X-RAY PRODUCTION
References:
Webster chapter 3
Christensen ( 3rd edition ) pages 10-22
Professor VanLysel’s notes
Attix chapter 9
Objectives: To understand;
1
Mechanisms for electron energy loss and associated equations
2
Derivation of X-ray energy distribution for thin and thick target bremsstrahlung
3
Effect of target angle on heat loading, effective focal spot and heel effect
4
Characteristic radiation and its dependence on kVp and atomic number
5
Definitions of exposure quantities
6
Exposure estimates
Wilhelm Conrad Roentgen
Site of the discovery, the Physical Institute of the University of Wurzburg,
taken in 1896. The Roentgens lived in apartments on the upper story, with
laboratories and classrooms in the basement and first floor.
Department of Radiology, Penn State University College of Medicine
Roentgen’s Laboratory
Department of Radiology, Penn State University College of Medicine
Radiograph of
Mrs. Roentgen’s hand
Department of Radiology, Penn State University College of Medicine
I-10 Radiograph of coins made by A.W. Goodspeed (1860- 1943) and William Jennings
(1860-1945) in 1896, duplicating one they had made by accident in Philadelphia on 22
February 1890. Neither Goodspeed nor Jennings claimed any priority in the discovery,
as the plates lay unnoticed and unremarked until Roentgen's announcement caused them
to review the images.
Department of Radiology, Penn State University College of Medicine
X-Ray Studios, like this one in
New York, opened in cities large
and small to take "bone portraits,"
often on subjects who had no
physical complaints.
Department of Radiology, Penn State University College of Medicine
The necessary apparatus was easily acquired. An evacuated glass
tube with anode and cathode, and a generator (coil or static machine),
combined with photographic materials could set anyone up in
business as a "skiagrapher."
Department of Radiology, Penn State University College of Medicine
Public demonstrations, like this one by Edison in May 1896, gave
the average person the opportunity to see his or her bones.
Department of Radiology, Penn State University College of Medicine
So excited was the public that each newly radiographed organ or system brought
headlines. With everything about the rays so novel, it is easy to understand the
frequent appearance of falsified images, such as this much-admired "first radiograph
of the human brain," in reality a pan of cat intestines photographed by
H.A. Falk in 1896.
Department of Radiology, Penn State University College of Medicine
FIRST FILM ANGIOGRAM
Hascheck and Lindenthal 1896
Contrast
chalk
Patient Preparation
amputation
Exposure time
57 minutes
( Reimbursement Rejected
by Medicare )
FIRST ANGIOGRAM
VOL. 1, NO.1, MAY,
1897
Early fluoroscopy exam
Consider the rough schematic of an x-ray tube shown in Figure 1.
k = photon energy
filament
e-
target
Electron kinetic energy
= Te
-
kvp
t
+
Electron energy loss is due to two sources
1
ionization
This accounts for about 99%
of the electron energy loss in the
diagnostic energy range and
shows up as heat.
2
bremsstrahlung This accounts for about 1%.
This process involves the acceleration
of the electron in the field of the nucleus
with a resultant emission of an x-ray as shown
in Figure 2.
Figure 2
X-ray
electron
The electron energy loss curve as a function of electron energy
looks very roughly as shown in Figure 3.
1/ Te
Notice that in the energy range of conventional x-ray tubes
( up to about 140 kVp) the rate of electron energy loss with
thickness is given by
dTe / dt = -b / Te
(1)
where b is a constant which depends on the target material.
We will use this approximation below in our calculation of the
thin target bremsstrahlung spectrum.
Thin Target Bremsstrahlung Spectrum
Consider the case in which the target thickness is sufficiently
small that no appreciable change in electron energy occurs
in the target.
Let N equal the total number of produced bremsstrahlung
x-rays, integrated over all angles. It is well known that the
differential number d2N of produced bremsstrahlung x-rays
per unit energy range dk per unit target thickness dt is given by
d2N / dkdt = C / kTe
(2)
where
C ~ Z / me2
where me is the electron rest mass and Z is the
atomic number.
It is interesting to note that the factor of 1 / me2
would be several orders of magnitude smaller if
ions were used as the bombarding particle in the
x-ray tube.
For a thin target the electron energy is relatively
constant and equation 2 can be integrated over
thickness to provide
dN / dk = C ∆t / kTe
(3)
For monoenergetic beams with only one energy bin
the total intensity I and the number of x-rays N are
related by
I = kN
(4)
For polyenergetic beams this relationship still
holds within each small (monoenergetic) energy
bin giving
dI / dk = k dN / dk
(5)
Therefore from equation 3
dN / dk = C ∆t / kTe
(3)
we get,
dI / dk = C ∆t / Te
(6)
This spectrum is shown in Figure 4 and indicates that dI / dk
is constant for all x-ray energies up to the incident electron energy Te.
dI
dk
k
Te
Now let’s use this spectrum to predict the shape of the spectrum
produced by a thick target.
Consider the electrons incident on a target as shown in Figure 5
e-
T0
t
T(t)
T(R)=0
target
R
t = distance penetrated into the target,
R = the maximum range of the electrons in the target
T(t) = the electron energy at depth t.
Since x-ray production depends on the electron energy, we need to
derive an equation for energy versus depth. We had
dTe / dt = - b / Te
(1)
By applying the following boundary conditions,
Te = 0 at t = R
and
Te = T0 at t = 0
(7)
it is easy to show that
R = T02 / 2b
(problem 1a) (8)
and
Te(t) = [ T02 ( 1 - t / R ) ]1/2
(problem 1b)
(9)
For any photon energy k there is a maximum
depth ( tmax (k) ) that an electron can penetrate and
still have enough energy to create a photon of energy k.
Energy conservation requires that
Te ( tmax(k)) = k
(10)
In one of the homework problems it is shown that
tmax(k) = R ( 1 - k2 / T02 )
(problem 1c)
(11)
With these relationships we can now calculate the
thick target spectrum.
From equations 2
d2N / dkdt = C / kTe
and 4
I = kN
we can write
2
dI
/ dtdk = C / Te
(12)
We can integrate this to obtain the desired
spectrum dI / dk.
Substituting into the upper limit from equation 11
tmax(k) = R ( 1 - k2 / T02 )
and into the integrand from equations 9
and 12
we get
Te(t) = [ T02 ( 1 - t / R ) ]1/2
d2I / dtdk = C / Te
In one of the homework problems this integral is
performed giving,
dI(k) / dk = 2CR( T0 - k) / T02
(problem 2)
= C(T0 - k ) / b = C( e * kVp - k ) / b
where e is the electron charge.
(15)
So far we have neglected the attenuation of the x-rays on
the way out of the target and x-ray tube window. This
spectrum is often used as the starting point for spectral
optimization programs. The spectrum is shown in
Figure 6.
Figure 6 shows the total spectrum as a sum of
contributions from several constant thin target spectra.
The contributions from the first three thin targets are
shown.
Figure 6
Notice that since, according to equation 3, dN / dk = C ∆t / kTe
the contribution from each thin target is inversely proportional to the
incident electron energy, the product of the height and energy extent of each
target, i.e. its total contribution to the spectrum, is constant for all targets
(again neglecting x-ray attenuation by the target which will be most severe
for the deepest targets).
dI
dk
Target
X-ray energy
Alteration of Spectrum Due to Filtration
1
2
3
4
Inherent filtration
3 mm Al added
3 mm Al and 10 cm water x4.0
3 mm Al and 20 cm water x15.0
Relative
number
of
photons
150kVp
Integrating to obtain the total intensity we obtain
(16)
Therefore, it can be expected that the total intensity in a thick
target bremsstrahlung spectrum is proportional to the square of the
x-ray tube voltage.
It is important to note that the population of any
particular energy bin is given by
dI(k) / dk = C( e * kVp - k ) / b
(17)
For energies near the peak x-ray energy the fractional increase in the
population can increase more rapidly than (kVp)2.
At low energies the increase is linearly proportional to kVp.
The bremsstrahlung angular distribution is dependent on energy as
Sketched below in Figure 7 for the cases of 34 keV and 2 meV.
34 keV
e
-
Thin
target
2 MeV
60o
The distribution shown is for a single thin target.
For the thick target case there is considerable multiple
scattering of the
electrons.
This results in considerable angular broadening of the
distribution leading to a much slower angular dependence
and a relatively constant intensity distribution at 90 degrees
where most diagnostic imaging is done.
The surface of the rotating x-ray anode is tilted at an angle  relative
to the line perpendicular to the incident electrons as shown below.

eanode
cathode
X-rays
Effective focal spot
is called the target angle and is typically on the order of
6 to 20 degrees.
As the target angle is increased, the area of the anode
which is subjected to electron bombardment increases,
thus increasing the instantaneous heat loading capabilities
of the target.
However, increasing target angle also leads to an increase
in the effective focal spot which is the size of the focal
spot projected in the direction of the detector.
Tubes with small effective focal spots, such as those used
for mammography, have low tube current ratings. Tubes
with larger effective focal spots are used for high heat load
applications such as computed tomography.
Attenuation of the x-rays on the way out of the target does introduce
an angular variation in intensity because of the fact that radiation
detected closest to the anode (target) side of the tube will have passed
through a greater thickness of target material. This is illustrated
below.
anode
cathode
It is necessary to correct for this effect in
applications such as dual energy imaging where
quantitative manipulation of the detected
x-ray intensity is performed..
In addition to the bremsstrahlung spectrum, x-rays are produced by
atomic electron transitions to vacant states. These characteristic x-rays
occur when incident electrons eject bound electrons as shown in the
next slide in Figure 10. The x-rays arising from transitions to the K
shell are designated by K , Kß etc depending on which shell the
transition originated from. Similarly transitions terminating on the L
shell produce L x-rays.
e-
Kß
K
eK
L
L
e-
M
N
Lß
A rough sketch of a spectrum including characteristic radiation and
target absorption is shown below in Figure 11.
The threshold energy required to produce K
radiation depends on the target material. A few
examples are
Iodine 33 keV
Tungsten 69.5 keV
Molybdenum 20 keV
The angular distribution of characteristic x-rays is
isotropic.
This fact, coupled with the fact that high energy
bremsstrahlung is predominantly in the forward direction
led Motz to suggest that quasi-monoenergetic beams could
be produced using electrons in the 1 meV range and
looking at the spectrum at 180 degrees relative to the
incident electron direction.
This approach has been pursued by Manning and was
reported in the October 1991 issue of Medical Physics.
The percentage of characteristic x-rays produced as a
function of energy increases with increasing energy.
For example
kVp
Percentage characteristic
80
10
100
19
150
28
This can be explained using our considerations of the
thin targets within the anode of the x-ray tube.
Basically, as the tube voltage goes up, the depth of
penetration increases and the fraction of the total
number of thin targets where characteristic x-rays are
energetically possible increases.
As the last target for which characteristic radiation is
possible gets closer and closer to the last target in
which bremsstrahlung is produced, the ratio of
characteristic to bremsstrahlung increases. This model
fits the data fairly well. This is described in more
detail.
Dependence of Percent
Characteristic Radiation on kVp
In answer to the question "why does the fraction of
characteristic radiation increase with electron
energy?", here are some hand waving arguments
which will attempt to quantitate this.
One guess is that it is purely a threshold
phenomenon. For Tungsten the threshold for k shell
electron emission is 69 kev. Below this value no k
characteristic photons will be emitted. As the
electron energy increases there will be electrons at
least in the first fraction of the electron range
which will have sufficient energy to eject a k shell
electron.
Let's test the following hypothesis. Suppose the
probability of ejecting an electron from the k shell
depends weakly on energy as long as we are above the
threshold.
Then if we imagine the thick target to be made up of thin
targets, the intensity of k characteristic x-rays will be
proportional to the fraction of targets with energies above
69 kev.
Ichar  fraction > 69 * total # of targets
We have shown that the Bremsstrahlung
intensity is equal for all of the thin targets.
Therefore the total bremsstrahlung intensity Ib
will be given by
Ib  total # of targets
Therefore the ratio rc of characteristic to
bremsstrahlung intensity is given by
rc = Ichar/ Ib  fraction > 69
This predicts that if we plot rc vs. fraction > 69 we
should get a straight line.
Now we can calculate the fraction > 69.
From equation 9,
Te(t) =( T02 ( 1-t/R))1/2 =( T02 - 2bt)1/2
where b is the constant in the electron energy
loss equation 1 and R is the range. If we guess
a value of 2.5 * 105 kev2 cm2/gm, we get the
results shown in Figure 12.
Figure 12
From this we can calculate the fraction of targets with energy > 69
for each kVp. We can plot this vs the percent characteristic intensity
versus kvp given in Ter Pogossian Table II-2. This is shown below.
Since we do get a fairly good linear relationship it
looks like our guess is reasonable, even though we
have not actually looked up the exact energy
dependence of k-shell electron ejection following
collision with another electron. We have neglected
the contribution from L radiation around 10 kev,
but much of this is filtered out by the inherent
filtration of the tube.
Dependence of Characteristic Radiation on Z
The bremsstrahlung cross section is proportional
to Z(kVp2). According to Johns and Cunningham,
the electron ionization loss process, which
dominates the energy loss process and precedes
the emission of characteristic radiation, is
proportional to Z0. Therefore, the ratio of
characteristic to bremsstrahlung will decrease like
1/Z as Z increases. This explains the greater
percentage of characteristic in the Molybdenum
spectrum at mammographic energies.
Before going further here are a few definitions.
X -Ray fluence = # X-Rays per unit area = N
Energy fluence (intensity)
I
X-Ray flux (fluence rate) = d / dt
(t= time)
Roentgen (R) = 2.58 * 10- 4 C / Kg in air
= 1 esu / cc
1 R ~ 2 * 1010 x-rays / cm2 at 35 keV
The number of x-rays per R depends on energy is given by
# x-rays / cm2 per R ~ 1 / (k * µk)
where µk is the attenuation coefficient associated with
energy deposition in the patient. This coefficient, which
we will discuss later, is large at low energies and
decreases with energy. The net effect is that the number of
x-rays required per Roentgen of exposure increase with
energy because of the smaller probability of interaction (
smaller µk ) and then decreases again as the amount of
energy deposited per x-ray increases and becomes the
dominant term in the denominator.
The curve looks approximately as shown below in Figure 14.
In this section we will present a rule of thumb for
estimating the exposure from a conventional x-ray tube
filtered with 2mm of aluminum. Such a tube will provide
approximately
10 mR per mAs
where 1 mR = 10- 3 R
at 1 meter at 100 kVp
(17)
Assuming a point focal spot
Exposure E ~ 1 / r2
where r is the source to patient distance, and
Intensity ~ kVp2
Therefore
E = (10 mR/mAs) * (1 meter/r)2 * (kVp/100)2 * mAs
(18)
E = (10 mR/mAs) * (1 meter/r)2 * (kVp/100)2 *
mAs
exposure at 2 meters, 50 kVp, 5mAs =
10(1/2)2 * (50 / 100)2 * 5 = 3.1 mR
Some Typical Diagnostic Exposures
Chest film
Mammogram
20 mR
≈ 1R surface
exposure per views
2 views/breast (300
mrads average glandular
dose)
CT
2R
Background
( excluding Radon )
125 mR / yr
It took a heck of a lot of x-rays but we finally
discovered what is wrong with you. You are suffering from
excessive exposure to radiation.