5.1, 5.5-5.10: Ionization, Interactions of Photons with Matter, Coherent Scattering, Photoelectric Effect, Compton Effect, Pair Production, & Relative Importance of Various Types of Interactions
- As radiation pass through media, it may strip electrons from atoms. This kind of radiation is characterized as ionizing radiation, and ionization is the process by which neutral atom acquires a positive or negative charge.
- Directly ionizing radiation are charged particles that can produce ionizations by themselves, whereas indirectly ionizing radiation are uncharged particles that liberate charged particles (directly ionizing radiation) which can then produce ionizations.
- Electrons that result from ionization are produced primarily by the following processes: photoelectric effect, Compton effect, and pair production.
- These processes contribute to the attenuation of an incoming beam, and depending on varying factors, the incoming beam will attenuate by each interaction by a different amount.
- The fraction of photons that will be absorbed by that particular interaction is represented by an attenuation coefficient, and the sum of each coefficient make up the total attenuation coefficient.
- The total attenuation coefficient is given by the following equation: μ/ρ = τ/ρ + σ_compton/ρ + π/ρ + σ_coherent/ρ (total attenuation coefficient = attenuation coefficient for: photoelectric effect + Compton scatter + pair production + coherent scatter).
- During the photoelectric effect, the incident photon is absorbed by an atom and an orbital electron is ejected in the process possessing energy equal to h * v (incident photon energy) – E_b (electron binding energy). The orbital shell vacancy will leave the atom in an excited state, and characteristic x-rays will emit as electrons transition down from the outer shells to fill the vacancy.
- The incident photon must have an energy that is equal to or greater than that of the electron binding energy to undergo the photoelectric effect. As a result, the incident photon may be limited in photoelectric interaction depending on its energy.
- However, the mass photoelectric attenuation coefficient (τ/ρ), or the probability of the photoelectric effect, is proportional to Z^3/E^3. Thus, an increase in the incident photon energy will not necessarily increase interaction by the photoelectric despite more orbital shells becoming available for interaction.
- A line graph comparing the τ/ρ (y-axis) with the photon energy (x-axis) will trend predominantly as a downward slope.
- Only when the incident photons are of enough energy to “unlock” new orbital shells, will the slope momentarily become positive before trending downwards again.
- This sudden change in slope upwards is referred to as an absorption edge which corresponds to the electron binding energies.
- The photoelectric effect is the predominant method of photon interaction when using low energy photons such as in imaging.
- The difference in the quantity of photons that reach the image receptor is what ultimately provides us an image.
- The differences in Z of different tissue types result in varying probabilities of photoelectric interaction by the incident photons, and this will correlate to the number of photons that will reach the image receptor to generate an image.
- The dependence of τ/ρ on Z^3 is taken advantage of in imaging by using high-Z contrast materials to further distinguish between different tissue types.
- An incident photon that has an energy far greater than the electron binding energy can eject the orbital electron by the Compton effect.
- The interaction resembles a collision between two marbles where the photon and the Compton electron are scattered by the angles θ and Φ respectively relative to a line extended from the incident path of the photon.
- The energy of the photon (E_p) is transferred to the Compton electron in the form of kinetic energy, and the Compton electron energy (E_c) can be calculated for by the following equation: E_c = E_p * {[α * (1 – cos Φ)] / [1 + α * (1 – cos Φ)]}.
- α is equal to E_p / E_e, where E_e is equal to the rest energy of the electron (0.511 MeV), and so α = E_p / 0.511 MeV.
- A photon that makes a direct hit with an electron will eject a Compton electron that will travel forwards (θ = 0°), and the scattered photon will travel backwards (Φ = 180°).
- Such a scenario will result in the Compton electron receiving the maximum energy possible (E_max) and the scattered photon left with the minimum energy possible by that interaction.
- A photon that makes a grazing hit will emit the Compton electron at a right angle (θ = 90°) receiving the minimum energy possible by that reaction, while the photon continues nearly unperturbed in its incident direction (Φ = 0°).
- The greater the energy of the incident photon, the greater will be the amount of energy transferred onto the Compton electron relative to the incident photon energy.
- This difference can be seen in the comparison of two photons of energies 51.1 keV and 5.11 MeV. The maximum energy transfer results when a photon interacts by a direct hit and thus, θ = 0° and Φ = 180°.
- α_0.0511 MeV = E_p / 0.511 MeV; α_0.0511 MeV = 0.1.
- E_cmax 0.0511 MeV = E_p * {[α * (1 – cos Φ)] / [1 + α * (1 – cos Φ)]}; 0.0511 MeV * {[0.1 * (1 – cos 180)] / [1 + 0.1 * (1 – cos 180]}; 0.0511 MeV * (0.2 / 1.2) = 0.00852 MeV.
- 0.00852 MeV / 0.0511 MeV = 16.7%.
- α_5.11 MeV = E_p / 0.511 MeV; α_0.511 MeV = 10.
- E_cmax 5.11 MeV = E_p * {[α * (1 – cos Φ)] / [1 + α * (1 – cos Φ)]}; 5.11 MeV * {[10 * (1 – cos 180)] / [1 + 10 * (1 – cos 180]}; 5.11 MeV * (20 / 21) = 4.87 MeV.
- 4.87 MeV / 5.11 MeV = 95.3%.
- This difference can be seen in the comparison of two photons of energies 51.1 keV and 5.11 MeV. The maximum energy transfer results when a photon interacts by a direct hit and thus, θ = 0° and Φ = 180°.
- The Compton effect not dependent on the atomic number (Z) of the material, but is instead dependent its electron density. Thus, materials of equal density thickness (g/cm^2) will result in approximately the same amount of attenuation of the beam.
- This is not true for tissues in the body as, for example, 1 cm of bone will attenuate the beam more than 1 cm of soft tissue because bone is more electron dense.
- The Compton effect becomes the main mode of interaction as the photon energy enters the megavoltage range, but will also decrease in interaction if the energy becomes too high.
- The interaction resembles a collision between two marbles where the photon and the Compton electron are scattered by the angles θ and Φ respectively relative to a line extended from the incident path of the photon.
- An incident photon may interact with the electromagnetic field of a nucleus by giving up all of its energy to create a negative electron and a positive electron referred to as pair production.
- Thus, the threshold energy for pair production is 1.02 MeV , and the excess energy beyond the threshold energy will be shared between the pair of electrons as kinetic energy.
- As a positron nears the end of its range, it will combine with a free electron in the medium to produce two annihilation photons that are ejected in opposite directions.
- Pair production increases in probability by Z^2 and with the logarithm of the incident photon energy beyond the threshold energy.
- Coherent scattering involves an incident electromagnetic wave that sets an orbital electron into oscillation which will emit an energy at the same frequency as the incident electromagnetic wave.
- This is an example of elastic scattering, but because coherent scattering only occurs at very low photon energies and high-Z materials, this attenuation coefficient is omitted in the total attenuation coefficient for therapeutic energies.
- In a graph comparing the total attenuation coefficient (y-axis) versus the energy (x-axis) of a high-Z and low-Z material, the following are notable points:
- The attenuation coefficient in the photoelectric range of energies are greater for a high-Z material versus that of a low-Z.
- The attenuation coefficient in the Compton range of energies do not differ much between high-Z and low-Z materials.
- The attenuation coefficient trends downward until the threshold energy of 1.02 MeV for pair production.
- These processes contribute to the attenuation of an incoming beam, and depending on varying factors, the incoming beam will attenuate by each interaction by a different amount.
5.2-5.4: Photon Beam Description, Photon Beam Attenuation, & Coefficients
- If a source is emitting radiation isotropically, taking a section of the sphere of radiation at a given distance from the source would be the fluence.
- Fluence is the number of incoming photons per centimeters squared, and flux density is the fluence rate which describes how fast the radiation is incoming.
- Increasing the dose rate for a beam from 600 MU/min to 1400 MU/min will increase the flux density or the number of incoming photons per centimeters squared per minute.
- Fluence is the number of incoming photons per centimeters squared, and flux density is the fluence rate which describes how fast the radiation is incoming.
- As Roentgen is to quantity as Gray is to quality, fluence is to quantity as energy fluence is to quality.
- Energy fluence is the amount of energy that is imparted into a material per centimeters squared, and energy flux density is the energy fluence rate.
- μ is the linear attenuation coefficient and is the fraction of the photons which interact in a thickness of x divided by a thickness of x. Units of μ are cm^-1, and a μ of 0.10 cm^-1 refers to a material that will absorb 10% of the incoming photons per centimeter of thickness.
- If 100 photons pass through a material with a μ of 0.10 cm^-1, how many photons will be absorbed in a thickness of 1 cm, 2 cm, and 3 cm?
- The following equation can be used to calculate for the absorbed photon number: dN = -μ * Ndx.
- With every added cm of thickness, the number of photons that become attenuated are reduced, and this phenomenon is referred to as exponential attenuation which only applies to monoenergetic beams.
- The equation mentioned above can be stated in terms of the transmitted intensity: dI = -μ * Idx.
- When expressed as a differential equation: I(x) = I_0 * e^-μ * x.
- If 100 photons pass through a material with a μ of 0.10 cm^-1, how many photons will be absorbed in a thickness of 1 cm, 2 cm, and 3 cm?
- The half-value layer (HVL) is the amount of thickness of a material required to reduce the intensity of a beam by half.
- In the differential equation above, when I(x)/I_0 is set to 1/2, it can be simplified to the following equation: HVL = ln 2 / μ.
- For a polyenergetic beam, the HVL will increase in thickness with every subsequent HVL.
- The filter will preferentially absorb the lower energy photons in the beam spectrum, and “harden” or increase the average energy of the beam.
- Whereas, the HVL of a monoenergetic beam will remain a constant thickness regardless of the number of HVLs the beam has passed through.
- The linear attenuation coefficient is not the best unit of measurement to compare the attenuation characteristics of materials as the amount of attenuation will depend on the filtration construct which may differ between manufacturers.
- Despite creating filters from the same material, the degree to how tightly packed the atoms are for a given thickness may differ.
- Thus, the mass attenuation coefficient (μ/ρ) is used to factor out the density and compare materials based on their atomic composition.
- Though μ (cm^-1) / ρ (g/cm^3) will yield units of cm^2/g, the units for μ/ρ are frequently seen as g/cm^2.
- In the differential equation for intensity, the thickness x (cm) is multiplied by ρ (g/cm^3) for the effective thickness, and when multiplied to μ (cm^-1) will yield units of g/cm^2.
- When a photon collides with an electron, it will transfer a fraction of its energy to the electron in the form of kinetic energy.
- The energy transfer coefficient (μ_tr) represents the fraction of the average energy transferred to an electron (E_tr) per unit thickness of a material and is given by the equation: μ_tr = (E_tr / h * v) * μ.
- Some of the energized electrons will undergo bremsstrahlung interactions and deposit their energy outside of the local volume.
- Thus, the electrons that do should be excluded during the calculation of local energy absorption or dose deposition.
- The energy absorption coefficient (μ_en) is the fraction of energy that is deposited locally per unit thickness of material, and is given by the equation: μ_en = μ_tr * (1 – g).
- g is the fraction of the energized electrons that will undergo bremsstrahlung interactions in the material.
Related Content: