X-RAY PRODUCTION

1
X-RAY PRODUCTION References Webster
chapter 3 Christensen ( 3rd edition ) pages
10-22 Professor VanLysels notes Attix
chapter 9 Objectives To understand 1 Mechanism
s 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
MEDICAL PHYSICS 567 Part II
2
Wilhelm Conrad Roentgen
3
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
4
Roentgens Laboratory
Department of Radiology, Penn State University
College of Medicine
5
Radiograph of Mrs. Roentgens hand
Department of Radiology, Penn State University
College of Medicine
6
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
7
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
8
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
9
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
10
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
11
FIRST ANGIOGRAM

FIRST FILM ANGIOGRAM
Hascheck and Lindenthal 1896
Contrast chalk Patient Preparation
amputation Exposure time 57 minutes


( Reimbursement Rejected by Medicare )

12
VOL. 1, NO.1, MAY, 1897
13
Early fluoroscopy exam
14
Consider the rough schematic of an x-ray tube
shown in Figure 1.
k photon energy
filament
target
e-
t
Electron kinetic energy Te
kvp

-
15

• 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.
• 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.

16
Figure 2
X-ray
electron
17
The electron energy loss curve as a function of
electron energy looks very roughly as shown in
Figure 3.
1/ Te
18
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.
19
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)
20
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.
21
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)
22
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)
23
For polyenergetic beams this relationship still
holds within each small (monoenergetic) energy
bin giving dI / dk k dN / dk (5)
24
Therefore from equation 3 dN / dk C
?t / kTe (3) we
get, dI / dk C ?t / Te (6)
25
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
Te
k ?
Now lets use this spectrum to predict the shape
of the spectrum produced by a thick target.
26
Thick Target Bremsstrahlung Spectrum
Consider the electrons incident on a target as
shown in Figure 5
T0
T(t)
T(R)0
t
e-
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.
27
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)
28
it is easy to show that R T02 / 2b
(problem 1a) (8) and
Te(t) T02 ( 1 - t / R ) 1/2 (problem 1b)
(9)
29
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)
30
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.
31
From equations 2 d2N / dkdt C / kTe and 4
I kN we can write d2I / dtdk C /
Te (12)
32
We can integrate this to obtain the desired
spectrum dI / dk.
33
Substituting into the upper limit from equation
11 tmax(k) R ( 1 - k2 / T02 ) and into
the integrand from equations 9 Te(t) T02 (
1 - t / R ) 1/2 and 12 d2I / dtdk C /
Te we get
34
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 (15)
where e is the electron charge.
35
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.
36
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).
37
dI dk
Target
X-ray energy
38
Alteration of Spectrum Due to Filtration

1. Inherent filtration
2. 3 mm Al added
3. 3 mm Al and 10 cm water x4.0
4. 3 mm Al and 20 cm water x15.0

150kVp
Relative number of photons
39
Integrating to obtain the total intensity we
obtain (16)

40
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.
41
The bremsstrahlung angular distribution is
dependent on energy as Sketched below in Figure
7 for the cases of 34 keV and 2 meV.
Angular Distribution of Bremsstrahlung
34 keV
Thin target
e -
2 MeV
60o
42
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.
43
Effective Focal Spot / Heat Loading
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.
?
e-
anode
cathode
X-rays
Effective focal spot
44
??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.
45
The Heel Effect
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
46
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..
47
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.
Characteristic Radiation
48
e -
Kß
Ka
e -
K
L
e -
La
M
Lß
N
49
A rough sketch of a spectrum including
characteristic radiation and target absorption
is shown below in Figure 11.
50
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
51
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.
52
The percentage of characteristic x-rays produced
as a function of energy increases with
increasing energy. For example kVp Perc
entage characteristic 80 10 100 19
150 28
53
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.
54
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.
55
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.
56
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 gt 69
total of targets
57
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
58
Therefore the ratio rc of characteristic to
bremsstrahlung intensity is given by rc
Ichar/ Ib ? fraction gt 69 This predicts that
if we plot rc vs. fraction gt 69 we should get a
straight line.
59
Now we can calculate the fraction gt 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.
60
Figure 12
61
From this we can calculate the fraction of
targets with energy gt 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.
62
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.
63
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.
64
Some Definitions
Before going further here are a few
definitions. X -Ray fluence X-Rays per
unit area N Energy fluence (intensity)
X-Ray flux (fluence rate) d? / dt (t time)
65
Exposure
Roentgen (R) 2.58 10- 4 C / Kg in air
1 esu / cc 1 R 2 1010 x-rays
/ cm2 at 35 keV
66
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.
67
The curve looks approximately as shown below in
Figure 14.
68
Estimating Exposure
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)
69
Assuming a point focal spot Exposure E 1 /
r2 where r is the source to patient
distance, and Intensity kVp2
70
Therefore E (10 mR/mAs) (1 meter/r)2
(kVp/100)2 mAs (18)
71
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
Example
72
Some Typical Diagnostic Exposures
Chest film 20 mR Mammogram 1R
surface exposure per views 2
views/breast (300 mrads average
glandular dose) CT
2R Background ( excluding Radon ) 125 mR
/ yr
73
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.