1. THE EXTERNAL HAZARD

2. 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

3. 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)

4. 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

5. 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?

6. 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

7. 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

8. 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

9. 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

10. X-ray generators

11. 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

12. 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

13. 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)

14. 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

15. Minimising the External Hazard

16. ALARP PRINCIPAL
USE LEAST ACTIVITY è

17. 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

18. ALARP PRINCIPAL
USE LEAST ACTIVITY è
USE LEAST TIME è

19. 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

20. 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

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

22. 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

23. Demonstration

24. ALARP PRINCIPAL USE LEAST ACTIVITY è USE LEAST TIME è
USE DISTANCE PROTECTION è
USE SHIELDING è

25. 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

26. 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

27. Demonstration

28. Remember the adage:
Time!
Distance!
Shielding!