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# THE EXTERNAL HAZARD.

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THE EXTERNAL HAZARD

External Hazard ‘That (biological) hazard arising from the immersion
of a body in a radiation field’ Source types exhibiting an external hazard Sealed sources Unsealed sources Electrical equipment generating em radiation Natural sources

Current Annual Occupational Limits
Skin (incl hands/feet) – 500 mSv public 50 mSv Eye – 150 mSv public 15 mSv Abdomen of female – 13 mSv in any 3 mnth period These are equivalent dose limits Equivalent dose (HT) = Absorbed dose (DT) X WR Adequate Shielding Level = 7.5 μSvh-1 (unclassified radiation workers)

Estimating the External Hazard - Calculation
Estimate of biological damage – i.e. Absorbed dose Type of isotope - α/β/γ Radiation generators – e.g. X-ray Geometry of source – point sources are isotropic Activity of source Distance from the source Exposure time Include natural radiation? e.g aircraft crew

Alpha emitters Not generally considered to be an external hazard Penetrate less than 4 cm in air Generally considered an Internal Hazard Bremstrahlung a problem?

Beta emitters Dose depends on number of beta particles per unit area
Independent of beta energy The range of 14C (low energy) betas in 1 mm The range of 32P (high energy) betas in 1 cm The dose rate Db in mSv/hr produced by a point source of beta activity M MBq at distance 0.1m is given by Db = 1000 M mSv/hr at 0.1m This translates to 1 beta particle cm-2 s-1 ~ 1 mSv/hr

Dβ = 1000 M μSv/hr at 0.1m 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 mSv/hr at 0.1m DOSE RATE = 5000 mSv/hr at 0.3m (inverse square law) 9 DOSE RECEIVED IN 0.25 HR = 5000 x 0.25 = 139 mSv

Gamma emitters – e.g. Cr51, Co60
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 Eg (Mev) produces a dose rate Dg = MEg mSv/hr at 1.0m (>0.1MeV) 7 To find the dose rate at other distances we apply the inverse square law

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 = = 2.5 MeV Dose rate at 1m = 50 x 2.5 = 17.9 mSv/hr 7 Dose rate at 0.5m = 17.9 x 4 = 71 mSv/hr

X-ray generators

X-ray tube X-rays produced by the bremstrahlung effect Dose rate is dependent on: Target material (atomic number) Applied tube voltage (kV) Tube current (mA) distance from the source (mm) Filtration? Rule of thumb formula: (with 1 mm Be filter) D = 670 ZVI mGy/s d2

Example Calculate dose rate at 50 mm from an X-ray tube Using a copper target and tube voltage 50 kV at 10 mA Z = 29 for copper D = 670 x 29 x 50 x 10 / 2500 D = 4 Gy/s i.e. finger dose limit of 500 mSv reached in 125 ms

In context: Dose 1 m for various (unshielded) sources 37 MBq (1 mCi) C14 – 0 mSv/h 37 MBq P32 – 0.5 mSv/h 400 GBq Cs137 – 34 mSv/h X-ray 75 kV and 10 mA – 3.5 Sv/h (2 mm Al filter)

ESTIMATION OF DOSE RATE BY MONITORING
SIGN ON SIDE OF MONITOR GIVES RESPONSE TO 10Sv hr-1 THUS BY USING MINI MONITOR CPS CAN BE APPROXIMATELY CONVERTED TO DOSE RATE ie, IF 20 CPS = 10Sv hr-1 15 CPS = 7.5 Sv hr-1 THE ADEQUATE SHIELDING LEVEL

Minimising the External Hazard

ALARP PRINCIPAL USE LEAST ACTIVITY è

LEAST 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 è USE LEAST TIME è

LEAST 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 up to 20 mSv per year ~ 400 Sv per week. How many hours of each week can he spend in an area having an average dose rate of 100 µSv/hr ? Dose = Dose Rate x Time  400 µSv = 100 µSv x T  T = 4h

ALARP PRINCIPAL USE LEAST ACTIVITY è USE LEAST TIME è
USE DISTANCE PROTECTION è

DISTANCE 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 è USE LEAST TIME è
USE DISTANCE PROTECTION è USE SHIELDING è

SHIELDING 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 plastic
For beta emitters – plastic / perspex thickness dependent on energy For gamma emitters – lead shielding or leaded glass For high activity sources bremsstrahlung may be an issue

Demonstration

Remember the adage: Time! Distance! Shielding!

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