gantry angles, all MV

Answer 37

scattering phantom at iso (simulates patient)

Answer 38

when door open/closed

Answer 39

surveys outside bunker

Answer 40

should have both rate and integration mode with a sensitivity in the rage 0.01 mR/hr to 5 R/hr

Answer 41

linac >/= 10 MV using n-survey meter, eg rem ball, bubble detectors, BF3 detectors). Geiger-Mueller, Farmer, and conventional Si diodes are not suitable for neutron detection due to very low neutron cross sections. However, 10B (used in BF3 detectors) has a high cross section for thermal neutrons

Answer 42

TVL = 1.5 cm Pb 15 cm concrete either a direct shielded door or a short maze typical walls are 4-5 cm of Pb or 35-61 cm of concrete mazed door is 3.2 mm of Pb

Answer 43

B = Pd^2/WT B= barrier transmission factor P= max permissible dose equivalent (= 0.1 mSv/wk controlled areas = 0.02 mSv/wk uncontrolled areas) d= distance from source to point protected T= occupancy factor W= workload at 1 m from source W = A K Np Tp A = activity K= dose rate constant Np = number of patients per week Tp = time/ patient tx

Answer 44

15 cm concrete and 1.5 cm lead

Answer 45

210 cm concrete, 40 cm lead

Answer 46

65-120 cm nominal is 80 cm Shielding calculation assumes 80 cm nominal and workloads are normalized to 1 m for shielding calculations. The recommended workload per treatment session is 12.5 Gy at the nominal treatment distance of 80 cm from the x-ray target

Answer 47

6 MV beam thus no neutrons IMRT factor is 15 and use factor is 0.05- very small tx beamlets aimed at many directions workload: ex 8 pt/day * 20 Gy/pt, 5 days/week- 800 Gy/week scatter is negligible compared to leakage since leakage is high and U is low, need to calculate both primary and secondary on every wall and add the results to get total dose rate Ceiling is secondary barrier as beam will not point higher than 22° above horizontal. Also, all beams will pass within 12 cm of room isocenter (92.1 cm above the floor).

Answer 48

primary = 150 cm concrette secondary (leakage) = 90 cm concrete

Answer 49

6 MV - no neutron shielding required SAD is 85 cm workload similar to linac Primary barrier: limited to narrow strip of wall typically 1 TVL more than linac due to focused beam beam stoppers in newer units ↓ shielding requirement (ie. no primary) Secondary barrier: many MU → IMRT factor 16 2 TVL more than linac most leakage & scatter along table axis (┴ to doughnut) conclusions likely only apply to serial (not helical) tomo since helical makes better use of the beam and has more internal shielding

Answer 50

no neutrons (1.25 MeV Co-60 source) 6500 Ci from about 200 sources use max activity in calculation because the work load stays constant- as source decays, the tx time lengthens Vendor provides an iso-kerma (Gy) maparound machine to aid in calculations. Workload= (supplied iso-kerma value) x (your weekly patient load) Typical shielding: 20 cm concrete (inherent shielding ↑↑, 40 cm cast iron) -barriers are only for scatter and leakage; primary doesn't exit unit -manufacturer supplies dose rate plots (i.e. matrix of dose rates at 1/2 m intervals at different heights) -dose distribution with gamma-helmet door open and couch fully retracted yield max dose rate and determine shielding -concrete thickness needed is read from curves in NCRP 49

Answer 51

-in Sv dose from radiation type R is weighted by radiation weighting factor Wr and summed over all radiation types Wr is roughly based on RBE Wr= 1 for photons 20 for alpha particles 5-20 for neutrons 2 for protons

Answer 52

equivalent dose to tissue T is weighted by tissue weighting factor Wt and summed over all irradiated tissues

Answer 53

Relative Biologic Effectiveness is defined as the ratio of the reference dose (250kVP x-ray) to test dose where both doses have the same biological effect (ie same cell survival fraction).

Answer 54

Ht = sum of wr Dr (equivalent dose) E = sum of wt Ht (effective dose)

Answer 55

time averaged dose equivalent rate barrier attenuated dose equivalent rate averaged over 1 week Rw = IDR Wpri Upri/Do Rw= TADR over 1 week (Sv/week) IDR = instantaneous dose-equivalent rate measured 30 cm beyond barrier with linac beam on at dose rate averaged over 20-60 s Do = absorbed dose rate at 1 m Wpri= primary-barrier weekly workload Upri= primary barrier use factor

Answer 56

Previous barrier shielding assumes workload is evenly distributed during the year, thus the barrier equation takes 1/50 of annual P. Additional shielding design goal must be implemented with very low workloads but with very high dose rates. e.g. a dedicated SRT machine (with very high dose rate) may be used only a few times per day. With exceedingly low workload, the barrier thickness would be low, therefore NRC requires additional shielding design goal defined as 20-micro-Sv “in-any-one-hour”.

Answer 57

IDR = (dose rate) x (barrier transmission) / (distance)^2 Also IDR = P(dose rate)/(WUT)

Answer 58

dose rate appears in IDR equation and is then divided by Do to get Rw

Answer 59

similar equation to primary barrier

Answer 60

Rh = Nmax Hpt Rh = TADR in any one hour Nmax = max number of patients in any one hour, considering patient set-up time Hpt= average dose equivalent per patient treatment 30 cm beyond barrier Rh = (Nmax/Nh) * Rw/40 Nh= average number of patient tx per hour Rw= average TADR over 1 week Here instead of proportionally scaling Rw to one hour, we multiply it by a number >1 US nuclear regulatory commission (NRC) requires dose equivalent in uncontroled area not exceed 20 μSv in-any-one-hour.(based on calculated Rh)

Answer 61

thermal, E = 0.025 eV at 20C, E < 0.5 eV intermediate, 0.5 eV< E< 10 keV fast: E > 10 keV

Answer 62

elastic collision with nuclei inelastic collision with nuclei neutron capture

Answer 63

total Ek before and after collision for head-on collision, Ef = Ei ((M-m)/(M+m))^2 Ef= kinetic energy of scattered neutron Ei = kinetic energy of incident neutron M= mass of targeted nucleus m= mass of neutron = 1 fraction of neutron Ek transferred to nucleus deltaE = (Ei-Ef)/Ei deltaE= 4M/(1+M)^2

Answer 64

1- for hydrogen (M=1) materials with high hydrogen content have efficient energy transfer (i.e. parraffin, wax, polyethylene) materials with heavier nucleus don;t have efficient energy transfer - poor shielding against neutrons (ex. Pb) therefore fat dose for a neutron beam is 20% higher than muscle dose!

Answer 65

recoil proton

Answer 66

high enetrgy neutron is absorbed by nucleus (usually high Z)- excited nucleus emits neutron and gamma (n, n gama) (n,p) and (n, alpha) also possible

Answer 67

(n, gamma) emitted gamma ray is called neutron capture gamma ray

Answer 68

1/(neutron velocity)^2 n spends more time in the vicinity of nucleus thus thermal n have higher cross section for capture.

Answer 69

boron borated polyethylene is used for n shielding

Answer 70

created for linacs with >10 MV Photon creates a (gamma, n) nuclear reaction other reactions (gamma, 2n), (gamma, pn) with lower yield

Answer 71

direct neutron: Eave a few MeV, forward peaked, yield only 15 % Evaporation neutron: Eave 1-2 MeV, n-spectra independent of Ephoton, isotropic, dominant process

Answer 72

> 2 orders of magnitude less than photons

Answer 73

put higher E treatments at end of day physics QA end of day (allow overnight decay) no repair near head within 40 min of a long test run use low E for IMRT, VMAT, since these use more MU than 3DCRT vendors should use materials with low photon and n activation yield

Answer 74

concrete (TVL for neutrons is smaller than that for photons)- thus if shielded for photons it is also adequate for neutrons heavy concrete (TVL n > TVL photons) for heavy concrete, TVLn > TVLp earth polyethylene BPE (borated polyethylene)

Answer 75

tenth value distance distance required for photon fluence to be reduced by 10X 5.4 m for 18-25 MV, 3.9 m for 15 MV for neutrons, TVD = 2.06* square root of surface area of maze hallway

Answer 76

direct n + scattered n + thermal n depend on Qn, n source strength emitted from head per Gy of x-ray dose at iso (tabulated) depend on total surface area of room (S) depend on distance from iso to pt A (d). A is at centerline of maze entrance where radiation from head would go to depend on beta- transmission factor for n penetrating head shield (1 for Pb, 0.85 for W) direct n fluence = beta * Qn/ (4pi d^2) scattered n fluence = 5.4 beta Qn/(2 pi S) thermal n fluence = 1.3 Qn /(2 pi S) 1/2pi is fraction of n that enter the maze

Answer 77

Wl * hphi Wl= workload for leakage hphi = dose equivalent from n-capture gamma-rays at maze entrance per unit x-ray dose at iso hphi is determined from scatter, thermal, and direct neutrons

Answer 78

dose equivalent from n-capture gamma-rays at maze entrance per unit x-ray dose at iso hphi = K phia 10 ^(-d/TVD) d is dstance from pt A in maze to entrance of maze K is ratio of n-capture ɣ-ray dose equivalent to total n fluence at A =6.9×10^–16Sv m2 per n fluence φA (phia) is total n fluence [m^–2] at A per unit dose (Gy) of x-rays at iso (i.e. scattered, thermal. direct)

Answer 79

neutron dose at maze door Hn=WlHn,D Wl= workload for leakage Hn,D = n dose equivalent at maze door per unit x-ray dose at iso Hn,D is calculated using modified Kersey's method

Answer 80

Hw = Htot + Hcg + Hn Htot= sum of leakage + scatter photons dose equivalent. If maze > 2.5 m, Htot is negligible Hcg= dose equivalent at door due to n-capture-gamma rays from concrete maze Hn= neutron dose equivalent at maze door

Answer 81

3.5 mm of Pb, then 54 mm of BPE, then 3.5 mm of Pb The Pb makes En decrease with inelastic scatter BPE thermalizes and captures the neutrons 2nd Pb layer stops the n-capture gamma rays that arise in BPE

Answer 82

Hcg gamma energy about 3.6 MeV, TVL = 6.1 cm of Pb Hn neutron energy about 100 keV, TVL = 4.5 cm of BPE

Answer 83

establish ALARA by reducing both Hcg and Hn to P/2 Hcg and Hn cannot be added to one equation so P is halved for both

Answer 84

7.6cm Pb + 28cm BPE + 7.6cm Pb). In this case door is also a secondary barrier. first PB layer slows down neutrons with inelastic collisions (remember elastic collisions in Pb with neutrons are essentially 0) and also attenuates photons BPE layer thermalizes and captures neutrons. It also produces gammas. These gammas are attenuated in 2nd Pb layer.

Answer 85

>/= 10 MV when space is needed

Answer 86

metal layer can become photoneutron source -for primary barriers only; for secondary barriers, scatter photons have lower energy and leakage and photoneutrons are negligible -equation is given for Hn - weekly dose equivalent produced in metal

Answer 87

-transmitted x-ray dose equivalent (Htr) is multiplied by 2.7 Htot = Hn + Hph = Hn + 2.7 Htr because Hph consisted of Htr and H-gamma capture

Answer 88

this produces max neutrons...

Answer 89

Steel is better choice because photo-neutron cross section in steel is 10X less than lead

Answer 90

ensure regulations are followed establish radiation protection program ensure complicance of: -NEW designation -initial and refresher training -occupational doses -active licenses and amendments -authorized users and physicists -radioactive source storage and inventory -patient release surveys -radioactiuve material waste disposal records -investigate rad safety problems, medical events, and emergencies

Answer 91

obtain a license to -operate -service -commission -decommission

Answer 92

provide orthogonal views of patient

Answer 93

must be submitted to CNSC at end of every calendar year

Answer 94

30% QA + research 70% clinical 1000 Gy/week

Answer 95

DON'T just assume thicker barrier apply 2 source rule twice- for leakage vs scatter and then for one energy vs the other

Answer 96

US likely not accepted by CNSC

Answer 97

BPE captures thermal neutrons captured gamma from BPE is lower energy than those from the other materials

Answer 98

Not for >/= 10 MV because of neutron production For < 10 MV, yes it can

Answer 99

at larger angles, scatter fraction decreases with energy at smaller angles, scatter fraction increases with energy for all the energies, scatter fraction decreases with increasing angle

Answer 100

dose equivalent, H (Sv) includes qlity factor for radiation type

Answer 101

35% with 95% CI for dose ratres < 0.02 mSv/h

Answer 102

admittance in the area is under supervision of someone in charge of radiation protection

Answer 103

numerical value of the quality factor is determined by the values of the stopping powers for the spectrum of the charged particles at the point in tissue where the energy is absorbed

Answer 104

the type and energy of the ionizing radiation that is incident on the body.

Answer 105

It is not practical to base shielding design directly on E. Deter- mination of E is complex, and depends on the attenuation of pho- tons and neutrons in the body in penetrating to the radiosensitive organs and hence on the energy spectra of the photons and neu- trons, and also on the posture of the recipient with respect to the source. Rotational exposure is most likely, since it is probable that an individual is moving about and would not be exposed from one direction only. shielding design goals will ensure that E in NCRP 147 is met

Answer 106

allows preganant women to still work there limits monthly equivalent dose to 0.5 mSv for fetus

Answer 107

-patient attnuates the beam- this is not considered -calculations often assume perpendicular incidence of radiation -leakage radiation is assumed to be max value allowed -recommended occuancy factors are high -min distance from barrier to occupied area is 0.3 m - usually people are more than 0.3 m from door, for example -2 source rule is used: . This has been shown to be a conserva- tively safe assumption since the tenth-value layer (TVL) and half-value layer (HVL) of the more penetrating radiation is always used. The two-source rule is even more conservatively safe when applied to dual-energy machines, even though the individual energies cannot be used simultaneously.

Answer 108

at 1 m from source

Answer 109

U will be higher in direction of TBI treatments

Answer 110

thickness and density of concrete; • thickness of metal shielding and polyethylene used for neu- tron shielding; • thickness of metal behind recesses in the concrete (e.g., laser boxes); • HVAC shielding baffle (Section 4.4) if used; • location and size of conduit or pipe used for electrical cable of any type; and • verification that the shielding design has been followed.

Answer 111

• shielding design report, including assumptions and specifi- cations; • construction, or as-built, documents showing location and amounts of shielding material installed; • post-construction survey reports; • information regarding remedies, if any were required; and • more recent reevaluations of the room shielding relative to changes (e.g., in utilization) which have been made or are still under consideration.

Answer 112

brem occurs in target neutron production occurs in both walls of room and linac head

Answer 113

neutrons pair production

Answer 114

room shielding consisting of high-Z material such as lead and steel only, or laminated barriers with insufficient hydrogenous material.

Answer 115

If the material used in the primary barrier is concrete (whether ordinary or heavy; see Sections 4.3.1 and 4.3.2), then experience has shown that the barrier will adequately absorb all photo- neutrons and neutron capture gamma rays and no additional bar- rier is required. This is due to the relatively high hydrogen content of concrete and its resultantly high neutron absorption cross sec- tion. If, on the other hand, materials other than concrete are used in the primary barrier, then special considerations are required -didn't think this was the case for heavy concrete... maybe depends on design

Answer 116

generally yes, to be conservative -calculated for perpendicular beam and held constant -can be tapered with obliquity of beam if space is a concern

Answer 117

ts= t/cos(theta) -since there is scatter in the barrier, the thickness required could be > t. - usually, the effect is small- slightly increase t However, if the required attenuation is orders of magnitude, and the angle of obliquity is large (>45 degrees), the increase for concrete barriers is ~2 HVL for low-energy photons and ~1 HVL for high-energy photons. the obliquity is usually taken into consideration only for primary radiation beams since the leakage and scattered-radiation sources can be too diffuse to apply a specific angle of incidence.

Answer 118

recess is usually 1 HVL use steel or other metal to add 1 HVL of shielding- also use this to mount the laser

Answer 119

size of diagonal of largest beam + 30 cm on each side , for scattered radiation at 20 degrees or less (since scatter fractions increase rapidly with accelerating voltage and scattered-beam energies approach the primary-beam energy), the 30 cm margin may not be adequate for the higher primary- beam energy if the barrier does not intercept at least the 20 degree scattered radiation.

Answer 120

multiply B factors together However, this does not take into account the attenuation and production of photoneutrons and neutron capture gamma rays that must be con- sidered if the primary-beam accelerating voltage is above 10 MV. In such high-energy cases, if a composite barrier design (e.g., steel or lead plus concrete) is not carried out correctly, the metal layer can become a photoneutron source potentially resulting in an increased exposure problem beyond the shield.

Answer 121

No, because energy of scattered radiation is below 10 MV and the leakage radiation intensity is so low that it doesn't produce many neutrons

Answer 122

U=1 and dL is measured from the isocenter if it can be assumed that the accelerator gantry angles used are, on average, symmetric. If this is not the situation, then the distance to the individual barriers should be taken from the closest approach of the accelerator head to each barrier and specific use factors should be used

Answer 123

monte carlo

Answer 124

usually ignored because it is insignificant compared to the leakage scatter

Answer 125

use E = 0.5 MeV to be conservative

Answer 126

~ 0.25 for 6-10 MV

Answer 127

room design 2 < dzz / square root (maze width x height) < 6 1< maze height/maze width<2 gantry use factors are uniform

Answer 128

No, because shielding for avg E 3.6 MeV photons from neutron capture will be adequate (usually)

Answer 129

closed collimators ie. most photoneutrons originate in the treatment head

Answer 130

100 keV TVL in polyethylene of 4.5 cm- use this conservative estimate to calculate borated polyethylene required thickness BPE is a little less effective in fast neutron shielding but much more effective in thermal neutron shielding compared to polyethylene

Answer 131

typical maze door may be very heavy, expensive, motorized other options: 1. reduce the opening at the inside maze entrance; 2. add a light-weight door containing a thermal neutron absorber (boron 9 % by weight) at the inside maze entrance; and 3. place a BPE (5 % boron) door at the inside maze entrance.

Answer 132

sometimes the maze isn't used and instead a heavy direct shielded door is used -mnust have same shielding as secondary barrier

Answer 133

incomplete shielding at door overlap -have to increase overlap (make door wider) or make a shielded door stop -Pb and BPE may need to be added on the surface of the concrete wall -easier to shield on jamb side than operator side- design so leakage radiation goes to jamb side

Answer 134

- face the gantry away from operator console and include a shielding wall behind it. Reduces door thickness by 50%

Answer 135

calculate leakage shielding required and add 1 HVL -conservative approach -do this because difficult to calculate neutron gamma rays in treatment room accurately anywways

Answer 136

1000 Gy/wk for E < 10 MV 500 Gy/wk for E > 10 MV

Answer 137

disregard unless the machine is electrons only

Answer 138

W is measured at 1 m and since TBI deliver dose at extended distance, the dose at 1 m will be high also have to consider additional scatter at the TBI wall usually scatter is determined separately for non-TBI and for TBI

Answer 139

because dose absorbed by the patient is similar to that with conventional

Answer 140

90 degree angle and U= 1 conservatively safe result, since the increased intensity of small angle scatter relative to 90 degree scatter is generally offset by the much smaller use factor for the gantry angles producing the small angle scatter

Answer 141

the use of a measured instantaneous dose-equivalent rate (IDR), with the accelerator operating at maximum output, does not properly repre- sent the true operating conditions and radiation environment of the facility. It is more useful if the workload and use factor are considered together with the IDR when evaluating the adequacy of a barrier. For this purpose, the concept of time averaged dose- equivalent rate (TADR) is used (american thing)

Answer 142

barrier attenuated dose-equivalent rate aver- aged over a specified time or period of operation. TADR is propor- tional to IDR, and depends on values of W and U. There are two periods of operation of particular interest to radiation protection, the week and the hour.

Answer 143

IDRscatter = IDRtotal - IDR leakage measure IDR leakage as dose 30 cm beyond a barrier with no phantom at iso and IDR total with a phantom at iso

Answer 144

IDR Wpri Upri/Do

Answer 145

Rh= Nmax * Hpt Nmax = maximum number of patient treatments in-any- one-hour with due consideration to procedure set-up time Hpt= average dose equivalent per patient treatment at 30 cm beyond the penetrated barrier Hpt is also equal to the time averaged dose equivalent per week (Rw) divided by the average number of patient treatments per week

Answer 146

more high Z material added- attenuates more photons however doesn't attenuate more neutrons heavier and more expensive but can keep barrier thinner

Answer 147

toxic, needs to be covered good against photons transparent to neutrons but slows down neutrons through inelastic scattering steel is more expensive but not toxic it is intermediate photon attenuator between concrete and lead

Answer 148

consider like concrete, but with density of 1.5 g/cm3 hard to define!

Answer 149

improves photon shielding and neutron shielding steel form ties also not concerning

Answer 150

lead amount of radiation scattered from lead is less than that from lighter materials and the scattered radiation is more readily attenuated in lead thickness of lead where required should be at least equal to equivalent thickness of displaced concrete (ratio of TVLs)

Answer 151

through shielding above the door, where photon and neutron fluence is lowest ducts for low E linacs typically won't require shielding but those for high E linacs may depending on the length of the maze (< 2.2 m)

Answer 152

along walls parrallel to gantry rotation

Answer 153

usually none

Answer 154

toxic needs to be held in place by steel or concrete produces neutrons and doesn't have hgh cross section for neutron absorption

Answer 155

more significant than through concrete because the % thickness of the Pb that the duct takes up is larger... dimension of the opening relative to the width of the barrier determines the absorp-tion of x rays that are diagonally incident on the barrier

Answer 156

-groundshine little shield- ing is provided by the concrete floor slab when the beam is aimed at the junction between the wall and the floor. To rectify this prob- lem it will be necessary to add steel or lead to the floor in order to reduce the scattering path length under the wall. Alternatively, the lead and polyethylene wall can be extended into the floor. -neutrons not an issue because they are attenuated by concrete in floor

Answer 157

can now make primary barriers into secondary barriers -usually attenuates scatter with angle up to 30 degrees

Answer 158

side-scattered photon radiation from ceiling barrier -neutrons produced in roof or oblique incidence aren't considered, but there are dominated by the scattered photon radiation

Answer 159

Linear accelerators that produce only electron beams are used within operating suites in which direct access to the tumor can be achieved. Shielding assessment of such a mobile electron accelera- tor was considered by Daves and Mills (2001) and they found that these IORT units could be used in standard operating rooms without added shielding if the machine on-time is restricted to ~30 min week–1. This results from: (1) the very low beam currents used for electrons only, (2) the low leakage radiation because no bending-magnets are employed, (3) the low bremsstrahlung pro- duction from the low-Z materials in the beam path, (4) the use of a compact beamstop beyond the tumor volume, and (5) low energy to eliminate neutron production

Answer 160

have to ensure that source always returns to "safe" position since it is always on

Answer 161

ensure barrier thicknesses are adequate ensure IDR is acceptable search for linac head hot spots with film search for hot spot, cracks, at lasers, ducts etc primary barriers are surveyed without phantom in beam, secondary barriers are surveyed with phantom in beam, max field size check for skyshine, groundshine

Answer 162

Neutron measurements inside the treatment room of a radio- therapy facility are fraught with difficulties because of photon interference from the primary and leakage photon beam and the fact that neutron detection is spread over many decades of energy ranging from thermal energies (0.025 eV) to several million elec- tron volts. No single detector can accurately measure neutron flu- ence or dose equivalent over the entire energy range. inside the room have to use passive detector because photon fluence overwhelms active detector outside the room, can use passive or active detector

Answer 163

Active neutron monitoring usually relies on slowing down fast neutrons or moderating them until they reach thermal energies. A thermal detector is then used to detect the thermal neutrons

Answer 164

Rem-Meters. Active detectors such as neutron rem- meters are useful in radiation fields for which the neutron spec- trum is not well characterized, since their response is designed to be proportional to the dose equivalent and therefore independent of neutron energy. Thus, in principle, no knowledge of the neutron spectrum is required. uses ICRP factors Typically, most rem-meters have a very large over-response in the intermedi- ate energy region, and give an adequate measure of dose equivalent between 100 keV and 6 MeV (Rogers, 1979). Therefore, it is impor- tant to know the spectrum, at least roughly, before any reliance can be placed on the instrument readings. usually consists of moderator (polyethylene) that slows down neutrons and a thermal neutron detector (ex. BF3, 3He) In the BF3 detector, the thermal neutrons are captured in the boron via the 10B(nth,α)7Li reaction. The alpha particle and recoil 7Li nucleus each produces a large pulse in the proportional counter. The large pulses are orders of magnitude higher than the pulses produced by photon interactions, and therefore can be discrimi- nated from the small pulses produced by photons in mixed fields

Answer 165

thermal neutron detectors in spheres of varying size to deterine neutron spectrum or use scintiallation spectrometer with hydrogeneous scintillator- but photons will interfere or time of flight spectrometer A signal is produced at the point of creation of the neutron or when it first enters the detection system, and the time is measured until that neutron gets to a detector some distance away. The energy of the neutron can be determined by knowing the time taken to travel a given distance.

Answer 166

-inside treatment rooms The various types of passive monitors are: moderated thermal-neutron activa- tion detectors, threshold activation detectors, TLDs, solid-state nuclear track detectors, and bubble detectors. Passive ther- mal-neutron monitors such as activation foils and TLDs can also be used inside a series of hydrogenous spheres of varying diameters to determine the neutron spectrum

Answer 167

Dosimeters called passive are dosimeters that do not need an external source of energy to operate. They are integrating dosimeters : they give only an estimate of an overall cumulated dose. They do not measure instantaneous doses, unlike active dosimeters that are able to follow the variations of the exposure.

Answer 168

provides thermal neutron fluence that is proprtional to fast neutron fluence often used with activation foils (gold and indium) -neutron absorption in foil produces a radioactive nucleus

Answer 169

detect neutrons mainly by sub-microscopic damage trails from the recoil nuclei of its constitu- ent atoms, namely hydrogen, carbon and oxygen. The damage trails or tracks can be revealed by a suitable etching proces track density is related to neutron dose

Answer 170

A bubble detector consists of tiny super- heated droplets that are dispersed throughout a firm elastic poly- mer contained in a small sealed tube. The detector is sensitized by unscrewing the cap. Secondary charged particles are produced when the neutrons strike the droplets. The energy deposited by the charged particles causes the droplets to vaporize, producing bub- bles which remain fixed in the polymer

Answer 171

yes very conservative since therapists do clinical and physicists do QA so nobody actually sees 1000 Gy/week

Answer 172

divide CNSC regulations by 1/20

Answer 173

no predominantly forward but not exclusively this is where leakage comes from (through head shielding i.e. primary collimator)

Answer 174

at 1 m from source

Answer 175

acceptance testing from vendor

Answer 176

add scraps of high Z material to concrete

Answer 177

barrier transmission factor

Answer 178

leakage and scatter are always there (isotropic)

Answer 179

leakage, due to higher E

Answer 180

strange constraint remote possibility someone exceeds their occupancy somewhere and exceeds limit?

Answer 181

for 2 MeV beam for neutrons- for 100 keV neutrons and 3.6 MeV photons

Answer 182

0.1-0.2 uSv/h i.e. high E machine returns to background after 48 h from beam-on (0.8 mGy/h 2 min after 30 min beam on)

Answer 183

-ensure it is constructed per design -can also take density audit of material -also have to report shielding results to CNSC

Answer 184

reduce dose at door by 3 X

Answer 185

-lower electron current (X3 order of magnitude) in electron mode vs current mode (since Bremstrahlung inefficient) -electrons have lower cross section for neutron production

Answer 186

-GM counter to look for hot spots -large volume ion chamber (i.e. survey meter) to measure dose rate at points of interest -do both energies -30 cm away from barrier -max FS, 45 degree colli -for primary, beam directed at all and no phantom -for secondary, phantom, and gantry in 4 angles- leakage highest where head is closest to wall

Answer 187

no isocenter no "primary"

Answer 188

high Z high pressure high voltage

Answer 189

<10 uSv/h at 10 cm, once source is shielded

Answer 190

like a ping pong ball hitting bowling balls the neutrons bounce off and don't transfer energy to the material

Answer 191

enriched B10 -cross section is 3840 barns -in comparison B11 is 0.005 barns (i.e. 800,000X less) -concrete is around 5 barn? -Fe about 2 barns

Answer 192

10B(n,alpha)7Li alpha is trapped in door gamma can escape

Answer 193

slower is easier to capture...

Answer 194

concrete = 3.6 MeV boron = 476 keV

Answer 195

nuclear arrangement changes in some way- i.e. neutron, proton, electron released kinetic energy of system not preserved

Answer 196

similar to photons -that is why we can use TVL for neutrons

Answer 197

polyethylene does the moderating Pb is only for photons thus can have Pb only on outside of the door if you want (but has to be on outside to get the ncapture gammas) -however putting some Pb in front does reduce high-energy neutrons in energy through inelastic interactions

Answer 198

-materials made radioactive by neutron activation -half life on order of minutes or hours -some products include 56 Mn (2.5 h), 24 Na (15 h), 28 Al (2 min), 62Cu (10 min), 64Cu (12 h), 187W(24 h), 57Ni (37 h)

Answer 199

aX(gamma,n)a-1X -produce evaporation neutrons and direct neutrons

Answer 200

No, they fall off as 1/d because Pb is a large volume source (not pt source)

Answer 201

neutron dose becomes even worse because Pb creates neutrons but doesn't have high cross section for capturing them -replacing Pb with steel will lower neutron production in metal by 10

Answer 202

-assume there is concrete and Pb laminate -evaluate neutron dose outside shield and photon dose -if neutron/photon dose is higher than protection level, add BPE or concrete above Pb -x-ray ateneuation outside BPE or concrete is evaluated -multiply x-ray level by 2.7 to account for neutron-capture gammas -if photon dose is high, add another Pb layer -measure neutron dose. Add BPE or concrete if needed

Answer 203

2.3 for Pb 1.2 for steel

Answer 204

-thickness and density of concrete, metal, BPE -shielding behind recesses (lasers) -HVAC shielding -location of pipe for physics cables -inspection to determine if required shielding design as followed

Answer 205

-time of first beam- preliminary survey to ensure no health concerns -once linac completely operational- complete shielding survey

Answer 206

-large volume ion chamber - rate and integration mode- 0-5 R/h, ie 0-0.05 Sv/h -geiger-muller (check for hot spots) -neutron survey meter -scattering phantom -film for photon head leakage measurement -moderated foil activation dosimeter and associated foil counting system if neutron head leakage is measured -ion chamber to spot check accelerator output

Answer 207

75% over fast neutron range

Answer 208

yes -does this in patient

Answer 209

fast: 1H(n,n)1H proton recoil (elastic) slow: 1H(n,gamma)2H neutron capture and 14N(n,p)14C nuclear reaction

Answer 210

relates instrument response to neutron dose equivalent -slows down neutrons and slow neutrons are detected -response RATE of detector per unit neutron flux as a function of neutron energy has a shape

Answer 211

photons are too intense- make an intense pulse when accelerator operated won't be able to measure neutrons

Answer 212

activation detectors -don't suffer from interference from pulse photons -gold or indium -induced radioactivity produced by slow neutron capture in foil is measured after irradiation -can also use moderator with activation detector (ex parraffin) -expose room to indium with moderator (get fast neutrons) and without (get thermal neutrons) -must correct for neutrons generated by photon interactions in cadmium shield

Answer 213

-junction of ceiling and floor with primary walls -door and HVAC duct (especially for neutrons)

Answer 214

yes but not in main beam, because photon interactions in dosimeter will generate neutrons that will be detected by the device

Answer 215

-date of survey, person doing it etc -methods- survey techniques used, W, T, U etc -instruments used for survey-serial number and date of calibration -results-table of max dose eqivalent to be expected a various points -conclusions and recommendations -floor plan and correlating survey points

Answer 216

40 cm of cast iron

Answer 217

-4 targets per patient, 1 fraction -120 s to open door to move patient in and out of unit -1320 s to deliver four dose fractions -100x4 + 1320 = 1800 s door open time per patient

Answer 218

avg of 2.77 nSv/s

Answer 219

calculate dose per week at point when door is open (i.e. dose rate X time open) calculate leakage dose as leakage dose X 40 hour work week -add these 2 together

Answer 220

B= Pd^2/(0.8f gamma A) 0.8 is transmission through patient for 137Cs radiation gamma is gamma factor for radium A is mg Ra equivalent in patient f is factor for 137Cs

Answer 221

-no, falls off slower (not pt source, maze not long enough?)

Answer 222

gamma factor * f factor * activity * time

Answer 223

-manufacturer provides isodose distributions (dose per scan or dose per mA min at the pt being evaluated) -manufacturer supplies scatter plots -no primary beam can strike walls, only consider leakage and scatter -use W, kV and mA, T -W is ecxpressed as mA s per week -12000 mA minutes/wk is typical -Dose per week at a point is given by W * dose per scan * T -transmission value TR = P/D and is also Xs/Xo Xs is R per mA min at 1 m for shield thickness s Xo is R per mA min at 1 m for no shielding Xs= Xo *TR -shielding thickness required is read off a curve using Xs

Answer 224

basically treat as linac except W is in mA min -thickness based on transmission is determined from curves -seems like the same kV energy is used for primary, scattered, and leakage radiation (conservative)

Answer 225

0.05 to 0.0015

Answer 226

for short maze with d < 2 m -can use HVAC baffle, duct wrap, concrete baffle

Answer 227

14N(gamma, n)13N, half life = 600s 16O(gamma,n)15O, half life = 122 s each of these is positron emitter - exposes techs to positrons and annihilation photons (0.5 MeV) -presents minimum hazard because max risk is dose to skin from positrons, and this was calculated to be well below 0.5 Sv/yr- limit

Answer 228

-dose equivalent on 3rd floor was higher than on 1st floor

Answer 229

water filled with boric acid much cheaper than typical Pb and BPE door

Answer 230

surround moderator with cadmium because slow neutrons cannot penetrate cadmium

Answer 231

thin= higher detection efficiency for low energy neutrons thick= higher detection efficiency for high energy neutrons

Answer 232

possible to design a moderated detector whose counting efficiency (counts per neutron) varies with neutron energy in the same way that the dose equivalent per unit neutron fluence varies with neutron energy

Answer 233

photons and neutrons only in maze

Answer 234

No, only on primary, not necessary for secondary laminate on inside part of linac room (so neutrons get absorbed in concrete)

Answer 235

6 mSv/year imaging (mostly CT) added 3 mSv/y to background

Answer 236

same for everyone Started Jan 1, 2001 and is 5 year increments from there

Answer 237

-badge a subset of workers -area monitors

Answer 238

Not as of 2021

Answer 239

biweekly dosimetry put on lower energy linac (high energy linac has activation products near linac head, where therapist would work)

Answer 240

-the licensee defines the action levels in the license. If exceeded, have to report to CNSC immediately and submit report in 21 days -At NSHA, action level is if a NEW gets 10% of annual limit

Answer 241

-circuit logic, intercom, CCTV system, beam on/off, enable/disable

Answer 242

6 MV photons have average energy around 2 MeV- less energetic

Answer 243

2.35 g/cm3 regular 5 g/cm3 heavy

Answer 244

radiation emitted in all directions DON'T say isotropic source

Answer 245

yes, both types can

Answer 246

-ask CNSC to increase workload- apply for license amendment -not considered incident since you are informing them in compliance report

Answer 247

bubble dector don't forget to talk about moderator!

Answer 248

license amendment because RSO is named in license

Answer 249

well counter

Answer 250

-internal or external -essentially do CNSC audit but its not from CNSC

Answer 251

code is recommendation regulation is law

Answer 252

20-100 mSv/year in planning

Answer 253

C(surface) =( (cpm(gross)- cpm(background))/ efficiency )* area of wet wipe/detector area cpm is counts/min efficiency 33 % for scintillation counter -wipe is usually 100 cm2 -answer is given as cpm/100 cm2 usually -dpm is disintegration/min

Answer 254

used to estimate hazard from surface contamination by transmission of the contamination from the surface into the body via inhalation or digestion

Answer 255

2 Gy: transient erythema (hours) 6 Gy: erythema (1-2 weeeks) 10 Gy: dry desquamation > 15 Gy: moist desquamation hair loss -deterministic for 3-5Gy -onset after 2-3 weeks -can be permanent for dose > 7 Gy

Answer 256

0.4/Gy at 8-15 weeks -four times lower at 16-25 weeks -treshold of 0.3 Gy

Answer 257

< 16 weeks

Answer 258

-congenital anomalies, neonatal death, temporary growth retardation for 1-6 weeks -permanent growth retardation > 6 weeks

Answer 259

hematopoetic= fever, chills, fatigue GI = nausea, diarrhea, vomiting, anorexia brain = disorientation, loos of coordination, seizure, coma