Radiography and Fluoroscopy Physics Review Questions

Take these questions as a:

Magnification and Collimation

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  1. Skin radiation burns are best measured with what metric?
    1. Tube current (mA)
    2. Tube voltage (kV)
    3. Tube air kerma (Ka)
    4. Dose-area product (DAP)

    Skin injuries are a result of the dose to the skin. Dose represents the energy deposited per gram of tissue, which can also be measured as Kerma (kinetic energy released in matter). Kerma in air can be converted to dose in soft tissue using numerical constants. Dose-area product represents the dose times the area exposed. Kerma depends on mA and kV, but each alone does not give the dose.

  2. As opposed to geometric magnification, electronic magnification does not change the:
    1. Dose
    2. Field-of-view
    3. Focal spot blur (penumbra)
    4. Resolution

    The focal spot blur does not change with electronic magnification but does change (is magnified) with geometric magnification. Resolution increases with magnification (that's the whole point), although this is degraded in geometric magnification by focal spot blur; typically, in magnification mode the device will switch to a smaller focal spot. Dose increases to maintain the same level of exposure (noise characteristics) on the magnified image. Field of view is typically smaller in a magnified image of any type.

  3. Motion is a problem in magnification mammography because
    1. Focal spot blur increases
    2. Motion blur increases
    3. Imaging time increases
    4. Resolution increases
    5. B and C

    In magnification mammography, in order to improve resolution a smaller focal spot is used. To prevent anode overheating, the current is decreased, and thus the exposure takes longer. The patient has to stay still for longer. Motion blur is also magnified with the geometric magnification.

  4. Collimation directly reduces
    1. Resolution
    2. Dose
    3. Dose-area product
    4. Tube voltage
    5. Magnification

    Collimation reduces the area of the patient that is exposed to radiation, thereby reducing DAP. The dose (energy per mass) is unchanged, although in reality since scatter is decreased, the dose actually will go down. Resolution and magnification are unchanged by collimation.

  5. Magnification (geometric or electronic) always results in
    1. Increase in Dose
    2. Increase in DAP
    3. Decrease in Dose
    4. Decrease in DAP
    5. No change in Dose

    In order for the magnified image to have the same noise as a full-view image, more dose is needed because the pixels are smaller. In fact, the dose will go up with the square of the magnification. This results in increasing risk of skin injury. However, the area irradiated is similarly smaller, so DAP will typically not change. (In the case of flat-panel detector electronic magnification, the DAP may actually decrease.)

X-Ray Interaction with Matter

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  1. In order to obtain higher-energy x-rays, we need to change
    1. Tube current (mA)
    2. Exposure time (s)
    3. Tube voltage (kV)
    4. Filament voltage
    5. Source to object distance (SOD)

    The maximum x-ray energy produced from an x-ray tube depends on the voltage across the tube. (Filament current and voltage, not discussed in the text, generate the electrons in the first place but do not accelerate them.) Tube current affects the number of electrons - and thus number of photons - produced.

  2. Deposition of x-ray energy in the patient is
    1. Always bad and should be eliminated in an ideal system
    2. Never bad and should be increased in an ideal system
    3. Necessary and represents a trade-off between diagnostic information and patient harm

    As silly as this question is, it is here to remind us that if x-rays did not interact with and deposit energy in the patient, there would be no image. X-rays going right through the patient produce no information. The image is generated by differences in how many x-rays are stopped by different tissues.

  3. The intensity of an x-ray beam passing through a patient
    1. Increases linearly
    2. Decreases linearly
    3. Decreases quadratically
    4. Decreases exponentially

    By the Beer-Lambert law, x-ray intensity drops off exponentially as it passes through tissue. The rate of decrease is determined by the linear attenuation coefficient.

  4. X-ray dose is deposited
    1. Preferentially near the skin
    2. Preferentially in the middle of the patient
    3. Preferentially towards the detector
    4. Relatively uniformly throughout

    Because of the exponential beam attenuation, almost all of the beam is attenuated in the first few centimeters of the patient. This attenuation is what is responsible for depositing dose - thus, in the first few centimeters of the patient. Remember that this represents the skin and subcutaneous tissues closest to the x-ray tube, of course.

  5. Larger patients require higher tube currents (more radiation, increased air kerma) in order to improve image
    1. Contrast
    2. Noise
    3. Resolution
    4. Dose

    Since larger patients attenuate the x-ray beam more, fewer photons would be available to form the image. Since noise is related to the number of photons available (SNR is proportional to 1/sqrt(N) ), we need to increase the number of x-rays to create a less noisy image. Tissue contrast is typically worse in large patients because of increased scatter, but that cannot be easily fixed.

Tissue Contrast and Mammography

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  1. Tissue contrast is generated by what type of interaction
    1. Coherent (Rayleigh) scatter
    2. Incoherent (Compton) scatter
    3. Photoelectric effect
    4. Overall attenuation
    5. Beam hardening effect

    The predominant means by which tissues differ in their attenuation - and thus contrast is generated - is with the photoelectric effect, which is highly dependent on the elemental composition of the tissue. Compton scatter depends on the electron density, which is similar in most biologic materials. Rayleigh scattering is only a minor effect at diagnostic energies.

  2. In order to decrease dose in large patients:
    1. Decrease kV and improve image noise
    2. Increase kV and improve image noise
    3. Increase kV and improve image contrast
    4. Decrease kV and improve image contrast

    Increasing kV decreases dose in large patients (well, in any size patients) because of less attenuation at higher energies. The trade-off is worse contrast.

  3. Iodine is an effective contrast agent because
    1. It has a high likelihood of photoelectric interaction at diagnostic x-ray energies
    2. It has a high likelihood of Compton interaction at diagnostic x-ray energies
    3. It has a high electron density
    4. It has a high mass density

    The k-edge of iodine around 30 keV strongly increases its likelihood of photoelectric interaction, thereby increasing the attenuation.

  4. In order to distinguish cancer from glandular tissue, mammography relies on
    1. Intravenous contrast agents
    2. High tube currents (mA)
    3. High x-ray energies to increase photoelectric effect
    4. Low x-ray energies to increase Compton scattering
    5. Low x-ray energies to increase photoelectric effect

    Again to re-emphasize the point: The photoelectric effect predominates at low x-ray energies and is the best way to distinguish different tissues.

  5. Molybdenum was traditionally used as an x-ray anode in mammography because of its
    1. Good heat capacity
    2. Characteristic x-rays
    3. Efficient Bremsstrahlung radiation
    4. Intrinsic filtration

    Molybdenum is an ideal anode for producing low-energy x-rays because it has characteristic x-rays at ~17 and 20 keV, which are in a very good range for photoelectric interactions in glandular tissue.

  6. Switching from a molybdenum to a tungsten anode demonstrates what trade-off
    1. Lower dose, worse contrast
    2. Higher dose, better contrast
    3. Higher dose, better heat capacity

    Tungsten does not have characteristic x-rays in the relevant spectrum (they are around 60-70 keV). Overall, the average x-rays end up being higher with tungsten anodes but therefore give lower dose. Also, tungsten's heat capacity is much better, allowing for longer exposures / multiple images such as for tomosynthesis.


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  1. The most important characteristic of a diagnostic image is
    1. It must look nice
    2. It must be fast to acquire
    3. It must contain sufficient information
    4. It must be high resolution

    Images can look terrible but contain diagnostic information and vice versa. In medicine, where there are always trade-offs between benefits and harms of any procedure or test, it is critical to obtain diagnostic information rather than focusing on how nice an image looks. We must employ as little radiation as possible in our imaging (ALARA principle).

  2. In x-ray imaging, noise is determined by
    1. Energy of x-rays hitting the detector
    2. Number of x-rays hitting the detector
    3. Tissue contrast
    4. Amount of scatter hitting the detector

    Noise is principally determined by the number of photons received by the detector (more photons, less noise). While scatter could be considered a type of structured or background noise, its principal effect is to worsen image contrast.

  3. With all other parameters unchanged, decreasing kV in a contrast angiogram has what effects?
    1. Improved contrast, improved noise
    2. Worse contrast, worse noise
    3. Improved contrast, worse noise
    4. Worse contrast, improved noise

    Decreasing kV decreases the x-ray penetration [and also photon flux], thus giving worse noise. (Typically the automated exposure compensation will increase the mA to compensate.) However, contrast is improved at low kV, especially with iodine compared to soft tissue, given its k-edge near 30 keV. This may allow one to actually use a lower dose overall, since SNR will still be higher (because the signal is greater).

  4. In x-ray imaging, improving SNR requires increased dose.
    1. True
    2. False

    Improving SNR can involve increased dose, but there are other interventions. For example, one can use pixel binning to trade resolution for dose. Alternatively, one can use a contrast agent to increase the signal itself. Changes in kV, as noted above, can also improve contrast at the expense of dose.


Back to section.

  1. In x-ray imaging, scatter:
    1. Contributes to tissue contrast
    2. Degrades tissue contrast
    3. Has no effect on tissue contrast

    Scatter degrades tissue contrast by adding a large amount of 'background' x-rays to every pixel. Scatter also does not distingush well between different tissue types and thus does not contribute to the generation of contrast.

  2. Scatter is a big factor in which of the following imaging examinations
    1. Pediatric PICC placement
    2. Dialysis fistulagram
    3. Mesenteric angiogram
    4. Knee arthrogram

    Scatter is a big factor in thick body parts such as the abdomen. Pediatric patients (non-obese ones) and adult extremities are much less affected by scatter.

  3. The major disadvantage of using grids is
    1. Worse resolution
    2. Increased dose
    3. Worse contrast
    4. Longer imaging time

    Because grids attenuate primary beam x-rays, they decrease signal; they also decrease the scatter background and thereby increase noise (since there are many fewer photons hitting the detector now). The combination of these effects requires an increased dose to obtain the same image noise.

  4. Collimation results in
    1. Increased radiation exposure to the patient
    2. Increased scatter within the patient
    3. Improved tissue contrast
    4. Improved quantum mottle

    By reducing scatter from elsewhere in the body, collimation improves the apparent contrast. Quantum mottle is unchanged. There is decreased radiation to the patient because a smaller area is irradiated (and there is less scatter).

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