Physics of Diagnostic X-rays For English please visit: http://www.thefreedictionary.com/
For More information about X-rays: http://www.youtube.com/watch?v=7hwzG0YI9ms http://www.youtube.com/watch?v=Bc0eOjWkxpU http://www.youtube.com/watch?v=I3s5HFQ2YME http://www.youtube.com/watch?v=megO38kZG8g (interaction of X-rays) http://www.youtube.com/watch?v=n9FkLBaktEY http://www.youtube.com/watch?v=_ERp1blPOsU http://www.youtube.com/watch?v=0vUQgKNpy2Y
Objectives: -
Electromagnetic wave spectrum Principles of x-rays production Atomic generation of x-rays X-rays spectrum and absorption Interaction of x-rays with matter Making an x-ray image (Rӧntgenogram) Dosimetry units Biological effects of x-rays
References: 1- Medical Physics textbook by Cameron 2- Physics in Biology and Medicine, Third Edition by Paul Davidovits 3- Physics of the Human Body, by Irving P. Herman
Electromagnetic radiation: Electromagnetic radiation (EM radiation or EMR) consists of two electric (horizontal plane) and magnetic (vertical plane) fields perpendicular to each other and perpendicular to the propagation direction as shown in figure.
Electromagnetic radiation is characterized by: 1- In space, it travels with the speed of light and undergo refraction, attenuation, diffraction, and reflection. 2- Wavelength (λ Lambda, measured in Length unit (nm) ) is the distance between any two points have the same phase. 3- Frequency (f) is the number of cycles or vibrations undergone during one unit of time (Hertz (Hz) or S-1) ( Speed of light C = λ (Wavelength) x f (frequency)) 4- Period (T) is the time for one complete cycle (S). 5- Energy of electromagnetic radiation is calculated by (E=hxf) (h is Planck’s constant, h= 6.5821 × 10-16 eV s)
X-rays were discovered by Wilhelm Conrad Röntgen , which have a wavelength in the range of 0.01 to 10 nm, corresponding to energies in the range of 120 eV to 120 keV (please Calculate E (eV) = 1240/λ(nm)) . Electromagnetic spectrum is shown in the underneath Figure and its possible applications as a function of X-ray energy.
X-rays from about 0.12 to 12 keV are classified as “Soft" x-rays (low energy), and from about 12 to 120 keV as “Hard" x - rays, due to their penetrating abilities. Soft X-rays hardly penetrate matter at all
X- rays production X-rays is produced in the traditional X-ray tube, See underneath figure when high speed (High energetic ) electrons collide with a high atomic number element (Target). (Cathode)
Focal spot (Anode) Target
Evacuated Electron beam The elements of X-rays tube are: 1- Evacuated Glass Envelop: to avoid the production of collision between high energetic electrons and air molecules, which give rise to production of undesired radiation. 2- Heated filament (Cathode) emits electrons by thermoionic emission, the number of electrons is controlled by the electric current in the filament (Control the X-rays intensity). The higher the electric current, the higher the number of emitted electrons, the higher the X-rays intensity.
3- High voltage to accelerate the emitted electrons (Control both the X-rays energy and intensity), Explain the effect of potential deference on the intensity and energy of X-rays.
4- Anode: it is the component in which the x-rays are produced in a very small area on the surface of the anode known as the focal spot (where the accelerated electrons collide with the target).
4-1- Most x-ray tubes use tungsten (18474W110), which has an atomic number (number of protons within its nucleus) of 74, ability to maintain its strength at high temperatures. Atomic number is defined as “The number of protons found in the nucleus of the atom”.
4-2 More than 99% of energetic electron that striking the anode is converted into heat which may damage the anode. The material for the anode is selected to enhance these functions. Furthermore X-ray tubes are equipped with a cooling system and rotating anode to overcome the overheating of the anode.
- Efficiency of x- rays production is defined as “The total X-rays energy expressed as a
fraction of the total electrical energy imparted into the surface of the anode”. - The two factors that determine x-rays production efficiency are the voltage applied to the tube (KVp) and the atomic number of the anode (Z) as given by
-
Efficiency = KVp x Z 10-6
In x-ray tube the number (quantity) of electrons accelerated toward the anode depend on the temperature of the filament. The intensity (number of X-rays photons) of the produced
x-rays when electrons strike the anode depends on the anode material. In general, the higher the atomic number (Z) and a high melting point of the target, the more efficiently x-
rays are produced. KeV is “The energy an electron gains in going across a potential difference of 1000 volt = 1kilo volt”. KeV = 1.6 x 10-16 (joules) = 1.6 x 10-9 (erg)
Atomic generation of X - rays In X-ray tube, X-rays are produced by two different mechanisms; these are: A- Bremsstrahlung (Continuous or white radiation): It is produced when energetic electrons loss some or all of their kinetic energy (1/2mv2, where v is the velocity of electron)because the attraction force between electrons (negative Charge) and the nucleus of the target atom (Positive Charge), the difference in energy will emitted as X-rays with wide spectrum (Why?) 𝐸2 𝐸1>𝐸2 Let the energy of incident electron is 𝑬1 and energy after and deflection is E2, 𝐸1>𝐸2, then the emitted X-rays has Energy : 𝑬 = 𝒉𝒇 = 𝑬1-𝑬2
𝐸1
The maximum X-rays energy is emitted when an energetic electron is completely stopped with in the target, 𝑬2 = 0. All kinetic energy of the electron will be converted into X-rays
𝒉𝒇 X-ray photon
𝑬 = 𝒉𝒇 = 𝑬1 = eVp (e is the charge of electron) The minimum wavelength of maximum X-ray energy, min (nm)
hc 1 1240 (C = 3x108 m/S. , is Velocity of light) e Vp Vp(V )
The Bremsstrahlung radiation is known as white radiation since it contains a wide range of wavelengths (Frequencies and energies).
•
The Bremsstrahlung intensity produced for a given number of electrons striking the anode depends upon two factors:
•
(1) Atomic number (Z) of the target (the higher the atomic number, the greater the deceleration of the electrons).
•
(2) Kilovolt peak (KVp) (The faster the electrons, the more they penetrate into the region of the anode’s nucleus), The higher the electron energy, the higher the energy can release.
λmin
2- Characteristic radiation involves a collision between high-speed (energetic) electrons and the orbital electrons in the target’s atoms. The interaction can occur only if the incoming electron has a kinetic energy greater than the binding energy of the electron within the atom. When a primary fast moving electron strikes (knocks) an inner bound (K) electron in a target atom and knocks it out of its orbit, the vacancy in the K shell is filled almost immediately by an electron from an outer shell either L or M which, cause energy levels in the target to change. In this process, a Characteristic K x-ray photon is emitted which have discrete energies.
For Tungsten, an L shell electron filling K shell vacancy results in characteristic X-ray radiation of energy hf = EK-shell - EL-Shell = 69.5 KeV-10.2 keV = 59.3 keV The same held for M-Shell
Example: A K-shell electron of binding energy of 69.5 keV is removed from a tungsten atom and is replaced by an shell electron of binding energy 12.1 keV. What is the energy of the characteristic x-ray that is emitted? Answer: The characteristic x-ray emitted has energy of: E = 69.5-12.1 = 57.4 keV
Example: X-ray tube is operated at a potential of 62000 V. Find the minimum wavelength limit of the x rays emitted. 1240 Min (nm) 1240 / V (Volt )
62000
0.02nm
Example: An energetic electron of initial energy of 50 KeV and final energy of 20 KeV, Calculate the emitted photon energy after deceleration? E = 50-20 = 30 KeV
Taif University
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An x-ray photon emitted when an electron falls from the L shell to replace the vacancy in the K shell is called: Kα x-ray, and that emitted when an electron falls from the M shell to the K shell is called Kβ x-ray. Characteristic x- rays photon (either α or β) are produced with energy equal to the difference in the energies of the outer and inner shell electrons. Characteristic x-rays are of little use at present except in mammography machine dedicated to breast images. Electron binding energy is a measure of the energy required to free electrons from their atomic orbits
Graph plotting of x-rays produced by a modern x-ray generator (wavelength versus relative intensity) LK
MK (Broad smooth curve) (80-90%) [Multiple(wide range) wavelengths]
6
(10-20%)
Interaction of X-rays with matter X-rays interaction with matter is strongly correlated with both the matter structure (atomic number) and Xrays energy, see the opposite figure. Photoelectric effect (PE): Occurs when the incoming xray encounters and transfers all of its energy to a K
electron and remove it from the atom. Compton effect (CE): Occurs when x-ray encounters an electron from an outer M - shell Pair production (PP): Occurs when a high energy x-ray photon enters the intense electric field of the nucleus.
Photoelectric effect (PE)
Low x-ray energy and high atomic number materials
Photoelectric effect (PE): when the incoming x-ray (of energy = hf = hc/λ), encounters and transfers all of its energy to an electron of an inner K shell electron [therefore disappears (absorbed)]. This inner bound electron of mass (me) will escape from the atom. The ejected electron is called photoelectron (PE), its binding energy must be lower than that of the incoming x-ray photon. Vacancy created in K-shell will be filled by electron from the L-shell, producing characteristic radiation (Energy of characteristic photon is equal to energy difference between the two shells). PE is more useful than the Compton one, because it permits to see bone and other heavy materials.
Compton effect (CE): High x-ray energy and low atomic number materials Recoil electron
Higher energy shorter wavelength
Lower energy Longer wavelength Compton effect (CE): Occurs when x-ray encounters an electron from an outer M shell (loosely bound electron). This electron is ejected and the remaining energy of the incident x-ray will be re-emitted immediately in the form of a (Compton) scattered x-ray photon. The direction of the emitted x-ray can be anywhere other than that of the original x-ray. Compton scattering results in: 1) Ionization events, 2) recoil electron, 3) positive ion and 4) lower energy scattered photon.
In Compton interaction the incoming x-ray photon will be scattered with less energy and longer wavelength (Compton x-ray).
Pair production (PP): x-ray photon energy must exceed (more than) 1.022 MeV. PP occurs in high atomic number materials. It leads to formation of two charged particles from a single high-energy photon. Formation of two charged particles from a single high-energy photon E=(me- + me+)C2 = 1.022 (MeV)
X-rays > 1.022 MeV
Pair production (PP): When the energy of a high energy x-ray photon (exceeds 1.022 MeV) enters the intense electric field of the nucleus, this energy will be converted into two particles e- (electron) & e+ (positron) (i.e., energy of x-ray photon is converted into mass). The remaining energy will be given to the particles as kinetic energy. When e+ spent its K.E in ionization it annihilates (collide) with an electron and produce annihilation photons. Both then vanish and their mass energy appears as two photons of 511 KeV each called annihilation radiation, which go in opposite directions.
Attenuation of X-rays is the Reduction in the intensity of x-ray due to absorption and scattering of some of the photons out of the x-ray beam”. The attenuation of x-ray is shown in the underneath figure. It contains: (1) A narrow beam of x-rays produced with an x-ray source this beam goes through a collimator (a lead plate with hole in it) (2) Absorber (sheets of aluminum introduced into the beam pass) (3) X-ray detector to measure the beam’s intensity. Scattered x-rays
x
I Io X- ray source
(Un-attenuated Incident)
Narrow beam Collimator
(Attenuated) Transmitted)
X-ray detector
(T%) Absorber (attenuating material)
I = Ioe-μx
An exponential equation describes the attenuation curve for a monoenergetic x-ray beam (photons of a same energy).
[Graph of the transmitted intensity versus the thickness of attenuator]
Intensity of x-ray decreases exponentially [intensity decreases with increasing thickness of absorbing material].
-Linear attenuation coefficient (cm-1) is the probability of interactions per unit length NA
μ a σ
A
ρ
= mass density [g/cm 3], NA = 6.02 x1023 [1/mol] (Avogadro’s number), A = Atomic Weight [g/mol] and a = Atomic cross section [cm2]. Linear attenuation coefficient is linearly related to density of the matter.
Since I = Io e-μx I/Io= e-μx ln I/Io= -μx ln Io/I= μx
Example: If the intensity of the un-attenuated x-ray beam is 150 (count/sec) and the intensity after attenuation is 30 (count/sec). Find the thickness of the attenuator, knowing that the linear attenuation coefficient of the attenuator is 0.5 (cm-1). Answer: ln Io/I = μx ln (150/30) = 1.6094 x = 1.6094/0.5 = 3.2188 (cm)
Example: Calculate the thickness of the lead (μ = 0.58 cm-1) used to attenuate the x-ray intensity to tenth its original value. Answer:
ln Io/I= μx I/Io =1/10 Io/I = 10 ln 10 = 0.58 x = 2.3025 Thickness= x = 2.3025/0.58 = 3.9699 ≈ 4 (cm)
The half value layer (HVL) for an x-ray beam is defined as the thickness of a given material that will reduce the beam intensity by one-half. HVL is related to the linear attenuation coefficient (μ) by the following equation:
HVL = ln2/μ = 0.693/µ
(cm)
The equivalent energy of an x-ray beam is determined by its half-value layer; it is the energy of a monoenergetic x-ray beam with the same half-value layer. Example: For a certain material if the linear attenuation coefficient is 0.562 (cm-1). Calculate the half value layer. Answer: HVL= 0.693/0.562= 1.233 (cm) Example: If the linear attenuation coefficient is 0.05 (cm-1). Find the HVL Answer: HVL = ln2/0.05 = 13.8629 ≈ 13.9 (cm)
Example: What is the linear attenuation coefficient (μ) of a given material. if you know that the
thickness of this material reduced the x-ray beam intensity to its half value is 15 (cm). Find the thickness (x) needed from this material to reduce the beam intensity (I) to the tenth of its original value (Io).
Answer: HVL = 15 (cm) then Io/I = 10 I/Io = 1/10 HVL = ln 2/μ then μ = ln 2 / HVL = 0.693 /15 = 0.0462 (cm-1)
ln (Io/I) = μx then x = ln (Io/I) / μ = ln (10)/0.0462 =2.3025/0.0462 = 49.839 (cm)
Mass attenuation coefficient (μm) Mass attenuation coefficient (μm) is used to remove the effect of density (ρ) when comparing attenuation in several materials. μm equals to the linear attenuation coefficient (µ) divided by density (ρ= g/cm3) of the material
µm = µ/ρ
(cm2/g)
µm emphasizes that the mass is primarily responsible for attenuating the xrays. μm is independent of physical state of material. Example: For a certain material if the linear attenuation coefficient is 0.562 (cm-1) and its density is 320 (g/cm3). Calculate the mass attenuation coefficient of this material.
Answer: µm =µ/ρ=0.562/320=0.001756=1.756x10-3
(cm2/g)
Making an x-ray image
Processing X-ray source Collimator
Object
Image
Film (wrapped in black paper )
X-rays images are shadows cast on film by various structures in the body. (Different type tissues attenuate X-rays differentially)
The x-ray imaging chain is a fairly simple system consisting of the source, collimator (which blocks the stray x-rays), object which the x-rays interact with, and the detector (photographic film in this case).
X- ray images
Soft tissues Bone Hand X-rays
Head X-ray
Bone is very dense and absorbs or attenuates a great deal of the x-rays. The soft tissues
around the bones is much less dense than bone and that is why it attenuates or absorbs less x-ray. To obtain a satisfactory x-ray image of thick parts such as the abdomen and hips
it is necessary to reduce the scattered radiation at the film. The amount of scattered radiation at the film depends on the energy of the x-ray and the thickness of the tissue that the x-ray beam passes through. The thicker the tissue the greater the scatter.
Geometric Unsharpness (penumbra) Penumbra: is lack of sharpness of the film, It is the blurred edge of the shadow. The width of the Penumbra can be calculated from this equation:
P = (f × b)/a Penumbra (P) can be reduced by increasing the distance from the x-ray tube to the patient, and minimized the distance between patient and film (detector).
The most effective parameter is the focal spot, which should be minimized to, minimized the penumbra (p), See opposite figure.
Improve the image quality by using Grid to eliminate the scattered X-rays photons
X-rays source
X-rays photons
Patient
Lead stripes (absorb )scattered X-rays
Grid
Film (detector) Grid used to reduce the amount of scattered radiation. It consists of alternating thin lead stripes and wide plastic stripes. The un-scattered x-rays pass through the plastic stripes, while most of the scattered x-rays are absorbed by the lead stripes.
Radiation Dosimetry The effect of ionizing radiation on matter (especially living tissue) is more closely related to the amount of energy deposited rather than the charge. This is called absorbed dose. Gray (Gy=J/Kg) is the Standard International (SI) unit of absorbed dose. [1 gray=100 rad]. Gray is defined as” The amount of radiation required to deposit energy of 1 Joule in 1 Kilogram of any kind of matter”. Rad [Radiation absorbed dose] It is the traditional unit used to measure radiation dose = 0.01 Joule of energy deposited per kilogram of mass (J/Kg) = 100 erg of energy per gram of mass. Equivalent dose is the measure of the biological effect of radiation on human tissues [it is measured in Sievert (Sv)]. For x-rays, equivalent dose is equal to absorbed dose. Rem [Roentgen equivalent man] it is the traditional unit used for measuring effective (or “equivalent”) dose of radiation. For x-rays rem = rad or 0.01 (J) of energy deposited per Kg. [1 Sv =100 rem]
Biological effects of radiation Direct action
Indirect action
Radiation damage to cells may occur directly from a radiation hit on the critical target, this results in ionization and excitation of the molecules inside the nucleus of cells
Creation
of
free
radicals
(reactive chemicals) due to
interaction of radiation (xrays) with water and leads to
formation of H2O2 which is a Somatic (Body cells)
Genetic (Germinal cells)
genetic effects in children Cancer is the somatic of irradiated parents. effect of most concern in radiation risk assessment. For chronic exposure to radiation over many years, the late somatic effects may occur during the irradiation period.
very
powerful
oxidative
agents causing damaging to DNA
or
undesired
modification in DNA.
Biological effects of radiation - Most Human Body cells can be damaged by ionizing radiation - Different cell can receive different amounts of radiation before damage can occur “Immature cells (cells that undergo rapid cell division) have a higher
sensitivity to radiation than mature cells” - Lymphocytes (white blood cells) are the most radiosensitive - Reproductive cells (sperm and ova) are highly sensitive - Linings and covers of body organs are moderately sensitive - Muscle and nerve cells are least sensitive.
