ATTENUATION, Lesson 5, MMed Physics

ATTENUATION, Lesson 5, MMed Physics

Attenuation:

ATTENUATION IS THE REDUCTION IN THE INTENSITY OF AN X-RAY BEAM AS IT TRAVERSES MATTER BY EITHER ABSORPTION OR DEFLECTION OF PHOTONS FROM THE BEAM.

BEAM Characteristics:

1)      Quantity: number of photons in beam

2)      Quality: Energy distribution of the photons in beam

Intensity is the weighted product of the number and energy of photons. Intensity depends on quantity and quality of the photons.

Beam Intensity can be measured in terms of the ions created in air by beam. Valid for monochromatic or polychromatic beam.

Monochromatic radiation. All Photons in the beam have the same energy.

Attenuation results in change in beam quantity, no change in beam quantity.

The number of photons and total energy of beam changes by same fraction.  Percentage change (reduction) remains consistent – “exponential attenuation”.  

FIG 5.1 IS NB. Attenuation of monochromatic radiation. ONLY ONE PEAK VS POLYCHROMATIC BEAM -

Attenuation Coefficients:

·         Parameter indicating the fraction of radiation attenuated by a given thickness.

·         Attenuation coefficient is a function of absorber and photon energy.

An attenuation coefficient is a measure of the quantity of radiation attenuated by a given thickness of an absorber.

Linear attenuation coefficient: 

Why called linear?

•          distance expressed in linear dimension “x”

Formula

N = N_oe^-µx

N = number of transmitter photons,

N_o = number of incident photons,

e = base of the natural logarithm (2.718...)

µ = linear attenuation coefficient (1/cm); property of energy and material.

x = absorber thickness in centimetres

Larger Coefficient = More Attenuation

v  Units: cm^-1 or 1/cm (or 1/distance)

v  Properties:

Ø  Reciprocal of absorber thickness that reduces beam intensity by e (~2.718 ...)

§  63% reduction

§  37% of original intensity remaining

Ø  As photon beam energy increases

§  Penetration increases / attenuation decreases

§  Attenuating distance increases

§  Linear attenuation coefficient decreases

Again for emphasis Linear Attenuation Coefficient

Linear attenuation coefficient is the quantitative measurement of attenuation per centimetre of absorber, so it tells us how much attenuation we can expect from a certain thickness of tissue.

Useful because we measure our patients in centimetres.

Linear Attenuation coefficient is only for the monochromatic radiation

The size of the coefficient changes with changing energies and it is specific for the type of material

Increasing the energy of radiation is proportionate to more penetrating beams, attenuation decreases

Decreasing the energy is proportionate to less penetrating beams, attenuation increases

Similar to radionucleotide decay N = N_oe^-λt

Mass Attenuation Coefficient

Mass attenuation coefficient = linear attenuation coefficient divided by density

v  normalizes for density

v  expresses attenuation of a material independent of physical state

coefficient (µ)

Ø  Mass attenuation Coefficient  cm ² / g  or rather {linear attenuation coefficient / density}

§  µ/ρ;               ρ =  g/cm ²;

Ø  absorber thickness(x) mass g / cm²   {linear distance X density}

H2O = 1g/cm3 = density or 1cm thick is 1 gram.

Aluminium =2,7 g/cm² or 1 gram is 0.37cm thick

Ø  Units for the absorber:

§  Linear Attenuation Coefficient = cm,

§  Mass Attenuation Coefficient = 1 grams / cm2

v  Units for the coefficient

Ø  LAC = cm-1

Ø  MAC = cm2/gram

v  Units for coefficient is cm-1 and cm2/g.  MAC = LAC / Density = µ/ρ ; Mac is consistent but densities differ.

Polychromatic Radiation

•          X-Ray beam contains spectrum of photon energies

–        highest energy = peak kilovoltage applied to tube

–        mean energy 1/3 - 1/2 of peak energy

•          depends on filtration

               

100 kVp beam has mean energy of 40kV

1)      Polychromatic X-ray Beam Attenuation

a)      Reduction in beam intensity by:

i)        Absorption (photoelectric)

ii)      Deflection (scattering)

b)      Attenuation alters beam

i)        Quantity

ii)      Quality

(1)    Higher fraction of low energy photons are removed

(2)    Beam Hardening

The lower energy beams are more readily attenuated then the higher energy beams. The mean energy of the remaining photons increases. This will continue until the mean energy will approach the peak energy

Half value layer

N = N_oe^-µx

Ø  absorber thickness required to reduce the intensity of the original beam by one half.

Ø  value of “x” which makes N equal to N_o / 2

N_o/2 = N_o e^-λt/2

1/2 = e-λt1/2

λ = 0.693

Ø  HVL = 0,693 / µ.

v  Indication of beam quality

v  Valid concept for all beam types

Ø  Mono-energetic

Ø  Poly-energetic

v  High HVL mean

Ø  More penetrating beam

Ø  Lower attenuation coefficient

A beam with high HVL is a more penetrating beam than one with a low half-value layer.

Polychromatic Attenuation

v  Yields curved line on  semi-log graph

Ø  Straight line with increasing attenuation

Ø  Slope approaches that of monochromatic beam peak energy

v  Increasing the HVL of radiation is proportionate to more penetrating beams

·         Decreasing the HVL is proportionate to less penetrating beams

§  N = N_oe^-µx

^·          800=1000 e^-µx

·         800=1000 / e^µx

·         e^µx =1000/800 = 1.25

·         where x= 1cm µ = .22/cm

·         HVL = 0.693/µ = 0.693/.22 = 3.15 cm

v  Mean energy increases with attenuation

Ø  Beam hardening

FACTORS AFFECTING ATTENUATION or BEAM QUALITY

v  Higher energy

Ø  More penetrating

Ø  Less attenuation

v  Matter

Ø  Density

Ø  Atomic Number

Ø  Electrons per gram

Ø  High density, atomic number or electrons per gram increases attenuation.

1)      4 FACTORS

NATURE OF RADIATION (1)	Composition of Matter (3)
ENERGY	Density Atomic Number Electrons per gram

ENERGY

a)      Increase radiation energy α to the number of photons transmitted

b)      Increasing Energy α decrease Attenuation

Density /Atomic Number / Electrons per gram

c)       Increase Density /Atomic Number / Electrons per gram α Decrease in number of photons

d)      Increase Density /Atomic Number / Electrons per gram α Increase in attenuation

2)      Density and Atomic Number Z

i)        Increase Density α increase in atomic number; except gold and lead

ii)       Au: Z=79, and ρ=19,3 g/cm2

iii)     Pb: Z=82, and ρ=11,0 g/cm2

iv)     Atomic number for water is consistent at 7.4 regardless of state. Solid, Liquid or Gas

3)       Density Ρ and Electrons per gram (e-/gram)

a)      P = weight / volume

b)      E- / gram = volume not relevant

4)      Atomic number and electron per gram

a)      Electrons per gram is a function of the number of neutrons in the atom

i)        Hydrogen has 6.0 x 10 ²³  No neutrons and double electrons per gram

ii)       Oxygen has 3.01 x 10 ²³

iii)     Z α 1/ (e- per gram). Element with low atomic numbers have more electrons per gram than those with high atomic numbers

5)      Effects of energy and Atomic Number

a)      As the energy increases the amount of photoelectric reactions decrease and Compton reactions increase (water/bone/sodium iodide)

b)      Compton % = (1-Photoelectric reactions %)% . Coherent scattering has minimal contribution and can be ignored.

c)       Low energy range (20 keV) the Photoelectric reactions predominate. As the energy increases the photoelectric effect becomes smaller and is replaced by the Compton effect. With high atomic number absorbers like sodium iodide, the photoelectric effect is the predominant interaction throughout the diagnostic energy range.

d)      Linear attenuation coefficient is the sum of the contributions from coherent scattering, photoelectric reactions, and Compton scattering. µ = µ-coherent + µ-pe + µ-compton

i)        Attenuation is

(1)    greater when the photoelectric effect predominates (concrete sidewalk) and

(2)    is less when Compton attenuation predominate (ice pond)

ii)       Attenuation is much more rapid when the photoelectric effect predominates.

6)      Energy

a)      Increasing energy increases transmission and decreases attenuation

i)        Low energy (<20keV), most of the interactions are photoelectric and few photons are transmitted

ii)         As energy increases, photoelectric attenuation becomes less important and completely ceases at 100 keV

iii)     More photons are transmitted with 150 keV than 100 keV

7)      Atomic Number

a)      Usually increasing radiation energy increases the number of transmitted photons and a decrease in the attenuation.

b)      BUT, with high atomic number absorbers, transmission may decrease with increasing beam energy.

i)        K edge

ii)      Comparison of the mass attenuation coefficients for tin and lead. Gram for gram, tin is a better absorber of x rays than lead between 29 and 88 keV. A lighter tin apron gives the same protection as the standard lead apron. Most photons in a polychromatic beam are less energetic than 88keV.

(1)    Tin is more expensive but lighter

(2)    Figure 5.5 (Mass Attenuation Coefficient vs Photon Energy (keV)

iii)     When maximum x-ray absorption is desired, the k-edge of an absorber should be closely matched to the energy of the x-ray beam.

(1)    Xeroradiography used a selenium plate with a K-edge of 12.7 keV as the x ray absorber. Excellent absorber for low energy radiation (30 to 35 kVp) used in mammography

(2)    High energy CXR uses 350 kVp with a field emission unit, needs a high energy absorber like tungsten with a K-edge of 59,5 keV

(3)    Rare earth intensifying screens and cesium iodide image intensifier is in a large part related to the excellent matching of their K edges to the beam energies

8)      Effect of Density on Attenuation

Tissue density is one of the most important factors in x-ray attenuation, and a difference in tissue densities is one of the primary reasons why we see an x-ray image

(1)    Density determines the number of electrons per given thickness, so it determines the tissue stopping power.

(2)    Linear relationship between density and attenuation

9)       Effect of electron per gram

a)      The number of Compton reactions depends on the number of electrons per given thickness. Absorbers with many electrons are more impervious to radiation then electrons with fewer electrons. Expressed in electrons per gram (but who is going to count?).

i)        If we know the density, we can workout the electrons per gram and then work out the x-ray attenuation.

ii)      e-/gram x gram / cm3 = e/cm3

(1)    Bone attenuates well because it has more electrons per centimetre

iii)    N_{o = NZ/ A;}

(1)    where No = number of electrons per gram; N = Avogadro’s number; Z= Atomic  weight; A= Atomic weight

(2)    Simply stated Z/A or number of electrons / weight of the atom

Application to Diagnostic radiology

v  Differential Attenuation

Ø  Uniform x-rays enter a patient. Some are attenuated (White) and some are transmitted (black). The mix between the two gives an image.  Image formation depends of the differential attenuation between tissues.

Ø  Remember the higher the attenuation coefficient, the greater the attenuation (more white from bone).

v  Energy

Ø  The greater the energy the less the difference in attenuation between tissue types, and less the contrast. The difference at low energies (20keV) is largely because of the difference in photoelectric effects, and

Ø  At higher energies (100 keV) the difference in contrast is because of the difference in Compton scattering

v  When Compton reactions predominate, differential attenuation entirely depends on differences in density

v  low energy photoelectric attenuation techniques

Ø  Fat and water are difficult to differentiate radiologically – so we need low energy photoelectric attenuation techniques. At higher energy Compton attenuation predominates, and the difference between fat and water would be determined by density and on the number of electrons per gram. The number of electrons per cm3, are nearly the same (2%), for fat and water.

Scatter Radiation

v  Scatter radiation has nothing useful to offer. It detracts from film quality and contributes no useful information.

Ø  50 to 90 % of the total number of total photons emerging from the patient

Ø  Contributes to film blackening

Ø  Compton scattering is the only scattering of significance

v  Three factors affecting scatter radiation

Ø  Kilovoltage (kVp)

Ø  Part thickness

Ø  Field size

v  Scatter radiation is maximum with high kVp techniques, thick parts and large field sizes. The only variable that is controllable is the kVp

Ø  Field size: small field size, “narrow beam”, irradiates a small volume of tissue, so it generates a small number of scatter photons. The smaller the angle of escape the less the scatter

.

§  As the x-ray beam is enlarged the scatter radiation increases rapidly and reaches saturation.

§  A field size of 30cm circle, most of the scatter photons don’t have sufficient energy to penetrate 15cm of tissue between the field margin and center.

§  30cm by 30cm is the saturation point for diagnostic radiology.

Ø  Part thickness saturation point – no control except to use a compression band or compression of beast in mammography

Ø  Kilovoltage (kVp) –

§  low energy (20kVp) photoelectric effect predominates and little scatter radiation is produced.

§  As energy is increased the percentage Compton effect increases and so does the production of Scatter. Managed by use of a grid.

BASIC INTERACTIONS BETWEEN X RAYS AND MATTER, MMED Physics, Lesson 4

BASIC INTERACTIONS BETWEEN X RAYS AND MATTER

X-ray photons may interact either with orbital electrons or with the nucleus of atoms. In the diagnostic energy range, the reactions are always with orbital electrons.

If the electron disrupted is an electron used to bind atoms together to form molecules, then the molecular structure of the tissue may be disrupted or altered. An atom will stop the same number of incident photons in a solid, liquid or gaseous state – it does not matter if oxygen is in ice, water or steam – neither does it matter if oxygen is in air or bound to hydrogen in water.

There are 5 basic ways that an x-ray photon can interact with matter: (LIST THE 5 WAYS)

1.       Coherent Scattering

2.       Photoelectric effect

3.       Compton Scattering

4.       Pair Production

5.       Photodisintegration

Absorbed – photons completely removed from the x-ray beam and crease to exist

Scattered – photons are deflected into a random course, and no longer carry useful information.

Noise – scattered photons carry no useful information and add noise to the system. It destroys image quality. Noise is also referred to as “film fog”. Noise Covers valid information with distracting or obscuring “garbage.”

v  COHERENT SCATTERING

Coherent scattering occurs when interactions undergo a change in direction without a change in wavelength.  Both can be described in terms of the wave-particle interaction or “classic scattering”

1.       Thomson scattering – photons interact with a single electron.

2.       Rayleigh scattering – photons interact with all the electrons of an atom.

When low energy radiation encounters an atom there is absorption of radiation, vibration of the atom, and emission of radiation as the atom returns to its undisturbed state.

·         Change in direction.

·         No change in energy, frequency or wavelength. 

·         There is NO IONISATION.  

·         Contributes to scatter as film fog.

·         Contributes less than 5% of all radiation, not significant but does contribute, in a small way, to “film fog” throughout the diagnostic energy range.

v  PHOTOELECTRIC EFFECT

K-shell electrons are at a lower energy level than electrons in the L shell. The outer electrons are “free”, and the inner electrons are in energy debt and energy debt is greatest when close to the nucleus in an element with a high atomic number. (Sounds like my bank balance).

The energy produced when electrons drop from a higher energy shell (L-shell) to a lower energy shell (K-shell) is characteristic for each element, and the radiation produced is called “characteristic radiation.”

Three end products of photoelectric effect:

1.       Characteristic radiation;

2.       A negative ion (the photoelectron – that is quickly absorbed and has obviously poor penetrating power); and

3.       A positive ion (an remaining atom that is deficient of one electron)

v  Probability of Occurrence of Photoelectric Interaction Probability – Three rules

1.       The Incident photon must have sufficient energy to overcome the binding energy of the electron

2.       A photoelectric reaction is most likely to occur when photon energy and electron binding energy are nearly the same

Ø  Low energy event

Ø  Photoelectric effect is proportionate to  1/(energy) ³ 

3.       The tighter an electron is bound in its orbit, the more likely it is to be involved in a photoelectric reaction

Ø  Photoelectric effect ~ (Atomic Number) ³

Photoelectric effect is most likely to occur with low energy photons and elements with high atomic numbers provided that the photons have sufficient energy to overcome the forces binding the electrons in their shells.

v  Characteristic Radiation

The method used is different from the high speed electrons used – here we are using the energy of an incident photon to eject the K-shell electron.

Iodine and Barium emit enough characteristic radiation to leave the patient and fog and reach the x-ray film screen – so called “secondary radiation”.

Applications to Diagnostic Radiology

Photoelectric effect enhances natural tissue contrast = good quality

Photoelectric effect does not produce scatter radiation.

Photoelectric interactions deposit most beam energy that ends up in tissue

¨       always use highest kVp technique consistent with imaging contrast requirements

Contrast is greatest when the difference in tissue absorption between adjacent tissues is greatest. Photoelectric effect ~ (Atomic Number) ³ – It magnifies the difference in tissues composed of different elements, such as bone and soft tissue.

Patient exposure is undesirable – patients receive more exposure from the photoelectric reactions than from any other type of interaction. All the energy from the incident photon is absorbed by the patient in a photoelectric absorption.

v  COMPTON SCATTERING

Virtually  all scatter radiation in diagnostic radiology comes from Compton Scattering.

A high energy incident photon strikes an outer shell free electron, ejecting it from its orbit. The photon, retaining part of its original energy, is deflected and travels in a new direction leaving an ion pair – a positive atom and a negative “recoil” electron. (ionization)

Two factors determine the amount of energy retained by the photon:

1)      Its initial energy

The angle of deflection off the recoil electron

Photons have momentum and the higher the energy of the photon the more difficult they are to deflect.

Ø  Change in wavelength Δλ = 0.024 (1 – cos Θ)

Ø  Δλ = change in wavelength Ǻ

Ø  Θ  = angle of photon deflection

But we don’t think in wavelength by keV

Ø  keV = 12.4 / λ

Ø  Δλ = change in wavelength Ǻ

Ø  keV  = energy of photon

Really very little energy is transferred to the recoil electron. At narrow angles of deflection, scattered photons retain most of their original energy. They have an excellent chance of reaching the x-ray film and producing fog:

Ø  as they can’t be removed by filters – they are too energetic; and

Ø  can’t be removed by grids because the angle of deflection is too small

Compton Scattering is a major Safety hazard – as it retains most of its initial energy. Both for the patient and the operator – especially during fluoroscopic examination.

v  Probability of Occurrence

Compton scattering is independent of the atomic number – as all elements contain the same number of electrons per gram, regardless of their atomic number. It is dependent on the energy of the radiation and the density of the absorber. The number of reactions diminish as the energy diminishes.

v  PAIR PRODUCTION AND PHOTODISINTERGRATION

Don’t occur in the diagnostic energy range.

Pair production – the high energy incident photon interacts with the nucleus and disappears – and its energy is converted into matter in the form of two particles, a positron and an electron – both with energies of 0.51 MeV – (Minimum energy required is 0.51 x 2 = 1.02 MeV).

v  Photodisintegration – part of the nucleus is ejected by high energy photon. The ejected part may be a neutron, a proton or an alpha particle, or a cluster of particles. Enough energy to overcome the nuclear binding energy is required. Typically 7 to 15 MeV. Threshold above diagnostic energies – does not occur in diagnostic radiology.

v  RELATIVE FREQUENCY OF BASIC INTERACTIONS