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ABR DMP practice questions

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Question 1Diagnostic Radiological Physics

For a given radiographic technique, the absorbed dose to tissue at a depth is reduced primarily by which interaction process that removes photons from the beam without producing a new scattered photon?

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

Explanation

In photoelectric absorption the incident photon transfers its full energy to a bound electron and disappears, so no scattered photon emerges and the beam is attenuated. Compton scattering redirects part of the photon energy as scatter, which still deposits dose elsewhere. Coherent scattering redirects without energy transfer, and pair production requires energies above 1.022 MeV.

Question 2Diagnostic Radiological Physics

The half-value layer (HVL) of a diagnostic x-ray beam is best described as which quantity?

  1. The energy at which half the photons are absorbed
  2. The thickness of material that reduces the air kerma rate to half its original value Correct answer.
  3. The distance at which the beam intensity falls to half its original value
  4. The thickness of material that stops all photons in the beam

Explanation

HVL is a thickness of attenuating material that halves the measured intensity (air kerma rate) of the beam. It is a measure of beam quality or penetrability, not a distance or an energy. Complete stopping of all photons would require an impractically large thickness.

Question 3Diagnostic Radiological Physics

As tube potential (kVp) increases at fixed mAs, which change occurs in the emitted x-ray spectrum?

  1. The maximum photon energy and the mean energy both increase Correct answer.
  2. The maximum photon energy decreases while the mean energy increases
  3. The maximum photon energy increases while the mean energy stays constant
  4. The maximum photon energy stays constant while the mean energy increases

Explanation

The peak of the spectrum is set by the tube potential, so the maximum photon energy rises with kVp. Because filtration removes proportionally fewer low-energy photons at higher kVp, the mean energy of the beam also increases, raising the HVL.

Question 4Diagnostic Radiological Physics

In a fluoroscopic system, the automatic brightness control (ABC) typically responds to a change in patient thickness by adjusting which parameter?

  1. Tube potential and/or tube current Correct answer.
  2. Collimator aperture
  3. Display gamma only
  4. Detector refresh rate

Explanation

ABC senses the signal at the image receptor and adjusts kVp and/or mA to maintain constant brightness as attenuation changes. It does not change frame rate, display gamma, or field size as its primary control. These adjustments alter patient and staff dose, so understanding ABC behavior matters for dose management.

Question 5Diagnostic Radiological Physics

The main purpose of added beam filtration in a diagnostic x-ray tube is to do which of the following?

  1. Reduce the maximum photon energy of the beam
  2. Remove low-energy photons that would contribute dose without improving the image Correct answer.
  3. Narrow the focal spot size
  4. Increase the total number of photons reaching the detector

Explanation

Filtration preferentially attenuates low-energy photons that are absorbed in the patient but contribute little to image formation, thus reducing dose while hardening the beam. It does not change the peak energy set by kVp, increase photon count, or affect focal spot dimensions.

Question 6Diagnostic Radiological Physics

Which quantity expresses the absorbed dose to tissue in SI units?

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

Explanation

The gray (joule per kilogram) is the SI unit of absorbed dose. The sievert is the unit of equivalent and effective dose, the becquerel measures activity, and the coulomb per kilogram is the unit of exposure.

Question 7Diagnostic Radiological Physics

The inverse-square law describes how the intensity of radiation from a point source varies with distance. If the distance from a point source is tripled, the intensity becomes which fraction of its original value?

  1. One third
  2. One sixth
  3. One ninth Correct answer.
  4. One half

Explanation

Intensity varies as the inverse of distance squared, so tripling distance reduces intensity by a factor of 3 squared, or nine. This relationship underlies distance-based radiation protection for scatter sources.

Question 8Diagnostic Radiological Physics

Which factor most directly determines the spatial resolution limit imposed by the geometry of an x-ray tube and detector?

  1. Tube current and exposure time
  2. Beam filtration and HVL
  3. Detector pixel pitch alone
  4. Focal spot size and object-to-detector distance Correct answer.

Explanation

Geometric unsharpness depends on the focal spot size and the ratio of object-to-detector distance to source-to-object distance. Larger focal spots and greater object-detector distances increase blur. mAs, filtration, and pixel pitch affect other image properties or are secondary to geometry.

Question 9Diagnostic Radiological Physics

A radiograph of a thick body part is underexposed. Which change would increase the number of photons reaching the detector while keeping the beam penetrability approximately unchanged?

  1. Add filtration at the same kVp and mAs
  2. Double the kVp at the same mAs
  3. Reduce the source-to-image distance
  4. Double the mAs at the same kVp Correct answer.

Explanation

Increasing mAs raises the number of photons emitted without changing the spectrum shape, so it increases detector signal while preserving penetrability. Raising kVp changes beam quality as well as quantity, and adding filtration reduces photon count.

Question 10Diagnostic Radiological Physics

A chest radiograph is acquired at 120 kVp and a second at 70 kVp with mAs adjusted to maintain detector signal. Compared with the 70 kVp image, the 120 kVp image will generally show which characteristic?

  1. Reduced scatter reaching the detector
  2. Lower subject contrast between bone and soft tissue Correct answer.
  3. Increased geometric unsharpness
  4. Higher subject contrast between bone and soft tissue

Explanation

Higher kVp increases beam penetrability, so the differential attenuation between bone and soft tissue is smaller, lowering subject contrast. Scatter production increases at higher kVp, and geometric unsharpness is independent of kVp.

Question 11Diagnostic Radiological Physics

The photoelectric effect dominates in which energy range and for which material?

  1. High photon energies and high atomic number absorbers
  2. Low photon energies and high atomic number absorbers Correct answer.
  3. Low photon energies and low atomic number absorbers
  4. High photon energies and low atomic number absorbers

Explanation

Photoelectric probability scales strongly with atomic number and falls steeply as photon energy rises, so it dominates at low energies in high-Z materials such as bone or contrast agents. Compton scattering dominates at higher energies and in low-Z soft tissue.

Question 12Diagnostic Radiological Physics

In CT, the quantity CTDIvol represents which of the following?

  1. The dose-area product for the complete examination
  2. The average dose within a single reconstructed slice from a series of rotations Correct answer.
  3. The total dose to the patient over the entire scan volume
  4. The peak skin dose at the scan isocenter

Explanation

CTDIvol is the volume-weighted average dose to a slice phantom per rotation and reflects scanner output normalized to a standard phantom. It is not the total patient dose; dose-length product combines CTDIvol with scan length to estimate that.

Question 13Diagnostic Radiological Physics

A CT protocol uses pitch greater than 1 (table moves faster than the collimated width per rotation). What is the primary effect on patient dose per unit of scanned length?

  1. Dose increases because of greater tube loading
  2. Dose is unchanged because pitch does not affect output
  3. Dose increases because of increased scatter
  4. Dose decreases because each volume is sampled with fewer rotations Correct answer.

Explanation

For a fixed tube output, increasing pitch spreads the same number of rotations over a longer distance, so the dose per unit length decreases. Pitch does not change the scanner's output per rotation but changes how densely the volume is sampled.

Question 14Diagnostic Radiological Physics

Which detector material property is most important for achieving high detective quantum efficiency (DQE) in a digital radiography detector?

  1. High x-ray absorption and efficient conversion of absorbed energy to signal Correct answer.
  2. Large pixel pitch for reduced noise
  3. Low atomic number for minimal attenuation
  4. Slow readout for longer integration

Explanation

High DQE requires that a large fraction of incident photons be absorbed and that the absorbed energy be converted efficiently into a measurable signal with low added noise. Low atomic number, larger pixels, and slow readout do not directly improve DQE.

Question 15Diagnostic Radiological Physics

In fluoroscopy, the air kerma rate at the patient entrance is most strongly influenced by which relationship?

  1. It increases approximately with the square of the tube potential Correct answer.
  2. It is determined only by collimation
  3. It decreases as tube current increases
  4. It is independent of tube potential

Explanation

Entrance air kerma rate rises steeply with kVp, roughly as the square of tube potential within the diagnostic range, in addition to scaling with mA. Collimation alters field size and integral dose but not the entrance air kerma rate at a point.

Question 16Diagnostic Radiological Physics

The modulation transfer function (MTF) of an imaging system describes which property?

  1. The ability to preserve contrast of objects as a function of spatial frequency Correct answer.
  2. The conversion gain of the detector
  3. The absorbed dose per unit exposure
  4. The number of photons required per pixel

Explanation

MTF quantifies how well the system transfers signal modulation (contrast) at each spatial frequency, with values falling toward zero at high frequency. It is a resolution metric, distinct from dose or conversion gain.

Question 17Diagnostic Radiological Physics

A radiographer moves the image receptor farther from the patient while keeping source-to-image distance constant. What is the effect on geometric unsharpness?

  1. It decreases because the beam becomes harder
  2. It is unchanged because focal spot size is constant
  3. It increases because object-to-detector distance increases Correct answer.
  4. It decreases because magnification decreases

Explanation

Geometric unsharpness grows with the object-to-detector distance, so increasing that distance while keeping source-to-image distance fixed increases blur. Magnification also increases in this arrangement, which is why contact positioning is preferred when sharpness matters.

Question 18Diagnostic Radiological Physics

Which statement correctly compares equivalent dose and absorbed dose?

  1. Absorbed dose is always larger than equivalent dose for photons
  2. Equivalent dose is numerically identical to absorbed dose for all radiations
  3. Equivalent dose is measured in gray
  4. Equivalent dose applies radiation weighting to absorbed dose averaged over tissue Correct answer.

Explanation

Equivalent dose is the absorbed dose multiplied by a radiation weighting factor that reflects biological effectiveness. For photons the weighting factor is one, so the values coincide numerically only for that case. Equivalent dose is expressed in sieverts.

Question 19Diagnostic Radiological Physics

In a pulsed fluoroscopy system, reducing the pulse rate from 15 to 7.5 pulses per second while keeping other factors constant has what primary effect?

  1. Patient dose is approximately halved Correct answer.
  2. Spatial resolution is halved
  3. Beam quality (HVL) is halved
  4. Patient dose is doubled

Explanation

Dose in pulsed fluoroscopy scales with the number of pulses per second, so halving the pulse rate roughly halves the dose. Spatial resolution and beam quality are not directly changed by pulse rate.

Question 20Diagnostic Radiological Physics

The heel effect causes which variation in beam intensity across the field?

  1. Lower intensity at the center of the field
  2. Uniform intensity with no variation
  3. Lower intensity on the anode side of the field Correct answer.
  4. Lower intensity on the cathode side of the field

Explanation

X-rays produced deeper in the anode target are attenuated more before exiting, so intensity is lower toward the anode side of the field. Positioning thicker body parts toward the cathode side helps compensate.

Question 21Diagnostic Radiological Physics

A lead apron is specified as 0.5 mm lead equivalent. This specification means the apron provides which attenuation?

  1. Complete attenuation of photons below 100 keV
  2. The same attenuation as 0.5 mm of lead at the specified beam quality Correct answer.
  3. Attenuation equal to 0.5 mm of aluminum
  4. Exactly 50 percent attenuation of all photons

Explanation

Lead equivalence states the thickness of lead that would produce the same attenuation at a stated beam quality. It is not a fixed percentage and depends on photon energy.

Question 22Diagnostic Radiological Physics

In CT, the dose-length product (DLP) is computed as which product?

  1. CTDIvol multiplied by the scanned length Correct answer.
  2. CTDIvol multiplied by pitch
  3. Effective dose multiplied by patient mass
  4. Tube current multiplied by rotation time

Explanation

DLP equals CTDIvol times the scan length along the patient, giving a length-integrated dose index. Effective dose is then estimated by multiplying DLP by a conversion coefficient.

Question 23Diagnostic Radiological Physics

A 2 mm thick lead shield reduces the intensity of a monoenergetic beam to 1/8 of its original value. What is the half-value layer of that beam in lead?

  1. 0.25 mm
  2. 2.0 mm
  3. 1.0 mm
  4. 0.67 mm Correct answer.

Explanation

Reducing intensity to one eighth requires three half-value layers, since 2 cubed equals 8. Dividing the 2 mm thickness by three HVLs gives about 0.67 mm per HVL.

Question 24Diagnostic Radiological Physics

A fluoroscopic procedure uses a source-to-skin distance of 40 cm. If the source-to-skin distance is increased to 60 cm at constant output, the entrance skin air kerma rate changes by approximately what factor?

  1. 1.50
  2. 0.44 Correct answer.
  3. 2.25
  4. 0.67

Explanation

The inverse-square law gives the ratio (40/60) squared, which equals 0.44. Increasing source-to-skin distance reduces entrance dose rate substantially, which is why a minimum distance is specified.

Question 25Diagnostic Radiological Physics

For a CT acquisition, increasing the reconstructed slice thickness while keeping other parameters fixed has which effect on image noise?

  1. Noise decreases because more photons are averaged per voxel Correct answer.
  2. Noise decreases because spatial resolution improves
  3. Noise is unchanged because CTDIvol is constant
  4. Noise increases because of greater partial volume

Explanation

Thicker slices average more detected photons into each voxel, reducing statistical noise. Partial volume effects blur small structures but do not increase noise. CTDIvol remains constant because it reflects scanner output, not reconstruction.

Question 26Diagnostic Radiological Physics

Which quantity is defined as the mean energy imparted by ionizing radiation to matter per unit mass?

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

Explanation

Absorbed dose is the energy deposited per unit mass, expressed in gray (J/kg). It is the fundamental dosimetric quantity for tissue effects.

Question 27Diagnostic Radiological Physics

Which interaction process produces scattered photons that degrade image contrast in radiography?

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

Explanation

Compton scattering deflects incident photons and sends them in new directions, producing scatter that reaches the detector as unwanted signal and reduces contrast.

Question 28Diagnostic Radiological Physics

Which statement correctly describes the relationship between x-ray beam energy and subject contrast?

  1. Beam energy has no effect on subject contrast because contrast depends only on detector type
  2. Higher beam energy generally increases subject contrast because more photons penetrate thick parts
  3. Higher beam energy generally reduces subject contrast because attenuation differences between tissues diminish Correct answer.
  4. Lower beam energy always reduces subject contrast by increasing scatter

Explanation

As beam energy rises, the photoelectric contribution falls and tissue attenuation coefficients converge, so differences in transmitted intensity between tissues shrink. This lowers subject contrast.

Question 29Diagnostic Radiological Physics

A digital radiography system reports a detector quantum efficiency (DQE) of 50% at 0.5 cycles/mm. What does this value primarily express?

  1. The efficiency with which the detector converts incident quanta into useful signal-to-noise ratio Correct answer.
  2. The ratio of detector pixel size to the focal spot size
  3. The fraction of scattered radiation rejected by the anti-scatter grid
  4. The fraction of incident x-ray quanta that are converted to light in the scintillator

Explanation

DQE compares the squared signal-to-noise ratio at the detector output to that at the input, so it measures how efficiently incident quanta become useful image signal. It incorporates both absorption and noise transfer.

Question 30Diagnostic Radiological Physics

A monoenergetic beam has a half-value layer of 3 mm of aluminum. What thickness of aluminum reduces its intensity to one-sixteenth of the incident value?

  1. 9 mm
  2. 48 mm
  3. 6 mm
  4. 12 mm Correct answer.

Explanation

Reducing intensity to 1/16 requires four half-value layers because (1/2)^4 = 1/16. Four times 3 mm equals 12 mm.

Question 31Diagnostic Radiological Physics

An x-ray tube operates at 100 kVp with a tungsten target. Which change would most increase the efficiency of x-ray production and the average photon energy?

  1. Increasing the tube current from 100 to 200 mA
  2. Increasing the added filtration to 5 mm aluminum
  3. Increasing the kVp from 100 to 120 Correct answer.
  4. Decreasing the focal spot size

Explanation

Raising kVp increases both the kinetic energy of incident electrons and the fraction converted to bremsstrahlung, shifting the spectrum to higher energies. Tube current changes quantity, not the energy spectrum.

Question 32Diagnostic Radiological Physics

A CT scanner is operated with a pitch of 0.5. How does this pitch affect dose compared with a pitch of 1.0 for the same tube current and rotation time?

  1. Dose is unchanged because pitch does not influence dose
  2. Dose is halved because the table moves more slowly
  3. Dose is reduced by a factor of four
  4. Dose is approximately doubled because the same anatomy is scanned twice Correct answer.

Explanation

Pitch is inversely related to dose for fixed tube current-time product; halving pitch means more rotations per unit length, so dose roughly doubles. This is why low pitch increases patient dose.

Question 33Diagnostic Radiological Physics

A radiograph is produced with a grid of 12:1 ratio. Compared with a 6:1 grid, what is the primary consequence?

  1. Elimination of the heel effect
  2. Reduced scatter rejection with lower patient dose
  3. Improved scatter rejection with a requirement for higher mAs Correct answer.
  4. Improved spatial resolution without any change in dose

Explanation

Higher grid ratio removes more scattered photons, improving contrast, but also removes more primary photons, so mAs must increase to maintain detector signal. This raises patient dose.

Question 34Diagnostic Radiological Physics

In fluoroscopy, which feature most reduces patient dose while preserving image brightness on the monitor?

  1. Increasing the field of view to include more tissue
  2. Using pulsed fluoroscopy with last-image hold Correct answer.
  3. Removing the copper filtration from the beam
  4. Increasing the fluoroscopic mA continuously

Explanation

Pulsed fluoroscopy exposes the patient only during acquisition pulses, and last-image hold lets the operator study a stored frame without additional exposure. Together they reduce exposure time substantially.

Question 35Diagnostic Radiological Physics

A radiologist wants to reduce the dose from a CT examination while keeping image noise acceptable. Which approach best reflects the principle of automatic exposure control in CT?

  1. Increase pitch to the maximum allowed value regardless of protocol
  2. Reduce the reconstruction kernel sharpness only
  3. Modulate tube current according to patient attenuation along the scan Correct answer.
  4. Fix the mAs at the maximum value for all patients

Explanation

Automatic exposure control adjusts tube current in response to measured attenuation, delivering lower mAs where the patient is thin and higher mAs where attenuation is greater. This maintains consistent noise at reduced dose.

Question 36Diagnostic Radiological Physics

A point source of radiation produces an exposure rate of 100 mR/h at 1 m. What is the approximate exposure rate at 4 m, ignoring attenuation?

  1. 400 mR/h
  2. 6.25 mR/h Correct answer.
  3. 25 mR/h
  4. 12.5 mR/h

Explanation

The inverse-square law states that intensity varies as 1/d^2. Quadrupling distance reduces intensity by a factor of 16, so 100/16 equals 6.25 mR/h.

Question 37Diagnostic Radiological Physics

A technologist notices that a radiograph shows a blurred, magnified heart shadow while the ribs appear sharp. Which geometric factor best explains this observation?

  1. A small focal spot combined with a short object-to-image distance
  2. A high grid ratio
  3. A large focal spot combined with a long object-to-image distance Correct answer.
  4. Excessive beam filtration

Explanation

Magnification and geometric unsharpness increase with larger focal spot size and greater object-to-image distance. Structures far from the detector, such as the heart, are magnified and blurred more than structures close to it.

Question 38Diagnostic Radiological Physics

A CT image reconstructed with a sharp (bone) kernel compared with a soft-tissue kernel will demonstrate which combination of properties?

  1. Lower spatial resolution and lower image noise
  2. Higher spatial resolution and lower image noise
  3. Lower spatial resolution and higher image noise
  4. Higher spatial resolution and higher image noise Correct answer.

Explanation

Sharp kernels emphasize high spatial frequencies, improving edge and detail resolution but amplifying noise. Soft kernels smooth the image, reducing noise at the expense of resolution.

Question 39Diagnostic Radiological Physics

A radiograph is acquired at 80 kVp and 10 mAs. To maintain the same detector signal while increasing kVp to 100, approximately what mAs is required if the beam is roughly 15% more penetrating?

  1. About 14 mAs
  2. About 20 mAs
  3. About 10 mAs
  4. About 7 mAs Correct answer.

Explanation

Increasing kVp raises photon penetration, so fewer photons are needed to produce the same detector signal. Reducing mAs by roughly the increase in penetration yields about 7 mAs.

Question 40Diagnostic Radiological Physics

A shielding calculation shows that a barrier transmits 1/1000 of the incident primary beam. How many tenth-value layers does this barrier represent?

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

Explanation

A tenth-value layer reduces intensity by a factor of ten. Reaching 1/1000 requires three factors of ten, so the barrier is three tenth-value layers thick.

Question 41Nuclear Medicine Physics

The physical half-life of technetium-99m is approximately how long?

  1. 73 hours
  2. 2 days
  3. 6 hours Correct answer.
  4. 13 hours

Explanation

Tc-99m decays with a physical half-life of about 6 hours, which is convenient for same-day imaging. The 6-hour value reflects the nuclear decay constant of the metastable state, independent of biological clearance.

Question 42Nuclear Medicine Physics

In a nuclear medicine image, the radiotracer distribution primarily reflects which property of the tissue?

  1. Proton relaxation time
  2. Electron density
  3. Physiologic uptake and clearance Correct answer.
  4. Acoustic impedance

Explanation

Nuclear medicine images map the biodistribution of a radiopharmaceutical, which depends on perfusion, metabolism, and clearance. Signal intensity therefore reflects function rather than anatomy alone.

Question 43Nuclear Medicine Physics

The radioisotope generator most commonly used to supply technetium-99m in hospitals produces Tc-99m from the decay of which parent?

  1. Xenon-133
  2. Strontium-90
  3. Molybdenum-99 Correct answer.
  4. Iodine-131

Explanation

Mo-99 decays by beta emission to Tc-99m, which is eluted from the generator column. The parent's longer half-life (about 66 hours) allows the generator to be milked repeatedly.

Question 44Nuclear Medicine Physics

A radionuclide has a physical half-life of 6 hours. Approximately what fraction of activity remains after 18 hours?

  1. One eighth Correct answer.
  2. One sixteenth
  3. One quarter
  4. One half

Explanation

Eighteen hours equals three physical half-lives, so activity falls by a factor of 2^3 = 8. The remaining fraction is therefore one eighth of the initial activity.

Question 45Nuclear Medicine Physics

In a scintillation camera, the function of the collimator is to:

  1. Restrict the direction of incident photons Correct answer.
  2. Convert gamma photons into light
  3. Correct for energy peak drift
  4. Amplify the photomultiplier signal

Explanation

A collimator is a lead septal structure that only admits photons traveling roughly parallel to the holes, thereby defining the image projection. Without it, spatial information would be lost because gamma rays cannot be focused.

Question 46Nuclear Medicine Physics

Which detector material is most commonly used in modern gamma cameras to convert gamma photons into visible light?

  1. Bismuth germanate
  2. Cesium iodide
  3. High-purity germanium
  4. Thallium-doped sodium iodide Correct answer.

Explanation

Thallium-activated sodium iodide, NaI(Tl), offers high light output and good stopping power for the gamma energies used clinically. It has long been the standard scintillation crystal in Anger cameras.

Question 47Nuclear Medicine Physics

The purpose of a pulse-height analyzer in a gamma camera is to:

  1. Reject events outside a selected energy window Correct answer.
  2. Increase the crystal's light yield
  3. Focus the incoming photons
  4. Synchronize the camera with the ECG

Explanation

The pulse-height analyzer accepts only signals whose energy falls within the chosen photopeak window, rejecting scattered and background events. This improves contrast and quantitative accuracy.

Question 48Nuclear Medicine Physics

A radionuclide has a physical half-life of 6 hours and a biological half-life of 6 hours. What is its effective half-life?

  1. 6 hours
  2. 3 hours Correct answer.
  3. 1.5 hours
  4. 12 hours

Explanation

Effective half-life combines physical decay and biologic removal as 1/Teff = 1/Tphys + 1/Tbio. With equal 6-hour components, Teff equals 3 hours.

Question 49Nuclear Medicine Physics

A parallel-hole collimator with longer septa and smaller holes (relative to another collimator of the same field of view) will generally produce:

  1. Higher sensitivity and lower resolution
  2. Lower sensitivity and lower resolution
  3. Lower sensitivity and higher resolution Correct answer.
  4. Higher sensitivity and higher resolution

Explanation

Resolution improves as hole diameter decreases and septal length increases, but these changes reduce the solid angle accepted, lowering sensitivity. This trade-off is fundamental to collimator design.

Question 50Nuclear Medicine Physics

Which collimator type is designed for imaging higher-energy photons such as those from iodine-131?

  1. Pinhole
  2. Medium-energy parallel-hole
  3. High-energy parallel-hole Correct answer.
  4. Low-energy high-sensitivity

Explanation

High-energy collimators use thick septa to prevent penetration by higher-energy photons, preserving spatial accuracy. Using a low-energy collimator for I-131 would cause septal penetration and image degradation.