1. 4/2003 Rev 2 I.3.5 – slide 1 of 30 Session I.3.5 Part I Review of Fundamentals Module 3Interaction of Radiation with Matter Session 5Attenuation IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

2. 4/2003 Rev 2 I.3.5 – slide 2 of 30  In this session we will discuss the process of attenuation including:  linear attenuation coefficients  mass attenuation coefficients  We will also discuss half value layer Overview

3. 4/2003 Rev 2 I.3.5 – slide 3 of 30 Attenuation vs Absorption When photons interact with matter three things can occur. The photon may be:  Transmitted through the material unaffected  Scattered in a different direction from that traveled by the incident photon  Absorbed by the material such that no photon emerges

4. 4/2003 Rev 2 I.3.5 – slide 4 of 30 Attenuation vs Absorption Attenuation of the photon beam can be considered a combination of scattering and absorption. Attenuation = Scattered + Absorbed If the photons are scattered or absorbed, they are no longer traveling in the direction of the intended target.

5. 4/2003 Rev 2 I.3.5 – slide 5 of 30 Attenuation vs Absorption a c b d RadiationSource Detector

6. 4/2003 Rev 2 I.3.5 – slide 6 of 30 Attenuation 100 90 81 73 66 90% 90% 90%90%

7. 4/2003 Rev 2 I.3.5 – slide 7 of 30 Exponential Attenuation dIdx = -  I = -  dx dII I0I0I0I0 IdII  0 x ln(I) – ln(I o ) = -  (x – 0) ln = -  x I IoIoIoIo I = I o e -  x = e eI IoIoIoIo ln -  x = e I IoIoIoIo -  x  represents the fractional linear attenuation coefficient

8. 4/2003 Rev 2 I.3.5 – slide 8 of 30 A half value layer of any material will permit only 50% or ½ of the incident radiation to pass. A second half value layer will permit ½ of the incident radiation (already reduced by ½) to pass so that only ¼ of the initial radiation (½ x ½) is permitted to pass. If “n” half value layers are used, (½) n of the initial radiation is permitted to pass. “n” may be any number. Half Value Layer

9. 4/2003 Rev 2 I.3.5 – slide 9 of 30 Half Value Layer - Example The half value layer (HVL) of a material is 2 cm. A researcher has a piece of the material which is 7 cm thick. What fraction of the initial radiation will pass through the piece? 7 cm 2 cm HVL = 3.5 HVL (½) 3.5 = 0.0883 (use a calculator y x ) Self Check – the answer must be between: (½) 3 = 1/8 = 0.125 and (½) 4 = 1/16 = 0.0625

10. 4/2003 Rev 2 I.3.5 – slide 10 of 30 The half value layer of material “A” is 2 cm and the half value layer of material “B” is 5 cm. A researcher has a piece of some material which is composed of 3 cm of “A” and 4 cm of “B”. What fraction of the initial radiation will pass through the piece? “A”: 3 cm = 1.5 HVL (2 cm/HVL) (2 cm/HVL) “B”: 4 cm = 0.8 HVL (5 cm/HVL) (5 cm/HVL) [(½) 1.5 ] x [(½) 0.8 ] = 0.354 x 0.574 = 0.203 Half Value Layer - Example 100 AB 3520 57%35%

11. 4/2003 Rev 2 I.3.5 – slide 11 of 30 The initial intensity is 192. It is desired to reduce the intensity to 12. How many HVL do we need? Irrespective of the material’s properties, a half value layer of ANY material passes ½ of incident photons. Going from 192 to 12 means that the initial intensity is reduced by a factor of 192/12 = 16. Or we could say that the final intensity is 1/16 of the initial. How many HVL do we need? (½) n = 1/16 or 2 n = 16 This one is easy. Since 2 4 is 16, we need 4 HVL Half Value Layer - Example

12. 4/2003 Rev 2 I.3.5 – slide 12 of 30 Suppose “n” was not a nice round integer. How do we solve for “n”? You need to remember that ln(y x ) = x ln(y) Looking at the previous problem: 2 n = 16 Take the natural logarithm of both sides ln(2 n ) = ln(16) n ln(2) = ln(16) n x 0.693 = 2.77 n = 2.77/0.693 = 4 Half Value Layer - Example

13. 4/2003 Rev 2 I.3.5 – slide 13 of 30 Given a specific material: For monoenergetic radiation, the HVL never changes. For polyenergetic X-ray beams, the HVL increases as more material is inserted into the beam due to the preferential removal of the lower energy X-rays (hardening of the beam). This effectively increases the energy resulting in more penetration and thus the need for more material to stop it. Half Value Layer

14. 4/2003 Rev 2 I.3.5 – slide 14 of 30 Energy P1P1P1P1 E1E1E1E1 E2E2E2E2 P2P2P2P2 E3E3E3E3 P3P3P3P3 E4E4E4E4 P4P4P4P4 E max As the amount of filtration increases, the effective energy also increases and so does the HVL since it takes more material to stop the higher energy radiation remaining. Half Value Layer

15. 4/2003 Rev 2 I.3.5 – slide 15 of 30 1000 500 250 125 62 HVL HVL HVLHVLmono-energeticpoly-energetic* 1000 500 300 200 155HVLHVLHVLHVL * Effective energy of the initial polyenergetic beam is the same as the energy of the monoenergetic beam above E1E1E1E1 E1E1E1E1 E1E1E1E1 E1E1E1E1 E1E1E1E1 E5E5E5E5 E4E4E4E4 E3E3E3E3 E2E2E2E2 E1E1E1E1 Half Value Layer

16. 4/2003 Rev 2 I.3.5 – slide 16 of 30 Half Value Layer (HVL) (mm) PhotonEnergy(keV) Half Value Layer

17. 4/2003 Rev 2 I.3.5 – slide 17 of 30 There are two types of attenuation coefficients:  Linear Attenuation Coefficient (LAC) provides a measure of the fractional attenuation per unit length of material traversed  Mass Attenuation Coefficient (MAC) provides a measure of the fractional attenuation per unit mass of material encountered Attenuation Coefficients

18. 4/2003 Rev 2 I.3.5 – slide 18 of 30  and HVL are functions of the energy of the photon radiation and the material through which it passes I = I o e (-  x) when x = HVL, then I = (½)I o (½)I o = I o e (-  HVL) ½ = e (-  HVL) ln(½) = ln(e (-  HVL) ) ln(½) = (-  HVL) ln(2) = (  HVL) ln(2)HVL Linear Attenuation Coefficient LAC =  M,E = ln 2 HVL M,E = 

19. 4/2003 Rev 2 I.3.5 – slide 19 of 30 LAC = MAC x density Mass Attenuation Coefficient 1 = cm 2 x g 1 = cm 2 x g cm g cm 3 The relationship between LAC and MAC is:   = x 

20. 4/2003 Rev 2 I.3.5 – slide 20 of 30

21. 4/2003 Rev 2 I.3.5 – slide 21 of 30 Mass Attenuation Coefficient PhotonEnergy Material

22. 4/2003 Rev 2 I.3.5 – slide 22 of 30 To express the attenuation of radiation as it passes through some material we can use either of two equations:  I = I o e (-  x)  I = I o (½) n These two equations are identical! Here’s how: I = I o e (-  x) = I o e { [-ln(2)/HVL] x} = I o e {-ln(2) *[ x/HVL] } let n = x/HVL ln(½) = -ln(2) = I o e { ln(½) * n} = I o e { n * ln(½)} = I o e { ln[(½) n ]} (n)ln(½) = ln(½) n = I o e { ln(½) * n} = I o e { n * ln(½)} = I o e { ln[(½) n ]} (n)ln(½) = ln(½) n = I o (½) n so I o e (-  x) = I o (½) n = I o (½) n so I o e (-  x) = I o (½) n Attenuation Equations

23. 4/2003 Rev 2 I.3.5 – slide 23 of 30 The dose rate is reduced from 300 mSv/hr to 100 mSv/hr using 5 cm of some material. The material has a mass attenuation coefficient of 0.2 cm 2 /g. What is the density of the material? Sample Problem #1

24. 4/2003 Rev 2 I.3.5 – slide 24 of 30 (½) n = 100/300 = 1/3 ln(½) n = ln(1/3) n = ln(1/3)/ln(½) = -1.0986/-0.693 = 1.585 HVL 5 cm/1.585 HVL = 3.2 cm/HVL LAC = lnproduction (2)/HVL = (µ/  ) x  = MAC x   = = = = 1.09 g/cm 3 Solution to Sample Problem #1 ln(2)HVL MAC0.693 3.2 cm 0.2 cm 2 /g 0.217 cm -1 0.2 cm 2 /g

25. 4/2003 Rev 2 I.3.5 – slide 25 of 30 One HVL of lead is used to shield a 60 Co source. What would be the dose reduction if the same amount of lead is used for a 137 Cs source? Sample Problem #2

26. 4/2003 Rev 2 I.3.5 – slide 26 of 30 60 Co energy = 1250 keVMAC for lead = 0.06 cm 2 /g 137 Cs energy = 660 keVMAC for lead = 0.1 cm 2 /g  for lead = 11.35 g/cm 3 NOTE: One might assume that since the energy of the 137 Cs photons is less than the energy of the 60 Co photons, then the amount of lead which reduces the 60 Co by 50% would reduce the 137 Cs by a greater amount. For 60 Co HVL = ln(2)/  and  = MAC x  = LAC LAC = 0.06 cm 2 /g x 11.35 g/cm 3 = 0.681 cm -1 HVL = 0.693/0.681 cm -1 = 1.02 cm Solution to Sample Problem #2

27. 4/2003 Rev 2 I.3.5 – slide 27 of 30 For 137 Cs LAC = 0.1 cm 2 /g x 11.35 g/cm 3 = 1.135 cm -1 HVL = 0.693/1.135 cm -1 = 0.61 cm 1.02 cm of lead represents (1.02/0.61) = 1.67 HVL for 137 Cs (½) 1.67 = 0.31 so 31% of initial 137 Cs dose is transmitted or, alternatively, there is a 69% reduction of the 137 Cs dose. AS one HVL implies a 50% reduction so 1.67 HVL implies more than a 50% reduction. Solution to Sample Problem #2

28. 4/2003 Rev 2 I.3.5 – slide 28 of 30 An X-ray beam is evaluated by sequentially placing thicknesses of aluminum in the beam path and measuring the amount of radiation transmitted. The results are: Al (mm)(μSv/hr)Al (mm)(μSv/hr) 0350041700 1290051500 2240061400 32000101000 Determine the effective energy of the radiation emitted by this X-ray unit. Sample Problem #3

29. 4/2003 Rev 2 I.3.5 – slide 29 of 30 HVL approximately 3.9 mm = 0.39 cm µ = ln(2)/HVL = ln(2)/0.39 cm = 1.78 cm -1  for aluminum = 2.7 g/cm 3 MAC = LAC/  = 1.78 cm -1 /2.7 g/cm 3 = 0.66 cm 2 /g Looking up the MAC for aluminum yields an effective energy somewhere between 35 and 40 keV Solution to Sample Problem #3

30. 4/2003 Rev 2 I.3.5 – slide 30 of 30 Where to Get More Information  Cember, H., Johnson, T. E., Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2008)  Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012)  Attix, F. H., Introduction to Radiological Physics and Radiation Dosimetry, Wiley and Sons, Chichester (1986)  Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, New York (1999)