Radiation Shielding

Agen-689Advances 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 shieldingTime

The short the time spent near the source, the smaller the dose

DistanceThe greater the distance the smaller the dose

ShieldingUse of materials to absorb the radiation dose

Shielding materialAny material provides some shieldingIron, 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−particlescan 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 shieldingTransmission of photons thru matter under conditions of ‘good’ geometrySince γ-rays exhibit a log relation between thickness and intensity, only partial reduction of the radiation can be obtained

Narrow beam

R

d

R>>d

xoeIxI µ−=)(

x

Gamma-ray shielding

Narrow beam

R

d

R>>d

xoeII µ−=

x

The particle flux for this situation is:

The intensity from a point source radiation can be decreased by increasing the distance r from the source or the x of the absorberAn absorber with higher µ can reduce the thickness needed

xernA µ

πφ −= 24

Broad beamThe measured intensity is greater than that of the good geometryScattered photons will also be detectedSo, including a constant B:

B = the building factors (B>1)Tables give values of B for different materials

xdetector

oΨ& Ψ&

hvo Φ=Ψ &&

x

xoeBII µ−=

Relaxation lengthThe thickness of a shield for which the photon intensity in a narrow beam is reduced to 1/e of its original valueOne relaxation length = 1/µ, the mean free pathDependence of B in tables on shield thickness is expressed by variation with number of relaxation lengths, µx

Building factor for concreteCan be obtained from tables as the average of values for Al and Fe:

[ ]FeAlconcrete BBB +=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 2.5 mR/h.

Answer

β− β−K-42

Ca-42

3.52 MeV82%

2.00 MeV18%

γ

1.52 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 µx is:

hRCEX /37.1)52.118.0(105.05.0 =×××==&

31.6

1082.11370

5.2 3

=

×== −−

x

e x

µ

µ

Answer, cont.The energy of the photons emitted by K-42 is 1.52 MeVFrom Table 15.1 (point source), for photons of this energy in lead and the thickness of 6.31 RLs, B = 3To keep the required reduction (1.82x10-3) the same when the buildup factor is used, the number of RLs in the exponential must be increasedThe number y of added RLs that compensate a B=3 is:

10.13ln31

==

=−

y

e y

Answer, cont.Added to the initial value, the estimated shield thickness becomes:

Inspection of Table 15.1 shows that B has increased to 3.5Thus a better guess is y= ln3.5 = 1.25, with an estimated sheild thickness of 6.31 + 1.25 = 7.56 RLsIt remains to verify a final solution numerically by try and error

RLs41.710.131.6 =+

Answer, contFor µx = 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.

4.483.662.0

4.313.533.351.52

3.743.021.0

107.567

356.7 1084.153.3 −−− ×== eBe xµ

Answer, contThe mass attenuation coeff. is 0.051 cm2/g (Fig 8.8)With ρ = 11.4 g/cm3 for lead, µ = 0.581 1/cmThe required thickness of lead shielding is:

A shield of this thickness can be interposed anywhere between the source and the point of exposureUsually, shielding is placed close to a source to realize the greatest solid-angle protection

cmx 13581.06.76.7

===µ

Photon with different energies

Up to now, we have discussed monoenergetic photonsWhen photons of different energies are present, separate calculations at each energy are usually neededSince the attenuation coefficient and buildup factors are different

ExampleA 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

lengths relaxation 62.5550020:mR/h 20 to thisreduce to

50536

7521445050m 6 da of MeV2.75 from rate exposure

MeV)1.37(for 061.0

;2.75MeV)(for 043.0

1

2

12

11

1 =⇒=

===

==

=

−

−

−

xe

R/h..**.dCE.X

cm

cm

x µ

µ

µ

µ

&

Sol, cont.

RLsofnumber larger evena need weso6 Bshows Table 7, for RL

41.779.162.5

79.1)6ln(is buildup

ofamount for this compensate that RLof # The RL)5.62 thickness thisof shieldwater

afor photons MeV2.75for 6 (B 15.1 tableFrom

1

1

>>=+=

==

=

xsoy

µ

Sol, 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 447

gives Table in ioninterpolat theso ,70.7 try slet' so added, be toneeded still

photonsenergy -lower thefrom rate exposure The

2

2

70.72.75

1

1

.Bx

hmRe.B

x

==×=

=×=

==

−

µ

µ

&

Sol, cont.

hmR

hmRXXX

hmReX

hR..X

.

.

/22080)19.8(600/1:is surface at the level exposure the

cm 1807.70/0.043: thenis water of depth needed the

mR/h) (20 figure design the toclose/8.1925.15.18

:is rate exposure totalthe/25.127408.24

:24.8)(B shield with

/74.26

37114450:is shieldingwithout

photons for these m 6at rate exposure The

2

37.175.2

9.10371

2

2371

=

=

=+=+=

=×=

=

=××

=

−

&&&

&

&

Shielding in X-Ray installations

SecondaryProtective barrier Leakage

Radiation

LeakageRadiation

ScatteredRadiation

PrimaryProtective barrier

Usefulbeam

subjectX-raytube

Primary protective barrierLead-lined wallDirection of the beamReduces exposure rate

Other locations exposed to photons

Leakage radiation from X-ray housingScattered photons from exposed objects in primary beamFrom walls, ceilings, etc

Secondary protective barriers needed to reduce exposure rates outside the X-ray area

Shielding in X-Ray installations

SecondaryProtective barrier Leakage

Radiation

LeakageRadiation

ScatteredRadiation

PrimaryProtective barrier

Usefulbeam

subjectX-raytube

Structural shielding designed to limit average dose equivalent to individuals outside and X-ray room

to 1 mSv/wk in controlled areasTo 0.1 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)

Primary Protective BarrierAttenuation of primary X-ray beams thru different thickness of various materials have been measuredThe primary beam intensity transmitted thru a shield depends strongly on the peak operating voltage but very little on the filtration of the beamThe total exposure per mA min is independent of the tube operating current itselfSo X-ray attenuation data for a given shielding material can be presented as a family of curves at different kVp valuesMeasurements are referred to a distance of 1 m from the target of the tube with different thicknesses of shield interposed

Primary Protective Barrier

10-5

10-4

10-3

10-2

10-1

100

101

K(R

/mA

min

) at 1

m

0 1 2 3 4 5 6 7Sheilding Thickness (mm)

#kVp

Attenuation curves for different peak voltages (kVp) are plotted with the ordinate as K (the exposure in R/mA min) and the abscissa gives the shield thicknessSee Fig.15.3 – for 2 mm of lead, the exposure 1m from the target of an X-ray machine operating at 150 kVp is 0.001 R/mA minIf the machine operates with beam current of 200 mA for 90 s, I.e., 200*1.5= 300 mA min, so the exposure will be 300*0.001 = 0.3 R behind the 2 mm lead shield

Primary Protective BarrierThe 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 individualThe 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 BarrierThe value of K can be computed as:

With P [R/wk], d [m], W [ mA min/wk], so K[ R /mA min] at 1 atm

WUTPdK

2

=

Secondary Protective BarrierDesigned to protect areas not in the line of the useful beam from leakage and scattered beamShielding requirements are computed separated for leakage and scattered radiationsThe final barrier thickness is the summ of each oneAssume that leakage and scattered radiations are isotropic (so U = 1)

Leakage radiationThe 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 kV1 R/h for X-ray machines not exceeding 500 kV1 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 radiationWhen 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:

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:

260dYtP =

260dYtTBP =

Leakage radiationSolving for B and let t = W/I where W [ mA min/wk] and I is the beam current [mA]:

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:

YWTIPdB

260=

693.0ln

2lnln BBN −=−=

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 beamIf 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 radiationThe value of K for scattered radiation can be calculated as:

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

fWTPdK

21000=

Protection from Beta emittersPresents two potential for external radiation hazards

Beta rays themselvesBremsstrahlung they produce

Beta particles can be stopped in a shield surrounding the source if it is thicker than their rangeTo 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 emittersThe bremsstrahlung shield thickness can be calculated as:

Calculate the radiation yield, Y, letting T = Tmax be the max beta-particle energyThis 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 thicknessSince 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

Design a suitable container for 3.7x1011

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.

SolutionConsider 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

The range for Tmax = 1.71 MeV is about 0.80 g/cm2

The thickness should be at least 0.8/0.93 = 0.86 cmTo estimate the bremsstrahlung yield we need ZeffMost energy will will be lost in water, so Zeff water = 7.22

SolutionThe fraction of beta-particle energy that is converted into bremsstrahlung is:

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

34

4

104.771.122.71061

71.122.7106 −−

−

×=×××+

×××≅Y

sMeVE /1059.270.0107.3 1111 ×=××=β&

SolutionThe rate of energy emission in the form of photons is:

The exposure rate from the unshielded bremsstrahlung is treated as coming from a point source at a distance of 1.5 mWe use the mass absorption coeff of air for 1.71 MeV photons, that is, 0.026 cm2/gThe intensity at a distance r = 1.5 m = 150 cm is:

sMeVEY /1092.11059.2104.7 9113 ×=×××= −β&

scmMeV

rEY

I 23

9

2 1079.6)150(4

1092.14

×=×

==ππ

β&

SolutionThe dose rate is:

Converting units and remembering that 1 R = 0.0088 Gy in air, we find for the exposure rate:

gsMeVID en 1771079.6026.0 3 =××==

ρµ&

hmR

sRX

GyR

JGykg

MeVkgJg

gsMeVX

6.111022.3

0088.011106.1177

6

13

=×=

×××

×=

−

−

&

&

SolutionLead is a convenient material for the bremsstrahlung shieldIgnoring the buildup and using the linear attenuation coeff for photon of energy Tmax to compute the bremsstrahlung shield thicknessThe attenuation for 1.71MeV photons in lead is 0.048 cm2/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:

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

cmxe x

5.46.111 55.0

== −