2
RADIATION GENERATING EQUIPMENT
2.3 PARTICLE AND SOURCE-BASED RADIATION
2.1 PRODUCTION OF X-RAYS
What are the two different mechanisms by which X-rays are produced?
Question 2
Describe how bremsstrahlung and characteristic X-rays are produced.
Question 3
What is the energy spectrum of bremsstrahlung and characteristic X-rays?
X-rays are produced by bremsstrahlung and characteristic X-ray emission. Useful X-ray beams in imaging and therapy are typically all bremsstrahlung, except in mammography where characteristic X-rays are desirable.
Answer 2
Bremsstrahlung X-rays result from the Coulomb interaction between the incident electron and the nuclei of the target material. The incident electron is decelerated, losing kinetic energy in the form of bremsstrahlung photons (radiative loss).
Characteristic X-rays result from Coulomb interactions between the incident electrons and atomic orbital electrons of the target material (collision loss). The orbital electron is ejected from its shell and an electron from a higher level shell fills the vacancy. The energy difference between the two shells may either be emitted from the atom in the form of a characteristic photon (characteristic X-ray) or transferred to another orbital electron that is ejected from the atom as an Auger electron.
Answer 3
Bremsstrahlung X-rays have a spectrum of energies. The maximum X-ray energy is equal to the energy of the incident electron. The energy of the electron corresponds to the peak accelerating voltage. The most probable X-ray energy is about one third of the maximum energy.
Characteristic X-rays have discrete energies, corresponding to the energy level difference between atomic shells involved in the electron transition.
X-ray photons produced by an X-ray tube are the combination of the two. It has a continuous distribution of energies for the bremsstrahlung photons superimposed with characteristic radiation at discrete energies.
Question 5
Relative to the incoming electron beam, at what angle are bremsstrahlung X-rays produced?
Question 6
How does fluorescent yield change with Z?
Question 7
How efficient is X-ray production?
X-rays used in radiation oncology are usually classified as:
(a) Grenz ray therapy: treatment that uses very low energy (below 20 kV) X-rays. No longer in use.
(b) Contact therapy: operates at 40 to 50 kV and facilitates irradiation of accessible lesions at very short source to surface distance (SSD) (2 cm or less). A filter of 0.5- to 1.0-mm-thick aluminum is usually interposed in the beam to absorb the very soft component of the energy spectrum. It is useful for tumors not deeper than 1 to 2 mm.
(c) Superficial therapy: treatment with X-rays ranging from 50 to 150 kV. Varying thicknesses of filtration (usually 1–6 mm aluminum) are added to harden the beam to a desired degree. The SSD typically ranges between 15 and 20 cm. Useful for irradiating tumors confined to about 5 mm depth.
(d) Orthovoltage therapy: treatment with X-rays ranging from 150 to 500 kV, and filtered with 1 to 4 mm copper. The SSD is typically 50 cm. Useful for tumor less than 2 to 3 cm deep.
(e) Supervoltage therapy: X-ray therapy in the range of 500 to 1,000 kV, filtered with 4 to 6 mm copper.
In contrast, diagnostic X-rays are usually in the 10 to 150 kV range.
Answer 5
In the kilovoltage energy range, X-rays are produced uniformly with regards to direction, shielding around the target produces an X-ray beam at 90°. In the megavoltage energy range (1–50 MV) most photons are produced in the direction of electron acceleration (forward direction 0°).
Answer 6
The fluorescent yield ω gives the ratio of fluorescent (characteristic) photons emitted per vacancy in a shell to the number of Auger electrons. It ranges from zero for low-Z atoms through 0.5 for copper (Z = 29) to 0.96 for high-Z atoms with K shell vacancies, which are the most prominent sources of characteristic X-rays.
Answer 7
Efficiency of X-ray production is proportional to the atomic number (Z) of target and energy of the electrons. Efficiency is less than 1% for X-ray tubes operating at 100 kVp (99% of input energy is converted into heat). Efficiency improves considerably for megavoltage accelerator beams (30%–95%, depending upon energy).
Question 9
Why is tungsten the chosen material for the cathode and anode of an X-ray tube?
Question 10
What are thin and thick targets, respectively, for an X-ray tube?
Question 11
What is the purpose of the added filtration placed externally to the X-ray tube?
A typical X-ray tube consists of a highly evacuated glass envelope, a cathode (negative electrode, tungsten filament), that produces the electrons, and an anode (positive electrode, tungsten target attached to a thick copper rod).
Answer 9
The choice of tungsten for filament (cathode) and target (anode) is based on its high melting point (3,370°C), due to the heat absorbed, and a high atomic number (Z = 74) to boost efficiency of X-ray production.
Answer 10
A thin target has a thickness much smaller than R, the range of electrons of a given energy, while the thickness of a thick target is of the order of R.
Answer 11
The purpose of the added filtration is to increase the proportion of high-energy X-rays in the beam by absorbing the lower energy components of the spectrum. Thus, the transmitted beam “hardens” (ie, it achieves higher average energy and therefore greater penetrating power). Another way of improving the penetrating power of the beam is by increasing the voltage across the tube. Since the total intensity of the beam decreases with increasing filtration and increases with voltage, a proper combination of voltage and filtration is required to achieve desired hardening of the beam as well as acceptable intensity.
Question 13
If a 2 mm thickness of material transmits 25% of a monoenergetic beam of photons, calculate the half-value layer (HVL) and μ of the beam.
Question 14
What is beam hardening?
Question 15
What is heel effect?
HVL is defined as the thickness of material required to reduce the intensity of a beam to one half of its initial value. A related quantity is the linear attenuation coefficient μ, where HVL = 0.693/μ.
Answer 13
Answer 14
The lower energy photons of the polyenergetic X-ray beam will be preferentially removed from the beam while passing through matter. The shift of the X-ray spectrum to higher effective energies as the beam transverses matter is called beam hardening. For polyenergetic beams, this causes the second half-value layer (HVL) to be greater than the first HVL.
Answer 15
Since the X-rays are produced at various depths in the target, they suffer varying amounts of attenuation in the target. There is greater attenuation for X-rays coming from greater depths than those from near the surface of the target. Consequently, the intensity of the X-ray beam decreases from the cathode to the anode direction of the beam. This variation across the X-ray beam is called the heel effect. The effect is particularly pronounced in diagnostic tubes because of the low X-ray energy and steep target angles. The problem can be minimized by using a compensating filter to provide differential attenuation across the beam to compensate for the heel effect and improve the uniformity of the beam.
Question 17
The actual anode focal area for a 20° anode angle is 4 mm (length) by 1.2 mm (width). What is the projected focal spot size at the central axis?
Question 18
For an X-ray beam, what is inherent filtration?
Question 19
What is the difference between tube current and filament current?
Question 20
How do tube current (mA) and tube voltage (kVp) affect X-ray intensity?
The size of the target area from which the X-rays are emitted is called the focal spot. The size of the focal spot depends on the size of the tungsten filament of the cathode. In diagnostic radiology, the focal spot should be as small as possible to produce sharp radiographic images. But smaller focal spots generate more heat per unit area of target and thus limit currents and exposure. Diagnostic tubes usually have two separate filaments to provide “dual focus,” a small and large filament for small and large images.
Answer 17
An angled target is used to reduce the effective size of the focal spot. The effective focal spot width is equal to the actual focal spot width since it is unaffected by the angle of the anode. The effective focal spot length = actual focal spot length × sin θ, where θ is the anode angle.
So for this problem, effective length = 4 mm × sin 20 = 1.36 mm.
Effective focal spot size = 1.36 mm × 1.2 mm = 1.632 mm2
Thus, a large electron beam can be used, to spread the heat on the target, but a small focal spot can be produced for sharp images.
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