the external hazard. that (biological) hazard arising from the immersion of a body in a radiation...

Radiation Protection ServiceRadiation Protection ServiceUniversity of GlasgowUniversity of Glasgow

THETHE EXTERNAL EXTERNAL HAZARDHAZARD


‘‘That (biological) hazard arising from the immersionThat (biological) hazard arising from the immersionof a body in a radiation field’of a body in a radiation field’

External HazardExternal Hazard

Source types exhibiting an external hazardSource types exhibiting an external hazard• Sealed sourcesSealed sources• Unsealed sourcesUnsealed sources• Electrical equipment generating em radiationElectrical equipment generating em radiation• Natural sourcesNatural sources


Current Annual Occupational LimitsCurrent Annual Occupational Limits

Skin (incl hands/feet) – 500 mSv Skin (incl hands/feet) – 500 mSv public 50 mSvpublic 50 mSvEye – 150 mSv Eye – 150 mSv public 15 mSvpublic 15 mSvAbdomen of female – 13 mSv in any 3 mnth periodAbdomen of female – 13 mSv in any 3 mnth period

These are These are equivalentequivalent dose limits dose limits

Equivalent dose (HEquivalent dose (HTT) = Absorbed dose (D) = Absorbed dose (DTT) X W) X WRR

Adequate Shielding Level = 7.5 Adequate Shielding Level = 7.5 μμSvhSvh-1-1

(unclassified radiation workers)(unclassified radiation workers)


Estimating the External Hazard - CalculationEstimating the External Hazard - Calculation

• Estimate of biological damage – i.e. Absorbed doseEstimate of biological damage – i.e. Absorbed dose• Type of isotope - Type of isotope - αα//ββ//γγ• Radiation generators – e.g. X-rayRadiation generators – e.g. X-ray• Geometry of source – point sources are isotropicGeometry of source – point sources are isotropic• Activity of sourceActivity of source• Distance from the sourceDistance from the source• Exposure timeExposure time• Include natural radiation? e.g aircraft crewInclude natural radiation? e.g aircraft crew


Alpha emittersAlpha emitters

• Not generally considered to be an external hazardNot generally considered to be an external hazard• Penetrate less than 4 cm in airPenetrate less than 4 cm in air• Generally considered an Internal HazardGenerally considered an Internal Hazard• Bremstrahlung a problem?Bremstrahlung a problem?


• Dose depends on number of beta particles per unit area

• Independent of beta energy

• The range of 14C (low energy) betas in tissue 1 mm

• The range of 32P (high energy) betas in tissue 1 cm

• The dose rate D in Sv/hr produced by a point source of

beta activity M MBq at distance 0.1m is given by

• D = 1000 M Sv/hr at 0.1m

• This translates to 1 beta particle cm-2 s-1 1 Sv/hr

Beta emittersBeta emitters


EXAMPLE:

ESTIMATE THE RADIATION DOSE RECEIVED AT THE EYES OF A WORKER USING A SMALL UNSHIELDED 32P SOURCE OF ACTIVITY 5 MBq FOR A PERIOD OF 15 MINUTES. ASSUME EYE-SOURCE DISTANCE IS 0.3M

Dβ = 1000 M μSv/hr at 0.1m

DOSE RATE = 5000 Sv/hr at 0.1m

DOSE RATE = 5000 Sv/hr at 0.3m (inverse square law) 9

DOSE RECEIVED IN 0.25 HR = 5000 x 0.25 = 139 Sv 9


The radiation dose delivered by a flux of gamma radiation

depends upon the number of photons incident on unit area

and it also depends upon photon energy, ie the dose

increases with photon energy.

A point source of gamma radiation of activity M MBq with a

total gamma photon energy per disintegration E (Mev)

produces a dose rate D = ME Sv/hr at 1.0m (>0.1MeV) 7

To find the dose rate at other distances we apply the inverse

square law

Gamma emitters – e.g. CrGamma emitters – e.g. Cr5151, Co, Co6060


EXAMPLE

Find the gamma dose rate at a distance of 0.5m from a 60Co source of activity 50 MBq. Each disintegration of 60Co results in the emission of two gamma ray photons of energy 1.17 and 1.33 MeV respectively

It follows that E = 1.17 + 1.33 = 2.5 MeV

Dose rate at 1m = 50 x 2.5 = 17.9 Sv/hr 7

Dose rate at 0.5m = 17.9 x 4 = 71 Sv/hr


X-ray generatorsX-ray generators


X-ray tubeX-ray tube

• X-rays produced by the bremstrahlung effectX-rays produced by the bremstrahlung effect• Dose rate is dependent on:Dose rate is dependent on:

• Target material (atomic number)Target material (atomic number)• Applied tube voltage (kV)Applied tube voltage (kV)• Tube current (mA)Tube current (mA)• distance from the source (mm)distance from the source (mm)• Filtration?Filtration?

Rule of thumb formula: (with 1 mm Be filter)Rule of thumb formula: (with 1 mm Be filter)

D = 670 ZVID = 670 ZVI mGy/smGy/sdd22


ExampleExample

Calculate dose rate at 50 mm from an X-ray tubeCalculate dose rate at 50 mm from an X-ray tubeUsing a copper target and tube voltage 50 kV atUsing a copper target and tube voltage 50 kV at10 mA10 mA

Z = 29 for copperZ = 29 for copper

D = 670 x 29 x 50 x 10 / 2500D = 670 x 29 x 50 x 10 / 2500D = 4 Gy/sD = 4 Gy/si.e. finger dose limit of 500 mSv reached in 125 msi.e. finger dose limit of 500 mSv reached in 125 ms


In context:In context:Dose rates @ 1 m for various (unshielded) sourcesDose rates @ 1 m for various (unshielded) sources

37 MBq (1 mCi) C37 MBq (1 mCi) C1414 – 0 mSv/h – 0 mSv/h37 MBq P37 MBq P3232 – 0.5 mSv/h – 0.5 mSv/h400 GBq Cs400 GBq Cs137137 – 34 mSv/h – 34 mSv/h

X-ray source @ 75 kV and 10 mA – 3.5 Sv/hX-ray source @ 75 kV and 10 mA – 3.5 Sv/h(2 mm Al filter)(2 mm Al filter)


ESTIMATION OF DOSE RATE BY MONITORINGESTIMATION OF DOSE RATE BY MONITORING

SIGN ON SIDE OF MONITOR GIVES RESPONSE TO 10Sv hr-1

THUS BY USING MINI MONITOR CPS CAN BE APPROXIMATELY CONVERTED TO DOSE RATE ie,

IF 20 CPS = 10Sv hr-1

15 CPS = 7.5 Sv hr-1

THE ADEQUATE SHIELDING LEVEL


Minimising the External HazardMinimising the External Hazard


ALARP PRINCIPAL

USE LEAST ACTIVITY


LEAST ACTIVITYLEAST ACTIVITY

• Use the least activity required to get good results

• Incorporation

• Counting efficiency / error / noise level etc

• Standard deviation varies as 1/√n

• i.e. doubling activity only improves statistical error by √2

• 95% confidence level (2σ level) ~ 1000 counts above b/g


ALARP PRINCIPAL

USE LEAST ACTIVITY

USE LEAST TIME


LEAST TIMELEAST TIME

• Conduct a dummy experiment to pinpoint deficiencies in the technique before using radioactivity

• All apparatus required should be ready prior to starting experiment

• Remember dose = dose rate x time


Example:A classified radiation worker is permitted to receive upto 20 mSv per year ~ 400 Sv per week. How many hoursof each week can he spend in an area having an averagedose rate of 100 µSv/hr ?

Dose = Dose Rate x Time 400 µSv = 100 µSv x T T = 4h


ALARP PRINCIPAL

USE LEAST ACTIVITY

USE LEAST TIME

USE DISTANCE PROTECTION


DISTANCE PROTECTIONDISTANCE PROTECTION

• If we double the distance from a point source of radiation, the number of particles/unit area in a given time is reduced by a factor of 4. This is an example of the inverse square law ie…

• if the dose rate produced by a small radioactive source is 1µSv/hr at 1 m then the dose rate at hand distance (0.1 m) = 100 µSv/hr

• finger distance (0.01m) = 104 µSv/hr = 10 mSv/hr


Demonstration


ALARP PRINCIPAL

USE LEAST ACTIVITY

USE LEAST TIME

USE DISTANCE PROTECTION

USE SHIELDING


SHIELDINGSHIELDING

• It is always best and most economical to shield the source not the personnel.

• Placing a shield round the immediate vicinity of the source requires a small amount of shielding material and also obviates the possibility of irradiation of hands etc

• Use the correct shielding material for the isotope concerned


• For alpha emitters – thin sheet of paper or plasticFor alpha emitters – thin sheet of paper or plastic

• For beta emitters – plastic / perspex thickness dependent on energyFor beta emitters – plastic / perspex thickness dependent on energy

• For gamma emitters – lead shielding or leaded glassFor gamma emitters – lead shielding or leaded glass

• For high activity sources bremsstrahlung may be an issueFor high activity sources bremsstrahlung may be an issue


Demonstration


Remember the adage:Remember the adage:

Time!Time!

Distance!Distance!

Shielding!Shielding!

the external hazard. that (biological) hazard arising from the immersion of a body in a radiation...

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