1 interaction of radiation with matter

Lecture 6

Shahid Younas

INTERACTION OF RADIATION WITH MATTER

INTERACTIONS

Lecture 6

At low photon energies (<26 keV), photoelectric effect dominates in soft

tissue.

When higher energy photons interact with low Z materials, Compton

scattering dominates

Rayleigh scattering comprises about 10% of the interactions in

mammography and 5% in chest radiography

PAIR PRODUCTION

Lecture 6

Can only occur when the energy of the photon exceeds 1.02 MeV

Photon interacts with electric field of the nucleus.

Energy transformed into an electron-positron pair of no

consequence in diagnostic x-ray imaging because of high energies

required

INTERACTIONS

Lecture 6

PAIR PRODUCTION

Lecture 6

A: Process

B: Annihilation

ATTENUATION OF X-AND GAMMA RAYS

Lecture 6

Attenuation is the removal of photons from a beam of x- or gamma

rays as it passes through matter.

Caused by both absorption and scattering of primary photons.

Linear Attenuation Coefficient

Lecture 6

Fraction of photons removed from a mono-energetic beam of x- or

gamma rays per unit thickness of material is called linear attenuation

coefficient ().

Typically expressed in cm-1

pairComptonphotoRayleigh


Lecture 6

n = number removed from beam.

N = number of photons incident on the material.

very small thickness x

Number of photons removed:

xNn


Lecture 6

Do you think that the linear equation clearly depicts the attenuation.

xNn


Lecture 6

100 keV photons, 1 mm of

thickness.

µ is 0.016/mm.

16% removed for every 1,000

photons.

100 keV photons, 6 cm of

thickness.

µ is 960/mm.

96% removed for every

1,000 photons.


Lecture 6

Attenuation is a continuous process from the front surface of the

attenuating material to the back exiting surface.


Lecture 6

For monoenergetic beam of photons incident on either thick or

thin slabs of material, an exponential relationship exists between,

Number of incident photons (N0) and those transmitted (N)

through thickness x without interaction:

xeNN 0


Lecture 04

Linear attenuation coefficient is the sum of the individual linear

attenuation coefficients for each type of interaction,

In diagnostic energy range, decreases with increasing energy except

at absorption edges (e.g., K-edge).

pairComptonphotoRayleigh


Lecture 6

Find the value of attenuation coefficient (µ) for Aluminum if 0.45cm

thickness of Aluminum reduces the radiation level by 50%.

Step 1:

Step 2: N/N = e-µx

Step 3: 50 % = e-µ(0.45 cm)

Step 4: Taking ln of both sides

Step 5: ln 0.5 = ln e-µ(0.45 cm)

Step 6: -0.693 = -0.45 µ

Step 7: u = 1.54 / cm

xeNN 0


Lecture 6

Could you guess the effect of density on linear attenuation coefficient?


Lecture 6

For given thickness of material, probability of interaction depends

on number of atoms the x- or gamma rays encounter per unit

distance.

Density () of material affects this number.

Linear attenuation coefficient is proportional to the density of the

material:

urwater vapoicewater

Mass Attenuation Coefficient

Lecture 6

For given thickness, probability of interaction is dependent on

number of atoms per volume.

Dependency can be overcome by normalizing linear attenuation

coefficient for density of material:


Lecture 6

)( Material ofDensity

)(t Coefficienn AttenuatioLinear

)/(t Coefficienn Attenuatio Mass


Lecture 6

Do you know the units of density and Mass Attenuation Coefficient?

Density: g / cm3

Mass attenuation Coefficient: cm2/g


Lecture 6

Mass attenuation coefficient is independent of density

For a given photon energy:

urwater vapourwater vapoiceicewaterwater ///


Lecture 6

Using the mass attenuation coefficient to compute attenuation:

Multiplying and dividing u with desired density.

x

eNN

0

xeNN 0


Lecture 6

Do you know about areal thickness or mass thickness?

x

eNN

0


Lecture 6

Half Value Layer

Lecture 6

Half value layer (HVL) defined as thickness of material required to

reduce intensity of an x- or gamma-ray beam to one-half of its initial

value.

Narrow Beam Geometry or “Good Geometry”- Quality of Beam.

Broad Beam Geometry or “Bad Geometry” – Overestimated HVL

Half Value Layer

Lecture 6

Half Value Layer

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The probability of attenuation remains the same for each additional

HVL for narrow beam.

Reduction in beam can be expressed as (1/2)n

“n” is number of HVLs.

Half Value Layer

Lecture 6

The relation between HVL and µ is,

HVL = 0.693/ µ

Half Value Layer

Lecture 6

Find HVL of a metal whose µ is 0.35 / cm.

HVL = 0.693 / 0.35 = 1.98 cm

Find µ for a metal whose HVL is 0.25 cm.

µ = 0.693 /0.25 = 2.8/cm

Half Value Layer

Lecture 6

Do you know Tenth Value Layer (TVL) and its usage?

Half Value Layer

Lecture 04

If a 2-mm thickness of material transmits 25% of a monoenergetic beam of

photons, calculate the HVL of the beam.

Step 1: N / No = e-µx

Step 2: 25 % = e-µ (0.2 cm)

Step 3: ln 0.25 = - µ(0.2 cm)

Step 4: µ = -(ln0.25) / (0.2 cm) ln 0.25 = -1.38

Step 5: = 1.38 /0.2 = 6.93 cm

Step 6: HVL = 0.693 / µ = 0.693 / 6.93 = 0.1 cm

Half Value Layer

Lecture 6

Image taken from: Johns & Cunningham, The Physics of Radiology, 4th Edition

HALF VALUE LAYER (mm)

Photon Source Tissue Aluminum Lead

Tl 201 37 11 0.2

99mTc 44 18 0.3

I131 6.3 cm N.A. 0.23 cm

Mean Free Path

Lecture 6

Range of a single photon in matter cannot be predicted.

Varies from zero to several centimeters.

Average distance traveled before interaction is referred to as Mean free

path (MFP) of photon beam.

Mean Free Path

Lecture 6

Mean free path (MFP) of photon beam is

HVL 44.1HVL/693.0

11MFP

ABSORPTION OF ENERGY

Lecture 6

Fluence

Number of photons (or particles) passing through unit cross-sectional area is called fluence.

(expressed in units of cm-2)

Area

Photons


Lecture 6

FluxFluence rate (e.g., rate at which photons or particles pass through a

unit area per unit time) is called flux .

expressed in units of cm-2sec-2

Time Area

Photons


Lecture 6

Energy FluenceAmount of energy passing through a unit cross-sectional area is called the

energy fluence.

Units of are energy per unit area (e.g., keV per cm2)

E

Photon

Energy

Area

Photons

KERMA

Lecture 6

A beam of indirectly ionizing radiation (e.g., x- or gamma rays or

neutrons) deposits energy in a medium in a two-stage process:

1. Energy carried by photons (or particles) is transformed into kinetic

energy of charged particles (such as electrons).

2. Directly ionizing charged particles deposit their energy in the medium

by excitation and ionization

KERMA

Lecture 6

Kerma (K) is an acronym for kinetic energy released in matter.

Defined as the kinetic energy transferred to charged particles by

indirectly ionizing radiation

1 interaction of radiation with matter

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