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.
x
N(0)
N(x)
N(x + ∆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)
= 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)
(1)
The linear attenuation coefficient µL is defined as,
µL = total  D (c m - 1)
Then from equation 1
∆N = - µL ∆x N ( x )
For small ∆x
dN = - µL dx N(x)
dN/N = -µL dx
(2)
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 -
µ x
L
(3)
This can also be expressed as
N(x) = N(o) e - ( µL/ )( x )
(4)
Where
( µL /  ) = µm
and
px = t
mass attenuation coefficient (cm2 / gm)
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)
I(x) = I(o) e - µLx
(6)
For that case,
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(0)
Slope = - L
Figure 3
ln N(x)
x
Contributions to UL
There are five 5 possible photon interactions:
1. Rayleigh (coherent scattering)
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
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.
bout 5% of cross-section in the Dx region
Elastic event, photon loses essentially none of its energy
Photon interacts with the entire atom
Atom recoils just enough to conserve momentum
Photon redirected through a small angle
Photons are only removed from beam in narrow geometry
Contributes nothing to dose
Contribution to µm ~ Z / k2 (cm2 / g)
(9)
The basic interaction is shown in Figure 4
ek
•
e-
+
kc
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).
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
e-
k
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’ =
with
k / {1 + (k / moc2) (1 - cos )}
moc2 = 0.511 MeV
(12)
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
Primary
intensity
Figure 7B
Detected Primary
Intensity + scattered
intensity
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)
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.
Monoenergetic slope = - L
polyenergetic
ln N(x)
Figure 9
x
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
x=0
Figure 10
dI/dk
Large x
k
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)
µ
e eff(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 that
µeff(t) ≠
(effective)
µ(t)
( “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.
C
dI
dk
Figure 13
k
km
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
giving
e-
µ(km)t
I(t) = Ce-µ(km)t ∫ e-
[µ(k) - µ(km)] t
µ (k) = +  = C 1 + C2 / k3
where C is fairly independent of energy.
dk
µ(k) -µ(km) =C2 (1/k3 - 1/km3)
= C2 ((km3 - k3) / k3km3)
The term ek ≈ km
[µ(k) - µ(km)]t
is negligible except at
Therefore, let
k = 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 kedges 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
and
µpl = cP + dC
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.