4/2003 rev 2 i.3.5 – slide 1 of 30 session i.3.5 part i review of fundamentals module 3interaction...
TRANSCRIPT
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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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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
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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
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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.
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4/2003 Rev 2 I.3.5 – slide 5 of 30
Attenuation vs Absorption
a
c
b
d
RadiationSource Detector
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4/2003 Rev 2 I.3.5 – slide 6 of 30
Attenuation
100 90 81 73 66
90% 90% 90% 90%
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4/2003 Rev 2 I.3.5 – slide 7 of 30
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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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
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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 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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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)
“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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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 2n = 16
This one is easy. Since 24 is 16, we need 4 HVL
Half Value Layer - Example
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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(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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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
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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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4/2003 Rev 2 I.3.5 – slide 15 of 30
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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4/2003 Rev 2 I.3.5 – slide 16 of 30
Half Value Layer (HVL) (mm)
PhotonEnergy(keV)
Half Value Layer
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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
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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 = 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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4/2003 Rev 2 I.3.5 – slide 19 of 30
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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4/2003 Rev 2 I.3.5 – slide 20 of 30
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4/2003 Rev 2 I.3.5 – slide 21 of 30
Mass Attenuation Coefficient
PhotonEnergy Material
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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 = 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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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 cm2/g. What is the density of the material?
Sample Problem #1
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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/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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4/2003 Rev 2 I.3.5 – slide 25 of 30
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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4/2003 Rev 2 I.3.5 – slide 26 of 30
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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4/2003 Rev 2 I.3.5 – slide 27 of 30
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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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)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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4/2003 Rev 2 I.3.5 – slide 29 of 30
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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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, 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)