Radiation ShieldingFactors that affect radiation doseShielding materialTypes of radiation and shieldingGamma-ray shieldingGamma-ray shieldingBroad beamRelaxation lengthBuilding factor for concreteExample#1AnswerAnswer, cont.Answer, cont.Answer, contAnswer, contPhoton with different energiesExampleSolutionSol, cont.Sol, cont.Sol, cont.Shielding in X-Ray installationsShielding in X-Ray installationsPrimary Protective BarrierPrimary Protective BarrierPrimary Protective BarrierPrimary Protective BarrierSecondary Protective BarrierLeakage radiationLeakage radiationLeakage radiationScattered radiationScattered radiationProtection from Beta emittersProtection from Beta emittersExampleSolutionSolutionSolutionSolutionSolutionRadiation ShieldingAgen-689Advances in Food EngineeringFactors that affect radiation dose Regulations and procedures have been developed and implemented to limit radiation dose by regulating the use, storage, transport, and disposal of radioactive material by controlling time,distance and shielding Time The short the time spent near the source, the smaller the dose Distance The greater the distance the smaller the dose Shielding Use of materials to absorb the radiation doseShielding material Any material provides some shielding Iron, concrete, lead, and soil.  Shielding ability of a material is determined by the thickness of the material required to absorb half of the radiation  This thickness of the material is called the half-thickness  Radiation that has passed through one half-thickness will be reduced by half again if it passes through another half-thickness (HT) The HT depends on the characteristics of the material and type and radiation energyTypes of radiation and shielding α−particles can be stopped, or shielded, by a sheet of paper or the outer layer of skin. β−particles  can pass through an inch of water or human flesh.  can be effectively shielded with a sheet of Al 1/25 of an inch thick. γ−rays  can pass through the human body like x - rays. dense materials such as concrete and Pb can provide shieldingGamma-ray shielding Transmission of photons thru matter under conditions of ‘good’ geometry Since γ-rays exhibit a log relation between thickness and intensity, only partial reduction of the radiation can be obtainedNarrow beamRdR>>dxoeIxIµ−=)(xGamma-ray shieldingNarrow beamRdR>>dxoeIIµ−=x The particle flux for this situation is: The intensity from a point source radiation can be decreased by increasing the distance rfrom the source or the xof the absorber An absorber with higher µ can reduce the thickness neededxernAµπφ−=24Broad beam The measured intensity is greater than that of the good geometry Scattered photons will also be detected So, including a constant B:B= the building factors (B>1) Tables give values of Bfor different materialsxdetectoroΨ&Ψ&hvoΦ=Ψ&&xxoeBIIµ−=Relaxation length The thickness of a shield for which the photon intensity in a narrow beam is reduced to 1/e of its original value One relaxation length = 1/µ, the mean free path Dependence of Bin tables on shield thickness is expressed by variation with number of relaxation lengths, µxBuilding factor for concrete Can be obtained from tables as the average of values for Al and Fe:[]FeAlconcreteBBB +=21Example#1 Calculate the thickness of a lead shield needed to reduce the exposure rate 1 m from a 10-Ci point source of K-42 to 2.5 mR/h.Answerβ−β−K-42Ca-423.52 MeV82%2.00 MeV18%γ1.52 MeV With no shielding, the exposure rate at r=1 m is: An initial estimate of the shielding required is based on narrow-beam geometry. The number of relaxation lengths µxis:hRCEX /37.1)52.118.0(105.05.0 =×××==&31.61082.113705.23=×==−−xexµµAnswer, cont. The energy of the photons emitted by K-42 is 1.52 MeV From Table 15.1 (point source), for photons of this energy in lead and the thickness of 6.31 RLs, B= 3 To keep the required reduction (1.82x10-3) the same when the buildup factor is used, the number of RLs in the exponential must be increased The number yof added RLs that compensate a B=3 is:10.13ln31===−yeyAnswer, cont. Added to the initial value, the estimated shield thickness becomes: Inspection of Table 15.1 shows that B has increased to 3.5 Thus a better guess is y= ln3.5 = 1.25, with an estimated sheild thickness of 6.31 + 1.25 = 7.56 RLs It remains to verify a final solution numerically by try and errorRLs41.710.131.6=+Answer, cont Forµx= 7.56, a 2-D linear interpolation in Table 15.1: The reduction factor with buildup included is: Which is the same value given before.4.483.662.04.313.533.351.523.743.021.0107.567356.71084.153.3−−−×== eBexµAnswer, cont The mass attenuation coeff. is 0.051 cm2/g (Fig 8.8) With ρ= 11.4 g/cm3for lead, µ = 0.581 1/cm The required thickness of lead shielding is: A shield of this thickness can be interposed anywhere between the source and the point of exposure Usually, shielding is placed close to a source to realize the greatest solid-angle protectioncmx 13581.06.76.7===µPhoton with different energies Up to now, we have discussed monoenergetic photons When photons of different energies are present, separate calculations at each energy are usually needed Since the attenuation coefficient and buildup factors are differentExample A 144-Ci point of Na-24 is to be stored at the bottom of a pool. The radionuclide emits 2 photons per disintegration with energies 2.75 MeV and 1.37 MeV in decaying by β-emission to stable Mg-24. How deep must be the water if the exposure rate at a point 6 m directly above the source is not to exceed 20 mR/h? What is the exposure rate at the surface of water right above the source?Solutionlengths relaxation 62.5550020:mR/h 20 to thisreduce to505367521445050m 6 da of MeV2.75 from rate exposure MeV)1.37(for 061.0;2.75MeV)(for 043.01212111=⇒=======−−−xeR/h..**.dCE.Xcmcmxµµµµ&Sol, cont. RLsofnumber larger evena need weso6 Bshows Table 7, for RL41.779.162.579.1)6ln(is buildupofamount for this compensate that RLof # The RL)5.62 thickness thisof shieldwater afor photons MeV2.75for 6 (B 15.1 tableFrom11>>=+====xsoyµSol, cont.824 Table, From9.10)0430.0/061.0(70.7:coef. att. of ratio by thelarger is RLsin thickness thephotons, MeV1.37for /5.18550044.7Xphotons 2.75for 447gives Table in