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THE EXTERNAL HAZARD

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**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

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**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)

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**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

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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?

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**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

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**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

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**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

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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

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X-ray generators

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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

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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

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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)

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**ESTIMATION 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

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**Minimising the External Hazard**

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ALARP PRINCIPAL USE LEAST ACTIVITY è

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**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

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ALARP PRINCIPAL USE LEAST ACTIVITY è USE LEAST TIME è

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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

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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

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**ALARP PRINCIPAL USE LEAST ACTIVITY è USE LEAST TIME è**

USE DISTANCE PROTECTION è

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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

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Demonstration

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**ALARP PRINCIPAL USE LEAST ACTIVITY è USE LEAST TIME è**

USE DISTANCE PROTECTION è USE SHIELDING è

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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

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**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

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Demonstration

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Remember the adage: Time! Distance! Shielding!

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Radiation Safety Rules of Thumb. TR-2 Alpha Particle An alpha energy of at least 7.5 MeV is required to penetrate the protective layer of the skin (0.07mm).

Radiation Safety Rules of Thumb. TR-2 Alpha Particle An alpha energy of at least 7.5 MeV is required to penetrate the protective layer of the skin (0.07mm).

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