xray theory

8/2/2019 Xray Theory

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HET408 Medical Imaging

An Introduction to ProjectionRadiography

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1 The X-ray Spectrum

X-rays form part of the electromagnetic spectrum extending from

wavelength() : 109 6 1012 m

frequency(f) : 3 1017 5 1019 Hz

The principle physical processes upon which X-ray generation rest are

Thermionic Emission

Bremsstrahlung and Characteristic Emission

When an electromagnetic wave interacts with a charged particle the amounts of energy andmomentum which are exchanged in the process are those corresponding to a photon.

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ultraviolet

infrared

microwave

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Figure 1: The Electromagnetic Spectrum

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2 Thermionic Emission

The energy band corresponding to the uppermost atomic shell, occupied by the valence elec-trons, primarily determines the bulk electrical properties.

If the uppermost band is not completely full it is called the conduction band, if it is full it iscalled the valence band and the empty band just above it is called the conduction band.

If the total number of electrons in the conduction band is less than the total number of energylevels available in the band, then electrons will occupy all energy states up to a maximumenergy called the Fermi Energy, F, if the metal is in its ground state.

The uppermost electrons in the conduction band can be thermally excited. The states occupiedby thermally excited electrons fall in an energy region of the order of 20 kT above F.

At high temperatures, T, the occupation of electronic states extends over energies well aboveF. To extract an electron from a metal it is necessary to give a conduction band electron atleast the energy

e (1)

where

e = elementary electronic charge

= work function and is the energy required to extract an electron from the highestoccupied energy level.

Thus at high temperatures some electrons will escape from the metal to form an electrongas around the solids (metal/filament) surface.

Based on the Fermi-Dirac distribution it can be shown that the thermo-electric currentdensity, j, coming form the surface of a solid is given approximately by the Richardson-Dushman equation,

j =4me

h3k2T2 exp[e/kT] (2)

where

me = electron rest mass T = temperature in Kelvin

k = Boltzmanns constant ( 1.38 1023 JK1)

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0

20 kT

EmaxF

dn

dE

Figure 2: Diagrammatic illustration of the energy states of free electrons in a solidat absolute zero (dashed line) and at 20 kT above absolute zero (solid line).

Thermionic emission can be implemented by passing an electric current through a piece ofwire (filament) with a very high melting point e.g tungsten. In the production of X-rays thisresultant electron gas is accelerated through an electric field, E. The kinetic energy acquiredafter an electron has traveled a distance x is

1

2me v

2 = E e x (3)

or

Energy = V e (4)

maxE

approx. 20kT

F

Figure 3: The distribution of free electrons between energy levels in the conductionband.

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This last expression motivates the definition of the electron volt as a unit of energy. 1electron volt is equal to the work done on a particle of charge e when it moves through a

potential difference of 1 V. Thus 1 eV is

eV = (1.6021 1019 C)(1 V)

= 1.6021 1019 J

3 Bremsstrahlung and Characteristic X-ray emission

X-rays are produced by the impact of fast electrons against the anode material of an X-raytube 1. The emitted X-ray spectrum will be composed of two main components dependent onthe energy of the incident photons;

Bremsstrahlung or deceleration radiation

Characteristic X-ray emission

For large anode potentials the emitted X-ray spectrum will be dominated by energy peakscorresponding to characteristic emission.

4 Bremsstrahlung

It is well known that an accelerated charge radiates electro-magnetic energy. The rate of energyradiation by a charge, q, moving with a velocity v and acceleration a, when v c is

dEdt

= q2

a2

60c3(5)

where

0 = permittivity of free space

c = velocity of light

If a particle is decelerated instead of being accelerated the previous equation still holds. Thuswhen a fast charge such as an electron hits a target and is stopped a substantial part of its

1Most of the energy absorbed from the electrons will appear in the form of heat. Only a small proportion,less than 1%, will appear in the form of X-rays

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10 20 300

characteristic

radiation

25 kV

20 kV15 kV

10 kV

5 kV

wavelength (nm)

0

1

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3

X-rayintensity(arbitraryunits)

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continuous

radiation

(bremsstrahlung)

Figure 4: The intensity distribution for the X-ray spectrum of Molybdenum as afunction of the applied voltage. Note the appearance of the K-series as excitationpotential is increased. The K-series excitation potential is 20.1 kV.

total energy is released as electro-magnetic radiation. This is called deceleration radiationor bremsstrahlung.

The energy of an electron may be radiated off as

the result of successive collisions such that several photons are produced.

the result of a single collision such that just one photon is produced.

Photons emitted by just one electron-anode collision will be the most energetic and thus havethe shortest wave-length. Thus the wavelengths of the X-rays so produced will be equal to orlonger than a threshold wavelength satisfying the relationship

E = h

=h c

Thus

=h c

V e

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=(6.6 1034 Js)(2.99 108 m s1)

V 1.6021 1019 C

= 1.24 106

Vm

Example

If electrons are accelerated by a potential difference of 18 kV, the minimum wavelength ofX-rays produced is

min =

1.24 106

1.8 104

= 6.9 1011 m

5 Characteristic X-ray Emission

Electrons filling the inner complete electron shells constitute the core or kernel. The bindingenergy of the kernel electrons is much higher than that of the valence or conduction electrons.These kernel electrons remain practically undisturbed in most of the processes in which the

atom participates. However if the energy of the incident electrons is sufficiently large it ispossible to knock one of the electrons out of the kernel.

For example when the energy of incident electrons is increased hitting a Molybdenum (Mo)target certain well defined peaks appear in the emitted X-ray spectrum. These peaks are calledcharacteristic X-rays. The wavelengths of these peaks are independent of the appliedvoltage, whereas the relative amplitude of these peaks is a function of the applied voltage.

The dynamical state of each electron can be described by four quantum numbers

n = energy

l = angular momentum

ml = quantization of orientation of angular momentum

ms = electron spin

States with n = 1, 2, 3, 4 constitute the K,L,M,N,.. shells.

Consider that an ejected electron is from the K-shell. When such a K-electron is removed andempty state (or hole) is left in the K-shell. Another electron in a higher energy level of thekernel (or even a valence or free electron) may fall into the vacant state in the K-shell. Theradiation emitted by the electron falling into this vacant state lies in the X-ray region of the

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spectrum. Such an electron may have proceeded from the L,M,N,... etc shells and thus a seriesof X-ray lines may be produced designated as K, K, K... etc. If the vacant state produced

by electron ejection is in the L-shell then transitions may only arise from the M,N,... etc shells.

L-series

K-series

K K

(M-shell)

4 (N-shell)

(K-shell)

(L-shell)

K

L L

infinity

n

3

Labsorptionedge

Ka

bsorptionedge

1

2

Figure 5: X-ray transitions in an atom with atomic number Z 36.

In many cases the photon emitted in an X-ray transition is absorbed by another electron withinthe same atom, which is therefore ejected as a result of an internal photo-electric effect. Thisprocess of the internal conversion of X-rays into photo-electrons is called the Auger effect andthe emitted photo-electrons are called auger electrons.

6 Interactions of EM radiation with tissue

When a beam of electro-magnetic radiation passes through an object the energy of this beamis gradually absorbed by various processes

photo-chemical reactions

atomic photo-electric effect (or photo-ionization)

Compton scattering

pair production

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photo-nuclear reactions

In the energy range used in diagnostic radiology (17 - 150 keV) the important photon interac-tions with tissue are the photo-electric effect and Compton scattering.

7 Photoionization

When an incident photon has enough energy, its absorption by an molecule or an atom mayresult in the ejection of an electron i.e

A + h = A+ + e (6)

This process is called photoionization or the atomic photo-electric effect. Thus when abeam of ultraviolet, X- or - radiation passes through matter, it produces ionization along itspath.

Th energy required to extract an electron from an atom or a molecule is called the ionizationpotential, I. The kinetic energy of the ejected electron is given by

Ek = h I (7)

Listed below are some typical values for the ionization potentials for some molecules

Ionization

Molecule Potential eV

H2O 12.6CO2 13.8O2 12.1

C9H10O3 8.4

Table 1: Ionization potentials for some common organic molecules.

A photo-electric interaction is followed by the ejection of a photo-electron, one or more char-acteristic X-rays and Auger electrons.

The highest energy electron that will be emitted in this process will be of the order of 150 keVand would have a range in water typically of 0.03 cm. Thus for most purposes the electronscan be considered as being locally absorbed. These emitted electrons will greatly contribute to

the patient dose and to the energy absorbed in the image receptor.

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8 Compton Scattering

When the energy of a photon is much larger than the binding energy of the electron in an atomor a molecule, the electron can be considered free. In this case Compton scattering is a moreprobable process than the photo-electric effect.

Analysis of electromagnetic radiation that has passed through a region in which free electronsare present shows in addition to the incident radiation another radiation of different frequency.This new radiation is interpreted as the radiation scattered by free electrons.

The frequency of the scattered radiation is smaller than the frequency of the incident radiation.Accordingly the wavelength of the scattered radiation is longer than the wavelength of the

incident radiation. This interesting effect is called the Compton effect.

9 The Attenuation of X-rays

Given that I0 is the intensity of radiation before it enters the substance, its intensity after ithas traversed a thickness x of the substance is given by the familiar equation

I = I0e

x

(8)

where (in say cm1) is a quantity characteristic of each substance and each process and isvariously called the coefficient of linear absorption, coefficient of linear attenuation orthe macroscopic cross-section.

For each substance there is one macroscopic cross-section for each possible process that wehave outlined. The total cross-section of a substance is the sum of all partial cross-sections.These cross-sections will also be a function of the energy of the incident photons. For the twoimportant attenuating processes occurring in diagnostic X-ray we have

Attenuation Process Attenuation

incident energy(E) atomic number(Z)

Photoionization E3 Z4

Compton Scattering > E3 Z

Table 2: Attenuation processes as a function of incident electromagnetic energy andatomic number

This table illustrates that scattering will will be less important in providing contrast between

tissues with differing average Z.

For soft tissue the photo-electric cross-section is larger than the scatter cross-section for energies 25 keV.

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I(Al)

II(Al)

III(Pb)

III(Al)

Al

Pb

II(Pb)

-1 0 1 2

photon energy (MeV)

K absorption edge

I(Pb)

10

(m

)-1

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Figure 6: Macroscopic cross sections for the absorption of photons in aluminium(Al) and lead (Pb). Curves labelled I are the partial cross sections due to photo-electric absorption. Curves labelled II are the partial cross sections due to Comptonscattering. Those curves labelled III are partial cross sections due to pair production(not important for the energies used in diagnostic radiography).

10 Image Formation

The projected radiographic image can be considered as the distribution of absorbed energy. Inwhat follows we will assume the following

a mono-chromatic X-ray source emits photons of energy E.

the X-ray source is sufficiently far away from the patient such that the photon beam canbe considered parallel to the normal of the the surface of the image receptor.

each photon reacts with the image substrate locally, the response of this imaging substratebeing linear such that the image so formed can be considered as the distribution of the

absorbed energy.

IfN photons are incident at the surface of the image receptor (or substrate) and I(x, y) dxdy isthe energy absorbedin the image receptor plane dxdy then we can write the following expression

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0.8

1.0

relativenumberofphotons

(a) (b)

0.6

Energy (keV)

incident transmitted

0.2

0.4

60 80 100 20 404020 60 80 100

Figure 7: X-ray spectra for a tungsten anode (100 kV accelerating potential with2.5 mm aluminium added). (a) X-ray spectra before and (b) after attenuation by 18.5cm soft tissue plus 1.5 cm bone. Note that the lower energy components have beenattenuated more heavily than the higher energy components after passage throughthe human tissue. Thus the average photon energy increases and the beam becomesharder. Such beam hardening can be troublesome in the context of tomographicimage reconstruction as the exponential law of linear attenuation no longer holds.(databased on Birch R, Marshall M and Andran G M (1979). Catalog of Spectral Data forDiagnostic X-rays. Hospital Physicists Association, London.).

I(x, y) = (E, 0)NEexp(

(x,y,z) dz)

+

(Es, )EsS(x,y,Es, ) d dEs

= primary + scatter

where

S(x,y,Es, ) is the number of scattered photons in the energy range Es to Es + dEs inthe solid angle range to + d. This is known as the scatter function.

(E, ) is the energy absorption efficiency coefficient of the image substrate as afunction of photon energy and angle .

(x,y,z) is the linear attenuation (made up of all the partial cross-sections).

The scatter function will in general be a complicated function of position and distribution ofattenuation within the tissue. For this reason it is often considered as a slowly varying functionof x and y. It can be shown that

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10

150100500

0

Compton scattering

photoelectric

-2

-1

10

10

photon energy (keV)

-1

linearattenuationcoefficient(cm

)

Figure 8: The variation of the partial cross sections (photo-electric and Compton)with incident photon energy for soft tissue.

I(x, y) = N(E, 0)Eexp(

(x,y,z) dz) + S(E)E (9)

= N (E, 0)Eexp(

(x,y,z) dz)(1 + R) (10)

where

R =scattered

primary(11)

The first equation essentially says that we consider the scatter proportional to the energy ofthe emitted X-rays. i.e as we increase the energy of the source X-rays so does the amount of

scatter.

10.1 Contrast and Unsharpness

Referring to the above figure 9 we can define contrast as

C =(I1 I2)

I1(12)

where I1 and I2 give the energy absorbed per unit area. Using the previous definitions forI(x, y) it can be shown for the simple model of contrast that the contrast, C, is given by

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12 x X

I1I2

Figure 9: A simple model for the definition and estimation of image contrast inplanar radiography.

C = {1 exp[(2 1)x]}/(1 + R) (13)

Thus from the above equation we can see that the factors affecting contrast will be

the thickness of the target (i.e x).

the difference in the linear attenuation coefficients between different structures that wewould like to visualize.

the scatter-to-primary ratio R i.e a greater proportion of scatter will act to degrade imagecontrast.

The linear attenuation coefficients will be a function of the incident photon energy. The scatter-

to-primary ratio will also be a function of the incident photon energy. Thus we expect contrastto vary with the mono-chromatic energy of the incident X-rays.

Increasing the energy of the emitted X-ray photons will result in reduced linear attenuationcoefficients and hence reduced patient dose. However increased source photon energy willresult in reduced contrast. Thus the energy of the X-ray spectrum chosen for radiographicalprocedures will be a compromise between patient does and contrast 2

Problem 1: By using equations (12) and (9) (why not equation 10 ?) andreferring to figure 9 derive equation (13)

2an aluminium plate interposed between the X-ray source and the patient is often used to filter out thelow-energy component of the emitted X-ray spectrum. As the low energy components of the spectrum will beheavily attenuated and thus contribute little to image formation the effect of such a filter is to remove the lowenergy component that would otherwise unacceptably contribute to patient dose.

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10.2 Relationships between noise and dose

Even during a simple plain-film radiogram there will be sources of noise that will conspire todegrade contrast and image quality. The contributions to this noise will in general arise fromtwo major sources

quantum noise due to statistical fluctuations in the number of photons detected perunit area by the imaging substrate.

fluctuations due to spatial and temporal inhomogeneities in the properties of the imagereceptor (substrate) and display system

Quantum noise, because it is a statistical process, can be reduced by increasing the numberof incident photons. However this will be at the expense of an increased dose to the patient.Therefore a central consideration is

What surface dose is required to achieve a contrast C over an area A against a backgroundnoise arising from the effects of quantum noise ?

It can be shown that the minimum surface dose, Sd, required to differentiate two areas withdifferent linear attenuation coefficients is

Sd = N Et (14)

where t is the mass energy absorption coefficient for tissue ( 0.004 m2 kg1). t =

En/ where En is the tissue mass energy cross-section or linear attenuation coefficient and is the tissue density. E the photon energy and N, the number of incident photons per unitarea

N = k2t (1 + R) exp(1X)/[(E, 0)(2 1)2x4] (15)

The constant kt is the signal-to-noise ratio for which an object just becomes detectable. Em-pirical evidence suggests this value is 5. From this equation note the following

the minimum dose required to visualize an object (against a background of 1) increasesas the inverse fourth power of the size of the object and as the inverse second powerof the difference in the linear attenuation coefficients. i.e larger doses will be required tovisualize thinner objects in the image plane thus ...

for a fixed dose and contrast there will be a minimum object size that can be visualized.

low contrast resolution will vary with object size.

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Example 1: Calculate the number of incident photons per unit area, N, andthe surface dose, Sd for imaging 1 mm

3 of tissue with 1% contrast assumingthe following values

E = 50 keV

= 0.3

x = 1 mm

kt = 5

1 = 22.6 m1

t = 0.004 m2kg1

R = 2

C(1 + R)/x (2 1) power series expansion of equation (13)

X = 0.2 m

N = k2t (1 + R) exp(1X)/[(E, 0)(2 1)2x4]

=52 (1 + 2) exp(22.6 m1 0.2 m)

0.3 9 102 m2 1012 m4

= 2.55 1013

photons m

2

thus the surface dose Sd is

Sd = NtE

= 22.55 1013 m2 0.004 m2 kg1 50 keV= 5.1 1012 keV kg1

= 0.817 mGy

where 1 Gy = 6.242 1015 keV kg1.

Problem 2: By assuming that the projected area of the object having alinear attenuation coefficient of 2 (figure 9) is A calculate the ratio of thetotal energy exiting the region A with respect to the expected variation in thetotal energy exiting a region of area A of thickness X and linear attenuationcoefficient 1. This is the signal-to-noise ratio (SNR). By assuming that theSNR must be of at least kt before an object is visible solve the expressionobtained for the SNR to obtain the number of photons per unit area thatmust be incident in order to obtain a contrast of C. Hint: Assume signal= (I1 I2)A = CI1A, noise =

I1A (i.e the number of photons detected per

unit area is a Poisson process) and C is given by equation (13). Also see pp124-127 Guy and ffytche

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11 Image Receptors and Substrates

The image receptor or substrate is what converts the transmitted X-rays into a visible image.All image receptors form an image by the absorption of energy from the transmitted X-raybeam. In general the transmitted spectra will consist of a distribution of photon energies andnumbers. Part of the performance of image receptors is measured by its response to particularphoton energies. The more restricted is its response, in general the better the resolution.

Below is an incomplete list of the major methods for visualizing the energy of the transmittedX-ray beam.

Direct exposure X-ray film

Screen-film combinations

Image-intensifiers

Xero-radiography

Ionography

Stimulated Luminescence (scintillation detector)

11.1 Direct Exposure X-ray Film

The direct exposure of X-ray film is not commonly used in radiography because it has a lowabsorption efficiency (). However the basic principles of operation are used in screen-filmcombinations. In general there are two film emulsions consisting of silver bromide suspendedin gelatin. This is illustrated below

protective layer

subbing layer

film base (200 micrometers)

film emulsion (20 micrometers)

Figure 10: Schematic outline of the construction of direct exposure X-ray film.

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When using X-ray film it is particularly important to know the relationship between the opticaldensity of the film as a function of exposure. Define the optical density of the film, D, as

D = log10(I0/I) (16)

where I0 is the intensity of a light beam (eg a light box) before passage through the film and Iis the intensity after passage through the film. It can be shown that the relationship betweenoptical density of the direct exposure film and the X-ray dose received by the film is given by

D = Dmax[1 exp(kB)] (17)

where k is a constant characterizing the response of the film and B is the dose applied to thefilm from transmitted X-rays. i.e the greater the film exposure the blacker the film. For amono-energetic beam of X-rays the dose B is related to N by

B = E N (18)

where is the photon absorption efficiency of the film. Curves that relate optical density to filmexposure are known as characteristic curves or H and D curves (after Hurter and Driffield).

It is common practice to plot film dose on a logarithmic scale. It is found that the filmcharacteristic curve will have a central portion for which the following linear relationship isvalid

D = 0.847 Dmax log10(1.883 k B)

= log10(B

B0)

where D is the optical density, B/B0 is the relative film exposure and is known as the filmgamma (typically between 2 and 3). For the case of an homogeneous solid of thickness x weexpect ln(B/B0) = x and it can be shown that we can calculate according to

log10(I) = xlog10(e) + log10

I0B

0

(EN0(ic)t)

(19)

where EN0(ic) is the emitted X-ray intensity as a function of the filament current. Thus thefilm can be estimated as the slope of log10(I) versus x log10(e).

Direct exposure X-ray film is excellent for high resolution studies where high resolution ismore important than dosing considerations eg hands and teeth. However in conventional plainfilm radiographic practice Screen-Film combinations are preferable as they are faster and

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thus minimize patient dose. The main drawback is that screen-film combinations do not haveas good inherent resolution as direct exposure film and for this reason are used when dose

minimization is more important than the preservation of fine detail.

opticaldensity

log (relative exposure)

0 0.5

2.0

1.0

A

B

3.0

1.0 1.5 2.0 2.5

Figure 11: Film characteristic curves for a direct-exposure X-ray film (curve A) andfor a film-screen combination (curve B). Note that the useful dynamic range for thelatter is less than the former.

12 References

Aird EGA (1988). Basic Physics for Medical Imaging. Heinemann, Oxford.

Alonso M and Finn EJ (1968). Fundamental University Physics, Volume III: Quantum andStatistical Physics. Addison-Wesley, Reading MA. pp 176-178,243-251.

Guy G and Dominic fftyche (2000). An Introduction to the Principles of Medical Imaging.Imperial College Press, London. pp 89-112,113-168,

Webb S(ed) (1992). The Physics of Medical Imaging. IOP, Bristol. pp 20-54.

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A International SI Units of radiation exposure and dose

A.1 Exposure

The index of exposure, iexposure, is defined by

iexposure = q/m (20)

where q is the total electric charge liberated by photoemmission in a mass m of air. There isno name for the SI unit of exposure ( 1 C kg1 of air). Thus the intensity of a beam of X-rays

can be specified in terms of the ionization per unit mass of air.

However in general strengths of X-ray sources are specified in terms of the energy deposited perunit mass as it is the energy deposited in tissue that determines the biological hazard. Thekinetic energy released per unit mass (kerma) of air is measured in units of Grays (Gy) suchthat

1Gy = 1 J kg1

Sources are also described in terms of air kerma rates i.e diexposure/dt.

A.2 Absorbed Dose

The absorbed dose will depend on the energy of incident radiation and the atomic compositionof the absorber. In general this is linearly related to iexposure by an empirical dimensionlessfactor that depends on the absorber and the incident radiation energy, fabsorber(E), i.e

doseabsorbed = fabsorber(E) iexposure (21)

Note that doseabsorbed will be in units of Gy3.

Absorbed dose X-ray dose can also measured in terms of its relative biological effectiveness(equivalent absorbed dose) such that it can be quantitatively compared to other forms ofradiation (e.g electron, proton or neutron radiation). The modern SI unit of equivalent absorbeddose is the Sievert (Sv) and for X-rays has the following units

1 Sv 1 J kg

1

= 1 Gy

4

3The older CGS unit of absorbed dose was the rad (1 erg s1). 1 Gy = 100 rad.4The old unit of absorbed equivalent dose was the rem (radiation equivalent man) with 1 Sv = 100 rem.

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A.3 Effective dose

The amount of energy deposited in a particular organ will depend on the what fraction of thetotal surface dose (contained in a weighting factor wR) is absorbed which can only be inferredfrom theoretical models and phantom studies. In general the wR is estimated assuming a 70kg standard man and will also incorporate information regarding the relative susceptibility ofan organ to radiation damage. Thus the effective organ dose will be given as

doseorgan = wR doseabsorbed

wR will range from 0.01 (dental, spine) to 0.2 (gonads).

xray theory

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