Radiation Shielding

1 Radiation ShieldingAgen-689 Advances 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 particles can be stopped, or shielded, by a sheet of paper or the outer layer of skin.

2 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. rays can pass through the human body like x -rays. dense materials such as concrete and Pbcan provide shieldingGamma-ray Shielding Transmission of photons thru matter under conditions of good geometry Since -rays exhibit a log relation between thickness and intensity, only partial reduction of the Radiation can be obtainedNarrow beamRdR>>dxoeIxI =)(xGamma-ray shieldingNarrow beamRdR>>dxoeII =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 can reduce the thickness neededxernA =24 Broad beam The measured intensity is greater than that of the good geometry Scattered photons will also be detected So, including a constant B.

3 B= the building factors (B>1) Tables give values of Bfor different materialsxdetectoro & &hvo = &&xxoeBII =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/ , the mean free path Dependence of Bin tables on shield thickness is expressed by variation with number of relaxation lengths, xBuilding factor for concrete Can be obtained from tables as the average of values for Al and Fe:[]FeAlconcreteBBB+=21 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 mR/h.

4 Answer MeV82% MeV18% 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 xis: ) ( ==& == xex Answer, cont. The energy of the photons emitted by K-42 is MeV From Table (point source), for photons of this energy in lead and the thickness of RLs, B= 3 To keep the required reduction ( ) 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 yeyAnswer, cont. Added to the initial value, the estimated shield thickness becomes: Inspection of Table shows that B has increased to Thus a better guess is y= = , with an estimated sheild thickness of + = RLs It remains to verify a final solution numerically by try and +Answer, cont For x= , a 2-D linear interpolation in Table : The reduction factor with buildup included is: Which is the same value given ==eBex Answer, cont The mass attenuation coeff.

5 Is cm2/g (Fig ) With = g/cm3for lead, = 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 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 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 MeV and MeV in decaying by -emission to stable Mg-24.

6 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 :mR/h 20 to thisreduce to505367521445050m 6 da of from rate exposure MeV) (for ; )(for ======= **. &Sol, cont. RLsofnumber larger evena need weso6 Bshows Table 7, for )6ln(is buildupofamount for this compensate that RLof # The RL) thickness thisof shieldwater afor photons 6 (B tableFrom11>>=+====xsoy Sol, Table, ) ( :coef. att. of ratio by thelarger is RLsin thickness thephotons, 447gives Table in ioninterpolat theso , try slet' so added, be toneeded stillphotonsenergy -lower thefrom rate exposure == === &Sol, ) (600/1:is surface at the level exposure thecm : thenis water of depth needed themR/h) (20 figure design the :is rate exposure )(B shield.

7 Is shieldingwithout photons for these m 6at rate exposure +=+== === = &&&&& Shielding in X-Ray installationsSecondaryProtective barrierLeakageRadiationLeakageRadiationS catteredRadiationPrimaryProtective barrierUsefulbeamsubjectX-raytube Primary protective barrier Lead-lined wall Direction of the beam Reduces exposure rate Other locations exposed to photons Leakage Radiation from X-ray housing Scattered photons from exposed objects in primary beam From walls, ceilings, etc Secondary protective barriers needed to reduce exposure rates outside the X-ray areaShielding in X-Ray installationsSecondaryProtective barrierLeakageRadiationLeakageRadiationS catteredRadiationPrimaryProtective barrierUsefulbeamsubjectX-raytube Structural Shielding designed to limit average dose equivalent to individuals outside and X-ray room to 1 mSv/wk in controlled areas To mSv/wk in uncontrolled areas Dose equivalent the product of absorbed dose Dand a dimensionless quality factor Q(fnc of LET) the unit is the siervet (Sv)

8 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 thicknessesof shield interposedPrimary Protective Barrier10-510-410-310-210-1100101K(R/mA min) at 1 m01234567 Sheilding Thickness (mm)#kVp Attenuation curves for different peak voltages (kVp) are plotted with the ordinate as K (the exposure in R/mAmin)

9 And the abscissa gives the shield thickness See for 2 mm of lead, the exposure 1m from the target of an X-ray machine operating at 150 kVp is R/mA min If the machine operates with beam current of 200 mA for 90 s, , 200* 300 mA min, so the exposure will be 300* = R behind the 2 mm lead shieldPrimary Protective Barrier The value of K in a specific application will depend on several other circumstances: The max permissible exposure rate, P(= R/wk or 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 d2is used to evaluate K)Primary Protective Barrier The value of K can be computed as.)

10 With P[R/wk], d[m], W[ mA min/wk], so K[ R /mA min] at 1 atmWUTPdK2=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 summ of each one Assume that leakage and scattered radiations are isotropic (soU= 1)Leakage Radiation The housing is construct so that the leakage exposure rate at a distance of 1 m from the target tube does not exceed R/h for X-ray tubes below 500 kV 1 R/h for X-ray machines not exceeding 500 kV 1 R/h or of the useful beam exposure rate for above 500 kV If Y is any one of these limits, the secondary barrier thickness for the leakage Radiation is computed as the number of half-value layers needed to restrict the exposure of individuals in other areas to allowed levelsLeakage Radiation When the tube operates tmin/wk, the week