Objectives: To understand; 1. Derivation of the attenuation law 2. Half value layer 3.Contributions...

Objectives: To understand;

1. Derivation of the attenuation law

2. Half value layer

3. Contributions to the attenuation coefficient and their energy dependence

4. Beam hardening calculations

5 . K-edge filters

6. Representation of attenuation coefficients as linear combinations of a basis set

Consider x-rays having an incident fluence N(0) which penetrate a distance x into a subject as shown in Figure 1.

N(0)

x

N(x)

∆x

N(x + ∆x)

The fluence at x + ∆x is related to that at x by

N( x + ∆x) = N(x) + ∆N where ∆N < 0.

The change in the fluence due to the attenuation in x is given by

∆N = - total( cm2 / el ) * ( D∆x) (el/cm2) * N(x)

(1) = prob. of attenuation per incident x-ray * N(x)

where total = total attenuation cross-section

= material density (gm/cm3)

D = electron density in (electrons/gram)

= NA (6x1023 atoms / mole) *Z (e l / atom) / A(grams/mole)

∆x = differential thickness traversed(cm)

The linear attenuation coefficient µL is defined as,

µL = total D (c m - 1) (2)

Then from equation 1

∆N = - µL ∆x N ( x )

For small ∆x

dN = - µL dx N(x)

dN/N = -µL dx

Then from equation 1

∆N = - µL ∆x N ( x )

For small ∆x

dN = - µL dx N(x)

dN/N = -µL dx

Integrating both sides

ln N = -µL x + C

elnN = N = e -(µL

x + C)

N = e c *e -µL

x

at x = 0 N = N(o) so

ec = N ( 0 )

N(x) = N(o) e - µLx (3)

This can also be expressed as

N(x) = N(o) e - ( µL

/ )( x ) (4)

Where

( µL / ) = µm mass attenuation coefficient (cm2 / gm)

and px = t thickness expressed in (gm / cm2)

It is useful to use g / cm2 when absorbing materials are mixed rather than separated in x.

For a monoenergetic beam, the intensity and fluence are related by,

I = k N (5)

For that case,

I(x) = I(o) e - µL

x (6)

The penetration of an x-ray beam is often defined in terms of the half value layer (HVL), the distance it must travel to decrease in intensity by a factor of two. This is illustrated in Figure 2 for the case of a monoenergetic beam.

Since the half value layer is the distance required to cut the initial number of photons in half,

e - (µL HVL1) = 1/2

e + ( µL HVL1 ) = 2

HVL1 = ln(2) / µL = 0.693 / µL (8)

For a monoenergetic beam

HVL1 = HVL2

and, as shown in Figure 3, µL is a constant

ln N(x)

x

Slope = - L ln N(0)

Figure 3

There are five 5 possible photon interactions:

1. Rayleigh (coherent scattering)

The fourth and fifth of these are not relevant for diagnostic x-ray imaging. We will just summarize the first three of these which are the only important interactions in the diagnostic x-ray energy range.

2. Photoelectric effect

3. Compton scattering Historically, low energy Compton scattering (~ < 20 keV) was known as Thompson or classical scattering.

4. Pair and triplet production

5. Photonuclear interactions

Contributions to UL

bout 5% of cross-section in the Dx region Elastic event, photon loses essentially none of its energyPhoton interacts with the entire atomAtom recoils just enough to conserve momentumPhoton redirected through a small anglePhotons are only removed from beam in narrow geometryContributes nothing to dose

Contribution to µm ~ Z / k2 (cm2 / g) (9)

The basic interaction is shown in Figure 4

The incoming photon is absorbed, the bound electron is ejected and the e- hole vacancy is filled with outer shell electrons producing characteristic photons and Auger electrons (electrons knocked out by characteristic x-rays).

k

e-

•e-

+

kc

The photoelectric cross section is sketched in Figure 5.

X-ray Energy k

L and K-edge discontinuities occur when the photon energy is just large enough to knock out an L or K shell electron respectively. Away from discontinuities,the photoelectric contribution to the mass attenuation coefficient behaves as

contribution to µm =

~ ( Z / k )3 ( cm2 / gm ) (10)

This is the dominant contribution to the attenuation coefficient at high Z and low energy.

__________________

The Compton interaction is shown in Figure 6. This is essentially a collision with the loosely bound outer orbital electrons. The incoming photon is deflected by an angle . Some of its energy is transferred to the “free electron”.The electron is absorbed almost immediately because of the short range of low energy charged particles. The scattered photon retains a large percentage of the incident energy and goes on to interact again.

Figure 6

k e-

e-T =electron kinetic energy

k’ =X-ray energy

The contribution m is usually called and behaves as

NA Z / A = D = electron density

(11) Z / A is unity for hydrogen and 0.4 - 0.5 for all other elements.

Compton scattering is the dominant effect at low Z and high energy.

The expression for the scattered x-ray energy is

k’ = k / {1 + (k / moc2) (1 - cos )} (12)

with moc2 = 0.511 MeV

Notice that it is very difficult to discriminate against scattered radiation on the basis of the x-ray energy loss since K / moc2 is on the order of 0.1 and 1 - cos is very small.

The electron energy is given by

T = k - k’

The electron scattering angle is given by

cot = (1 + k / moc2) tan ( / 2) (13)

Compton scattering decreases image contrast as illustrated in Figure 7 where the recorded image intensity for the case of primary (unscattered) x-rays (7A) and primary + scatter (7B) are separately shown.

Figure 7A

Detected Primaryintensity

Figure 7B

Detected PrimaryIntensity + scatteredintensity

Compton scattering decreases contrast by permitting x-rays from unrelated regions of the anatomy to be detected behind the contrast producing object. Scatter grids which preferentially transmit primary x-rays and attenuate predominantly scattered x-rays are used in a wide variety of clinical imaging situations.

The total mass attenuation coefficient µm is equal to the sum of the individual coefficients.

µm = µL/ p = R + (14)

The Raleigh scattering contribution is often ignored for the purpose of approximate calculations.

Once you determine the ratio of and for one element, you can predict the attenuation coefficient at another energy, e.g.,

if for substance x, = at 20 keV

and µ(20) = (20) +(2 0 ) = 2 (2 0 )

then at 40 keV

µ(40) =(40) + (40) ≈ (20) + (2 0 ) x (2 0 / 4 0 )3

=(20) + (20) / 8 = 9/8 (2 0 ) = 9 / 1 6 µ(20)

The mass attenuation coefficients in units of cm2 / gm are shown in Figure 8 for bone, muscle, fat and iodine.

Note that since bone has a higher atomic number, it provides higher attenuation at low energy. At higher energies, where the Compton effect is dominant, bone, fat and muscle provide the same attenuation for the same thickness measured in gm / cm2.

Since bone obscures chest images, chest radiography is typically done at 120-140 kVp in order to reduce the relative contrast of bone. For mammography where soft tissue contrast is important, energies in the 20 keV range are used. For angiography, the imaging of blood vessels, optimal image contrast is obtained when the beam energy is predominantly in the region of the iodine k-edge where, at the threshold for ejection of k shell electrons, there is an abrupt discontinuity in the iodine attenuation coefficient.

It is useful to remember that

= at 40 keV for bone 20 keV for tissue (15)

polyenergetic

Let us examine the changes in the x-ray spectrum for the case of polyenergetic beams. We saw previously that for a monoenergetic spectrum the log of the attenuation was a straight line. In the polyenergetic case the dependence of the x-ray fluence is shown in Figure 9.

ln N(x)

x

Monoenergetic slope = - L

Figure 9

The deviation from linearity is due to the fact that the attenuation coefficient decreases with energy. This is due to the preferential removal of low energy x-rays from the beam as sketched in Figure 10

dI/dk

k

Large x

x = 0

Figure 10

The value x indicated is the total amount of attenuating material in the beam. As x increases, intensity decreases, but average energy, k, increases and µeff decreases. This is "beam hardening." Because the photoelectric cross section varies as 1 / k 3, low energy x-rays are preferentially removed. This causes the second half value layer to be larger than the first as shown in Figure 11.

Figure 11

The homogeneity factor H is defined as

H = HVL1 / HVL2 < 1 (16)

A homogeneity factor H ≥ 0.6 is required for patient use to insure that the beam is adequately hardened.

Now let’s consider the mathematical behavior of the intensity as a function of thickness for a broad spectrum as shown in Figure 12.

dI(k,0)dk

k

Figure 12

The spectrum at depth t can be expressed in terms of the initial spectrum as

dI / dk (k,t) = dI / dk (k, 0) e-µ(k)t (17)

The integrated intensity is expressed as

I (t) = ∫ dI / dk (k,t) dk

= ∫ dI / dk (k, 0 ) e-µ(k)t dk (18)

Sometimes the intensity at depth t is expressed in terms of an effective attenuation coefficient µeff defined by the equation

I(t) = I(o) e-µeff(t)t (19)

µeff(t) is the coefficient you would have to use to predict the intensity if you were to pretend that the beam were monoenergetic.

Note thatµeff(t) ≠ µ(t)

(effective) ( “instantaneous” )

µeff(t) describes attenuation from 0 to t. µ(t) describes attenuation from t to t + ∆t.

The instantaneous coefficient, which is most important for estimating the contrast produced by a differential thickness placed in the patient, is defined by

∆I(t) / I(t) = -µ(t) ∆t (20)

Now let’s do an example of a beam hardening calculation.

Calculate µ(t) for very large t for the case of a thin target spectrum which is shown in Figure 13 at t=0.

dIdk

C

k kmFigure 13

I(t) = ∫ (dI (k, 0 ) / dk) e-µ(k)t dk

where the integral goes from 0 to km. Assuming very large t, the beam will be very narrow in energy

Pull out e- µ(km)t

giving I(t) = Ce-µ(km)t ∫ e- [µ(k) - µ(km)] t dk

µ (k) = + = C 1 + C2 / k3

where C1 is fairly independent of energy.

µ(k) -µ(km) =C2 (1/k3 - 1/km3)

= C2 ((km

3 - k3) / k3km3)

The term e- [µ(k) - µ(km)]t is negligible except at k ≈ km

Therefore, letk = km - ∆

Plug in, omit 2 terms, set km~k in the denominator, and do the algebra. This results in

µ(k) - µ(km) = 3C2(km - k) / km4 = (Km - K)

where = 3C2/km4.

Then I(t) I(t) = Ce-µ(km)t ∫ e - km - k)t dk

Let y = -(k m - k) t

dy = t d k

Then I(t) = [Ce- µ(km)t ∫ e ydy ] /

where the integral is performed from - km t to 0. This integrates to

I(t) ={ (C/) e-µ( km ) t} / t

at very large t.

The instantaneous µ(t) is obtained by taking the log and applying equation 20,

ln I(t) = ln (C/ - µ(km)t - ln(t)

Taking the differential

(t)∆t = ∆I(t) / I(t) = - ( µ (km) + 1/t ) ∆t

So we get µ(t) = µ(km) + 1/t

As shown in one of the homework problems for the case of thick target bremsstrahlung the result turns out to be

µ(t) = µ(km) + 1/t2

Several years ago Peter Joseph used this relationship along with attenuation measurements at very large thicknesses as a means for determining kVp. This is done by plotting (t) as a function of t and determining (km).

The absorption properties of various elements having k-edges in the diagnostic region can be used to alter spectra for special purposes. These k edge filters may decrease the average beam energy in some cases leading to “beam softening”. This is illustrated in Figure 14 for the case of iodine and cerium filters which have been used to provide spectra for dual energy imaging where it is desired to have spectra peaked above and below the k-edge of iodine, the element to be imaged in angiography.

Figure 14

Since all attenuation coefficients in the diagnostic x-ray range may be expressed in terms of the basic Compton and photoelectric cross sections, any two material coefficients can serve as a basis set from which all other attenuation coefficients may be formed by linear combination.

For example the bone attenuation coefficient may be expressed as

µb = aP + bC (21)

where P and C are the photoelectric and Compton cross sections respectively.

The attenuation coefficients for plastic and aluminum are given by

µpl = cP + dC and

µal = eP + fC

Clearly we could solve for P and C in terms of µpl and µal to obtain

µb = gµpl + hµal

Because the P and C have different energy dependences, P proportional to 1/k3 and C independent of energy, by making measurements at two different energies it is possible to solve for the separate contributions of P and C to each part of the image and then to recombine these into tissue, bone, aluminum and plastic images. The advantage of describing tissue and bone in terms of aluminum and plastic is that aluminum and plastic phantoms can be used to calibrate various system parameters or remove nonlinearities.

For example c,d, e and f can be forced to provide the right answers for the known plastic and aluminum coefficients. Using these corrected values, corrected values for g and h, which are functions of c,d,e,and f, can be used to accurately calculate µb which, for example, varies with the amount of bone mineral.

The same sort of process is used for tissue. The exact procedure involving the appropriate calibration constants has not been presented above and depends on the details of the image acquisition apparatus.