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Radiation Shielding Radiation Shielding Agen-689 Advances in Food Engineering Factors 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 dose Shielding 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 energy Types 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 shielding Gamma-ray shielding Transmission of I(x) = I e−µx photons thru matter o x under conditions of ‘good’ geometry Since γ-rays exhibit a Narrow beam log relation between d thickness and intensity, only partial reduction of the radiation can be R obtained R>>d Gamma-ray shielding The particle flux for this I = I e−µx situation is: o nA x φ = e−µx 4πr2 The intensity from a point source radiation can be Narrow beam decreased by increasing the d distance r from the source or the x of the absorber An absorber with higher µ can R reduce the thickness needed R>>d Broad beam The measured intensity is greater than that of the good geometry Scattered photons will also be detected So, including a constant B: Ψ& o x Ψ& detector −µx I = BIoe B = the building factors Ψ& o = Φ& hv (B>1) Tables give values of B for x different materials 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 B in tables on shield thickness is expressed by variation with number of relaxation lengths, µx Building factor for concrete Can be obtained from tables as the average of values for Al and Fe: 1 B = []B + B concrete 2 Al Fe Example#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. Answer With no shielding, the exposure K-42 rate at r=1 m is: − β− X& = 0.5CE = 0.5×10 × (0.18×1.52) = 1.37R / h β An initial estimate of the shielding 2.00 MeV 18% required is based on narrow-beam 3.52 MeV geometry. The number of 82% relaxation lengths µx is: γ −µx 2.5 −3 e = = 1.82 ×10 1.52 MeV 1370 µx = 6.31 Ca-42 Answer, 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 -3 To keep the required reduction (1.82x10 ) the same when the buildup factor is used, the number of RLs in the exponential must be increased The number y of added RLs that compensate a B=3 is: 1 e− y = 3 y = ln 3 = 1.10 Answer, cont. Added to the initial value, the estimated shield thickness becomes: 6.31+1.10 = 7.41RLs 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 error Answer, cont For µx = 7.56, a 2-D linear interpolation in Table 15.1: 7 7.56 10 1.0 3.02 3.74 1.52 3.35 3.53 4.31 2.0 3.66 4.48 The reduction factor with buildup included is: Be−µx = 3.53e−7.56 = 1.84 ×10−3 Which is the same value given before. Answer, cont 2 The mass attenuation coeff. is 0.051 cm /g (Fig 8.8) 3 With ρ = 11.4 g/cm for lead, µ = 0.581 1/cm The required thickness of lead shielding is: 7.6 7.6 x = = = 13cm µ 0.581 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 protection Photon 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 different Example 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? Solution −1 µ1 = 0.043cm (for 2.75MeV); −1 µ2 = 0.061cm (for 1.37 MeV) exposure rate from 2.75 MeV of a d = 6 m 0.5CE 0.5*144*2.75 X& = = = 5.50R/h d 2 36 to reduce this to 20 mR/h : −µ1x 20 = 5500e ⇒ µ1x = 5.62 relaxation lengths Sol, cont. From table 15.1 (B = 6 for 2.75 MeV photons for a water shield of this thickness 5.62 RL) The # of RL that compensate for this amount of buildup is y = ln(6) = 1.79 so µ1x = 5.62 +1.79 = 7.41 for RL > 7, Table shows B1 > 6 so we need a even larger number of RLs Sol, cont. The exposure rate from the lower - energy photons still needed to be added, so let's try µ1x = 7.70, so the interpolation in Table gives B1 = 7.44 for 2.75 photons −7.70 X& 2.75 = 7.44 ×5500e = 18.5mR / h for 1.37 MeV photons, the thickness in RLs is larger by the ratio of att. coef. : µ2 x = 7.70 × (0.061/ 0.0430) = 10.9 From Table, B2 = 24.8 Sol, cont. The exposure rate at 6 m for these photons without shielding is : 0.5×144 ×1.37 X& = = 2.74R / h 1.37 62 with shield (B2 = 24.8) : −10.9 X& 1.37 = 24.8× 2740e = 1.25mR / h the total exposure rate is : X& = X& 2.75 + X& 1.37 = 18.5 +1.25 = 19.8mR / h close to the design figure (20 mR/h) the needed depth of water is then : 7.70/0.043 = 180 cm the exposure level at the surface is : 19.8(600/180)2 = 220mR / h Shielding in X-Ray installations Primary protective barrier Lead-lined wall Secondary Protective Direction of the beam barrier Leakage Radiation Reduces exposure rate Primary Protective Other locations exposed to barrier photons subject X-ray Leakage radiation from X-ray tube housing Useful Scattered photons from Leakage beam exposed objects in primary Radiation beam From walls, ceilings, etc Secondary protective barriers Scattered needed to reduce exposure Radiation rates outside the X-ray area Shielding in X-Ray installations Structural shielding designed to limit average dose Secondary Protective equivalent to individuals barrier Leakage Radiation Primary outside and X-ray room Protective barrier to 1 mSv/wk in controlled subject areas X-ray tube To 0.1 mSv/wk in Useful Leakage uncontrolled areas beam Radiation Dose equivalent – the product of absorbed dose D and a dimensionless quality Scattered factor Q (fnc of LET) – the Radiation unit is the siervet (Sv) Primary Protective Barrier Attenuation of primary X-ray beams thru different thickness of various materials have been measured The primary beam intensity transmitted thru a shield depends strongly on the peak operating voltage but very little on the filtration of the beam The total exposure per mA min is independent of the tube operating current itself So X-ray attenuation data for a given shielding material can be presented as a family of curves at different kVp values Measurements are referred to a distance of 1 m from the target of the tube with different thicknesses of shield interposed Primary Protective Barrier Attenuation curves for different peak voltages (kVp) are plotted with 1 the ordinate as K (the exposure in 10 R/mA min) and the abscissa gives 0 the shield thickness 10 See Fig.15.3 – for 2 mm of lead, the m -1 10 exposure 1m from the target of an 1 t a X-ray machine operating at 150 kVp ) n i -2 is 0.001 R/mA min 10 m If the machine operates with beam A m / -3 current of 200 mA for 90 s, I.e., R 10 ( 200*1.5= 300 mA min, so the K #kVp exposure will be 300*0.001 = 0.3 R -4 10 behind the 2 mm lead shield -5 10 0 1 2 3 4 5 6 7 Sheilding Thickness (mm) Primary Protective Barrier The value of K in a specific application will depend on several other circumstances: The max permissible exposure rate, P (=0.1 R/wk or 0.01 R/wk) The workload, W (weekly amount of use of machine in mA min/wk) The use factor U (fraction of workload during which the useful beam is pointed in a direction considered) The occupancy factor T ( that takes into account the fraction of the time that an area outside the barrier is likely to be occupied by a given individual The distance d, in meters from the target of the tube to the location under consideration (for other distances than 1 m, a factor of d2 is used to evaluate K) Primary Protective Barrier The value of K can be computed as: Pd 2 K = WUT With P [R/wk], d [m], W [ mA min/wk], so K [ R /mA min] at 1 atm Secondary Protective Barrier Designed to protect areas not in the line of the useful beam from leakage and scattered beam Shielding requirements are computed separated for leakage and scattered radiations The final barrier thickness is the
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