4/2003 Rev 2 I.3.5 – slide 1 of 30

Session I.3.5

Part I Review of Fundamentals

Module 3 Interaction of Radiation with Matter

Session 5 Attenuation

IAEA Post Graduate Educational CourseRadiation Protection and Safety of Radiation Sources

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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

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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

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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.

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Attenuation vs Absorption

a

c

b

d

RadiationSource Detector

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Attenuation

100 90 81 73 66

90% 90% 90% 90%

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Exponential Attenuation

dIdx

= -I

= - dxdII

I0

I dII

= - dx 0

x

ln(I) – ln(Io) = - (x – 0)

ln = - xIIo

I = Io e- x

= eeIIo

ln - x

= eIIo

- x

represents the fractional linear attenuation coefficient

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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

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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 cm2 cmHVL

= 3.5 HVL

(½)3.5 = 0.0883 (use a calculator yx)

Self Check – the answer must be between:

(½)3 = 1/8 = 0.125 and(½)4 = 1/16 = 0.0625

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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)

“B”: 4 cm = 0.8 HVL (5 cm/HVL)

[(½)1.5 ] x [(½)0.8 ] = 0.354 x 0.574 = 0.203

Half Value Layer - Example

100

A B

35 20

57%35%

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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 2n = 16

This one is easy. Since 24 is 16, we need 4 HVL

Half Value Layer - Example

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Suppose “n” was not a nice round integer. How do we solve for “n”?

You need to remember that ln(yx) = x ln(y)

Looking at the previous problem: 2n = 16

Take the natural logarithm of both sides

ln(2n) = ln(16)n ln(2) = ln(16)n x 0.693 = 2.77n = 2.77/0.693 = 4

Half Value Layer - Example

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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

4/2003 Rev 2 I.3.5 – slide 14 of 30Energy

P1

E1E2

P2

E3

P3

E4

P4

Emax

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

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1000 500 250 125 62

HVL HVL HVL HVLmono-energetic

poly-energetic*

1000 500 300 200 155

HVL HVL HVL HVL

* Effective energy of the initial polyenergetic beam is the same as the energy of the monoenergetic beam above

E1 E1E1E1 E1

E5E4E3E2E1

Half Value Layer

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Half Value Layer (HVL) (mm)

PhotonEnergy(keV)

Half Value Layer

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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

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and HVL are functions of the energy of the

photon radiation and the material through which it

passes

I = Io e (- x) when x = HVL, then I = (½)Io

(½)Io = Io 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

=

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LAC = MAC x density

Mass Attenuation Coefficient

1 = cm2 x gcm g cm3

The relationship between LAC and MAC is:

= x

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Mass Attenuation Coefficient

PhotonEnergy Material

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To express the attenuation of radiation as it passes through some material we can use either of two equations:

I = Io e (- x)

I = Io (½) n

These two equations are identical! Here’s how:

I = Io e(-x) = Io e{ [-ln(2)/HVL] x} = Io e{-ln(2) *[ x/HVL] } let n = x/HVL

ln(½) = -ln(2)

= Io e{ ln(½) * n} = Io e { n * ln(½)} = Io e{ ln[(½)n]} (n)ln(½) = ln(½)n

= Io (½)n so Ioe(-x) = Io(½)n

Attenuation Equations

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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 cm2/g. What is the density of the material?

Sample Problem #1

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(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/cm3

Solution toSample Problem #1

ln(2)HVLMAC

0.6933.2 cm

0.2 cm2/g0.217 cm-1

0.2 cm2/g

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One HVL of lead is used to shield a 60Co source. What would be the dose reduction if the same amount of lead is used for a 137Cs source?

Sample Problem #2

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60Co energy = 1250 keV MAC for lead = 0.06 cm2/g137Cs energy = 660 keV MAC for lead = 0.1 cm2/g

for lead = 11.35 g/cm3

NOTE: One might assume that since the energy of the 137Cs photons is less than the energy of the 60Co photons, then the amount of lead which reduces the 60Co by 50% would reduce the 137Cs by a greater amount.

For 60Co

HVL = ln(2)/ and = MAC x = LACLAC = 0.06 cm2/g x 11.35 g/cm3 = 0.681 cm-1

HVL = 0.693/0.681 cm-1 = 1.02 cm

Solution toSample Problem #2

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For 137Cs

LAC = 0.1 cm2/g x 11.35 g/cm3 = 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 137Cs

(½)1.67 = 0.31 so 31% of initial 137Cs dose is transmitted or, alternatively, there is a 69% reduction of the 137Cs dose.

AS one HVL implies a 50% reduction so 1.67 HVL implies more than a 50% reduction.

Solution toSample Problem #2

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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)0 3500 4 17001 2900 5 15002 2400 6 14003 2000 10 1000

Determine the effective energy of the radiation emitted by this X-ray unit.

Sample Problem #3

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HVL approximately 3.9 mm = 0.39 cmµ = ln(2)/HVL = ln(2)/0.39 cm = 1.78 cm-1

for aluminum = 2.7 g/cm3

MAC = LAC/ = 1.78 cm-1/2.7 g/cm3 = 0.66 cm2/g

Looking up the MAC for aluminum yields an effective energy somewhere between 35 and 40 keV

Solution toSample Problem #3

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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, 6th 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 (8th Edition, 1999 update), Wiley, New York (1999)