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PRC RadTech practice questions

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Question 1Radiologic Physics, Equipment and Maintenance

Which component of the x-ray tube is the source of the electrons that form the tube current?

  1. Rotating anode focal track
  2. Collimator shutter assembly
  3. Tungsten filament in the cathode Correct answer.
  4. Tube housing dielectric oil

Explanation

Electrons are released by thermionic emission when the tungsten filament of the cathode is heated by filament current. These electrons are then accelerated toward the anode to produce the tube current. The anode is the target, not the source of electrons.

Question 2Radiologic Physics, Equipment and Maintenance

Increasing tube current (mA) while all other exposure factors are held constant changes the beam by altering the:

  1. average energy of the photons produced
  2. total number of photons produced Correct answer.
  3. half-value layer of the beam
  4. maximum photon energy in the beam

Explanation

Tube current determines how many electrons cross from cathode to anode per unit time, so it governs beam quantity. Photon energy, including the maximum and average energies, is set by kVp. Filtration and kVp determine the half-value layer.

Question 3Radiologic Physics, Equipment and Maintenance

Raising kVp while other factors are unchanged increases the beam's:

  1. spatial resolution
  2. penetrating ability Correct answer.
  3. focal spot size
  4. grid ratio

Explanation

kVp sets the kinetic energy of the electrons and therefore the energy of the emitted photons, which determines how readily the beam penetrates tissue. Higher kVp shifts the spectrum toward higher energies. Resolution, focal spot size and grid ratio are unrelated to kVp.

Question 4Radiologic Physics, Equipment and Maintenance

Bremsstrahlung radiation is produced when an incident electron:

  1. transfers energy to a second orbital electron
  2. ejects an inner-shell electron from a target atom
  3. is deflected and decelerated by a nuclear field Correct answer.
  4. is captured by the nucleus of a target atom

Explanation

Bremsstrahlung results from the electrostatic attraction between the passing electron and the nucleus of a target atom, which slows the electron and converts the lost energy into a photon. The emitted photon can carry any fraction of the electron's energy, giving a continuous spectrum. Nuclear capture does not occur at diagnostic energies.

Question 5Radiologic Physics, Equipment and Maintenance

Characteristic x-rays are emitted when:

  1. the filament is heated to thermionic emission
  2. the anode rotates at high rotational speed
  3. a free electron slows near a tungsten nucleus
  4. an outer-shell electron fills a vacancy in an inner shell Correct answer.

Explanation

Characteristic radiation occurs after an incident electron removes an inner-shell electron; the resulting vacancy is filled by an outer-shell electron that releases energy as a photon. Because shell binding energies are unique to each element, the photons have discrete energies.

Question 6Radiologic Physics, Equipment and Maintenance

Because of the anode heel effect, the thickest part of the patient should be positioned:

  1. beyond the edge of the collimated field
  2. toward the cathode end of the tube Correct answer.
  3. toward the anode end of the tube
  4. exactly at the central ray axis

Explanation

X-rays produced deeper in the anode angle must pass through more target material, so intensity is lowest on the anode side and highest on the cathode side. Placing the thicker anatomy under the cathode end equalizes receptor exposure. The beam is not usable outside the collimated field.

Question 7Radiologic Physics, Equipment and Maintenance

If the source-to-image distance is doubled, the beam intensity reaching the receptor becomes:

  1. four times the original value
  2. one quarter of the original value Correct answer.
  3. one half of the original value
  4. double the original value

Explanation

Beam intensity follows the inverse square law, varying as the reciprocal of the square of the distance. Doubling the distance multiplies intensity by 1/2², or one quarter. Intensity cannot rise as distance increases.

Question 8Radiologic Physics, Equipment and Maintenance

The primary purpose of a radiographic grid is to:

  1. reduce the patient's skin dose
  2. reduce scatter reaching the receptor Correct answer.
  3. increase the size of the collimated field
  4. remove low-energy photons from the beam

Explanation

A grid is placed between the patient and the receptor to absorb radiation that has been scattered in the patient, improving contrast. Because some primary photons are also absorbed, mAs usually must be increased. Field size and low-energy photon removal are functions of collimation and filtration.

Question 9Radiologic Physics, Equipment and Maintenance

A high-frequency generator differs from a single-phase generator mainly because it:

  1. eliminates the need for beam filtration
  2. requires no rectifiers in the output circuit
  3. produces a nearly constant potential waveform Correct answer.
  4. uses a rotating anode instead of a stationary one

Explanation

High-frequency generators convert the supply to a high-frequency alternating waveform that is rectified, resulting in minimal ripple and a nearly constant potential across the tube. This raises the effective kV and output for a given setting. Anode type and filtration requirements are unaffected.

Question 10Radiologic Physics, Equipment and Maintenance

The line-focus principle is applied in x-ray tube design to:

  1. make the effective focal spot smaller than the actual one Correct answer.
  2. shorten the response time of the exposure timer
  3. reduce anode heating during long exposures
  4. increase the tube's maximum kV rating

Explanation

Angling the anode target means the projected, or effective, focal spot is smaller than the actual area where electrons strike. A small effective focal spot improves recorded detail while the larger actual area spreads heat. Anode heating and kV rating are not the purpose of the principle.

Question 11Radiologic Physics, Equipment and Maintenance

A three-phase, 6-pulse exposure is made at 80 kVp, 300 mA and 0.1 s. How many heat units are generated?

  1. 2400 HU
  2. 3240 HU Correct answer.
  3. 2160 HU
  4. 3384 HU

Explanation

Heat units equal kVp × mA × time × the waveform factor, which is 1.35 for three-phase 6-pulse equipment. Here 80 × 300 × 0.1 × 1.35 = 3240 HU. Factors of 1.0 and 1.41 apply to single-phase and 12-pulse equipment respectively.

Question 12Radiologic Physics, Equipment and Maintenance

The half-value layer of an x-ray beam is the thickness of absorber that:

  1. doubles the beam intensity at the receptor
  2. reduces the field size by one half
  3. reduces beam intensity by one half Correct answer.
  4. halves the beam's maximum photon energy

Explanation

Half-value layer quantifies beam quality as the absorber thickness that attenuates beam intensity to 50 percent of its original value. Increasing filtration or kVp raises the HVL. HVL does not describe energy maxima or field size.

Question 13Radiologic Physics, Equipment and Maintenance

During an AEC exposure, the exposure is terminated when the:

  1. ionization chamber has received a preset charge Correct answer.
  2. anode reaches its thermal capacity
  3. backup timer reaches 50 milliseconds
  4. light field is fully covered by the part

Explanation

In automatic exposure control, a chamber placed between the patient and the receptor monitors radiation reaching it and signals the generator to end the exposure once a preset charge is collected. This produces consistent receptor exposure regardless of patient thickness. The backup timer is only a safety limit.

Question 14Radiologic Physics, Equipment and Maintenance

Replacing a 5:1 grid with an 8:1 grid while keeping receptor exposure constant requires the mAs to:

  1. remain unchanged
  2. decrease slightly
  3. change inversely with kVp
  4. increase Correct answer.

Explanation

A higher grid ratio absorbs a greater fraction of scattered and primary radiation, so more mAs is needed to maintain receptor exposure. The grid conversion factor rises with grid ratio. kVp does not change the grid's absorption characteristics directly.

Question 15Radiologic Physics, Equipment and Maintenance

The rotating anode of an x-ray tube is driven by:

  1. the electron beam striking the focal track
  2. bearings lubricated by the tube's oil bath
  3. a gear train inside the tube envelope
  4. a stator winding outside the tube envelope Correct answer.

Explanation

Rotation is produced by an induction motor: a stator coil assembly outside the glass envelope creates a rotating magnetic field that turns the anode assembly inside. There is no mechanical linkage through the envelope. Bearings support the rotor but do not drive it.

Question 16Radiologic Physics, Equipment and Maintenance

For a general-purpose x-ray tube operated above 70 kVp, the minimum total filtration is:

  1. 0.5 mm Al equivalent
  2. 2.5 mm Al equivalent Correct answer.
  3. 1.0 mm Al equivalent
  4. 1.5 mm Al equivalent

Explanation

Regulatory standards require a minimum total filtration of 2.5 mm Al equivalent for tubes that operate above 70 kVp, to remove low-energy photons and reduce patient dose. Units limited to 70 kVp or below require 1.5 mm Al equivalent.

Question 17Radiologic Physics, Equipment and Maintenance

In computed radiography, the latent image stored in the imaging plate is read by:

  1. heating the plate to release trapped electrons
  2. scanning with a laser and collecting the emitted light Correct answer.
  3. applying a high voltage across a selenium layer
  4. washing the plate in developer and fixer

Explanation

X-ray exposure traps electrons in metastable states within the photostimulable phosphor, forming a latent image. A scanning laser releases the stored energy as visible light, which is detected and converted to a digital signal. The plate is then erased with bright light for reuse.

Question 18Radiologic Physics, Equipment and Maintenance

An indirect-conversion flat-panel detector first converts incident x-ray energy into:

  1. a pattern of magnetic domains
  2. a charge pattern on a selenium layer
  3. a latent image in a europium foil
  4. light within a scintillator layer Correct answer.

Explanation

In indirect conversion, a scintillator such as cesium iodide converts x-rays to light, which then strikes photodiodes to produce charge stored at each thin-film transistor. Direct conversion skips the light step and uses a photoconductor such as selenium.

Question 19Radiologic Physics, Equipment and Maintenance

Decreasing the object-to-image distance while holding SID constant will:

  1. decrease magnification Correct answer.
  2. increase focal spot blur only
  3. increase magnification
  4. have no effect on magnification

Explanation

Magnification depends on the ratio of object-to-image distance to source-to-object distance; reducing the object-to-image distance brings the object closer to the receptor and reduces magnification. Geometric unsharpness also decreases. SID and OID together determine magnification.

Question 20Radiologic Physics, Equipment and Maintenance

Geometric unsharpness of the recorded image increases as the:

  1. source-to-image distance increases
  2. object-to-image distance increases Correct answer.
  3. focal spot size decreases
  4. object-to-image distance decreases

Explanation

Unsharpness from the focal spot grows as the object moves farther from the receptor and as the focal spot enlarges, and it shrinks as SID increases. Minimizing OID and using the small focal spot improve recorded detail.

Question 21Radiologic Physics, Equipment and Maintenance

An anode has a heat capacity of 200,000 HU and currently holds 120,000 HU. How many additional 90 kVp, 500 mA, 0.2 s exposures on a three-phase 6-pulse unit can be made before capacity is reached?

  1. 13 exposures
  2. 9 exposures
  3. 7 exposures
  4. 6 exposures Correct answer.

Explanation

Each exposure generates 90 × 500 × 0.2 × 1.35 = 12,150 HU. The remaining capacity is 80,000 HU, and 80,000 ÷ 12,150 = 6.58, so only 6 complete exposures fit. The seventh would exceed the anode's rating.

Question 22Radiologic Physics, Equipment and Maintenance

A digital detector with 200-micron pixels has a theoretical limiting spatial resolution of approximately:

  1. 2.5 lp/mm Correct answer.
  2. 1.25 lp/mm
  3. 10 lp/mm
  4. 5 lp/mm

Explanation

The Nyquist limit states that the highest resolvable frequency is one cycle per two pixels. With 0.2 mm pixels this gives 1 ÷ (2 × 0.2 mm) = 2.5 lp/mm. Smaller pixels raise the limit proportionally.

Question 23Radiologic Physics, Equipment and Maintenance

Compared with single-phase full-wave rectification, high-frequency generation produces a voltage waveform with:

  1. a larger ripple and a lower effective kV
  2. an identical ripple but a higher mA
  3. no ripple and no characteristic radiation
  4. a smaller ripple and a higher effective kV Correct answer.

Explanation

High-frequency generators produce a waveform with very low ripple percentage, so the potential stays near the peak kV throughout the exposure. This raises the effective kV and output compared with single-phase equipment. A small residual ripple remains, and characteristic radiation is still produced.

Question 24Radiologic Physics, Equipment and Maintenance

At the same exposure level, a detector with high detective quantum efficiency will, compared with a lower-DQE detector:

  1. have a lower conversion efficiency
  2. show more quantum mottle in the image
  3. produce a higher signal-to-noise ratio Correct answer.
  4. require a higher dose to form an image

Explanation

Detective quantum efficiency expresses how much of the available signal-to-noise ratio in the incident beam is preserved in the output image. Higher DQE means more absorbed photons contribute useful signal, yielding a better signal-to-noise ratio or the same image quality at lower dose.

Question 25Radiologic Physics, Equipment and Maintenance

The most direct method of verifying kVp accuracy on an installed generator uses:

  1. an ion chamber measuring output linearity
  2. a star pattern tool for resolution
  3. a noninvasive kVp meter placed in the beam Correct answer.
  4. a spinning top device on the receptor

Explanation

A noninvasive kVp meter measures the peak potential directly from the beam, so the displayed kVp can be compared with the measured value. Spinning tops assess timer accuracy, ion chambers assess output consistency, and star patterns assess resolution.

Question 26Radiologic Physics, Equipment and Maintenance

The filament in a rotating-anode x-ray tube is most commonly made of:

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

Explanation

Tungsten is used because it has a high melting point and can undergo thermionic emission at the temperatures needed to produce a useful tube current. It also has a high atomic number, which contributes to x-ray production at the anode. Copper and aluminum melt or vaporize too readily at filament temperatures.

Question 27Radiologic Physics, Equipment and Maintenance

Thermionic emission in the x-ray tube refers to the:

  1. release of electrons from a heated filament Correct answer.
  2. removal of electrons from the anode by high voltage
  3. production of x-rays at the focal spot
  4. acceleration of electrons by the anode

Explanation

Thermionic emission is the liberation of electrons from a heated filament when the filament current raises its temperature. The tube current (mA) is determined by the number of electrons emitted and attracted across the tube.

Question 28Radiologic Physics, Equipment and Maintenance

Changing the tube current (mA) while all other factors remain constant primarily alters the:

  1. energy of the characteristic x-rays
  2. number of x-ray photons produced Correct answer.
  3. penetrating power of the beam
  4. half-value layer of the beam

Explanation

Tube current controls the number of electrons crossing from cathode to anode per unit time, so it determines the quantity of x-ray photons. The energy of the photons is controlled mainly by kVp.

Question 29Radiologic Physics, Equipment and Maintenance

Increasing kVp while keeping mAs constant primarily increases the beam's:

  1. quantity of photons
  2. grid ratio
  3. penetrating ability Correct answer.
  4. spatial resolution

Explanation

Higher kVp gives electrons greater kinetic energy, producing x-rays with higher average energy and greater penetrating ability. Quantity is primarily controlled by mAs.

Question 30Radiologic Physics, Equipment and Maintenance

A smaller focal spot size will generally:

  1. increase geometric unsharpness
  2. increase anode heat capacity
  3. improve spatial resolution Correct answer.
  4. decrease beam intensity

Explanation

A smaller focal spot reduces the geometric penumbra, so detail is sharper and spatial resolution improves. Focal spot size does not directly determine beam intensity or anode heat capacity.

Question 31Radiologic Physics, Equipment and Maintenance

Decreasing the anode target angle while keeping the actual focal spot area constant will:

  1. reduce the tube voltage requirement
  2. decrease the effective focal spot size Correct answer.
  3. increase the heat capacity of the anode
  4. increase the effective focal spot size

Explanation

The effective focal spot is the projection of the actual focal spot and is proportional to the sine of the anode angle. A smaller target angle therefore produces a smaller effective focal spot, which improves spatial resolution.

Question 32Radiologic Physics, Equipment and Maintenance

The primary function of the rectifier in an x-ray generator is to:

  1. store electrical charge
  2. smooth the voltage waveform
  3. rectify alternating current Correct answer.
  4. increase voltage amplitude

Explanation

Rectifiers allow current to flow in only one direction, converting alternating current into unidirectional current for the x-ray tube. Voltage amplitude is changed by transformers, and waveform smoothing is done by capacitors or generator design.

Question 33Radiologic Physics, Equipment and Maintenance

A transformer with 100 turns in the primary and 1,000 turns in the secondary receives 220 V. The secondary voltage is:

  1. 2,200 V Correct answer.
  2. 1,000 V
  3. 220 V
  4. 22 V

Explanation

The transformer law states that secondary voltage divided by primary voltage equals secondary turns divided by primary turns. The turns ratio is 1,000/100 = 10, so 220 V × 10 = 2,200 V.

Question 34Radiologic Physics, Equipment and Maintenance

If the source-to-image distance is increased from 100 cm to 150 cm, the beam intensity at the image receptor becomes:

  1. 225% of the original
  2. 67% of the original
  3. 44% of the original Correct answer.
  4. 150% of the original

Explanation

Intensity follows the inverse square law, so it changes by the inverse of the square of the distance ratio. The new intensity is (100/150)² = 0.44, or 44% of the original.

Question 35Radiologic Physics, Equipment and Maintenance

The grid ratio is defined as the:

  1. lead strip height divided by interspace width Correct answer.
  2. number of lead strips per centimeter
  3. lead strip thickness divided by height
  4. ratio of primary to scatter radiation

Explanation

Grid ratio compares the height of the lead strips to the width of the interspace between them. Higher ratios remove more scatter but require more mAs.

Question 36Radiologic Physics, Equipment and Maintenance

A grid with 40 lines per centimeter has a grid frequency of approximately:

  1. 100 lines per inch Correct answer.
  2. 10 lines per inch
  3. 400 lines per inch
  4. 40 lines per inch

Explanation

There are 2.54 centimeters in an inch, so 40 lines/cm × 2.54 cm/inch ≈ 102 lines/inch, or about 100 lines per inch.

Question 37Radiologic Physics, Equipment and Maintenance

Moiré artifact on a digital image is most commonly caused by:

  1. incorrect collimation
  2. patient motion
  3. grid-pattern aliasing with the pixel matrix Correct answer.
  4. detector temperature drift

Explanation

Moiré is an aliasing artifact produced when the grid line pattern interacts with the discrete pixel sampling matrix. It appears as a repeating moiré pattern and is reduced by matching grid frequency to detector sampling.

Question 38Radiologic Physics, Equipment and Maintenance

In an automatic exposure control system, the ionization chamber is located:

  1. at the collimator exit
  2. between patient and image receptor Correct answer.
  3. inside the x-ray tube housing
  4. within the high-voltage transformer

Explanation

The ionization chamber measures the radiation that reaches the image receptor, so it is placed between the patient and the receptor. It terminates the exposure when the preset signal is reached.

Question 39Radiologic Physics, Equipment and Maintenance

The backup timer in an AEC system is designed to:

  1. end the exposure if AEC fails Correct answer.
  2. compensate for patient thickness
  3. increase kVp if exposure is too long
  4. reset the AEC density control

Explanation

The backup timer is a safety device that terminates the exposure after a preset maximum time if the AEC does not stop it. This prevents excessive patient dose from a failed AEC.

Question 40Radiologic Physics, Equipment and Maintenance

The photostimulable phosphor in a computed radiography imaging plate is typically:

  1. gadolinium oxysulfide
  2. cesium iodide
  3. barium fluorobromide Correct answer.
  4. amorphous selenium

Explanation

Computed radiography plates commonly use europium-doped barium fluorobromide as the photostimulable phosphor. Cesium iodide and amorphous selenium are used in different direct or indirect digital detectors.

Question 41Radiologic Physics, Equipment and Maintenance

After a computed radiography plate is read, residual signal is removed by:

  1. washing the plate in developer
  2. applying a strong magnetic field
  3. heating the plate to 100°C
  4. exposing the plate to bright light Correct answer.

Explanation

Reading the plate releases most trapped electrons, but some residual signal remains. A bright light erasure cycle clears the plate for reuse.

Question 42Radiologic Physics, Equipment and Maintenance

In a direct-conversion flat-panel detector, the thin-film transistor (TFT) array functions to:

  1. read out pixel charge Correct answer.
  2. store the image permanently
  3. convert x-rays directly into light
  4. amplify the x-ray beam

Explanation

The TFT array acts as a switching matrix that reads the charge collected at each pixel after x-ray exposure. Direct-conversion detectors use a photoconductor such as amorphous selenium to create charge.

Question 43Radiologic Physics, Equipment and Maintenance

Two detectors receive the same entrance exposure. The detector with the higher detective quantum efficiency will produce an image with:

  1. less quantum mottle Correct answer.
  2. higher contrast but more noise
  3. more quantum mottle
  4. lower spatial resolution

Explanation

Detective quantum efficiency (DQE) describes how efficiently a detector converts incident x-ray quanta into useful image signal. A higher DQE means more quanta contribute to the image, so quantum noise is lower at the same exposure.

Question 44Radiologic Physics, Equipment and Maintenance

The modulation transfer function (MTF) of an imaging system describes:

  1. the conversion efficiency of x-rays to light
  2. the radiation dose required for an exposure
  3. contrast preservation across spatial frequencies Correct answer.
  4. the contrast improvement provided by a grid

Explanation

MTF quantifies how well an imaging system preserves contrast at different spatial frequencies. A system with high MTF maintains detail better at higher frequencies.

Question 45Radiologic Physics, Equipment and Maintenance

A three-phase, 12-pulse generator is used for an exposure at 90 kVp, 400 mA, and 0.1 s. How many heat units are generated?

  1. 7,200 HU
  2. 5,076 HU Correct answer.
  3. 4,860 HU
  4. 3,600 HU

Explanation

For a three-phase, 12-pulse generator, heat units equal kVp × mA × s × 1.41. Here, 90 × 400 × 0.1 = 3,600, and 3,600 × 1.41 ≈ 5,076 HU.

Question 46Radiologic Physics, Equipment and Maintenance

If an anode cooling chart indicates that the anode loses 50% of its stored heat in 5 minutes, an anode currently holding 200,000 HU will hold approximately how much after 5 minutes?

  1. 200,000 HU
  2. 150,000 HU
  3. 50,000 HU
  4. 100,000 HU Correct answer.

Explanation

Cooling is exponential, and the chart states that half the stored heat is lost in 5 minutes. Therefore 200,000 HU becomes approximately 100,000 HU after one 5-minute interval.

Question 47Radiologic Physics, Equipment and Maintenance

Tube rating charts typically provide maximum allowable exposure factors as a function of:

  1. patient thickness and OID
  2. detector pixel size and fill factor
  3. grid ratio and collimation
  4. kVp, mA, and time for a given focal spot Correct answer.

Explanation

Tube rating charts relate kVp, mA, and exposure time for a specific focal spot and generator waveform. They are used to avoid exceeding the anode's thermal limits during a single exposure.

Question 48Radiologic Physics, Equipment and Maintenance

Increasing total filtration will:

  1. increase beam intensity
  2. decrease the average photon energy
  3. remove low-energy photons and raise HVL Correct answer.
  4. increase the characteristic x-ray peak

Explanation

Added filtration preferentially absorbs low-energy photons, so the beam becomes more penetrating and its half-value layer increases. The total number of photons is reduced, not increased.

Question 49Radiologic Physics, Equipment and Maintenance

Restricting the x-ray field with collimation primarily:

  1. reduces dose and scatter Correct answer.
  2. increases spatial resolution
  3. increases the beam's HVL
  4. increases the number of x-ray photons

Explanation

Collimation reduces the volume of tissue irradiated, which lowers patient dose and the amount of scatter produced. It does not change beam energy or the number of photons in the remaining field.

Question 50Radiologic Physics, Equipment and Maintenance

A 15% increase in kVp generally produces approximately:

  1. no change in receptor exposure
  2. a 50% decrease in receptor exposure
  3. a doubling of receptor exposure Correct answer.
  4. a 15% increase in receptor exposure

Explanation

The 15% rule states that increasing kVp by 15% approximately doubles receptor exposure, all else equal. The mAs is usually halved to maintain the same exposure.