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ABOMR practice questions

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Question 1Radiation physics and radiation biology

The SI unit of absorbed dose is the gray. One gray is defined as which of the following?

  1. One joule per kilogram Correct answer.
  2. One joule per coulomb
  3. One coulomb per kilogram
  4. One sievert per kilogram

Explanation

Absorbed dose is energy deposited per unit mass, so 1 Gy equals 1 J/kg. The coulomb per kilogram is the unit of exposure, not absorbed dose.

Question 2Radiation physics and radiation biology

A radiation worker receives 2 mGy from alpha particles and 2 mGy from gamma rays. Which statement best describes the equivalent doses from these two exposures?

  1. The alpha dose is weighted higher than the gamma dose Correct answer.
  2. The gamma dose is weighted higher than the alpha dose
  3. They are equal because absorbed doses are equal
  4. Neither exposure carries any equivalent dose

Explanation

Equivalent dose multiplies absorbed dose by a radiation weighting factor. Alpha particles carry a much higher weighting factor than photons, so equal absorbed doses yield a larger equivalent dose for alpha radiation.

Question 3Radiation physics and radiation biology

Which interaction of x-rays with matter predominates at diagnostic energies in soft tissue and produces both a photoelectron and characteristic photons?

  1. Coherent scattering
  2. Compton scattering
  3. Pair production
  4. Photoelectric absorption Correct answer.

Explanation

Photoelectric absorption predominates at lower diagnostic energies and high atomic number, ejecting an inner-shell electron and emitting characteristic photons. The photoelectron carries most of the incident energy.

Question 4Radiation physics and radiation biology

During Compton scattering, the incident photon interacts with which target?

  1. An inner-shell electron that is ejected
  2. The electric field of the nucleus
  3. The nucleus as a whole
  4. A loosely bound outer-shell electron Correct answer.

Explanation

Compton scattering involves a photon colliding with a loosely bound or free outer-shell electron, transferring part of its energy and scattering at an angle. The ejected electron is the Compton (recoil) electron.

Question 5Radiation physics and radiation biology

Why does the photoelectric effect produce higher subject contrast between bone and soft tissue than Compton scattering does?

  1. Compton scattering depends strongly on atomic number
  2. Compton scattering deposits no energy in tissue
  3. Photoelectric absorption is independent of photon energy
  4. Photoelectric absorption depends strongly on atomic number Correct answer.

Explanation

The photoelectric probability scales approximately with the cube of atomic number, so differences in Z between bone and soft tissue amplify attenuation differences. Compton scattering depends mainly on electron density, which differs far less between these tissues.

Question 6Radiation physics and radiation biology

In an x-ray tube, what is the primary role of the tungsten target?

  1. To convert electron kinetic energy into x-rays Correct answer.
  2. To focus the electron beam onto a small area
  3. To filter low-energy photons from the beam
  4. To remove heat from the anode assembly

Explanation

Electrons accelerated across the tube potential strike the target, where their kinetic energy is converted into bremsstrahlung and characteristic x-rays. Most of the energy becomes heat, which the anode must dissipate.

Question 7Radiation physics and radiation biology

Which change increases the maximum energy of photons produced in an x-ray tube?

  1. Increasing the peak kilovoltage Correct answer.
  2. Increasing the tube current
  3. Increasing the focal spot size
  4. Increasing the anode rotation speed

Explanation

The maximum photon energy equals the electron kinetic energy set by the tube potential, so raising kVp raises the high-energy limit of the spectrum. Tube current changes the number of photons, not their peak energy.

Question 8Radiation physics and radiation biology

Adding aluminum filtration to an x-ray beam primarily accomplishes which of the following?

  1. Narrows the focal spot to improve resolution
  2. Raises the peak photon energy above the kVp set
  3. Increases the number of photons in the beam
  4. Removes low-energy photons and reduces patient skin dose Correct answer.

Explanation

Filtration preferentially absorbs low-energy photons that would otherwise contribute to skin dose without reaching the receptor. It hardens the beam and shifts the mean energy upward without exceeding the kVp limit.

Question 9Radiation physics and radiation biology

The half-value layer of an x-ray beam is best described as which of the following?

  1. The thickness of material that stops all photons
  2. The thickness of material that reduces intensity to half Correct answer.
  3. The distance at which intensity halves in air
  4. The energy at which half of photons are absorbed

Explanation

Half-value layer quantifies beam quality as the absorber thickness that attenuates intensity to 50% of its original value. Larger HVL indicates a more penetrating, higher-quality beam.

Question 10Radiation physics and radiation biology

A monoenergetic beam has a half-value layer of 2 mm of aluminum. What thickness of aluminum reduces its intensity to one-eighth of the original value?

  1. 4 mm
  2. 8 mm
  3. 6 mm Correct answer.
  4. 16 mm

Explanation

Each half-value layer halves the intensity, so reducing to one-eighth requires three half-value layers because 2 cubed equals 8. Three times 2 mm gives 6 mm.

Question 11Radiation physics and radiation biology

Which process results in the production of annihilation photons?

  1. A positron combines with an electron Correct answer.
  2. A nucleus absorbs a thermal neutron
  3. An electron is ejected from an inner shell
  4. A photon scatters from a free electron

Explanation

When a positron loses its kinetic energy and combines with an electron, their masses convert into two 511 keV photons emitted in opposite directions. This is the basis of positron emission tomography.

Question 12Radiation physics and radiation biology

In a photoelectric interaction, the kinetic energy of the ejected photoelectron is best approximated by which expression, where E is photon energy and B is binding energy?

  1. E plus B
  2. E minus B Correct answer.
  3. B minus E
  4. E divided by B

Explanation

The photon must first overcome the electron binding energy, and the remainder appears as the photoelectron's kinetic energy. This is why photoelectric absorption requires photon energy at or above the binding energy.

Question 13Radiation physics and radiation biology

The linear attenuation coefficient of a material depends on which of the following?

  1. Focal spot size and filtration
  2. Photon energy and material density Correct answer.
  3. Tube current and exposure time
  4. Source-to-detector distance

Explanation

The linear attenuation coefficient reflects the probability of interaction per unit path length and varies with photon energy and the density and composition of the absorber. Geometry and exposure settings do not define it.

Question 14Radiation physics and radiation biology

Which statement correctly distinguishes the mass attenuation coefficient from the linear attenuation coefficient?

  1. Mass coefficient is normalized to material density Correct answer.
  2. Linear coefficient is expressed in units of area per mass
  3. Linear coefficient applies only to air
  4. Mass coefficient is independent of photon energy

Explanation

Dividing the linear attenuation coefficient by density yields the mass attenuation coefficient, which allows comparison between materials of different densities. It still varies with photon energy.

Question 15Radiation physics and radiation biology

Which quantity describes the probability of stochastic effects such as cancer induction from radiation exposure?

  1. The exposure measured in air
  2. The half-value layer of the beam
  3. The absorbed dose rate only
  4. The equivalent dose, which accounts for radiation type Correct answer.

Explanation

Stochastic risk is estimated using equivalent and effective dose, which incorporate radiation weighting and tissue sensitivity. Absorbed dose alone does not account for the differing biological effectiveness of radiation types.

Question 16Radiation physics and radiation biology

Which statement best describes deterministic effects of radiation?

  1. Their severity increases with dose above a threshold Correct answer.
  2. Their probability rises with dose with no threshold
  3. They occur only after a single high-energy photon
  4. They are independent of the dose received

Explanation

Deterministic effects have a threshold dose above which severity increases with dose, reflecting cell killing. Examples include erythema and cataract formation.

Question 17Radiation physics and radiation biology

According to the linear no-threshold model, how is the risk of stochastic effects treated at very low doses?

  1. Risk exists only above the deterministic threshold
  2. Risk decreases as dose increases
  3. Risk is assumed zero below 1 mGy
  4. Risk is assumed proportional to dose with no safe threshold Correct answer.

Explanation

The linear no-threshold model assumes stochastic risk increases linearly with dose and extrapolates through zero, with no threshold. It is a conservative basis for radiation protection.

Question 18Radiation physics and radiation biology

Which radiation-induced cellular effect is most closely associated with high-dose exposure and results in loss of reproductive capability of cells?

  1. Reversible membrane depolarization
  2. Reproductive cell death Correct answer.
  3. Point mutation without cell death
  4. Increased mitotic rate

Explanation

High radiation doses can cause reproductive cell death, in which cells lose the ability to divide. This underlies deterministic tissue responses such as skin desquamation.

Question 19Radiation physics and radiation biology

Which cell cycle phase is generally most radiosensitive?

  1. G2 and M phases Correct answer.
  2. Early G1 phase
  3. Late S phase
  4. G0 phase

Explanation

Cells are most radiosensitive during G2 and M, when chromosomes are condensed and DNA repair is limited. Late S phase is relatively radioresistant.

Question 20Radiation physics and radiation biology

The law of Bergonié and Tribondeau states that cells are most radiosensitive when they have which characteristics?

  1. Low differentiation and high mitotic activity Correct answer.
  2. High differentiation and low mitotic activity
  3. High differentiation and long lifespan
  4. Low mitotic activity and high specialization

Explanation

Radiosensitivity increases with mitotic activity and undifferentiated status, and decreases with specialization. Rapidly dividing, poorly differentiated cells such as those in bone marrow are most vulnerable.

Question 21Radiation physics and radiation biology

Which of the following tissues is generally considered most radiosensitive?

  1. Mature skeletal muscle
  2. Peripheral nerve
  3. Mature cartilage
  4. Bone marrow Correct answer.

Explanation

Bone marrow contains rapidly dividing hematopoietic stem cells and is among the most radiosensitive tissues. Mature muscle, cartilage, and nerve are highly differentiated and relatively resistant.

Question 22Radiation physics and radiation biology

Which DNA lesion is considered most important for radiation-induced cell killing and mutation?

  1. Single-strand break
  2. Double-strand break Correct answer.
  3. Base oxidation
  4. DNA-protein crosslink

Explanation

Double-strand breaks are difficult to repair faithfully and are strongly associated with cell death and chromosomal aberrations. Single-strand breaks are more readily repaired using the intact complementary strand.

Question 23Radiation physics and radiation biology

The oxygen enhancement ratio describes which of the following?

  1. The ratio of dose in oxygenated versus hypoxic conditions for equal effect Correct answer.
  2. The ratio of beam intensity to filtration thickness
  3. The ratio of photoelectric to Compton interactions
  4. The ratio of absorbed dose to equivalent dose

Explanation

The oxygen enhancement ratio compares doses needed to produce the same biological effect under oxygenated and hypoxic conditions. Oxygen increases the fixation of radiation damage, raising effectiveness.

Question 24Radiation physics and radiation biology

Which statement best explains why fractionating a radiation dose spares normal tissue more than tumor tissue in radiotherapy?

  1. Tumor cells are always hypoxic and resistant
  2. Normal tissues repair sublethal damage better between fractions Correct answer.
  3. Normal tissues lack DNA repair mechanisms
  4. Fractionation reduces the total dose delivered

Explanation

Normal tissues generally repair sublethal damage and repopulate between fractions better than tumors. Fractionation exploits these differences to widen the therapeutic ratio.

Question 25Imaging equipment and image acquisition techniques

Which radiographic density change results from doubling the milliampere-seconds (mAs) while all other exposure factors are held constant?

  1. The image noise decreases markedly
  2. The image becomes approximately half as dark
  3. The image contrast increases markedly
  4. The image becomes approximately twice as dark Correct answer.

Explanation

mAs is directly proportional to the number of x-ray photons produced, so doubling mAs roughly doubles receptor exposure, which increases radiographic density. Contrast and noise are governed mainly by other factors such as kVp and photon statistics.

Question 26Imaging equipment and image acquisition techniques

Increasing kilovoltage peak (kVp) while mAs is unchanged primarily produces which effect on the image?

  1. No change in density or contrast
  2. Increased penetration with a longer scale of contrast Correct answer.
  3. Reduced penetration and higher contrast
  4. A proportional decrease in beam collimation

Explanation

Higher kVp increases photon energy and penetrance, so more structures are penetrated and the histogram of transmitted intensities broadens, producing a longer scale (lower) contrast. Density also tends to rise because of greater transmission through the patient.

Question 27Imaging equipment and image acquisition techniques

What is the primary function of added filtration in the x-ray beam?

  1. To focus the beam onto the receptor
  2. To increase the beam intensity at the center
  3. To remove low-energy photons that would add patient dose without contributing to the image Correct answer.
  4. To increase the effective focal spot size

Explanation

Added filtration attenuates low-energy photons that are absorbed in tissue but do not reach the receptor, thereby reducing patient skin dose and hardening the beam. It does not focus the beam or enlarge the focal spot.

Question 28Imaging equipment and image acquisition techniques

Which device is used to restrict the x-ray field and reduce scatter reaching the receptor?

  1. Photostimulable phosphor plate
  2. Intensifying screen
  3. Grid
  4. Collimator Correct answer.

Explanation

A collimator contains adjustable lead shutters that define the field size, limiting the volume of tissue irradiated and reducing scatter production. Grids remove scatter after it is produced rather than limiting the field.

Question 29Imaging equipment and image acquisition techniques

In a dental panoramic unit, the receptor and x-ray tube rotate around the patient's head to produce an image of the curved dental arch. What is this motion called?

  1. Rotational panoramic movement Correct answer.
  2. Cephalometric translation
  3. Tomographic rotation
  4. Linear scanning

Explanation

Panoramic radiography uses coordinated rotation of the tube and receptor about a moving center of rotation to image the curved focal trough. This rotational panoramic movement blurs structures outside the trough.

Question 30Imaging equipment and image acquisition techniques

Which factor most directly determines the size of the effective focal spot in an x-ray tube?

  1. The filament temperature only
  2. The kVp selected
  3. The exposure time
  4. The angle of the anode target Correct answer.

Explanation

The effective focal spot is the projection of the actual focal spot onto the patient and is reduced by a steeper target angle (line-focus principle). Filament temperature and kVp affect output and spectrum, not the projected spot size.

Question 31Imaging equipment and image acquisition techniques

Which statement best describes the purpose of the focal trough in panoramic radiography?

  1. It sets the kilovoltage for the exposure
  2. It determines the radiation dose to the thyroid
  3. It defines the region where structures are imaged with acceptable sharpness Correct answer.
  4. It limits the field of view to the anterior teeth

Explanation

The focal trough is the three-dimensional curved zone in which structures are positioned to be imaged sharply; objects outside it are blurred by motion. It is a geometric concept, not a dose or exposure control.

Question 32Imaging equipment and image acquisition techniques

A clinician wants to reduce motion blur on a panoramic image of a nervous patient. Which adjustment is most appropriate?

  1. Use the shortest available exposure time Correct answer.
  2. Increase the exposure time
  3. Decrease the milliamperage
  4. Increase the focal spot size

Explanation

Shorter exposure times reduce the opportunity for patient movement to degrade the image, which is especially important for panoramic units with long rotational times. mA and focal spot size do not address motion.

Question 33Imaging equipment and image acquisition techniques

Which receptor type requires the lowest radiation exposure to produce a diagnostic image?

  1. Direct-exposure film
  2. Photostimulable phosphor plate
  3. Charge-coupled device sensor Correct answer.
  4. Screen-film with rare-earth screens

Explanation

Solid-state CCD/CMOS sensors convert x-rays directly to electronic signal with high sensitivity, allowing lower exposures than film or phosphor plates. Film and phosphor systems require more photons to reach a detectable signal.

Question 34Imaging equipment and image acquisition techniques

What is the main advantage of a charge-coupled device (CCD) intraoral sensor over photostimulable phosphor plates?

  1. It requires no cable connection to the computer
  2. It provides immediate image display without a scanning step Correct answer.
  3. It has a larger active area for the same size
  4. It can be sterilized by autoclaving

Explanation

CCD/CMOS sensors produce an electronic image immediately after exposure, eliminating the plate scanning and erasing cycle. They typically require a cable or wireless link and cannot be autoclaved.

Question 35Imaging equipment and image acquisition techniques

Which combination of exposure factors would most likely produce a quantum mottle (noisy) digital image?

  1. High mAs and high kVp
  2. Low mAs and high kVp Correct answer.
  3. Low mAs and low kVp
  4. High mAs and low kVp

Explanation

Quantum mottle arises when too few photons reach the receptor, which occurs with low mAs; high kVp further reduces the number of photons needed per unit density but does not compensate for an inadequate photon count, so the combination of low mAs and high kVp is most likely to be noisy. Adequate mAs is the key to reducing noise.

Question 36Imaging equipment and image acquisition techniques

A digital image appears grainy and the clinician increases mAs. What is the expected effect?

  1. Spatial resolution increases because pixels enlarge
  2. The image becomes lighter because of reduced exposure
  3. Contrast increases because kVp rises
  4. Noise decreases because more photons reach the receptor Correct answer.

Explanation

Increasing mAs raises the number of photons reaching the receptor, improving the signal-to-noise ratio and reducing graininess. It does not change kVp or pixel size.

Question 37Imaging equipment and image acquisition techniques

Which statement about the relationship between spatial resolution and pixel size is correct?

  1. Pixel size has no effect on resolution
  2. Larger pixels improve spatial resolution
  3. Resolution depends only on kVp
  4. Smaller pixels improve spatial resolution Correct answer.

Explanation

Spatial resolution is limited by pixel dimensions; smaller pixels sample finer detail and improve resolution. kVp affects contrast and noise, not the geometric sampling limit.

Question 38Imaging equipment and image acquisition techniques

A clinician notices that the same anatomical region appears elongated on a panoramic image. Which patient positioning error is most likely?

  1. The patient's chin is raised too high Correct answer.
  2. The patient's head is positioned too far forward relative to the focal trough
  3. The patient's lips are closed
  4. The patient's head is tilted to one side

Explanation

Raising the chin too high places the mandible and maxilla outside the focal trough and produces elongation of the image. Head tilt causes asymmetry, and forward positioning causes widening or blurring.

Question 39Imaging equipment and image acquisition techniques

Which exposure factor primarily controls the contrast of a digital image when the receptor is not the limiting factor?

  1. Source-to-receptor distance
  2. mAs
  3. Focal spot size
  4. kVp Correct answer.

Explanation

kVp determines the energy spectrum and thus the subject contrast transmitted to the receptor; higher kVp produces a longer scale of contrast. mAs controls quantity and noise, while geometry affects resolution and magnification.

Question 40Imaging equipment and image acquisition techniques

What is the effect of increasing the source-to-receptor distance while maintaining receptor exposure?

  1. Increased magnification and decreased sharpness
  2. No change in magnification
  3. Increased contrast and decreased noise
  4. Decreased magnification and increased sharpness Correct answer.

Explanation

Increasing source-to-receptor distance reduces the geometric magnification of structures and improves sharpness because the beam becomes more parallel. Exposure must be adjusted to compensate for the inverse-square loss.

Question 41Imaging equipment and image acquisition techniques

Which statement best describes the line-focus principle in dental x-ray tubes?

  1. The cathode focuses electrons onto a large area
  2. The anode is angled so the effective focal spot is smaller than the actual focal spot Correct answer.
  3. The receptor is positioned along a straight line
  4. The beam is collimated to a line shape

Explanation

Angling the target allows a larger actual focal spot to dissipate heat while projecting a smaller effective focal spot toward the patient, improving sharpness. It does not describe collimation or receptor geometry.

Question 42Imaging equipment and image acquisition techniques

A panoramic image shows the anterior teeth blurred while the posterior regions are sharp. Which adjustment is most likely needed?

  1. Decrease the exposure time
  2. Increase the kVp
  3. Increase the anode angle
  4. Move the patient posteriorly within the focal trough Correct answer.

Explanation

Anterior blurring often results from the patient being positioned too far forward or backward relative to the focal trough; repositioning along the trough brings the anterior teeth into the sharp zone. Exposure factors and anode angle do not correct focal trough misalignment.

Question 43Imaging equipment and image acquisition techniques

Which factor is most important for minimizing geometric unsharpness in intraoral radiography?

  1. Using the shortest possible source-to-receptor distance
  2. Increasing the object-to-receptor distance
  3. Using a small focal spot and long source-to-receptor distance Correct answer.
  4. Using the largest available focal spot

Explanation

Geometric unsharpness is reduced by a small focal spot and a long source-to-receptor distance, both of which narrow the penumbra. A short distance or large focal spot increases blur.

Question 44Imaging equipment and image acquisition techniques

A digital sensor produces an image with a horizontal band of reduced density. Which cause is most likely?

  1. Patient motion
  2. A defective column in the sensor Correct answer.
  3. A grid artifact
  4. Incorrect kVp selection

Explanation

A horizontal band of altered density on a solid-state sensor often indicates a defective row or column of pixels. Motion produces blur, and kVp affects overall contrast rather than a localized band.

Question 45Imaging equipment and image acquisition techniques

Which statement about the effect of collimation on image quality is correct?

  1. Collimation increases the effective focal spot size
  2. Collimation has no effect on contrast
  3. Tighter collimation reduces scatter and improves contrast Correct answer.
  4. Tighter collimation increases scatter reaching the receptor

Explanation

Restricting the field reduces the volume of tissue irradiated and thus the amount of scatter produced, which lowers fog and improves contrast. Collimation does not alter focal spot size.

Question 46Imaging equipment and image acquisition techniques

An intraoral image appears uniformly too light despite correct positioning. Which change is most appropriate?

  1. Decrease the kVp
  2. Increase the mAs Correct answer.
  3. Decrease the exposure time
  4. Increase the source-to-receptor distance

Explanation

A uniformly light image indicates underexposure, which is corrected by increasing mAs to raise the number of photons reaching the receptor. Decreasing kVp or exposure time would worsen underexposure.

Question 47Imaging equipment and image acquisition techniques

Which statement best explains why a grid is used with a large field of view in extraoral imaging?

  1. It reduces the focal spot size
  2. It absorbs scattered radiation before it reaches the receptor Correct answer.
  3. It increases magnification to improve detail
  4. It converts x-rays to visible light

Explanation

A grid consists of radiopaque strips that absorb obliquely directed scatter while transmitting primary photons, improving contrast when a large volume of tissue produces significant scatter. It does not magnify or convert photons.

Question 48Imaging equipment and image acquisition techniques

A clinician compares two panoramic units: one uses a moving center of rotation and the other a stationary center. Which statement is most accurate?

  1. The moving center of rotation allows the focal trough to better follow the dental arch Correct answer.
  2. The stationary center produces a sharper image of the entire arch
  3. The center of rotation has no effect on image geometry
  4. The center of rotation determines the kVp

Explanation

A moving center of rotation changes the focal trough shape during the scan so it more closely follows the curved dental arch, improving sharpness across the arch. The center of rotation is a geometric parameter and does not set kVp.

Question 49Image processing, quality assurance, and radiation safety

Which exposure factor change will reduce patient dose the most while still producing a diagnostic image, all else equal?

  1. Adding filtration to the beam
  2. Halving the mAs Correct answer.
  3. Doubling the kVp
  4. Increasing the source-to-skin distance

Explanation

Patient dose in radiography is directly proportional to mAs, so halving mAs halves the number of photons reaching the patient. The other changes affect beam quality or geometry in ways that do not simply reduce entrance dose by the same proportion.

Question 50Image processing, quality assurance, and radiation safety

In digital radiography, which factor primarily determines spatial resolution of the final image?

  1. The focal spot size alone
  2. The window and level settings applied
  3. Matrix size and pixel pitch of the receptor Correct answer.
  4. The mAs selected at the control panel

Explanation

Spatial resolution in digital imaging is governed by the receptor's matrix dimensions and pixel pitch, which set the smallest resolvable detail. Display processing and exposure factors affect contrast or noise but not the fundamental sampling limit.