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4/2003 Rev 2 I.4.12 – slide 1 of 10 Session I.4.12 Part I Review of Fundamentals Module 4Sources of Radiation Session 12Filtration and Beam Quality IAEA.

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Presentation on theme: "4/2003 Rev 2 I.4.12 – slide 1 of 10 Session I.4.12 Part I Review of Fundamentals Module 4Sources of Radiation Session 12Filtration and Beam Quality IAEA."— Presentation transcript:

1 4/2003 Rev 2 I.4.12 – slide 1 of 10 Session I.4.12 Part I Review of Fundamentals Module 4Sources of Radiation Session 12Filtration and Beam Quality IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

2 4/2003 Rev 2 I.4.12 – slide 2 of 10  In this session we will discuss X-ray spectra  We will also discuss filtration and it’s effect on the spectra  Finally we will discuss X-ray beam quality Overview

3 4/2003 Rev 2 I.4.12 – slide 3 of 10 Introduction The three topics in this section have already been introduced in Session I.3.5 where the following concepts were discussed:  exponential attenuation  linear attenuation coefficient and  mass attenuation coefficient A brief review will be useful in understanding the current topics

4 4/2003 Rev 2 I.4.12 – slide 4 of 10 X-Ray Spectra

5 4/2003 Rev 2 I.4.12 – slide 5 of 10 Attenuation 100 90 81 73 6690%90%90%90%

6 4/2003 Rev 2 I.4.12 – slide 6 of 10 1000 500 250 125 62 HVL HVL HVLHVLmono-energeticpoly-energetic* 1000 500 300 200 155 * 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

7 4/2003 Rev 2 I.4.12 – slide 7 of 10 Half Value Layer 500 250 125 6? mono poly

8 4/2003 Rev 2 I.4.12 – slide 8 of 10 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)(mSv/hr)Al (mm)(mSv/hr) 03.541.7 12.951.5 22.461.4 32.0101.0 Determine the effective energy of the radiation emitted by this X-ray unit. Sample Problem

9 4/2003 Rev 2 I.4.12 – slide 9 of 10 HVL approximately 3.9 mm = 0.39 cm µ = Ln2/HVL = Ln2/0.39 cm = 1.78 cm -1 ρ for aluminum = 2.7 g/cm 3 MAC = LAC/ ρ = 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

10 4/2003 Rev 2 I.4.12 – slide 10 of 10 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)  Turner, J. E., Atoms, Radiation and Radiation Protection, 3 rd Edition, Wiley VCH Verlag, Chichester (2007)  Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, New York (1999)

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