X-RAY PRODUCTION References: Webster chapter 3 Christensen ( 3rd edition ) pages 10-22 Professor VanLysel’s notes Attix chapter 9 Objectives: To understand;

X-RAY PRODUCTION

References: Webster chapter 3Christensen ( 3rd edition ) pages 10-22Professor VanLysel’s notesAttix 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 byH.A. Falk in 1896.

Department of Radiology, Penn State University College of Medicine

FIRST ANGIOGRAM

FIRST FILM ANGIOGRAM

Contrast chalk

Patient Preparation amputation

Exposure time 57 minutes

( Reimbursement Rejected by Medicare )

Hascheck and Lindenthal 1896

VOL. 1, NO.1, MAY, 1897

Early fluoroscopy exam

kvp +-

target

filament

t

e-

Electron kinetic energy= Te

k = photon energy

Consider the rough schematic of an x-ray tube shown in Figure 1.

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.

X-ray

electron

Figure 2

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.

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)

Thin Target Bremsstrahlung Spectrum

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 / me

2 would be several orders of magnitude smaller ifions 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.

Now let’s use this spectrum to predict the shape of the spectrum produced by a thick target.

dIdk

k Te

Consider the electrons incident on a target as shown in Figure 5

t = distance penetrated into the target, R = the maximum range of the electrons in the targetT(t) = the electron energy at depth t.

e-T0

target

t

R

T(t) T(R)=0

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

d2I / 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

Te(t) = [ T02 ( 1 - t / R ) ]1/2

and 12d2I / dtdk = C / Te

we get

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.

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 / kTethe 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).

X-ray energyTarget

dIdk

1 Inherent filtration2 3 mm Al added3 3 mm Al and 10 cm water x4.04 3 mm Al and 20 cm water x15.0

Relative number of photons

Alteration of Spectrum Due to Filtration

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 thex-ray tube voltage.

It is important to note that the population of anyparticular 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 thepopulation can increase more rapidly than (kVp)2.

At low energies the increase is linearly proportional to kVp.

34 keV

60o

The bremsstrahlung angular distribution is dependent on energy as Sketched below in Figure 7 for the cases of 34 keV and 2 meV.

e -Thin target

2 MeV

The distribution shown is for a single thin target.

For the thick target case there is considerable multiple scattering of theelectrons.

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.

cathode anode

Effective focal spot

e-

X-rays

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.

cathodeanode

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-raysoccur 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 -

e -

e -

K

Kß

K

L

M

N

L

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 keVTungsten 69.5 keVMolybdenum 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.

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.

Dependence of Percent Characteristic Radiation on kVp

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 =( T0

2 - 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.

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.

Dependence of Characteristic Radiation on Z

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)

I

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

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

Some Typical Diagnostic Exposures

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.