Leicester Course Mock

Question 1

2. Concerning properties of x-rays:

Beam intensity is the total energy per unit area per unit time. True.
True.
Beam intensity refers to the amount of energy the x-ray beam carries per unit area per unit time. It is typically measured in units like milliampere-seconds (mAs), which reflects the amount of energy produced by the x-ray tube. Intensity is related to both the number of photons and the energy of those photons.

The inverse square law applies to all x-ray exposures.
False. only point source in a vacuum

X-rays have lower linear energy transfer than alpha particles
X-rays are a form of high-energy electromagnetic radiation and generally have low linear energy transfer (LET) compared to particles like alpha particles, which are charged and have much higher LET.

Answer 1

All electromagnetic radiation can cause ionization
FALSE.
Ionizing radiation includes x-rays, gamma rays, and ultraviolet (UV) radiation with enough energy to ionize atoms (i.e., knock electrons out of atoms).

At equivalent energy, an x-ray cannot be distinguished from a gamma ray

True.
X-rays and gamma rays are both forms of electromagnetic radiation and, at the same energy level, they are indistinguishable. The distinction between x-rays and gamma rays lies in their origin:

X-rays are typically produced by the interaction of electrons with matter (such as in an x-ray tube).
Gamma rays are produced by nuclear processes or radioactive decay.

Question 2

3. Regarding the anode angle:

Is the angle that the target face makes with the x-ray beam. TRUE.

Is the only factor determining focal spot size. FALSE.
While the anode angle is a significant factor in determining the focal spot size, it is not the only factor. The focal spot size is also influenced by other factors such as:

Electron beam size: The size of the electron beam striking the anode target also impacts the focal spot.
Anode material: The type of material used in the anode (e.g., tungsten) can affect the resolution and heat dissipation, which in turn can impact the focal spot.
Focus design: The geometry of the x-ray tube and the way the electron beam is focused on the anode also play a role.

Is generally 20-35 degrees DK 7-20.
FALSE.
The anode angle typically ranges from 7 to 20 degrees, depending on the type of x-ray tube and its application. A common range for medical x-ray tubes is 10 to 20 degrees, but it is not 20 to 35 degrees. A larger anode angle may provide a larger focal spot and reduce the tube’s heat load, while a smaller angle creates a smaller focal spot but increases heat concentration.

Answer 2

Increases the tube rating if the angle is reduced.
False.
The tube rating (or heat capacity of the tube) is actually reduced if the anode angle is reduced. When the anode angle is small, the effective focal spot becomes smaller, which means the heat concentration on the anode increases, reducing the ability of the tube to dissipate heat effectively. A smaller angle leads to a higher heat load on the anode, potentially causing damage. Therefore, reducing the angle does not increase the tube’s rating; it typically decreases the thermal capacity.

The anode heel effect is greater if the angle is reduced (made more steaper).

True.
The anode heel effect refers to the variation in x-ray beam intensity across the field, with a higher intensity near the cathode side of the tube and a lower intensity near the anode side. This effect is more pronounced when the anode angle is reduced. A smaller anode angle causes the x-rays to be emitted more tangentially, increasing the gradient of intensity from the cathode to the anode side. In other words, the anode heel effect is more significant at smaller angles

Question 3

4. Radiation output from an x-ray tube increases with:

The addition of a filter
False.
The addition of a filter to the x-ray beam does not increase the radiation output. In fact, filters are used to remove low-energy, non-diagnostic x-rays (soft x-rays) from the beam. While this increases the beam quality by reducing the patient dose (as soft x-rays are less penetrating), it does not increase the total radiation output. In other words, filtering the beam may reduce the total number of x-rays hitting the patient but increases the average energy (beam quality).

Increasing kV (with all other factors kept constant)
True.
Increasing the kilovolt (kV) applied to the x-ray tube increases the energy of the x-rays produced. This increases both the penetrating power of the x-rays and the quantity of x-rays generated. Higher kV results in more efficient production of x-rays, leading to an increase in radiation output. Higher kV also increases the speed of electrons in the tube, leading to more interactions with the anode target and a higher overall x-ray yield.

Increasing mA (with all other factors kept constant)
True.
Milliamperes (mA) refers to the current applied to the x-ray tube, which controls the number of electrons flowing from the cathode to the anode. Increasing the mA increases the number of electrons striking the anode, thus increasing the quantity of x-rays produced. This directly increases the radiation output. Since mA determines the quantity (not the energy) of the x-rays, it is a key factor in controlling the intensity of the x-ray beam.

Answer 3

Decreasing cathode to anode distance
False.
Decreasing the cathode to anode distance would theoretically concentrate the x-rays in a smaller area, but it does not increase the overall radiation output. In fact, it would typically increase the beam divergence, leading to a reduction in intensity and potentially compromising image quality. The inverse square law also applies here: as the distance between the cathode and anode decreases, the intensity would increase in a localized area but not necessarily result in an overall increase in radiation output. The radiation output primarily depends on factors like mA, kV, and target material.

A single phase waveform
False.
A single-phase waveform produces less efficient radiation output compared to three-phase or high-frequency waveforms. In a single-phase system, the voltage fluctuates between zero and the peak value, which means the x-ray tube does not operate at full potential all the time, resulting in a lower average radiation output. In contrast, three-phase or high-frequency waveforms provide a more consistent and higher average voltage, leading to higher radiation output.

Question 4

5. Characteristic radiation

is produced when an electron interacts with the coulomb field of the nucleus. FALSE.
Characteristic radiation is actually produced when an electron interacts with the inner-shell electrons of the target atom (typically in the K-shell or L-shell), not with the Coulomb field of the nucleus itself. When an electron from the tube strikes an atom, it can eject an inner-shell electron, creating a vacancy. An electron from a higher energy level then falls into the vacancy, releasing energy in the form of characteristic radiation. The process involves interactions with electrons in the atom, not with the nucleus.

can be produced in an x-ray tube
True.
Characteristic radiation is indeed produced in an x-ray tube when high-energy electrons (from the cathode) strike the target material (usually tungsten). These electrons can dislodge inner-shell electrons from atoms in the target material, and as higher-energy electrons drop down to fill these vacancies, characteristic x-rays are emitted. The energy of the emitted x-rays is characteristic of the target material and the difference in energy between the two electron shells involved.

is always produced in a photoelectric interaction
False.
The photoelectric effect is a different type of interaction where an x-ray photon is completely absorbed by an atom, ejecting an inner-shell electron. Characteristic radiation can be produced as a secondary effect, but it is not always the result of a photoelectric interaction. In the photoelectric effect, the ejected electron leaves a vacancy, which may later be filled by an electron from a higher energy level, emitting characteristic radiation. However, the photoelectric interaction itself does not always result in characteristic radiation.

Answer 4

has energies that depends on NUCLEAR binding energy of the target material
False.
The energy of characteristic radiation depends on the binding energy of the electron shells of the target material, not the nuclear binding energy. Specifically, it depends on the difference in energy between the inner-shell electrons (e.g., K-shell, L-shell) and the higher-energy electron shells (e.g., L-shell, M-shell) in the target atom. These binding energies are specific to each element, but they are related to the electron shell energies, not the nuclear forces.

Definition: The nuclear binding energy is the energy required to hold the nucleus of an atom together. It is the energy needed to break up a nucleus into its individual protons and neutrons.

Nature of Energy: This energy is a result of the strong nuclear force that binds the protons and neutrons together inside the nucleus. It is typically on the order of MeV (million electron volts).

has energies that depend upon the kVp
False.
The energy of characteristic radiation is not directly dependent on the kVp (kilovolt peak) setting of the x-ray machine. The energy of characteristic radiation is determined by the difference in energy between the inner and outer electron shells in the target material. However, kVp influences the quantity and quality of the overall x-ray beam, and higher kVp can increase the likelihood of producing characteristic radiation by increasing the energy of electrons striking the target. But the specific energies of the characteristic x-rays themselves remain fixed by the atomic structure of the target material.

Question 5

6. Regarding scattered radiation affecting the image in diagnostic radiology:

Compton scatter is an important process in diagnostic radiology.
True.
Compton scattering is one of the primary interactions of x-ray photons with matter, especially in diagnostic radiology. It occurs when an x-ray photon interacts with an outer-shell electron of an atom, resulting in the photon being scattered in a different direction and losing some of its energy. Compton scatter contributes to image noise and radiation dose to the patient, making it an important consideration in radiographic imaging.

has a longer wavelength than the primary radiation
True.
Scattered radiation (such as from Compton scattering) typically has a longer wavelength than the primary radiation. This is because during Compton scattering, the scattered photon loses energy, resulting in a lower energy (and therefore longer wavelength) photon compared to the primary incident x-ray photon. The amount of energy loss depends on the angle of scattering, with a greater loss occurring at larger scattering angles.

Answer 5

is increased by compressing thick parts of the patient
False.
Compressing thick parts of the patient actually reduces scatter radiation, rather than increasing it. When thicker body parts are compressed, the overall patient thickness is reduced, which decreases the volume of tissue that the x-ray beam passes through, and as a result, there is less scatter generated. Additionally, compression can help reduce motion blur and improve image quality.

is greater in the forward direction than backwards
True.
Compton scatter tends to be greater in the forward direction (in the direction of the incident x-ray beam) compared to the backward direction. This is because the scatter angle influences the energy loss of the scattered photon: photons scattered at smaller angles (forward scatter) retain more energy than those scattered at larger angles (backward scatter). Therefore, forward scatter contributes more to the overall scattered radiation compared to backward scatter.

is independent of atomic number
True.
Compton scattering is independent of the atomic number (Z) of the material through which the x-rays are passing. Unlike photoelectric absorption, which is strongly dependent on the atomic number of the tissue (higher Z materials like bone interact more with x-rays), Compton scattering mainly depends on the electron density of the tissue, which is relatively similar across different materials in the body. Therefore, it does not significantly vary with the atomic number.

Question 6

7. An adequately filtered diagnostic x-ray beam:

reduces the relative amount of scatter reaching the film

False.
While beam filtration (typically through the use of aluminum filters) does remove low-energy x-rays from the x-ray beam (which are less penetrating), it does not directly reduce the amount of scatter reaching the film. Scatter radiation is produced primarily through interactions like Compton scattering in the body, and its reduction depends more on factors like beam collimation, patient positioning, and the use of grid devices rather than filtration alone.

can be produced by placing approx. 2mm of Aluminium in the beam
True.
Aluminum filtration is typically used to filter out low-energy x-rays from the x-ray beam. 2mm of aluminum is a common standard for adequate filtration in diagnostic radiology. This amount of aluminum can reduce the patient dose by removing low-energy x-rays that do not contribute to the diagnostic quality of the image but do contribute to unnecessary patient exposure.

Answer 6

increases the half value layer value, as compared with the unfiltered beam
True.
Half-value layer (HVL) refers to the thickness of a material (typically aluminum) required to reduce the intensity of the x-ray beam by half. Filtered x-ray beams have higher HVL values compared to unfiltered beams because the filtration process removes the lower energy, less penetrating x-rays, leaving a beam with higher energy and greater penetrating power. As a result, the HVL for a filtered beam is higher, indicating greater beam quality (penetration).

produces only marginally-smaller skin doses than an unfiltered beam
False.
Adequate filtration significantly reduces the skin dose by removing low-energy x-rays that would be absorbed by the skin. These low-energy photons contribute little to the image but significantly increase the dose to the skin. Therefore, filtration results in a substantial reduction in skin dose, not just a marginal one.

relies on an inverse cube relation between photo electric attenuation and photon energy
False.
The statement is not accurate in describing how photoelectric attenuation relates to photon energy. Photoelectric absorption is more directly related to the inverse square of the photon energy (or more specifically, the cube of the photon energy in some contexts) in its interaction with matter, but this is not the basis for how filtration works in diagnostic radiology. Filtration removes low-energy photons to improve beam quality, and the attenuation characteristics of the beam depend more on the material properties of the filter and the photon energy, not on an inverse cube relation.

Question 7

8. Photoelectric interactions between radiation and biological tissues

result in ionization
True.
Photoelectric interactions occur when an x-ray photon is completely absorbed by an atom, ejecting an inner-shell electron (often from the K-shell or L-shell). This ejected electron is called a photoelectron, and it leaves a vacancy in the atom’s electron shell, which can lead to ionization of the atom. Therefore, photoelectric interactions result in ionization of the atom from which the electron was ejected.

produce a deflected incident photon
False.
Photoelectric interactions involve the complete absorption of the incident photon, so there is no deflected photon as part of this interaction. This is in contrast to Compton scattering, where the incident photon is only partially absorbed and scattered in a different direction. In the photoelectric effect, the incident photon is entirely absorbed by the atom, leaving no scattered photon behind.

increase in probability of occurrence as the electron binding energy increases
True.
The probability of photoelectric interaction increases as the binding energy of the electron increases. In other words, atoms with higher atomic numbers (Z) have more tightly bound inner electrons. These tightly bound electrons are more likely to participate in photoelectric interactions, which is why photoelectric absorption is more common in high-Z materials (e.g., bone, iodine). High electron binding energy increases the likelihood of the photon being absorbed rather than scattered.

decrease in proportion to the fourth power of the incident photons energy .
FALSE. CUBED.

result in increased absorption at energies immediately below the electrons binding energy

Question 8

9. Subject contrast
10. Depends on the focal spot size
    False.
    Subject contrast is primarily determined by the tissues being imaged, the energy of the x-ray beam, and the thickness of the body part. The focal spot size does not directly affect the subject contrast, though it can influence the image sharpness (spatial resolution). A smaller focal spot improves image sharpness, but it does not directly alter the contrast of the subject being imaged.
11. Depends on the kVp
    True.
    Subject contrast is indeed influenced by the kilovolt peak (kVp). The kVp controls the energy of the x-ray photons. A higher kVp increases the penetration power of the x-ray beam, leading to lower subject contrast, because higher energy x-rays pass through tissues more easily and produce less differential attenuation between tissues. Conversely, a lower kVp results in a greater difference in attenuation between tissues, increasing subject contrast.
12. Depends on the mAs
    False.
    The milliamperage-seconds (mAs) affects the radiographic density (or exposure) but does not directly impact the subject contrast. mAs controls the quantity of x-rays produced, which influences the overall image brightness or darkness but not the contrast between different tissues. The subject contrast is more dependent on factors like tissue composition, beam energy (kVp), and thickness of the subject.
13. Varies linearly with patient thickness
    False.
    Subject contrast does not vary linearly with patient thickness. As the thickness of the patient increases, beam attenuation increases, and there can be more scattering and loss of contrast due to the additional tissue the x-rays must pass through. However, this effect is not linear; instead, subject contrast typically decreases as the patient thickness increases due to increased scattering and absorption.
14. Is independent of the film’s gamma
    False.
    The film’s gamma (or film contrast) is an important factor in determining the overall image contrast, and it is related to how the film responds to the exposure. Subject contrast and film contrast both contribute to the overall radiographic contrast. If the film has low gamma (low contrast), it will not show the differences in tissue attenuation well, even if the subject contrast is high. Conversely, a film with high gamma (high contrast) can enhance the subject contrast, making tissue differences more visible.

Question 9

10. Geometric unsharpness is reduced by using

Geometric Unsharpness Ug = f x b / a

f = x-ray focal spot size
b = OBJECT - DETECTOR distance
a = SOURCE - OBJECT distance

https://www.radiologycafe.com/frcr-physics-notes/x-ray-imaging/image-quality/

A smaller object film distance = b
True. Object being closer to detector reduces geometric unsharpness

A shorter exposure time
False. Not part of the equation.

Answer 9

A larger focus film distance.
FALSE.
Increasing SOURCE TO IMAGE distance INCREASES BOTH MAGNIFICATION and GEOMETRIC BLURRING.

An increased target angle technically.
FALSE.
LARGER anode angle = LARGER focal spot size = f
This will increase geometric unsharpness

A grid
False. No effect.

Question 10

In mammography

for an average breast a molybdenum anode is the anode of choice True.

1. For an average breast, a molybdenum anode is the anode of choice
   True.
   Molybdenum is commonly used as the anode material in mammography because it provides a low-energy x-ray spectrum, which is ideal for imaging the soft tissues of the breast. This energy range helps to highlight differences in tissue densities without over-penetrating the breast tissue. It also reduces the radiation dose compared to other materials like tungsten.
2. A 3mm thick molybdenum filter is usually used
   False.
   The typical filter used in mammography with molybdenum anodes is much thinner—around 0.03 mm (30-60 micrometers). This thin filter helps to remove lower-energy x-rays from the beam, which would otherwise increase patient dose without contributing to image quality. The filter ensures that the x-ray beam has a more appropriate energy range for optimal imaging of breast tissue.
3. A small focal spot improves geometric resolution but increases exposure times
   True.
   A small focal spot improves geometric resolution (or sharpness) because it reduces the size of the x-ray beam’s focus, resulting in less geometric unsharpness. However, a smaller focal spot also increases the heat concentration at the target and requires longer exposure times to produce adequate image density because it produces less x-ray output. This is why mammography typically uses a small focal spot for high-resolution images despite the trade-off in exposure time.

Answer 10

4. The typical skin dose is in the region of 50mGy
   False.
   The typical skin dose in mammography is generally much lower than 50mGy. In fact, the typical skin dose is usually in the range of 1-2 mGy per view. A dose of 50mGy would be considered excessively high and far beyond the typical exposure range for mammography. Low-dose imaging techniques are used to minimize radiation exposure to the skin and surrounding tissues.
5. An anti-scatter grid lowers the skin dose and improves film contrast
   False.
   Anti-scatter grids are used in mammography to reduce the amount of scatter radiation that reaches the film or detector, which helps to improve image contrast by preventing scattered radiation from degrading the image. However, grids also increase the radiation dose to the skin and tissues because they absorb some of the primary beam as well. Therefore, grids improve image quality but do not lower skin dose. In fact, they often result in a higher dose to the patient.

Question 11

12. In computed radiography

The x-ray absorbtion efficiency of the photostimulable phosphor plates (PSPs) is much higher compared to film screen

The latent image stored on a PSP can decay if not read promptly

The intrinsic resolution in CR is limited by the thickness of the phosphor layer

Spatial resolution of a standard CR system is significantly higher than that of a competing film screen combination

Quantum mottle is proportional to the square of photon fluence incident upon the image plate (N)

Answer 11

1. True
2. True. Over time, the latent image fades spontaneously by the process of phosphorescence
3. False. Matrix size?
4. False. A standard CR system generally has a significantly lower spatial resolution compared to a competing film-screen combination, meaning the film-screen combination can typically differentiate smaller details in an image than a standard CR system
5. False?

Question 12

13. In digital radiography

The charged couple device (CCD) converts photons into an electronic signal

Resolution on a flat panel array is limited by the width of the detector elements

The efficiency of signal recording in a solid state DR detector is increased with increasing fill factor

DR detectors have a narrow latitude

The modulation transfer function (MTF) of indirect conversion detectors is better than that of direct conversion detectors

Answer 12

1. True
2. True. spatial resolution is determined not only by the size of the thin film transistor matrix per unit area, but also by the control technique of the scattered light
3. True. In a solid-state digital radiography (DR) detector, a higher fill factor means a larger proportion of the detector’s area is actively sensing radiation, leading to a more efficient recording of the signal and improved image quality
4. NO, digital radiography (DR) detectors have a wide exposure latitude, which is the range of exposure settings that can produce a diagnostic image
5. The MTF of direct conversion detectors is considered better than that of indirect conversion detectors

Question 13

14. Dose in fluoroscopy

Automatic Brightness Control (ABC) is the same as Automatic Exposure Control (AEC)

The maximum entrance skin dose rate limit for a standard patient is 100mGy per minute

Patient entrance skin dose rate is reduced by increasing the thickness of a spectral filter

Entrance surface dose is increased by reducing the fluoroscopy pulse rate from 30 to 15 frames per second

Leakage from the tube housing is typically 5 uGy/hr at 1 m distance

Answer 13

A. True. Automatic Brightness Control (ABC) in fluoroscopy is essentially the same concept as Automatic Exposure Control (AEC) in radiography; both systems automatically adjust the radiation exposure to maintain a consistent image quality by adapting the kVp and mA based on the attenuation of the X-ray beam by the patient, with the key difference being that ABC is specifically used in fluoroscopy while AEC is used in standard radiography

B. True

C.

D.

E.

Question 14

15. Digital subtraction angiography

In DSA x-ray tubes with a lower rating can be used

Between the acquisition of the pre contrast and post contrast images the patient is allowed to move

A road map cannot be used in conjunction with live fluoroscopic images

Subtracted images have very high signal to noise ratio compared to non-subtracted images

Temporal frame averaging is used to decrease displayed image noise

Answer 14

A. True. In Digital Subtraction Angiography (DSA), X-ray tubes with a lower rating can be used because the technique allows for significantly lower radiation doses to be used while still achieving high image quality

B. False

C. False

D. FALSE. Subtracted images generally have a lower signal-to-noise ratio compared to non-subtracted images, meaning they appear noisier due to the subtraction process eliminating not only unwanted tissue information but also some signal information, leading to a reduced signal-to-noise ratio

E. If frame averaging is used, the image is improved; and the more frames averaged, the better the results for noise decreases with the square root of the number of frames averaged. That is, if 16 frames are averaged, noise reduction would be a factor of 4.

Question 15

16. Absorbed dose

is a measure of energy absorbed per unit thickness

is measured in joules per kilogram T

is stated in Sieverts F

is stated in Grey T

can be estimated by an ionisation chamber T

Answer 15

A. False. A measure of energy absorbed per unit thickness is called the linear absorption coefficient

Question 16

17. The 10 day rule:

is intended to reduce the incidence of leukaemia in children

does not apply to single women

refers to the 10 day interval following the onset of menstruation

should be applied for all x-ray exposures

does not apply if image intensification is used

Question 17

18. A 20 mSv dose to a member of the radiographic staff requires the following actions

The Health and Safety Executive needs to be informed

A medical examination must be carried out

The radiographer must cease work

An investigation must be launched

The radiographer automatically becomes a classified person

Question 18

19. A deterministic effect of ionising radiation is where

The probability of occurance is proportional to dose

The severity is related to dose

A threshold must be exceeded before an effect is observed

Radiation induced cataracts may be formed

Genetic effects are noticed in the 2nd generation of offspring

Question 19

20. A radiation dose of 5mSv corresponds to

About the dose a patient may receive from an abdominal x-ray

The current statutory dose limit for members of the public

The average annual background radiation received by a member of the UK public

annual radiation dose threshold above which a member of the staff must be “classified”

A lifetime excess risk of harm of approximately 1 in 100,000

Answer 19

A. False. Abdo XR 0.7mSv

B. False. 1mSv

C. False. 2.7 mSv

D. False. 6 mSv

E. 5 / 20,000 =1/4000 lifetime risk

Question 20

21. In computed tomography (CT)

the images have a higher spatial resolution than film/screen combinations

the noise in the image is quoted as the CT number of water

the partial volume effect is more noticeable at large slice thicknesses

the partial volume effect dose not affect spiral (helical) CT images

detector non-uniformities normally causes streaking artefacts

Answer 20

A. False. Computed radiography’s spatial resolution is inferior to conventional radiography; however, this limitation is considered clinically insignificant.

B. False. Instead, the noise level in a CT image is expressed as a percentage of image contrast in CT numbers. The standard deviation for noise should be +/-3.

C. True. the partial volume effect, where small structures are averaged with surrounding tissue due to the imaging resolution, becomes more prominent when using larger slice thicknesses. This means that thinner slices generally lead to a less pronounced partial volume effect and better visualization of small structures.

D. False?

E. False ? Ring Artifact

Question 21

22. Computed tomography

Most multi-slice scanners are based on the 3rd generation scanner geometry

The axis of rotation is the z-axis

The typical kVp value for an x-ray tube is about 80-140Kv

The main interaction in CT scanning is the Compton effect

Pixel size is determined by the number of detectors

Answer 21

A. True

B. When a CT scanner acquires data, the X-ray source and detector rotate around the patient, with the z-axis representing the center of this rotation.

C. True. The typical kVp value for an x-ray tube in CT is about 80-140Kv

D. True. The primary interaction that occurs during CT scanning is the Compton effect, where X-ray photons interact with electrons in the tissue, causing them to scatter and lose some of their energy, which is the dominant mechanism responsible for image formation in CT scans

E. Pixel size is determined by the number of detectors in a given area.

Question 22

23. In computed tomography

Fourth generation scanners use a rotate/stationary scan motion

Scintillators can be used as detectors

The resolution is normally limited by the focal spot size

Spatial resolution is limited by the number of pixels

Tube anode/cathode axis is parallel to the fan beam plane

Answer 22

A. True

B. Yes, scintillators are used as detectors in CT scanners. Scintillators are luminescent materials that convert X-ray photons into visible light, which is then converted into an electrical signal. This electrical signal is used to create a 3D image of the patient’s body.

C. As an example, if your CT system is equipped with an X-ray source that have a minimum focal spot size of 4 µm, even if you reach voxel size below 4 µm, you will not be able to resolve the details in the image as the spatial resolution is physically limited by the X-ray source.

D. Smaller focal spots give higher resolution, but the max mA is limited to prevent damage to the anode.

E. True. When a CT scanner’s tube anode/cathode axis is parallel to the fan beam plane, it means that the imaginary line connecting the anode to the cathode of the X-ray tube is aligned with the plane of the fan-shaped X-ray beam emitted, which helps minimize the “heel effect”

Question 23

24. In computed tomography

scans in the sagittal plane are produced by electronic reconstruction

the window level indicates the tissue density being examined

the radiation dose to the scalp from a brain scan is about 10 times that of a conventional lateral skull x-ray

Beam hardening is the same throughout the field of view for a given patient F

the absorbed radiation dose is measured in Hounsfield units T?

Answer 23

A. True.

B. True.

The window level in CT refers to the central value of the window width that determines the range of densities that will appear in the image. The window level corresponds to the specific tissue density being examined and allows the radiologist to adjust the image contrast for optimal visualization of structures. For example, the window level might be adjusted to focus on soft tissues, bones, or air, depending on the clinical need.

C. False. Skull AP/PA: 1.8 mGy vs CT Head 60 mGy. So about 30 TIMES MORE.

Chat GPT: As a result, the radiation dose to the scalp from a brain CT scan can be approximately 10 times higher than from a conventional lateral skull X-ray, depending on the specific scan protocols used.

D. False. Beam hardening refers to the phenomenon where lower-energy X-rays are absorbed more than higher-energy X-rays as the beam passes through tissue, leading to an increase in the average energy (or “hardness”) of the X-ray beam. This effect varies depending on the composition and density of the tissues in the field of view. For example, beam hardening may be more pronounced in areas with denser tissues (like bone) and less so in areas with soft tissues. Therefore, beam hardening is not the same throughout the field of view for a given patient.

E. False. Hounsfield units (HU) are a scale used in CT imaging to represent the relative density of tissues. They are not a measure of absorbed radiation dose. Hounsfield units are calculated based on the attenuation of the X-ray beam by tissues, with water set at 0 HU and air at around -1000 HU. The absorbed dose is typically measured in units like Gray (Gy) or milliGray (mGy), not Hounsfield units.

Question 24

25. CT dose measurement

CTDI varies with slice width

CTDIvol is directly proportional to pitch

DLP is directly proportional to scanned length

Conversion coefficients to derive effective dose from DLP are independent of scanner design

mA modulation leads to a lower patient dose compared to fixed mA techniques

Answer 24

A. True. The CT Dose Index (CTDI) typically varies with the slice thickness or slice width. A wider slice leads to more radiation exposure and a higher CTDI, as the x-ray beam covers a larger portion of the body. Conversely, narrower slices typically result in a lower CTDI.

B. False.

The CTDIvol (volume CT dose index) is inversely proportional to the pitch in spiral (helical) CT imaging. As the pitch increases (meaning the table moves faster relative to the x-ray beam), the dose per unit length decreases because the beam is spread over a larger area. Hence, a higher pitch leads to a lower CTDIvol, assuming all other factors remain constant.
CTDIvol = CTDIw / pitch SO INVERSELY proportional

C. True.
DLP calculation: DLP is calculated by multiplying the CTDIvol (dose per slice) by the scan length.

D. False.
The conversion coefficients used to estimate the effective dose from DLP may vary depending on factors such as scanner design and protocols used. These coefficients are based on average reference data, and scanner technology (such as tube design or type of detectors) can affect how radiation interacts with the patient, thus influencing the effective dose calculation.

E. True.

mA modulation (tube current modulation) adjusts the mA (milliamperes) based on the patient’s size, the region of interest, and other factors, optimizing the dose for each part of the body. This reduces the dose to areas that don’t require as much radiation (e.g., areas with less tissue density), compared to fixed mA techniques, which apply a constant radiation dose throughout the scan, regardless of the patient’s anatomy.

Question 25

26. Concerning MRI

a spin echo sequence shows arteries as a flow void

T1 is always greater than T2 T

T2 represents energy lost to the lattice F

the magnitude of free induction decay is related to nuclear concentration F?

Spin Echo techniques (as opposed to gradient echo) are used to increase the read out signal T

Answer 25

A. True. Flow voids is a term used when describing MRI studies and refers to signal loss occurring within moving fluids (usually blood but also frequently seen in CSF or urine) when the fluid is moving at a sufficient velocity relative to the MRI apparatus. Some MRI sequences are more susceptible to flow-induced signal voids than others as it is a combination of time-of-flight and spin-phase effects and thus usually seen in spin-echo techniques (such as T2-weighted images

D True.
Yes, in MRI, the magnitude of the Free Induction Decay (FID) signal is related to the nuclear concentration of the spins (protons) in the tissue being imaged.

E. True.

Spin Echo techniques are generally preferred when aiming to maximize the readout signal in MRI because they use a 180° refocusing pulse to counteract the dephasing of spins caused by magnetic field inhomogeneities, leading to a stronger and more consistent signal compared to Gradient Echo sequences which are more susceptible to signal loss from these inhomogeneities.

Question 26

27. Gadolinium DPTA

is largely excreted by glomerular filtration

reduces both the T1 and T2 relaxation time

Answer 26

A. True. Gadolinium chelates that extracellularly distributed and eliminated rapidly through renal glomerular filtration

B. True. Yes, Gadolinium DTPA (Gd-DTPA) reduces both the T1 and T2 relaxation times in magnetic resonance imaging (MRI), meaning it shortens the time it takes for tissue to return to its equilibrium state in both longitudinal and transverse magnetization, effectively enhancing the signal intensity on T1-weighted images while slightly decreasing it on T2-weighted images

C. Yes, Gadolinium DTPA (Diethylenetriaminepentaacetic acid) is a contrast agent commonly used in MRI (Magnetic Resonance Imaging) to enhance the visibility of blood vessels and tissues. It can increase the signal from arteries during imaging because it is a paramagnetic substance. When injected into the bloodstream, it causes a local shortening of the T1 relaxation time in the tissues, which results in a brighter signal on T1-weighted images

D. FALSE.
The half-life of Gadolinium DTPA in the blood typically ranges between 1 to 2 hours.

E. FALSE.
Paramagnetic substances, like Gadolinium, have unpaired electrons and exhibit a magnetic moment that aligns with an external magnetic field, but they do not retain magnetization when the external field is removed. This is what makes Gadolinium effective in enhancing MRI signals. It shortens the T1 relaxation time in tissues, increasing signal intensity in T1-weighted images.

Superparamagnetic materials, on the other hand, also have unpaired electrons, but they exhibit a stronger magnetic response and can retain some magnetization even after the external magnetic field is removed. Superparamagnetism typically occurs in nanoparticles or certain types of iron oxide particles used in imaging or drug delivery.

Question 27

28. In gradient echo sequences use in magnetic resonance imaging (MRI) :

A. A gradient replaces the 90° rephasing pulse used in spin echo imaging

B. Large flip angles (70° - 110°) give T1-weighted images:

C. Flow is sensitively detected and this is exploited for magnetic resonance angiography (MRA)

D. TR (Repetition Time) in gradient echo sequences is typically shorter compared to spin echo sequences.

E. Resultant images are affected by T2* information

Answer 27

A. False. 180 degree rephasing

B. True. Larger flip angles result in greater sensitivity to T1 relaxation effects,

C. True.

Gradient echo sequences are particularly sensitive to flow, which is exploited in magnetic resonance angiography (MRA). The flow sensitivity comes from the way gradient echo sequences interact with moving blood or other fluids, and this can be used to visualize blood vessels in MRA without needing contrast agents, or with specialized techniques using contrast.

D. True.

TR (Repetition Time) in gradient echo sequences is typically shorter compared to spin echo sequences. Gradient echo sequences are generally faster and can be performed with shorter TRs because they do not require multiple 180° refocusing pulses, which helps reduce scan time.

E. True.
True. T2 (T2-star)* is the effective transverse relaxation time in gradient echo sequences. Because gradient echo imaging is more sensitive to field inhomogeneities and magnetic susceptibility effects (e.g., in the presence of air, bones, or implants), the images produced are often influenced by T2* decay, which includes both T2 relaxation and the effects of field inhomogeneity. This means the resulting images are more susceptible to signal loss due to T2* effects.

Question 28

29. In magnetic resonance imaging the following are true?

A. A spin echo sequence with a short repeat time (TR) and a short echo time TE is T1 weighted

True.

In spin echo sequences, a short TR and short TE give T1-weighted images. This is because a short TR allows for limited longitudinal relaxation (favoring T1 effects), and a short TE reduces the contribution from T2 decay, which minimizes T2 effects, resulting in a strong sensitivity to T1 differences between tissues.

B. A spin echo sequence with a long repeat time TR and a long echo time TE is proton density weighted

False.

A spin echo sequence with long TR and long TE will result in T2-weighted images, not proton density (PD)-weighted. A long TR allows for full longitudinal relaxation, and a long TE maximizes T2 decay effects, making the image more sensitive to T2 differences between tissues. Proton density weighting typically requires a long TR but a short TE.

C. A spin echo sequence with a long repeat time TR and a short echo time TE is T2 weighted

False. A long TR with a short TE is actually proton density weighted rather than T2-weighted. A long TR ensures that all tissues have sufficient time to relax longitudinally, minimizing T1 effects, and the short TE minimizes T2 decay, making proton density (the number of hydrogen protons in the tissue) the dominant factor for contrast. For T2-weighted imaging, you need both a long TR and a long TE to highlight differences in T2 decay.

Answer 28

D. The contrast in a T1 weighted spin echo sequence is independent of alpha angle

True

True. In T1-weighted imaging, the flip angle (often denoted as alpha) influences the signal strength, but the contrast between tissues primarily depends on the T1 relaxation times of different tissues. A typical flip angle used in a T1-weighted spin echo sequence is around 90° (in spin echo), and it doesn’t significantly affect the T1 contrast because the T1 differences are what primarily determine the image contrast.

E. T2* includes effects due to magnetic field inhomogeneity

True. T2* (T2-star) relaxation time includes both T2 decay (transverse relaxation) and the effects of magnetic field inhomogeneity (such as imperfections in the magnetic field, susceptibility effects, or differences in tissue composition). T2* is a more complex decay process that combines the inherent T2 relaxation of tissues with additional signal loss due to inhomogeneities in the magnetic field.

In Spin Echo
T1 = SHORT SHORT
T2 = LONG LONG
Proton Density = shoRT TR and long TE

Question 29

30. In magnetic resonance imaging

Aliasing may be caused by a part of the patient outside the field of view

True.
Aliasing, also known as “wrap-around artifact,” occurs when anatomy outside the field of view (FOV) is mapped into the image. This happens because the scanner assumes the imaged region is continuous, leading to overlaps when data from outside the FOV is folded into the visible image.

Pulsation artefact occurs in the phase encode direction

True.
Pulsation artifacts, such as those caused by cardiac or vascular pulsations, typically occur in the phase-encoding direction due to the timing and encoding of phase data. Adjusting the phase-encoding direction or using gating techniques can help minimize these artifacts.

Chemical shift artefact increases with field strength

True.
The chemical shift artifact results from differences in resonance frequencies between fat and water. Since the frequency difference is proportional to the magnetic field strength, the artifact becomes more pronounced at higher field strengths.

Answer 29

Decreasing the FOV where there are artefacts caused by ferromagnetic objects improves the image

False.
Decreasing the FOV can increase susceptibility artifacts caused by ferromagnetic objects because these artifacts are related to field inhomogeneity. Using sequences like spin-echo or adjusting the imaging parameters (e.g., increasing bandwidth) is typically more effective.

On T1 flowing blood appears dark

True.
Flowing blood appears dark on T1-weighted images due to the phenomenon of “flow void,” where rapidly moving spins are not adequately excited or refocused, resulting in no signal.

Question 30

31. An ultrasonic beam passing through the body

Is attenuated T
Produces heating of the tissues T
Can be reflected T
Produces ionising radiation F
Travels relatively fast in gases F

Answer 30

velocity = square root of [Bulks or compressibility / density]

Air’s compressibility is much higher compared to liquids or solids, which significantly lowers the speed of sound. While air has low density (ρ), its much higher compressibility dominates the equation, resulting in a slower propagation speed compared to denser media like water or bone.

Question 31

32. In relation to an Ultrasound beam

The speed of propagation depends on the density of the material through which it passes.

True.
velocity = square root of [Bulks or compressibility / density]

The wavelength of a 1 MHz beam is shorter than that of a 20 MHz beam. False.

Answer 31

The attenuation of a beam in tissue depends on the frequency of the beam. True

The acoustic impedance of a medium depends on its density. True

20dB loss means that the beam has been reduced in intensity by a factor of 100 (2 Bells = 102)

Question 32

Ultrasound:

May be detected by a Piezoelectric crystal T

Voltage changes across the crystal cause expansion and contraction resulting in pressure waves T

Answer 32

The same crystal cannot be used for producing as well as receiving Ultrasound

False.
The same piezoelectric crystal can function as both a transmitter and a receiver in most modern ultrasound transducers. It generates ultrasound waves when voltage is applied and detects returning echoes by converting mechanical vibrations into electrical signals.

Continuous sound is used to detect Doppler effect
In pulsed Ultrasound the pulse width is shorter than the listening interval

True.
Continuous wave (CW) ultrasound is commonly used for Doppler applications to measure blood flow velocities. This method uses one crystal to continuously transmit sound and another to continuously receive it, allowing for precise detection of high velocities.

Question 33

34. Ultrasound modes

Foetal biparietal diameter can be measured using A-mode scanning

FALSE.
A-mode (amplitude mode) displays one-dimensional information about echo amplitudes and depths but does not provide two-dimensional images needed for measuring structures like foetal biparietal diameter (BPD). B-mode (brightness mode) is typically used for such measurements.

In B-mode scanning multiple pulses are required per line to build up a single image

TRUE.
In B-mode, multiple pulses are used to scan along different lines across the imaging plane. The echoes from these pulses are processed to create a 2D cross-sectional image of the structures.

Answer 33

In B-mode the amplitude of the echo is inversely proportional to the brightness of the corresponding point on the display

FALSE. DIRECTLY PROPORTIONAL.

In M-mode scanning the vertical coordinate represents depth

True.
In M-mode (motion mode), the vertical axis represents depth (distance from the transducer), while the horizontal axis represents time. This mode is used to analyze motion patterns, such as heart valve movements.

In M-mode scanning each transmission-reception sequence results in a new M-mode line being displayed alongside the other

True.
In M-mode, each ultrasound pulse-transmission and reception sequence generates a single line of data, which is displayed sequentially next to previous lines, forming a dynamic representation of motion over time.

Question 34

The doppler shift frequency is dependent on the velocity of Ultrasound in the propagation medium

True.
2f0 vcosθ /c
​Where:
c: Speed of sound in the propagation medium (e.g., 1,540 m/s in soft tissue) Hence, the Doppler shift depends on the propagation speed of sound in the medium.

There is an upper and lower limit on the blood velocities that may be measured using the Doppler effect which is due to aliasing

True.
Aliasing occurs when the Doppler shift frequency exceeds the Nyquist limit, which is HALF the pulse repetition frequency (PRF). Blood velocities that produce Doppler shifts outside this range cannot be measured accurately, leading to ambiguities in velocity detection.

The magnitude of the returning Doppler signal is dependent on frequency at the point at which the echo is produced

True.
While the transmitted frequency affects the Doppler shift, the magnitude of the returning signal is more significantly influenced by attenuation and scattering.

Answer 34

Doppler Ultrasound gives a direct measurement of volume blood flow rate

FALSE.
Doppler Ultrasound measures blood flow VELOCITIES, not volume blood flow rate directly. Calculating blood flow volume requires additional information, such as the cross-sectional area of the vessel

When the pulse repetition rate is 9 kHz a Doppler shift 5 kHz cannot be measured accurately due to aliasing.

TRUE.
For a PRF of 9 kHz, the Nyquist limit is 4.5 kHz. A Doppler shift of 5 kHz exceeds this limit, leading to aliasing and inaccurate measurements.

Question 35

Concerning radioactivity:

Activity of 1Mbq = 1 disintegration per sec
FALSE.
The unit MBq (Megabecquerel) is defined as 1 million disintegrations per second

The daughter always decays with the same activity as the parent when in equilibrium
True.
At radioactive equilibrium, the activity of the parent and daughter isotopes are equal. This occurs when the rate of decay of the daughter (due to the parent decay) equals the rate of the daughter’s decay, meaning the activity of both parent and daughter are constant over time.

The activity is independent of the half life of the daughter. False.
The activity of the daughter is dependent on its half-life, as the daughter’s decay rate (activity) is influenced by its own decay constant (
𝜆daughter), which is related to its half-life. Shorter half-lives result in higher decay rates (activity) for the daughter, and longer half-lives result in lower decay rates.

Or if it meant activity of parent…
The activity of the parent is determined solely by its own decay constant (which is related to its half-life), and it is independent of the half-life of the daughter.

Answer 35

The daughter decays exponentially when at equilibrium

True.
When at equilibrium, the daughter isotope decays exponentially, just like the parent. However, the exponential decay rate of the daughter depends on its own decay constant. At equilibrium, the activity of both parent and daughter are constant and match each other.

The activity is not related to the time between elusions

False.

The activity of a radionuclide is indeed related to the time between elutions, especially in cases where the radionuclide is used in medical imaging (like PET or SPECT). The longer the time between elutions, the more the parent isotope will decay, potentially decreasing the available activity for the daughter isotope.

Question 36

37. A scintillation detector

Detects by Compton effect only. FALSE.
A scintillator detector does not detect radiation solely through the Compton effect. The scintillator works by absorbing high-energy photons (like X-rays or gamma rays), which cause electrons in the scintillator material to become excited. When the material de-excites, it emits light (scintillation) that is then detected. The Compton effect can contribute to the interaction of gamma rays with the scintillator, but it is not the only mechanism by which scintillator detectors work.

Detects alpha particles. TRUE.
Scintillator detectors can indeed detect alpha particles. The scintillator material can interact with alpha particles, causing them to produce light (scintillation) that can be measured. However, the efficiency for detecting alpha particles is dependent on the type of scintillator used and the energy of the alpha particles.

Cannot count faster than 5,000 per second. FALSE.
Modern scintillator detectors are capable of counting much faster than 5,000 counts per second, depending on the type of scintillator and the detector electronics used. The limiting factor is often the electronics and the processing speed of the detector system, which can count in the range of tens of thousands or even millions of counts per second.

Answer 36

Cannot differentiate between two radioactive isotopes. FALSE.
Scintillator detectors, especially when used in conjunction with pulse shaping or energy discriminators, can differentiate between isotopes based on the energy of the detected radiation. The different isotopes emit gamma rays or other radiation at distinct energies, and the scintillator can detect these energies, allowing differentiation between isotopes.

Converts X radiation into light. TRUE.
This is one of the primary functions of a scintillator detector. When X-rays or gamma rays interact with the scintillator material, they excite the atoms in the material, and the energy is emitted as visible light (scintillation). This light is then detected by a photodetector (such as a photomultiplier tube) and converted into an electrical signal for measurement.

Question 37

38. A gamma camera has

A crystal of thallium-activated gallium iodide
True.
Many gamma cameras use a thallium-activated sodium iodide (NaI(Tl)) crystal, but some systems use gallium iodide (GaI) crystals, often thallium-doped, for certain applications. These crystals are used to detect gamma photons by converting them into visible light (scintillation) that can be detected by photomultiplier tubes (PMTs).

An array of 19 or 37 photomultiplier tubes. True.
Gamma cameras typically have an array of photomultiplier tubes (PMTs) that are arranged around the scintillation crystal. The exact number can vary, and it’s not limited to 19 or 37; the arrangement can depend on the design of the camera. More PMTs allow for better spatial resolution and efficiency in detecting gamma radiation.

A pulse height analyser. True.
Gamma cameras often incorporate a pulse height analyzer (PHA) to filter and select gamma photons based on their energy. This helps to discriminate between the energy of interest (from the radiopharmaceutical) and background radiation, improving the quality of the image by ensuring only the appropriate energies are used.

Answer 37

Physical characteristics well suited to imaging with photons above 250keV. True.
Gamma cameras are designed to handle a wide range of photon energies, including those above 250 keV. The crystal material and PMT setup can efficiently detect higher-energy photons, which are common in nuclear medicine imaging, such as with technetium-99m or iodine-131. However, higher-energy photons can reduce resolution due to scattering and attenuation.

Appreciable reduction in sensitivity at high count rates. True.
Gamma cameras do experience a reduction in sensitivity at high count rates due to pulse pile-up and dead time. When too many photons hit the detector in a short period, the system may miss some events or misinterpret them, reducing the overall sensitivity and image quality at high count rates.

Question 38

39. Radionuclides/radiopharmaceuticals

All radionulides for medical imaging are produced artificially. True.
All radionuclides used in medical imaging, such as technetium-99m (Tc-99m) and fluorine-18 (F-18), are produced artificially, typically in nuclear reactors or cyclotrons. Naturally occurring isotopes do not have the necessary properties for medical imaging applications

Tc-99m is the product of electron capture decay. FALSE.
Technetium-99m (Tc-99m) is actually produced as the decay product of molybdenum-99 (Mo-99), which decays by beta decay to produce Tc-99m. Tc-99m itself decays by gamma emission to stable Tc-99. Electron capture is not the decay mode of Tc-99m

Xenon-133 (13Xe) is used in lung ventilation imaging. True.
Xenon-133 (Xe-133) is indeed used in lung ventilation imaging. It is a gas that can be inhaled by patients, and it emits gamma radiation, which can be detected by a gamma camera to evaluate lung function and ventilation patterns.

Answer 38

Fluorine-18 (18F) is produced in a nuclear reactor.
False.
Fluorine-18 (F-18) is typically produced in a cyclotron, not a nuclear reactor. A cyclotron accelerates particles to bombard a target material (usually oxygen-18) to produce fluorine-18, which is then incorporated into radiopharmaceuticals, like FDG (fluorodeoxyglucose), for PET imaging.

Fluorine-18 (18F) has a half life of 110 minutes

Question 39

40. In positron emission tomography (PET)

Radionuclides that decay by beta+ emission are used. TRUE.

Annihilation photons have energy of 0.51MeV
TRUE.
The annihilation photons produced when a positron meets an electron have an energy of 0.511 MeV each. This is the energy corresponding to the rest mass of the electron and positron. Two 0.511 MeV photons are emitted in opposite directions (180 degrees apart).

The annihilation photons are emitted at 90 degrees to each other. FALSE.
The annihilation photons are emitted 180 degrees apart, not 90 degrees. The two photons travel in opposite directions to conserve momentum, and both are detected simultaneously in a PET scanner.

Answer 39

The spatial resolution is superior to single photon emission tomography (SPECT)
TRUE.
PET typically provides superior spatial resolution compared to SPECT. This is due to the high-energy annihilation photons (511 keV) and the advanced detector technology in PET systems, which allows for better localization and imaging of the tracer distribution.

Radiation dose of PET is significantly less than SPECT

FALSE.

PET generally involves higher radiation doses compared to SPECT. The use of shorter-lived radionuclides (like fluorine-18) in PET and the need for high-energy photons (511 keV) for detection contribute to a higher radiation exposure compared to SPECT, although the dose is still considered to be within acceptable levels for diagnostic imaging.

Radionuclides used in PET (such as fluorine-18 or carbon-11) undergo beta-plus decay (β⁺), which produces a positron. The positron then annihilates with an electron, producing two 511 keV gamma photons.

Radionuclides used in SPECT (such as technetium-99m, iodine-131, or gallium-67) decay via gamma emission (without the intermediate positron emission found in PET). These gamma photons tend to have lower energy compared to PET’s annihilation photons.