Topic 18

1. Topic 18 Reducing Occupational Exposure

2. Written and produced by Michael A. Thompson, Professor Nuclear Medicine Technology Program University of Alabama at Birmingham  Board of Trustees, University of Alabama, 1999

3. Occupational radiation exposure refers to that exposure received from radiation sources while inthe workplace. It does not include that exposure received from personal medical or dental procedures prescribed by your physician or dentist.

4. Personnel dosimetry devices (i.e., film badges and TLDs) are used to measure occupational exposure and should not be stored close to radiation sources or worn when having personally prescribed radiation procedures performed.

5. The basic philosophy of radiation protection proposes that • all procedures involving radiation exposure to an individual shouldalways produce a net benefit, • all occupational exposures to workers should not exceed the allowed established limits, and

6. Basic Radiation Protection Philosophy (con’t) • all radiation exposures should be kept as low as reasonablyachievable, social and economic conditions be taken into account. This last concept is referred to as ALARA.

7. As Low As Reasonably Achievable

8. Just how does one reduce their occupational radiation dose? Radiation exposure from any radiation source can be reduced by effective use of: • time • distance • appropriate shielding

9. Time Reduce the time spent near sources of radiation

10. Distance Increase the distance between you and the source of radiation. Recall theinverse square law: I1(d1)2 = I2(d2)2

11. Sample Calculation: A radiation source produces an exposure rate of 100 mR/hr at a distance of 1m. What is the exposure rate at a distance of 2m? I1(d1)2 = I2(d2)2 (100 mR/hr)(1m)2 = I2(2m)2 I2 = (100 mR/hr)(1/4) I2 = 25 mR/hr

12. Note that in this simple example radiation exposure rate was reduced to 25%of its originalvalue by simply doubling the distance from the source.

13. Appropriate Shielding Shielding can be used to effectively reduce one’s occupational radiation exposure but only when available and appropriate.

14. Recall from previous discussions that some materials are not always appropriate to use as shielding. Example:Lead (Pb) is not appropriate to use with high energy beta emitters since it will produce bremsstrahlung (x-rays). Example:Thin Pb shielding should not be used against high energy gamma emitters due to the production of scatter radiation.

15. Good shielding for high energycharged particles consists of lowZ-number materials such as: • Plexiglas • wood • cardboard or particleboard

16. Good shielding materials for photon radiation( i.e., x-rays and gamma rays) include sufficiently thickhigh Z-numbermaterials such as: • lead (Pb) • tungsten (W) • depleted uranium (U)

17. How does one predict the amount of photon radiation transmitted through a specific thickness of an absorber? To answer such a question, it is useful to use the concept of the Half-Value Layer (HVL)

18. Half-Value Layer (HVL) That thickness of an absorber required to attenuate or remove 50% of the incident radiation beam intensity. Note: This thickness will vary with photonenergy and the type of absorber.

19. The HVL layer thickness concept can be easily used to determine the reduction in intensity or exposure rate caused by an absorber of thickness “x” as illustrated: Absorber Detector Incident Beam Intensity ( I0 ) Transmitted Intensity (I) x Knowing I0, x, and the HVL, how does one find I?

20. To answer this question, we use an attenuation equation given by: I = I0 (0.5)N Where I = transmitted intensity I0 = initial intensity N =absorber thickness in HVLs = actual thickness/HVL

21. It should be noted that (0.5)Nin the attenuationequation is known as the transmission factorsince it represents the fraction or percent of the incident radiation that is transmitted through the absorber.

22. Transmission Factor (0.5)NEvaluated for Various Values of N: N ValuePercent Transmission 1 50% 2 25% 3 12.5% 3.3 10% 6.6 1% 10 0.1%

23. HVLs in Lead (Pb) for Commonly Used Medical Radionuclides 125I 0.02 mm 201Tl 0.051 mm 99mTc 0.25 mm 111In 0.72 mm 18F 4.0 mm 131I 2.4 mm

24. HVLs in Lead (Pb) for Commonly Used Medical Radionuclides 60Co 11.0 mm 99Mo 6.8 mm 137Cs 5.8 mm 133Xe 0.08 mm 57Co 0.19 mm

25. Sample Problem: A 137Cs source (HVL in Pb = 5.8 mm) produces an exposure rate of 100 mR/hr. If a Pb shield 1.4 cm thick is inserted between the source and the detector, what is the new exposure rate? Solution: First determine N N = actual thickness/HVL = 14 mm/5.8 mm  2.4 HVLs

26. Sample Problem (con’t): Then insert this value into the attenuation equation: I = I0 (0.5)N = (100 mR/hr)(0.5)2.4 = (100 mR/hr)(0.189) = 18.9 mR/hr Note for this case a transmission factor of 18.9%.

27. It is also important to realize that not all radiation sources require shielding. This is particularly true for certain pure alpha and pure, low energy betaemitters as illustrated in the next table.

28. Shielding Requirements for Commonly Used Radionuclides Radiation Type Shielding Maximum Radionuclide Emitted Energy (keV) Required 3H 18.6  none 14C 156  none 32P 1710  Plexiglas 35S 168  none 99Mo 740  ,  lead 131I 364  ,  lead 137Cs 662  ,  lead

29. Important Concept to Remember: Radiation dose can always be reduced by appropriate use of • time, • distance, and • shielding 