Quiz No. 2
What is the heel effect?
The heel effect refers the greater attenuation of x-rays produced on the anode side when compared to those produced on the cathode side. Generally x-rays are produced below the surface of the tungsten target. Because of the way it is positioned, x-rays produced on the anode side must travel through a greater thickness of the tungsten target before escaping. This means that there can be up to a 45% difference in x-ray intensity of the beam. Thus, the thicker body part to be image can be positioned on the cathode side, with the thinner body part on the anode side.
What is the heat unit?
The heat unit is a unit for thermal energy used by radiographers. It is defined on the basis of full-wave rectification and single phase power, while joules are defined on the basis of full-wave rectification and three phase power.
How is heat produced in the target of the X-ray tube?
99% of the kinetic energy of electrons is converted into heat energy. This means that an incoming electron excites an outer shell electron to a higher energy state. The electron immediately drops back down and loses energy in the form of infrared radiation. The emitted infrared radiation heats the target.
Heat production is directly proportional to tube voltage and tube current. This means doubling either one of these variables with double the amount of heat produced at the anode target.
How is characteristic radiation produced?
High energy incident electrons eject an inner shell electron (useful if K shell), thus ionising the atom. A higher shell electron then falls down into the vacancy and loses energy in the form of a characteristic x-ray. The energy of the characteristic radiation is equivalent to the difference in the electron binding energies for the various possible electron transitions to the K shell. The K shell binding energy for tungsten is 69.500keV. The incident electron must have a kinetic energy of equal or greater energy than the binding energy of the electron it will eject.
L characteristic photons are not useful for diagnostic imaging as they are only of around 10keV, which is normally absorbed by glass around tube and aluminium filters.
How is bremsstrahlung radiation produced?
When electrons are accelerated towards the anode target, they approach the nucleus of a target atom. The positive electric field of the nucleus causes the negative electron to slow down and change direction. The electron therefore loses energy when it changes direction, which is emitted as a bremsstrahlung x-ray photon. An electron can lose all of its energy in one interaction, or small amounts over a maximum of 1000 interactions. The energies of bremsstrahlung photons are continuous, unlike discrete characteristic x-ray photon energies. They do not depend on the target material.
The maximum bremsstrahlung photon energy is produced when an electron loses all its energy in a single interaction. Because the energy of the energy is equivalent to the tube voltage, this means that maximum bremsstrahlung photon energy is also equivalent to the tube voltage value.
Define the following terms:
The heel effect refers the greater attenuation of x-rays produced on the anode side when compared to those produced on the cathode side. Generally x-rays are produced below the surface of the tungsten target. Because of the way it is positioned, x-rays produced on the anode side must travel through a greater thickness of the tungsten target before escaping. This means that there can be up to a 45% difference in x-ray intensity of the beam. Thus, the thicker body part to be image can be positioned on the cathode side, with the thinner body part on the anode side.
What is the heat unit?
The heat unit is a unit for thermal energy used by radiographers. It is defined on the basis of full-wave rectification and single phase power, while joules are defined on the basis of full-wave rectification and three phase power.
How is heat produced in the target of the X-ray tube?
99% of the kinetic energy of electrons is converted into heat energy. This means that an incoming electron excites an outer shell electron to a higher energy state. The electron immediately drops back down and loses energy in the form of infrared radiation. The emitted infrared radiation heats the target.
Heat production is directly proportional to tube voltage and tube current. This means doubling either one of these variables with double the amount of heat produced at the anode target.
How is characteristic radiation produced?
High energy incident electrons eject an inner shell electron (useful if K shell), thus ionising the atom. A higher shell electron then falls down into the vacancy and loses energy in the form of a characteristic x-ray. The energy of the characteristic radiation is equivalent to the difference in the electron binding energies for the various possible electron transitions to the K shell. The K shell binding energy for tungsten is 69.500keV. The incident electron must have a kinetic energy of equal or greater energy than the binding energy of the electron it will eject.
L characteristic photons are not useful for diagnostic imaging as they are only of around 10keV, which is normally absorbed by glass around tube and aluminium filters.
How is bremsstrahlung radiation produced?
When electrons are accelerated towards the anode target, they approach the nucleus of a target atom. The positive electric field of the nucleus causes the negative electron to slow down and change direction. The electron therefore loses energy when it changes direction, which is emitted as a bremsstrahlung x-ray photon. An electron can lose all of its energy in one interaction, or small amounts over a maximum of 1000 interactions. The energies of bremsstrahlung photons are continuous, unlike discrete characteristic x-ray photon energies. They do not depend on the target material.
The maximum bremsstrahlung photon energy is produced when an electron loses all its energy in a single interaction. Because the energy of the energy is equivalent to the tube voltage, this means that maximum bremsstrahlung photon energy is also equivalent to the tube voltage value.
Define the following terms:
- Quantity: the total energy of an x-ray beam, i.e. the area under the curve
- Quality: the average energy of an x-ray beam, i.e. the horizontal position of the curve
- Half-value thickness: the thickness of material required to attenuate the intensity of a beam to half its original value
- Filtration: the process by which low energy or soft photons are removed, e.g. aluminium filters, can be inherent or added
- Wedge filter: a compensating filter which compensates for the unequal attenuation by matching thicker portions of the filter to thinner body parts
What is the 15% rule?
Quantity of an x-ray beam is proportional to kVp², which means a 40% increase in kVp will double quantity. Thus a 40% increase in kVp is equivalent to doubling mAs, meaning this is a much better option for patient dose. However, by increasing kVp, we are also increasing the energy of the x-rays, meaning they become more penetrating and go straight through to the film. Thus, it has become apparent that only a 15% increase in kVp is equal to doubling mAs. This means to maintain optical density on the film, we increase kVp by 15% and half mAs.
What are the different types of filtration?
Inherent filtration is the filtration which is embedded in the x-ray tube, e.g. the lead sheath in the tube housing. Added filtration is the filtration added external to the x-ray tube, e.g. aluminium filters. The total filtration is the sum of the inherent and added filtration.
How does quantity and quality vary with the following?
- mAs: ↑ in mAs ↑ quantity because more electrons are moving from cathode to anode
- kVp: ↑ in kVp ↑ quality and ↑ quantity because more electrons are attracted to the anode and increasing kVp increases the attraction between the electron and the positive anode, meaning they are accelerated faster and gain more kinetic energy, thus increasing quality
- Filtration: ↑ quality because low energy photons are removed but ↓ quantity because of the removed soft photons
- Distance: ↑ quantity but has no effect on quality, because increasing distance means the intensity of the beam is reduced by: I = 1/d²
- Target material: the higher the atomic number the greater the efficiency of bremsstrahlung radiation, thus ↑ quantity, also as atomic number increases, the characteristic radiation energy ↑, thus increasing quality (quality of bremsstrahlung not affected)
- Voltage waveform: the smoother the voltage waveform, ↑ quality and ↑ quantity
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