0751 - H122 - Basic Health Physics - 32 - Shielding Radiation.
Shielding RadiationAlphas,pBetas, Gammas and Neutrons7/05/20111
ContentsGeneralGeneralShielding Alpha EmittersShielding Alpha EmittersShieldingShildi BBetat PParticlesti lGeneralBeta Particle RangeBremsstrahlungExample CalculationShielding PositronsShielding Positrons2
ContentsShielding Gamma RaysGeneralBasic Equation – First Example CalculationBasic Equation – Second Example CalculationRelaxation LengthCalculating the Required Shielding ThicknessCalculating the Required Shielding Thickness – ExampleShielding Multiple Gamma Ray EnergiesShielding Multiple Gamma Ray Energies – ExampleLayered (Compound) ShieldingLayered (Compound) Shielding – ExampleHalf Value Layers3Half Value layers – Example
ContentsShielding Neutrons continuedMultipurpose Materials for neutron ShieldsPossible Neutron Shield OptionsNeutron Shielding Calculations – Fast NeutronsNeutron Shielding Calculations – Alpha-Beryllium SourcesNeutron Shielding CalculationsGamma and Neutron Shields – GeneralGeneralRadiation energyShielding MaterialShielding Material - Lead4
ContentsShielding Gamma Rays continuedTenth Value layersBuildupBuildup FactorBuildup Factor ExampleDetermining the Required shield Thickness with BuildupShielding NeutronsThe Three Steps1. Slow the Neutrons2. Absorb the Neutrons3. Absorb the Gamma rays5
ContentsGamma and Neutron Shields – General continuedShielding Material – WaterShielding Material – ConcreteStreamingStreaming – Door DesignStreaming – LabyrinthShineAppendicesAppendix A – Densities of Common MaterialsAppendix B – How to do a Linear InterpolationAppendix C – Range and Energy of Beta particlesAppendix D – Transmission CurvesAppendix F – Buildup Factors6
GeneralAugust 25, 20097
General Shielding calculations done “by hand” are often ballparkapproximations The most accurate shielding calculations are performed bycomputers ShiShieldsld are oftenft ““over-designed”d id” so thatth t ththey can hhandledlmore intense sources of radiation that might be used in thefuture The photon shielding calculations in this handout onlyapply to monoenergetic photons (e.g., gamma rays).Shielding x-rays, which have a continuous range ofenergies, is a specialized topic not considered here8
GeneralThe required shield thickness depends on three things:1. Radiation Energy. The greater the energy of theradiation (e.g., beta particles, gamma rays, neutrons) thethicker the shield must be.2. The shield material. For photons (x-rays, gamma rays)the lower the atomic number of the shield, the thicker itmust be. For neutrons, the general rule is that the lowerthe hydrogen content, the thicker the shield must be.3. Radiation intensity. The higher the radiation intensity,and the more it must be reduced, the thicker the shield.9
Shielding Alpha EmittersAugust 25, 200910
Shielding Alpha Emitters For external alpha emitting sources, the alpha particlesthemselves are not a radiation hazard. However, we mustconsider the possibility that other types of radiation mightbe present. This possibility must always be consideredwhen shielding any type of radiation! While we might assume that an alpha emitter doesn’t emitgamma rays, it might actually do so. Pu-239 for exampleemits a large number of gamma rays. Nevertheless, weoften consider Pu-239 a “pure alpha” emitter because thegamma intensity is low .11
Shielding Alpha Emitters Alpha emitters might also emit conversion electrons andlow energy x-rays. Furthermore, alpha sources often contain trace quantitiesof impurities some of which might emit beta particles. These kinds of issues are primarily a concern when largequantities of alpha emitters are handled, e.g., in glovebox facilities where curie and megacurie quantities mightbe encountered.12
Shielding Beta ParticlesAugust 25, 200913
Shielding Beta ParticlesGeneral The dose from an external source of beta particles isprimarily to the skin. The penetrating power of betas is low- approximately a few mm in tissue. Beta particles, like alpha particles, can be completelystopped by a shield. It rarely requires more than 5- 10 mmof material to do so. The material the shield is made of is not very important interms of its ability to stop the beta particles.14
Shielding Beta ParticlesBeta Particle Range Beta ranges are often expressed as a “density thickness”which has units of g/cm2. These ranges can beconverted to centimeters as follows: The easy way to estimate the range (in mg/cm2) of betaparticles is to obtain it from the graph on page 162 of thePTP Radiological Health Handbook. The energy specifiedon the graph is the maximum beta energy.15
Shielding Beta ParticlesBeta Particle RangeSeveral equations approximate the range of betas, i.e., thethickness of the shield required to stop all the betas. The equation below (Feather's Rule) applies to betas witha maximum energy above 0.6 MeV:Range (g/cm2) 0.542 E – 0.133E is the maximum β energy (MeV) The following equation applies to betas with a maximumenergy below 0.8 MeV:Range (g/cm2) 0.407 E1.3816
Shielding Beta ParticlesBremsstrahlung While it is easy to completely stop beta particles, the veryact of doing so produces bremsstrahlung, a morepenetrating type of radiation. Shielding bremsstrahlung canbe more of a problem than shielding gammas. A goodd shieldinghi ldi materialt i l ffor bbetast minimizesi i ibremsstrahlung i.e., low Z material such as plastic. Like beta particles, bremsstrahlung photons have a range ofenergies up to a maximum. The maximum energy of thebremsstrahlung photons is the same as the maximumenergy of the betas. When shielding bremsstrahlung, wecan take the very conservative approach that all the17bremsstrahlung photons have this maximum energy.
Shielding Beta ParticlesExample calculationA 1 Ci P-32 source (50 mls aqueous) is in a polyethylenebottle (0.93 g/cm3). The effective atomic numbers of thesolution and vial wall are assumed to be 7.5.a.How thick should the vial be?b.What is the exposure rate at 1 meter due tobremsstrahlung?18
Shielding Beta ParticlesExample calculationa. How thick should the vial be?The maximum beta energy for P-32 is 1.71 MeV. From thecurve (p 162) in the PTP Radiological Health Handbook, themaximum range of the P-32 betas is 800 mg/cm2 (0.8 g/cm2).Essentially the same result is obtained using the equation:Range (g/cm2) 0.542 E – 0.133 0.542 x 1.71 – 0.133 0.79 g/cm2As such, the vial should be at least 0.86 cm thick ( 0.8 g/cm2divided by 0.93 g/cm3).19
Shielding Beta ParticlesExample calculationb. What is the exposure rate at 1 meter due tobremsstrahlung?The fraction of the beta energy released in the form ofbremsstrahlungg from interactions in the solution ((and to alesser extent in the vial wall) is calculated as follows:20
Shielding Beta ParticlesExample calculationb. What is the exposure rate at 1 meter due tobremsstrahlung? - continuedThe beta energy emission rate from the 1 Ci P-32 source is: 3.737 x 1010 dpsdx 00.77MMeVV per ddecay2.59 x 1010 MeV/sThe bremsstrahlung energy emission rate is therefore: 0.0043 x 2.59 x 1010 MeV/s1.11 x 108 MeV/s21
Shielding Beta ParticlesExample calculationb.What is the exposure rate at 1 meter due tobremsstrahlung? - continuedThe energy fluence rate (R) through a 1 cm2 area at 1 meteris:22
Shielding Beta ParticlesExample calculationb. What is the exposure rate at 1 meter due tobremsstrahlung? - continuedThe dose rate to air is the energy fluence rate (MeV, cm-2, s-1)times the mass energy absorption coefficient for air (cm2/g).Absorbed dose rate R x uen/DMass energy absorption coefficients apply to single photonenergies but bremsstrahlung photons have a range ofenergies. A conservative assumption is that they all have anenergy of 1.71 MeV (the maximum energy for P-32).The mass energy absorption coefficient for 1.71 MeV photons23in air is approximately 0.025 cm2/g.
Shielding Beta ParticlesExample calculationb. What is the exposure rate at 1 meter due tobremsstrahlung? - continuedAbsorbed dose rate R x u/D 883 ((MeV cm-2 s-1) x 0.025 cm2/gg 22.1 MeV g-1 s-1 22.1 x 106 eV g-1 s-1Since one ion pair is produced for every 34 eV absorbed bythe air, the ion pair production rate is: 22.1 x 106 (eV, g-1, s-1) / 34 (eV/ion pair) 0.65 x 10 6 ion pairs per gram per second24
Shielding Beta ParticlesExample calculationb.What is the exposure rate at 1 meter due tobremsstrahlung? - continuedSince the charge produced in air for one member of eachion pair is 11.66 x 10-19 C,C the rate of charge production is : 0.65 x 10 6 (ion pairs g-1 s-1) x 1.6 x 10-19 C1.23 x 10-13 C g-1 s-11.23 x 10-10 C kg-1 s-14.44 x 10-7 C kg-1 hr-125
Shielding Beta ParticlesExample calculationb.What is the exposure rate at 1 meter due tobremsstrahlung? - continuedFinally, the answer:Since 2.58 x 10-4 C/kg equals 1 roentgen, the exposure ratein R/hr is 4.44 x 10-7 (C kg-1 hr-1)/2.58 x 10-4 (C/kg) 1.72 x 10-3 R/hr26
Shielding PositronsAugust 25, 200927
Shielding Positrons In general, positron energies (and the bremsstrahlung theyproduce) tend to be higher than beta energies. Even so, thepositrons themselves are easy to stop. It requires at most acm or so of plastic.The most important positron emitter, F-18, emits positronsth t are off ffairlythati l llow energy: maximumi633633.55 kkeV,V average249.8 keV. What we don’t want to forget is that each emitted positronwill produce two 511 keV annihilation photons that must beshielded. These annihilation photons are not alwaysmentioned in tables of decay data.28
Shielding Gamma RaysAugust 25, 200929
Shielding Gamma RaysGeneralX0 is the exposure rate in the absence of shielding.X0Sourced30
Shielding Gamma RaysGeneralWe want an equation that gives the exposure rate (X) whenshielding of a given thickness (x) is present.XSourcexd31
Shielding Gamma RaysBasic EquationThe following basic equation assumes a narrow beam ofradiation penetrating a thin shield (a situation referred to as"good geometry").X is the exposure rate with the shield in place (e.g., R/hr)Xo is the exposure rate without the shield (e.g., R/hr)x is the thickness of the shield (e.g., cm)u is the linear attenuation coefficient (e.g., cm-1).The linear attenuation coefficient (u) is the probability of anytype of interaction (PE, CS, PP) per unit path length.32
Shielding Gamma RaysBasic EquationListings of the linear attenuation coefficient (u) are hard tofind. However, listings of a related quantity, the massattenuation coefficient (u/D), are easy to find.The “classic”Th“ li ” listingli ti off mass attenuationttti coefficientsffi i t iis ththatt offHubbell. Hubbell’s tables can be found in the PTPRadiological Health Handbook beginning on page 90.The mass attenuation coefficient (u/D) is the probability of aninteraction (PE, CS, PP) per unit density thickness. Its unitsare cm2/g. In other words, u/D is the probability ofinteractions per g/cm2 travelled by the photon.33
Shielding Gamma RaysBasic EquationThe value of the mass attenuation coefficient depends on theshielding material and the energy of the photons.Nevertheless, mass attenuation coefficients for differentmaterials are very similar. For example, the following tableli t some mass attenuationliststtti coefficientsffi i t ffor 1 MMeVV photons.h tShield Materialu/D 9The mass attenuation coefficient is the linear attenuationdivided by the density of the material.34
Shielding Gamma RaysBasic EquationWhen using the mass attenuation coefficient instead of thelinear attenuation coefficient, the basic shielding equation is:μ/ρ is the mass attenuation coefficient (cm2/g)ρ is the density of the shielding material (g/cm3)Note, the units in the exponent must cancel out:cm2/g x g/cm3 x cm35
Shielding Gamma RaysBasic Equation – First example calculationThe exposure rate at a particular point is 100 R/hr due to1332 keV gamma rays from Co-60. What would be theresulting exposure rate if a 1 cm lead shield were employedbetween the source and the point?Warning!In this example, we will use the basic equation whichassumes we have “good” geometry, i.e., a narrow beam andthin shield. As we will see later, in the “real world” we rarelyhave good geometry and this equation would result in anunderestimate of the exposure rate.36
Shielding Gamma RaysBasic Equation – First example calculationu/ρ is approximately 0.057 cm2/g (I eyeballed this)ρactual is assumed to be the same as the theoretical density,11.35 g/cm337
Shielding Gamma RaysBasic Equation – Second example calculationThe exposure rate at a particular point is 100 R/hr due to 662keV gamma rays from Cs-137. What would be the resultingexposure rate if a 1 cm lead shield were employed?ρ is approximatelyppy 0.11 cm2/gg ((I eyeballedythis))u/ρ38
Shielding Gamma RaysRelaxation LengthThe relaxation length is the thickness of a shielding materialthat will reduce the intensity of the radiation to 1/e (37%) ofits original intensity.Relaxation length (cm) 1/uu is the linear attenuation coefficient (cm-1)39
Shielding Gamma RaysCalculating the required shielding thicknessThe preceding equation can be rewritten as follows so as tocalculate the shielding thickness (x) necessary to achieve thedesired reduction in the exposure rate:40
Shielding Gamma RaysCalculating the required shielding thickness - ExampleWhat thickness of a lead shield would be required to reducethe exposure rate due to Cs-137 from 100 R/hr to 1 R/hr?41
Shielding Gamma RaysShielding multiple gamma ray energiesSince the mass attenuation coefficient (u/D) depends on theenergy of the photons, the preceding shielding calculationsonly work if the exposure rate is due to a single gamma rayenergy.This works fine for Cs-137 which has a single gamma ray.How do we handle the situation where the exposure rate isdue to more than one gamma ray energy, e.g., the twogamma rays of Co-60 at 1173 keV and 1332 keV?42
Shielding Gamma RaysShielding multiple gamma ray energiesWe could take a conservative approach and do thecalculation only using the highest of the gamma ray energies.This overestimates the exposure rate and the requiredshielding.This approach might be reasonable for something like Co-60where the two gamma rays are of similar energy andcontribute nearly equally to the exposure rate.On the other hand, this approach could be a disaster for thesituation where the highest gamma ray energy contributesonly a little to the exposure rate (e.g., a high activity Am-241source combined with a low activity Cs-137 source.)43
Shielding Gamma RaysShielding multiple gamma ray energiesThe correct approach is to determine the exposure rate foreach gamma ray energy. Then we determine the attenuationof each gamma ray energy as an independent calculation.Here is an example problem of how this can be donedone.The exposure rate at a particular point is 100 R/hr due to the1173 and 1332 keV Co-60 gamma rays. What would be theresulting exposure rate if a 1 cm lead shield were employedbetween the source and the point? Once again, we assume“good” geometry, i.e., a narrow beam and thin shield.44
Shielding Gamma RaysShielding multiple gamma ray energies - ExampleThe 100 R/hr is the sum of the 1173 keV gamma ray (X1173)and 1332 keV gamma ray (X1332 ) exposure rates.X1173 andd X1332 can beb ddeterminedti d ffrom ththeiri specificifi gammaray constants: '1173 and '1332The specific gamma ray constant for Co-60 (1.3 R m2 hr-1 Ci-1 )is the sum of the specific gamma ray constants for the 1173and 1332 keV gamma rays:45
Shielding Gamma RaysShielding multiple gamma ray energies - ExampleThe specific gamma ray constant for a given gamma rayenergy can be calculated with the equation:' is the specific gamma ray constant in R hr-1 m2 Ci-1E is the gamma ray energy in MeV,I is the gamma ray intensity (abundance)uen/p is the mass energy absorption coefficient46
Shielding Gamma RaysShielding multiple gamma ray energies - ExampleThe specific gamma ray constants for the two gammas are:47
Shielding Gamma RaysShielding multiple gamma ray energies - Example48
Shielding Gamma RaysShielding multiple gamma ray energies - ExampleFirst we calculate the effect of 1 cm of lead on the 47 R/hrdue to the 1173 keV photons. The mass attenuationcoefficient is estimated to be 0.0625 cm2/g49
Shielding Gamma RaysShielding multiple gamma ray energies - ExampleSecond we calculate the effect of 1 cm of lead on the 53 R/hrdue to the 1332 keV photons. The mass attenuationcoefficient is estimated to be 0.057 cm2/g50
Shielding Gamma RaysShielding multiple gamma ray energies - ExampleFinally we add the exposure rates for the 1173 and 1332 keVgamma rays:As expected, the exposure rate (51 R/hr) is not muchdifferent from what we calculated when we assumed theexposure rate was only due to the 1332 keV gamma ray (52R/hr).51
Shielding Gamma RaysLayered (compound) ShieldingIt is often the case that the shielding material consists ofseveral layersXSourcex1x252
Shielding Gamma RaysLayered (compound) ShieldingXSourcex1x253
Shielding Gamma RaysLayered (compound) Shielding – ExampleThe exposure rate at a particular point is 100 R/hr due to 662keV gamma rays. Estimate the resulting exposure rate if ashield consisting of 1 cm of lead and 2 cm of iron wereemployed between the source and the point? Assumegood geometry,geometry ii.e.,e a narrow beam and thin shieldshield.“good”u/ρ is approximately 0.11 cm2/g for leadu/ρ is approximately 0.075 cm2/g for ironρ for the lead is assumed to be the same as itstheoretical density of 11.35 g/cm3ρ for the iron is assumed to be the same as itstheoretical density of 7.86 g/cm354
Shielding Gamma RaysLayered (compound) Shielding – Example55
Shielding Gamma RaysHalf Value LayersInstead of using attenuation coefficients to perform shieldingcalculations, we can use half (or tenth) value layers. A halfvalue layer is the thickness of material that reduces theradiation intensity by one-half.The following equation using HVL's looks very similar to theradioactive decay equation:56
Shielding Gamma RaysHalf Value LayersA mathematically equivalent form of this equation is:Half value and tenth value layers for some materials can befound in PTP’s Radiological Health Handbook on page 79.Don’t be surprised if you see slightly different values in theliterature for the half value layers of different nuclides, e.g.,0.6 cm or 0.65 cm for Cs-137 in lead.57
Shielding Gamma RaysHalf Value LayersThe following table gives an idea as to the approximate halfvalue layer thicknesses for several shielding materials.Energy(MeV)Approximate Half Value Layers in cm (TVL in 36.6013.970.481.274.50.65 (2.16)1.6 (5.3)4.8 (15.7)1.2 (4.0)2.1 (6.9)6.2 (20.6)1.66 (5.5)2.2 (7.4)6.9 2260.690.7958
Shielding Gamma RaysHalf Value Layers - ExampleThe exposure rate is 100 R/hr due to a Cs-137 source. Whatwould the exposure rate be if the source were shielded with 2cm of lead?59
Shielding Gamma RaysTenth Value LayersIn a similar manner, a tenth value layer (TVL) is defined asthat thickness which will reduce the radiation intensity to onetenth of its original value. The equations for using TVL's areas follows:60
Shielding Gamma RaysBuildupWhen the exposure rate at a given distance from a pointsource is calculated using the standard equation, we assumescattered radiation doesn’t contribute to the exposure rate.Only those gamma rays originally directed to the point ofinterest contribute.contributeSourceX61
Shielding Gamma RaysBuildupSimilarly, the shielding equations considered so farassume no contribution from radiation scattered by theshield towards the point of interest. This is the so-called“good geometry” only possible with a narrow (collimatedbeam) and a thin shield.CollimatedSourceXShield62
Shielding Gamma RaysBuildupFor a broad beam or thick shield situation like that below,radiation scattered in the shield contributes to the exposurerate. This is the real world (“bad geometry”). Scattering ofradiation towards the point of interest is referred to asbuildup.SourceXShield63
Shielding Gamma RaysBuildupTo account for buildup, we canincorporate a “fudge factor” known asthe buildup factor into our equations.SourceXShield64
Shielding Gamma RaysBuildup FactorThe buildup factor is a unitlessnumber greater than 1.The value we assign to the buildup factor depends on threethings: Shield materialEnergy of the photonsValue of the equation’s exponent (:/D) D x (i.e., :x)Buildup factors can be found on page 88 of PTP’sRadiological Health Handbook.65
Shielding Gamma RaysBuildup FactorThe equation’s exponent, the product ux , is sometimesreferred to as the number of mean free paths.The mean free path (mfp) is the average distance a photontravels before an interaction takes place. It is the reciprocaloff theth linearliattenuationttti coefficientffi i t (u).( )66
Shielding Gamma RaysBuildup Factor - ExampleThe exposure rate at a particular point is 100 R/hr due to1332 keV gamma rays. What would be the resultingexposure rate if a 1 cm lead shield were employed betweenthe source and the point? Assume “poor” geometry, i.e., abroad beam and thick shield.u/ρ is approximately 0.057 cm2/gρactual is assumed to be 11.35 g/cm3B is estimated to be 1.2, based on an energy of 1332, anda value of 0.647 for ux (number of mean free paths)67
Shielding Gamma RaysBuildup Factor - Example68
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupThe following equation gives the shield thickness (x)necessary to reduce the exposure rate from X0 to X:A problem with this equation is that the buildup factor (B)depends on the thickness of the shield (x), which is exactlywhat we are trying to determine.In other words, we can’t determine B before we know x.69
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupFirst approach - iterationWe could estimate B by guessing what x is likely to be.Then we could plug B and x into the following equation, andget to the requiredqexposureprate X.see how close we gIf X is too high, increase the shield thickness and redo thecalculation. Keep doing this until we find the shield thicknessthat achieves the desired exposure rate. This approach isreferred to as iteration. An unsavory sledgehammer70approach to be sure, but it works.
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupSecond approach – energy absorption coefficientWe could eliminate the buildup factor in our equations bysubstituting the mass energy absorption coefficient (:en/D) forthe mass attenuation coefficient (:/D) . Then we use thefollowing equation for x:Although not technically correct, this approach is simple .71
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupThird approach- computersSwallow our pride and use a computer.pcalculations can pproduce veryy accurate results forComputercomplicated shield and source geometries.Computer modeling typically uses the Discrete OrdinatesTechnique (DOT for short) or the Monte Carlo Technique.The DOT is a numerical integration technique used to derivesolutions of the Boltzmann Transport equation in certainsimplified cases.72
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupThird approach - computersThe Monte Carlo method uses extensive data tables and arandom number generator to mathematically "track"thousands of simulated emissions of radiation through ashield.Perhaps the best known shielding code is Microshield 73
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupFourth approach – transmission curvesOne way to determining the required shielding thickness (orreduction in exposure rate due to a given shield thickness) isto use a transmission curve that accounts for buildup.Transmission curves are plots of transmission (X/Xo) versusshield thickness for various radionuclides. Examples curvesare on page 82 of PTP’s Radiological Health Handbook.Alas, transmission graphs are only available for a limitednumber of radionuclides and shield materials and therefore74have somewhat limited applicability.
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupFourth approach – transmission curves exampleWhat thickness of lead shielding is required to reduce theexposure rate due to Co-60 from 100 R/hr to 22 mR/hr?75
Shielding Gamma RaysDetermining the Required Shield Thickness with BuildupFourth approach – transmission curves exampleFrom the curve on page 83 of PTP’s Radiological HealthHandbook, approximately 15 cm of lead correspond to atransmission factor of 2.2 x 10-476
Shielding Neutrons77
Neutron ShieldingThe Three StepsShielding neutrons involves three steps:1. Slow the neutrons2. Absorb the neutrons3. Absorb the gamma rays78
Neutron Shielding1.Slow the neutrons Neutrons are slowed to thermal energies withhydrogenous material: water, paraffin, plastic. Water can evaporate or leak, paraffin is flammable andplasticl ti iis expensive.i To slow down very fast neutrons, iron or lead might beused in front of the hydrogenous material.79
Neutron Shielding2.Absorb the neutrons Hydrogenous materials are also very effective atabsorbing neutrons - the cross section for neutroncapture by H-1 is 0.33 barns. Unfortunately, a difficult to shield 2.2Unfortunately2 2 MeV gamma ray isemitted when H-1 absorbs a neutron. Boron might be incorporated into the shield because ithas a large cross section for neutron absorption and onlyemits a low energy capture gamma ray. To slow down very fast neutrons, iron or lead might beused in front of the hydrogenous material.80
Neutron Shielding3.Absorb the gamma rays Gamma rays are produced in the neutron shield byneutron (radiative) capture, inelastic scattering, and thedecay of activation products.81
Neutron ShieldingMultipurpose Materials for Neutron Shields Concrete, especially with barium mixed in, can slow andabsorb the neutrons, and shield the gamma rays. good multipurposep pshieldinggPlastic with boron is also a gmaterial.82
Neutron ShieldingPossible Neutron Shield OptionsnHydrogenousmaterialnHi Z moderator for fastneutronsGammashieldnCombination material e.g.,borated plastic or concretewith barium83
Neutron ShieldingNeutron Shielding Calculations – fast neutronsNeutron shielding calculations are best done by computers.Nevertheless, in some limited situations, it is possible toemploy a simplistic exponential equation similar to that usedfor monoenergetic photons.The following equation (Schaeffer 1973) describes the effectof a given shielding material (e.g., steel) on fast neutron doserate. It only works if the there is at least 50 cm of water (orequivalent hydrogenous material) behind the shield.84
Neutron ShieldingNeutron Shielding Calculations – fast neutronstH2ODD is the dose rate with shieldD0 is the dose rate without shieldt is the shield thickness (e.g., cm)ER is the neutron removal cross section (e.g., cm-1)85
Neutron ShieldingNeutron Shielding Calculations– alpha-beryllium sourcesThe following equation (modified from one in from NBSHandbook 63) describes the effect of shield thickness onthe neutron dose rate associated with a radioactive neutronsource (e.g., AmBe).B is a buildup factor usually assumed to be 586
Neutron ShieldingNeutron Shielding CalculationsThe following values for removal cross sections are fromNBS Handbook 63.MaterialRemoval Cross SectionER (cm-1)water0 1030.103iron0.1576ordinary concrete0.0942barytes concrete0.0945graphite0.0785For what it is worth, the removal cross section isapproximately 2/3 to 3/4 of the total cross section.87
Neutron ShieldingNeutron Shielding CalculationsIn neutron shielding calculations, we might also use themass attenuation coefficient symbolized ER/DAccording to Schaeffer (1973), the mass attenuationcoefficient (ER/D) for fast neutrons can be approximated withyour choice of one of the following:ER/D 0.19 Z-0.743 cm2/g (Z 8) 0.125 Z-0.565 cm2/g (Z 8)ER/D 0.206 A-1/3 Z-0.294 0.206 (A Z)-1/3ER/D 0.21 A-0.5888
Neutron ShieldingNeutron Shielding CalculationsWhen the neutron mass attenuation coefficient (ER/D) isemployed instead of the neutron removal cross section, theprevious equation becomes:89
Neutron ShieldingNeutron Shielding CalculationsAnother simple approach that might be possible is toemploy the neutron transmission/attenuation curves ofAllen and Futterer which can be found beginning on page12 of PTP’s Radiological Health Handbook.These graphs provide dose reduction factors for neutronsof specified energies as a function of shield thickness. Theshield is indicated to be polyethylene, but the graphsprovide conversion factors for water, concrete and soil(NTS).90
Gamma and Neutron Shields General91
Gamma and Neutron Shields - GeneralGeneralConsider the possible use of temporary shielding (mobile orfixed) rather than, or in addition to, permanent shielding.Consider the consequences of the shield becomingcontaminated oror, if neutrons are presentpresent, activatedactivated.If permanent shielding is used, consider the cost and easeassociated with dismantling it – this might be necessarywhen the facility is decommissioned. Concrete shields aresometimes built with an inner (closest to the source) layerthat is intended to be removed at the end of the facility’s life.92
Gamma and Neutron Shields - GeneralGeneralIf a source emits gamma rays at a variety of energies, thehigh energy gamma rays sometimes determine theshielding requirements. For example, even though the1764 and 2204 keV gamma rays of radium sourcesnormally contribute very little to t
th t f f i l l i 633 5 k V Shielding Positrons 28 that are of fairly low energy: maximum 633.5 keV, average 249.8 keV. What we don’t want to forget is that each emitted positron will produce two 511 keV annihilation photons that must be shielded. These annihilation
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Objectives zTo review the need for statistics in health physics. zTo review measures of central tendency and dispersion. zTo review common statistical models (distributions). 2 To review common statistical models (distributions). zTo calculate the uncertain ty for a count and count rate. zTo review the statistics for a single count (rate) versus those for a series of counts of the same
RaBe (“raybee”) source, a mix of Ra-226 and Be-9 Yield: ca. 15 x 106 neutrons/sec. per Ci ca. 40 x 104 neutrons/sec. per GBq Alpha Neutron Sources 14 Half-life: 1,600 years Average neutron energy: 3.6 MeV (13.2 MeV max) Gamma exposure rates of these source
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