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Arab Board Radiology Part 1 study materials
Arab Board Radiology Part 1 study resources

Arab Board Radiology Part 1 practice questions

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Question 1Physics, Instrumentation, and Radiologic Technology

Which quantity is defined as the rate of flow of electric charge through a conductor?

  1. Electric resistance
  2. Electric current Correct answer.
  3. Electric potential
  4. Electric capacitance

Explanation

Electric current is the rate at which charge passes a point in a circuit, measured in amperes (coulombs per second). Potential difference drives the charge, resistance opposes its flow, and capacitance describes charge storage.

Question 2Physics, Instrumentation, and Radiologic Technology

A radiograph requires a mAs of 20 at a source-to-image distance of 100 cm. To maintain the same receptor exposure at 200 cm, what mAs is required?

  1. 80 mAs Correct answer.
  2. 40 mAs
  3. 160 mAs
  4. 10 mAs

Explanation

The inverse square law states that intensity falls with the square of distance. Doubling the distance reduces intensity to one quarter, so mAs must increase fourfold (20 × 4 = 80 mAs).

Question 3Physics, Instrumentation, and Radiologic Technology

In an X-ray tube, the purpose of the oil surrounding the tube housing is primarily to:

  1. Provide electrical insulation and heat conduction Correct answer.
  2. Convert electrons into X-ray photons
  3. Filter low-energy photons from the beam
  4. Collimate the useful beam to the field size

Explanation

Tube oil serves as a dielectric medium for high-voltage insulation and conducts heat away from the anode assembly. Filtration, collimation, and photon production occur at other components.

Question 4Physics, Instrumentation, and Radiologic Technology

Increasing the tube voltage across an X-ray tube while keeping tube current constant will:

  1. Increase the maximum photon energy Correct answer.
  2. Decrease the number of photons produced
  3. Decrease the maximum photon energy
  4. Leave the spectrum completely unchanged

Explanation

The maximum photon energy (in keV) is numerically equal to the peak kilovoltage applied across the tube. Raising kVp therefore raises the endpoint of the bremsstrahlung spectrum and shifts it to higher energies.

Question 5Physics, Instrumentation, and Radiologic Technology

The half-value layer of an X-ray beam is best described as the:

  1. Energy at which half the photons are absorbed
  2. Distance at which intensity falls to half
  3. Thickness of material that reduces beam intensity to half Correct answer.
  4. Time needed for tube output to halve

Explanation

Half-value layer (HVL) is a measure of beam quality expressed as the absorber thickness that attenuates intensity by 50 percent. It reflects the penetrating power of the beam.

Question 6Physics, Instrumentation, and Radiologic Technology

Which interaction predominates in the diagnostic energy range for soft tissue and produces scattered photons that degrade image contrast?

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

Explanation

Compton scattering dominates at diagnostic energies in soft tissue and deflects photons, producing scatter that reduces contrast. Photoelectric absorption dominates in bone and at low energies, while pair production requires energies above 1.02 MeV.

Question 7Physics, Instrumentation, and Radiologic Technology

A radiographer increases kVp from 70 to 80 while leaving mAs unchanged. What is the most likely effect on the image?

  1. Decreased density and increased noise
  2. No change in density or contrast
  3. Increased density with lower contrast Correct answer.
  4. Increased contrast and decreased density

Explanation

Raising kVp increases beam penetration and output, raising receptor exposure (density) while reducing subject contrast because more photons penetrate rather than being differentially absorbed.

Question 8Physics, Instrumentation, and Radiologic Technology

Which material is most commonly used as the anode target in a general-purpose diagnostic X-ray tube?

  1. Molybdenum
  2. Tungsten Correct answer.
  3. Aluminium
  4. Copper

Explanation

Tungsten is chosen for its high atomic number, high melting point, and good thermal conductivity, giving efficient X-ray production and heat tolerance. Copper and aluminium lack the atomic number, and molybdenum is reserved for specific applications.

Question 9Physics, Instrumentation, and Radiologic Technology

The line-focus principle in X-ray tube design is used to:

  1. Reduce off-focus radiation reaching the detector
  2. Increase the effective focal spot area for heat loading
  3. Spread heat over a larger area while keeping a small effective focal spot Correct answer.
  4. Align the collimator with the central ray

Explanation

Angling the target face allows a larger actual focal spot to project a smaller effective focal spot, improving spatial resolution while maintaining heat capacity.

Question 10Physics, Instrumentation, and Radiologic Technology

Which factor most directly limits the spatial resolution of a computed radiography image?

  1. Tube current and exposure time
  2. Laser spot size and plate pixel pitch Correct answer.
  3. Generator rectification type
  4. Collimator aperture size

Explanation

In CR, resolution is limited by the size of the laser readout spot and the sampling pitch of the photostimulable plate. Exposure factors and generator design affect dose and output, not the sampling limit.

Question 11Physics, Instrumentation, and Radiologic Technology

In digital radiography, the term detective quantum efficiency (DQE) describes:

  1. The variation in pixel values across a uniform field
  2. The ability of a detector to convert incident radiation into useful signal relative to an ideal detector Correct answer.
  3. The fraction of incident X-rays absorbed by the detector
  4. The maximum spatial frequency the detector can resolve

Explanation

DQE expresses how efficiently a detector converts incident radiation into output signal-to-noise ratio compared with an ideal detector. It combines absorption efficiency and noise characteristics into one metric.

Question 12Physics, Instrumentation, and Radiologic Technology

Which detector type directly converts X-ray energy into an electrical charge without a light conversion step?

  1. Photostimulable phosphor plate
  2. Caesium iodide scintillator with photodiode
  3. Rare-earth intensifying screen with film
  4. Amorphous selenium flat panel Correct answer.

Explanation

Amorphous selenium is a photoconductor that converts absorbed X-rays directly into electron-hole pairs, which are collected as charge. The other listed systems use a light intermediate step.

Question 13Physics, Instrumentation, and Radiologic Technology

A grid with a ratio of 12:1 is used with a beam of 80 kVp. What is the main consequence compared with using no grid?

  1. Higher patient dose and improved contrast Correct answer.
  2. Lower patient dose and higher scatter reaching the detector
  3. No change in dose or contrast
  4. Higher patient dose and reduced contrast

Explanation

A high-ratio grid removes scatter before it reaches the detector, improving contrast but requiring increased exposure to compensate for primary beam attenuation, thereby raising patient dose.

Question 14Physics, Instrumentation, and Radiologic Technology

Grid cut-off is most likely to occur when:

  1. The central ray is angled across the grid lines Correct answer.
  2. The grid ratio is decreased
  3. The collimated field is reduced
  4. The source-to-image distance is increased

Explanation

Grid cut-off occurs when primary photons are absorbed by the grid strips because of misalignment, such as tube angulation across the grid lines or decentring. Decreasing ratio or adjusting distance reduces this risk.

Question 15Physics, Instrumentation, and Radiologic Technology

The main purpose of added beam filtration in diagnostic radiography is to:

  1. Increase the maximum photon energy of the beam
  2. Reduce the effective focal spot size
  3. Remove low-energy photons that would increase patient dose without contributing to the image Correct answer.
  4. Increase the number of low-energy photons

Explanation

Added filtration preferentially absorbs low-energy photons that would be absorbed in the patient without reaching the detector, reducing skin dose while leaving the useful beam relatively intact.

Question 16Physics, Instrumentation, and Radiologic Technology

Automatic exposure control terminates the exposure when:

  1. A preset mAs value is reached
  2. A predetermined quantity of radiation reaches the detector Correct answer.
  3. The collimator light is switched off
  4. The tube anode reaches a set temperature

Explanation

AEC uses ionisation chambers or solid-state detectors to monitor radiation reaching the receptor and terminates exposure once a preset signal is achieved. This adjusts exposure time to patient thickness.

Question 17Physics, Instrumentation, and Radiologic Technology

In mammography, a molybdenum anode with a molybdenum filter is used primarily because it:

  1. Eliminates the need for a grid
  2. Produces a high-energy beam for dense breasts
  3. Reduces the need for breast compression
  4. Provides characteristic X-rays in the low-energy range optimal for breast tissue Correct answer.

Explanation

The Mo/Mo combination produces characteristic photons around 17 to 20 keV, which optimise contrast between breast tissues. This low-energy spectrum suits the small thickness and low attenuation of breast tissue.

Question 18Physics, Instrumentation, and Radiologic Technology

Which statement correctly describes the relationship between spatial frequency and the modulation transfer function (MTF)?

  1. MTF generally decreases as spatial frequency increases Correct answer.
  2. MTF increases as spatial frequency increases
  3. MTF remains constant across all spatial frequencies
  4. MTF is unrelated to spatial resolution

Explanation

The MTF describes how well a system preserves contrast at a given spatial frequency and typically falls off as frequency rises, reflecting loss of fine detail. It is a key indicator of spatial resolution.

Question 19Physics, Instrumentation, and Radiologic Technology

The principal advantage of pulsed fluoroscopy over continuous fluoroscopy is:

  1. Improved spatial resolution beyond detector limits
  2. Higher frame rate and smoother motion
  3. Reduced patient and staff dose from fewer exposure pulses Correct answer.
  4. Elimination of the image intensifier

Explanation

Pulsed fluoroscopy acquires images at discrete intervals rather than continuously, reducing the total exposure time and therefore dose while maintaining adequate visualisation.

Question 20Physics, Instrumentation, and Radiologic Technology

Which physical principle underlies the operation of an image intensifier's input phosphor?

  1. Photoelectric emission of electrons
  2. Conversion of X-ray photons into light photons Correct answer.
  3. Magnetic deflection of photons
  4. Thermionic emission of electrons

Explanation

The input phosphor (typically caesium iodide) absorbs X-ray photons and emits light, which then liberates electrons from the photocathode. The light conversion is the first step in brightening the image.

Question 21Physics, Instrumentation, and Radiologic Technology

The brightness gain of an image intensifier is the product of which two factors?

  1. Conversion gain and collimation gain
  2. Flux gain and minification gain Correct answer.
  3. Resolution gain and contrast gain
  4. Absorption gain and scatter gain

Explanation

Total brightness gain equals flux gain (increase in light photons per electron) multiplied by minification gain (reduction in image area from input to output). Together they determine the intensification.

Question 22Physics, Instrumentation, and Radiologic Technology

Which of the following best explains why the effective focal spot is smaller than the actual focal spot?

  1. The cathode filament is small
  2. The tube housing absorbs off-axis photons
  3. The anode rotates during exposure
  4. The target face is inclined at an angle to the electron beam Correct answer.

Explanation

Because the target is angled, the projected focal spot viewed along the central ray is foreshortened relative to the actual area struck by electrons. This improves resolution while maintaining heat dissipation.

Question 23Physics, Instrumentation, and Radiologic Technology

Which property of a photostimulable phosphor plate allows it to store a latent image before readout?

  1. Permanent chemical change in the phosphor
  2. Trapping of electrons in metastable energy states Correct answer.
  3. Immediate emission of light proportional to exposure
  4. Magnetic alignment of crystals

Explanation

X-ray exposure raises electrons to metastable traps in the phosphor, storing the latent image. Laser stimulation later releases this energy as light proportional to the original exposure.

Question 24Physics, Instrumentation, and Radiologic Technology

Which of the following correctly describes the relationship between noise and the number of detected photons in a radiographic image?

  1. Relative noise increases as the square root of detected photons increases
  2. Relative noise decreases as the square root of detected photons increases Correct answer.
  3. Relative noise decreases linearly with detected photons
  4. Relative noise is independent of detected photons

Explanation

Quantum noise follows Poisson statistics, so the signal-to-noise ratio rises with the square root of detected photons. Consequently relative noise falls as the square root of the number of photons increases.

Question 25Radiation Biology, Radiation Protection, and Safety

Which radiation quantity is defined as the absorbed dose multiplied by a dimensionless radiation weighting factor?

  1. Kerma
  2. Equivalent dose Correct answer.
  3. Exposure
  4. Absorbed dose

Explanation

Equivalent dose (in sieverts) accounts for the biological effectiveness of different radiation types by multiplying absorbed dose (in grays) by a radiation weighting factor (wR). Absorbed dose alone does not reflect biological harm.

Question 26Radiation Biology, Radiation Protection, and Safety

The SI unit of absorbed radiation dose is the:

  1. Coulomb per kilogram
  2. Gray Correct answer.
  3. Becquerel
  4. Sievert

Explanation

The gray (Gy) is defined as one joule of absorbed energy per kilogram of matter. It quantifies absorbed dose, not equivalent dose or activity.

Question 27Radiation Biology, Radiation Protection, and Safety

Which type of radiation has a radiation weighting factor (wR) of 1?

  1. Alpha particles
  2. X-rays Correct answer.
  3. Protons
  4. Neutrons

Explanation

X-rays, gamma rays, and electrons are assigned a radiation weighting factor of 1 because of their low linear energy transfer. Higher-LET radiations such as alpha particles and neutrons have higher weighting factors.

Question 28Radiation Biology, Radiation Protection, and Safety

The primary purpose of using a gonad shield during a radiographic examination is to:

  1. Shorten the exposure time
  2. Improve image contrast
  3. Reduce the risk of hereditary effects Correct answer.
  4. Decrease scattered radiation to the image receptor

Explanation

Gonads are highly radiosensitive with respect to hereditary effects, so shielding them reduces the genetic dose to future generations. Shielding does not affect image quality or exposure time.

Question 29Radiation Biology, Radiation Protection, and Safety

The annual occupational effective dose limit for radiation workers recommended by the ICRP is:

  1. 1 mSv per year
  2. 150 mSv per year
  3. 50 mSv per year
  4. 20 mSv averaged over 5 years Correct answer.

Explanation

The ICRP recommends an occupational effective dose limit of 20 mSv per year averaged over five years, with no single year exceeding 50 mSv. This replaces the older 50 mSv annual limit.

Question 30Radiation Biology, Radiation Protection, and Safety

A worker receives a whole-body absorbed dose of 10 mGy from gamma radiation. What is the equivalent dose?

  1. 10 mSv Correct answer.
  2. 10 mGy
  3. 100 mSv
  4. 1 mSv

Explanation

For gamma radiation, wR = 1, so equivalent dose (Sv) equals absorbed dose (Gy) numerically. Thus 10 mGy becomes 10 mSv.

Question 31Radiation Biology, Radiation Protection, and Safety

The effective dose from a chest radiograph is approximately:

  1. 0.1 mSv
  2. 0.02 mSv Correct answer.
  3. 1 mSv
  4. 10 mSv

Explanation

A standard PA chest radiograph delivers an effective dose of about 0.02 mSv, equivalent to a few days of natural background radiation. Higher values apply to CT or fluoroscopic procedures.

Question 32Radiation Biology, Radiation Protection, and Safety

The most radiosensitive phase of the cell cycle is:

  1. M phase Correct answer.
  2. G2 phase
  3. G1 phase
  4. S phase

Explanation

Cells are most radiosensitive during mitosis (M phase) and late G2, when chromosomes are condensed and DNA repair is limited. S phase is the most radioresistant because of DNA synthesis and repair.

Question 33Radiation Biology, Radiation Protection, and Safety

The law of Bergonié and Tribondeau states that cells are most radiosensitive when they are:

  1. Slowly dividing and well differentiated
  2. Undergoing apoptosis
  3. Mature and non-dividing
  4. Rapidly dividing and undifferentiated Correct answer.

Explanation

Bergonié and Tribondeau observed that immature, rapidly proliferating cells are more radiosensitive than mature, differentiated cells. This principle explains why bone marrow and intestinal epithelium are radiosensitive.

Question 34Radiation Biology, Radiation Protection, and Safety

Which subcellular structure is considered the most critical target for radiation-induced cell killing?

  1. Ribosomes
  2. Cell membrane
  3. DNA Correct answer.
  4. Mitochondria

Explanation

DNA is the principal target because radiation-induced damage (single- and double-strand breaks) can lead to loss of reproductive integrity. Other organelles are less critical for cell survival.

Question 35Radiation Biology, Radiation Protection, and Safety

A radiation worker's badge shows a dose of 5 mSv for the month. If this dose were received uniformly over the whole body, what additional risk of fatal cancer is estimated using the ICRP risk coefficient of 5% per sievert?

  1. 0.025% Correct answer.
  2. 0.25%
  3. 5%
  4. 2.5%

Explanation

Using the nominal risk coefficient of 5% per Sv, a dose of 5 mSv (0.005 Sv) gives 0.005 × 0.05 = 0.00025, or 0.025%. This is a small but non-zero stochastic risk.

Question 36Radiation Biology, Radiation Protection, and Safety

The dose-response relationship for stochastic effects of radiation is generally considered to be:

  1. Hormetic
  2. Linear, no threshold Correct answer.
  3. Non-linear with a threshold
  4. Linear with a threshold

Explanation

Stochastic effects (cancer, hereditary) are assumed to follow a linear-no-threshold (LNT) model for radiation protection purposes. This conservative model assumes risk increases proportionally with dose even at low doses.

Question 37Radiation Biology, Radiation Protection, and Safety

The main purpose of the ALARA principle in radiology is to:

  1. Increase image quality regardless of dose
  2. Eliminate all radiation exposure
  3. Keep radiation doses as low as reasonably achievable Correct answer.
  4. Reduce exposure time to zero

Explanation

ALARA (As Low As Reasonably Achievable) is a fundamental principle of radiation protection that requires optimizing protection to minimize doses while considering economic and social factors. It does not demand zero dose.

Question 38Radiation Biology, Radiation Protection, and Safety

A patient undergoes a CT scan of the abdomen with a dose-length product (DLP) of 500 mGy·cm. Using a conversion coefficient of 0.015 mSv/(mGy·cm), what is the estimated effective dose?

  1. 7.5 mSv Correct answer.
  2. 75 mSv
  3. 750 mSv
  4. 0.75 mSv

Explanation

Effective dose is calculated as DLP × conversion coefficient: 500 × 0.015 = 7.5 mSv. This is a typical effective dose for an abdominal CT.

Question 39Radiation Biology, Radiation Protection, and Safety

The radiation weighting factor for alpha particles is:

  1. 5
  2. 20 Correct answer.
  3. 10
  4. 1

Explanation

Alpha particles have a high linear energy transfer and are assigned a radiation weighting factor of 20, reflecting their greater biological effectiveness per unit absorbed dose. This is why internal contamination with alpha emitters is of concern.

Question 40Radiation Biology, Radiation Protection, and Safety

In a fluoroscopy procedure, the primary method to reduce patient dose while maintaining adequate image quality is to:

  1. Increase the source-to-skin distance
  2. Remove the grid
  3. Use pulsed fluoroscopy and last-image hold Correct answer.
  4. Increase the field size

Explanation

Pulsed fluoroscopy and last-image hold reduce the total exposure time and number of frames, thereby lowering dose. Other options may increase dose or degrade image quality.

Question 41Contrast Media, Pharmacology, and Adverse Reactions

Which iodinated contrast agent is classified as iso-osmolar relative to human plasma at approximately 290 mOsm/kg?

  1. Iohexol
  2. Iopamidol
  3. Iodixanol Correct answer.
  4. Ioxaglate

Explanation

Iodixanol is a non-ionic dimer formulated to be iso-osmolar with plasma at about 290 mOsm/kg. Iohexol and iopamidol are non-ionic monomers with higher osmolality, and ioxaglate is an ionic dimer with intermediate osmolality.

Question 42Contrast Media, Pharmacology, and Adverse Reactions

A patient with a documented prior moderate immediate reaction to an iodinated contrast agent requires a clinically indicated contrast-enhanced CT. Which premedication approach is most consistent with current guidance?

  1. Give a corticosteroid regimen beginning 6 to 13 hours before the study Correct answer.
  2. No premedication is needed because repeat reactions never occur
  3. Give intramuscular epinephrine 30 minutes before the study
  4. Give a single dose of antihistamine only, immediately before injection

Explanation

Elective premedication for prior immediate reactions uses a corticosteroid started several hours before contrast, often with an antihistamine, to reduce breakthrough reactions. Prophylaxis reduces but does not eliminate risk, so readiness to treat a reaction is still required.

Question 43Contrast Media, Pharmacology, and Adverse Reactions

Which laboratory finding is the most specific marker of contrast-induced acute kidney injury after intravascular iodinated contrast?

  1. Rise in serum potassium
  2. Fall in serum bicarbonate
  3. Rise in blood urea nitrogen
  4. Rise in serum creatinine Correct answer.

Explanation

Contrast-induced acute kidney injury is defined by a rise in serum creatinine from baseline, typically within 48 to 72 hours, reflecting a fall in glomerular filtration. Other electrolyte or acid-base changes are nonspecific and not diagnostic.

Question 44Contrast Media, Pharmacology, and Adverse Reactions

Which mechanism best explains the vasodilatory, heat-like sensation experienced during rapid intravenous injection of hyperosmolar iodinated contrast?

  1. Direct chemoreceptor stimulation of the medulla
  2. Complement activation by the contrast molecule
  3. Osmotically driven fluid shifts and vessel wall effects Correct answer.
  4. Direct histamine release from mast cells

Explanation

Hyperosmolality draws fluid and produces vessel wall and endothelial effects that generate warmth and discomfort during injection. This is a physiologic osmolality-related effect rather than an immunologic one.

Question 45Contrast Media, Pharmacology, and Adverse Reactions

Which patient factor is the strongest established risk factor for nephrogenic systemic fibrosis after gadolinium-based contrast exposure?

  1. Well-controlled type 2 diabetes with normal eGFR
  2. Older age with preserved renal function
  3. Severe chronic kidney disease with low eGFR Correct answer.
  4. Prior mild allergic reaction to iodinated contrast

Explanation

Nephrogenic systemic fibrosis is linked to gadolinium exposure in the setting of severely impaired renal clearance, particularly with less stable linear agents. Preserved renal function markedly reduces risk.

Question 46Contrast Media, Pharmacology, and Adverse Reactions

A patient develops urticaria, wheeze, and hypotension within minutes of an intravenous iodinated contrast injection. Which agent is the first-line treatment?

  1. Intravenous furosemide
  2. Oral prednisolone
  3. Intravenous diphenhydramine
  4. Intramuscular epinephrine Correct answer.

Explanation

Acute anaphylaxis with airway or hemodynamic compromise is treated first with intramuscular epinephrine, which reverses bronchospasm and vasodilation. Adjuncts such as antihistamines and steroids are secondary and never delay epinephrine.

Question 47Contrast Media, Pharmacology, and Adverse Reactions

Which statement best describes the physicochemical behavior of extracellular gadolinium-based contrast agents after intravenous administration?

  1. They remain intravascular and are excreted unchanged by glomerular filtration Correct answer.
  2. They accumulate in the reticuloendothelial system for weeks
  3. They are actively secreted by renal tubules after hepatic conjugation
  4. They distribute into total body water and are excreted hepatically

Explanation

Standard extracellular gadolinium chelates distribute mainly into the extracellular space and are cleared unchanged by glomerular filtration. They are not metabolized or actively secreted.

Question 48Contrast Media, Pharmacology, and Adverse Reactions

Which property distinguishes a high-relaxivity gadolinium chelate from a standard agent in clinical use?

  1. Greater T2 shortening at low concentration
  2. Greater T1 shortening per unit concentration Correct answer.
  3. Absence of any renal excretion
  4. Lack of chelation to the metal ion

Explanation

Relaxivity describes the ability of a contrast agent to shorten T1 or T2 relaxation times per unit concentration. Higher relaxivity means greater T1 shortening at a given dose, which can allow lower doses.

Question 49Contrast Media, Pharmacology, and Adverse Reactions

Which statement about the ionic versus non-ionic nature of iodinated contrast agents is accurate?

  1. Non-ionic agents dissociate into charged particles in solution
  2. Non-ionic agents cannot be used for intravascular injection
  3. Ionic agents generally have higher osmolality at equal iodine concentration Correct answer.
  4. Ionic agents produce fewer osmolality-related symptoms

Explanation

Ionic monomers dissociate into multiple particles, giving higher osmolality for the same iodine concentration than non-ionic monomers. Non-ionic agents are widely used intravascularly and generally cause fewer osmolality-related symptoms.

Question 50Contrast Media, Pharmacology, and Adverse Reactions

A patient with a documented shellfish allergy is scheduled for an iodinated contrast-enhanced study. What is the most appropriate approach?

  1. Proceed without special prophylaxis, as shellfish allergy is not a specific risk factor Correct answer.
  2. Cancel the study because shellfish allergy contraindicates iodinated contrast
  3. Give premedication because cross-reactivity is guaranteed
  4. Substitute gadolinium for the iodinated contrast

Explanation

Shellfish allergy reflects reactivity to tropomyosin and is not a specific predictor of iodinated contrast reaction. It should not by itself trigger prophylaxis or cancellation, though any prior contrast reaction history remains relevant.