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 summ of each one

„ Assume that leakage and scattered radiations are isotropic (so U = 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

„ 0.1 R/h for X-ray tubes below 500 kV

„ 1 R/h for X-ray machines not exceeding 500 kV

„ 1 R/h or 0.1% 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 levels Leakage radiation

„ When the tube operates t min/wk, the weekly exposure in R in a area at a distance of d meters from the tube target is, with no structural shielding: Yt P = 60d 2

„ If P is the individual weekly max. allowed exposure in the area, having an occupation factor T, so the required reduction B for the leakage is: YtT P = B 60d 2 Leakage radiation

„ Solving for B and let t = W/I where W [ mA min/wk] and I is the beam current [mA]: 60IPd 2 B = YWT

„ The number N of half-value layers that reduces the radiation to the factor B of its unshielded value is given by B=2-N, or:

ln B ln B N = − = − ln 2 0.693 Scattered radiation

„ If the tube operating potential is less than 500 kVp, then the barrier penetrating capability of the scattered beam is the same as the useful beam

„ If the potential is greater than 500 kVp, then the scattered photons are treated like primary photons in a useful beam of 500 kVp X rays Scattered radiation

„ The value of K for scattered radiation can be calculated as: 1000Pd 2 K = fWT „ f is an empirical factor (to accont for increased output of an X-ray machine), the 1000 factor is used to account for the snaller intensity of scattered radiation compared with the useful beam Protection from Beta emitters

„ Presents two potential for external radiation hazards

„ Beta rays themselves

„ Bremsstrahlung they produce

„ Beta particles can be stopped in a shield surrounding the source if it is thicker than their range

„ To minimize bremsstrahlung production, this shield should have low Z, and it can be enclosed in another material (high Z) that is thick enough to attenuate the bremsstrahlung intensity Protection from Beta emitters

„ The bremsstrahlung shield thickness can be calculated as:

„ Calculate the radiation yield, Y, letting T = Tmax be the max beta-particle energy

„ This assumption overestimate the actual bremsstrahlung intensity (most photons have energy much lower than Tmax) „ So, one ignores buildup in the shielding material and uses the linear attenuation coeff. for photons of energy Tmax to calculate shield thickness

„ Since the bremsstrahlung spectrum is hardned by passing thru the shield, the exposure rate around the source is calculated by using the air absorption coeff for photons of energy Tmax Example

11 „ Design a suitable container for 3.7x10 Bq source of P-32 in a 50mL aqueous solution, such that the exposure rate at a distance of 1.5 m will not exceed 1 mR/h. P-32 decays to ground state of S-32 emission of beta particles with an average energy of 0.70 MeV and a maximum energy of 1.71 MeV. Solution

„ Consider a bottle made of polyethylene (density = 0.93 g/cm3 and low Z)

„ It would be thick enough to stop beta particles of max energy

2 „ The range for Tmax = 1.71 MeV is about 0.80 g/cm „ The thickness should be at least 0.8/0.93 = 0.86 cm

„ To estimate the bremsstrahlung yield we need Zeff

„ Most energy will will be lost in water, so Zeff water = 7.22 Solution

„ The fraction of beta-particle energy that is converted into bremsstrahlung is: 6 ×10−4 × 7.22 ×1.71 Y ≅ = 7.4 ×10−3 1+ 6 ×10−4 × 7.22 ×1.71

„ The rate of energy emission by the source of beta particles with an average energy of 0.70 MeV:

11 11 E& β = 3.7 ×10 × 0.70 = 2.59 ×10 MeV / s Solution

„ The rate of energy emission in the form of photons is:

−3 11 9 YE&β = 7.4 ×10 × 2.59 ×10 = 1.92 ×10 MeV / s

„ The exposure rate from the unshielded bremsstrahlung is treated as coming from a point source at a distance of 1.5 m

„ We use the mass absorption coeff of air for 1.71 MeV photons, that is, 0.026 cm2/g

„ The intensity at a distance r = 1.5 m = 150 cm is:

YE& 1.92 ×109 MeV I = β = = 6.79 ×103 4πr2 4π (150) cm2s Solution

„ The dose rate is: µ MeV D& = I en = 0.026 × 6.79 ×103 = 177 ρ gs „ Converting units and remembering that 1 R = 0.0088 Gy in air, we find for the exposure rate: 177MeV 1.6 ×10−13 Jg 1Gykg 1R X& = × × × gs MeVkg J 0.0088Gy R mR X& = 3.22 ×10−6 = 11.6 s h Solution

„ Lead is a convenient material for the bremsstrahlung shield

„ Ignoring the buildup and using the linear attenuation coeff for

photon of energy Tmax to compute the bremsstrahlung shield thickness 2 „ The attenuation for 1.71MeV photons in lead is 0.048 cm /g, so µ = 0.048*11.4 = 0.55 cm-1

„ The thickness x needed to reduce the exposure rate to 1 mR/h is: 1 = 11.6e−0.55x x = 4.5cm

„ A lead container of this thickness could be used to hold the polyethylene bottle