iaea interaction of radiation with matter - 3 x and gamma rays day 2 – lecture 3 1

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Interaction of radiation with matter - 3

X and Gamma Rays

Day 2 – Lecture 3

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• To discuss photon interactions including:• Photoelectric effect• Compton Scattering• Pair Production

• To learn about:• Linear and mass attenuation coefficients• Half value layer

Objective

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Photoelectric Effect

When an incident photon ejects an electron, the process is called “photoelectric” effect. It is a photon-electron interaction rather than a charged particle interaction.

The photon transfers all its energy to the electron in a single cataclysmic event after which the photon ceases to exist and the ejected electron retains all the energy originally possessed by the photon except for the energy required to overcome the binding energy of the electron.

since an electron has been ejected, a vacancy exists in the K shell and thus one or more characteristic x-rays will be produced as the vacancy is filled by electrons from higher orbits.

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-Example: Eincident photon = 80 keV

Ebinding energy = 20 keVEphotoelectron = 60 keV

Ephotoelectron = Eincident photon – Ebinding energy

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The photoelectric effect is predominant for:

Low energy photons but higher than BE of the electrons

High atomic number “Z” materials

Probability is proportional to:Z4

E3

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Compton Scattering

Compton scattering occurs when the incident x-ray photon is deflected from its original path by an interaction with an electron.

The electron is ejected from its orbital position and the x-ray photon loses energy because of the interaction but continues to travel through the material along an altered path.

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Compton Scattering

incidentphoton (Eip)

scatteredphoton (Esp)

scatteredelectron (Ese)

loosely boundelectron (Eie)

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The energy change depends on the angle of scattering and not on the nature of the scattering medium.

Since the scattered x-ray photon has less energy, it has a longer wavelength and less penetrating than the incident photon.

Compton Scattering

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Pair Production

Photon converted into two particles

(energy into mass)

electron (-)positron (+)

The rest mass energy of a positive or negative electron is 0.511 MeV

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Pair Production

To create these two particles requires a minimum energy of 2 x 0.511 MeV = 1.02 MeV

If the photon has an energy of exactly1.02 MeV, all of it is used to create the rest mass energy of the two particles so that the particles remain at rest and soon recombine.

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Pair Production

A positron cannot exist at rest. It combines with an electron. The two particles annihilate each other converting mass back into energy.

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Pair Production

• If, the incident the photon has an energy 1.02 MeV, the excess energy is given to the particles as kinetic energy which permits them to travel at some velocity away from the point of creation.

• These then undergo the typical charged particle interactions losing energy during each interaction.

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Photon Energy (MeV)

PairProduction

Compton

Photoelectric

Combined

Pro

bab

ility

WATER

Photon Interactions

The Compton Scattering effect occurs over all energies but peaks in the midrange ;

While the Pair Production event occurs only above 1.02 MeV and dominates as the photon energy increases.

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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 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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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

I = Io e- x

represents the fractional linear attenuation coefficient unit is per cm.

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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?

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7 cm2 cmHVL

= 3.5 HVL = n

(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

Determine the number of “n” half value layers used

(½)n of the initial radiation is permitted to pass.

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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?


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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 = n (2 cm/HVL)

“B”: 4 cm = 0.8 HVL = n (5 cm/HVL)

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


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?

You don’t need to know anything about the material. A half value layer of ANY material passes ½.


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The initial intensity is 192. It is desired to reduce the intensity to 12. How many HVL do we need?

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


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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

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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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)

or when x = HVL

I = 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

Use I = Io (½) n

or I = Io e (- x)

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(½)n = 100/300 = 1/3 or 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 = ln(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

Using I = Io (½) n

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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) (mR/hr) Al (mm) (mR/hr)0 350 4 1701 290 5 1502 240 6 1403 200 10 100

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

Sample Problem #2

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HVL approximately 3.9 mm = 0.39 cmµ = ln(2)/HVL = 0.693/0.39 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 #2

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Where to Get More Information

Cember, H., Johnson, T. E, Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2009)

International Atomic Energy Agency, Postgraduate Educational Course in Radiation Protection and the Safety of Radiation Sources (PGEC), Training Course Series 18, IAEA, Vienna (2002)

iaea interaction of radiation with matter - 3 x and gamma rays day 2 – lecture 3 1

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