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ABSNM RP practice questions

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Question 1Radiation Physics and Instrumentation

Which unit expresses absorbed dose in the SI system?

  1. Becquerel
  2. Gray Correct answer.
  3. Roentgen
  4. Sievert

Explanation

Absorbed dose is energy deposited per unit mass, and its SI unit is the gray, equal to one joule per kilogram. The becquerel measures activity, the sievert measures equivalent dose, and the roentgen measures exposure in air.

Question 2Radiation Physics and Instrumentation

The half-life of a radionuclide is best defined as the time required for which change?

  1. Activity to fall to one tenth of its initial value
  2. Mass of the sample to double
  3. Activity to fall to one half of its initial value Correct answer.
  4. Number of emitted photons to reach zero

Explanation

Half-life is the interval over which radioactive activity decays to 50 percent of its starting value. Decay is exponential, so activity never reaches zero in a finite number of half-lives.

Question 3Radiation Physics and Instrumentation

A radionuclide has a half-life of 6 hours. What fraction of its initial activity remains after 18 hours?

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

Explanation

Eighteen hours equals three half-lives, and each half-life halves the remaining activity, so the fraction is (1/2)^3 = 1/8.

Question 4Radiation Physics and Instrumentation

In a photomultiplier tube, what is the primary role of the dynode chain?

  1. Converting light photons into heat
  2. Multiplying electrons through successive secondary emission Correct answer.
  3. Storing charge for later readout
  4. Collimating the incoming gamma rays

Explanation

Each dynode is held at a progressively higher potential so that an incident electron releases several secondary electrons, producing cascade amplification. The photocathode, not the dynodes, performs the initial light-to-electron conversion.

Question 5Radiation Physics and Instrumentation

Which detector type directly converts absorbed gamma energy into electron-hole pairs within a semiconductor crystal?

  1. Cadmium zinc telluride detector Correct answer.
  2. Geiger-Mueller counter
  3. Sodium iodide scintillator
  4. Ionization chamber

Explanation

Semiconductor detectors such as cadmium zinc telluride create electron-hole pairs whose number is proportional to deposited energy, giving direct high-resolution spectra. Scintillators instead emit light that must be converted by a photodetector.

Question 6Radiation Physics and Instrumentation

How does the energy resolution of a detector relate to the width of a photopeak?

  1. Resolution improves only when peak height increases
  2. Resolution is independent of photopeak width
  3. Wider photopeak means better resolution
  4. Narrower photopeak means better resolution Correct answer.

Explanation

Energy resolution is expressed as the full width at half maximum of the photopeak relative to its centroid energy, so a smaller width indicates better resolution. Peak height alone does not define resolution.

Question 7Radiation Physics and Instrumentation

What is the purpose of pulse-height analysis in a gamma camera or spectrometer?

  1. Reducing the detector's dead time
  2. Selecting events within an energy window Correct answer.
  3. Increasing the photopeak width
  4. Shortening the radionuclide half-life

Explanation

Pulse-height analysis sorts detected events by amplitude so that only those falling in a chosen energy window, typically around the photopeak, are accepted. This rejects Compton-scattered events and improves image contrast.

Question 8Radiation Physics and Instrumentation

Which process produces the dominant interaction of diagnostic-energy gamma rays with soft tissue?

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

Explanation

At energies typical of most diagnostic radiopharmaceuticals in soft tissue, Compton scattering is the most frequent interaction because of the low effective atomic number. Pair production requires energies above 1.022 MeV.

Question 9Radiation Physics and Instrumentation

A source gives a count rate of 400 counts per second with a background of 100 counts per second. What is the net source count rate?

  1. 500 counts per second
  2. 300 counts per second Correct answer.
  3. 400 counts per second
  4. 100 counts per second

Explanation

Net count rate is obtained by subtracting the background rate from the measured rate: 400 minus 100 equals 300 counts per second.

Question 10Radiation Physics and Instrumentation

A detector has a dead time of 10 microseconds and records an observed rate of 20000 counts per second. Using the paralysable model, what is the approximate true rate?

  1. 22000 counts per second
  2. 40000 counts per second
  3. 18000 counts per second
  4. 25000 counts per second Correct answer.

Explanation

For a paralysable system, n = m·e^(mτ). With m = 20000 and τ = 10 microseconds, mτ = 0.2, so e^0.2 ≈ 1.221 and n ≈ 24400, closest to 25000 counts per second.

Question 11Radiation Physics and Instrumentation

Two radionuclides are present and one has a much longer half-life. What is the practical consequence for counting?

  1. The long-lived component decays to zero within minutes
  2. The two components cannot be distinguished by decay
  3. The long-lived component contributes a nearly constant background Correct answer.
  4. The short-lived component decays faster than the long-lived one

Explanation

A nuclide with a long half-life shows little activity change over a typical counting period, so it behaves like a steady baseline. The short-lived component decays measurably and can be separated by following the decay curve.

Question 12Radiation Physics and Instrumentation

What does the sensitivity of a gamma camera describe?

  1. Energy resolution at the photopeak
  2. Counts detected per unit activity in the source Correct answer.
  3. Ability to resolve two nearby point sources
  4. Uniformity of the detector field

Explanation

Sensitivity is the fraction of emitted photons that are actually detected, expressed as counts per unit activity. Spatial resolution, uniformity, and energy resolution are separate performance measures.

Question 13Radiation Physics and Instrumentation

A parallel-hole collimator with longer holes and smaller diameter holes will produce which change in performance?

  1. Improved sensitivity with reduced resolution
  2. Improved spatial resolution with reduced sensitivity Correct answer.
  3. No change in resolution or sensitivity
  4. Improved both resolution and sensitivity

Explanation

Longer and narrower channels restrict the accepted angle of incoming photons, sharpening spatial resolution but rejecting more photons and lowering sensitivity. Resolution and sensitivity trade off in collimator design.

Question 14Radiation Physics and Instrumentation

Why is a high-energy general-purpose collimator used for iodine-131 imaging?

  1. It eliminates the need for pulse-height analysis
  2. It provides the highest possible sensitivity
  3. It improves energy resolution of the crystal
  4. Its thick septa limit penetration by high-energy photons Correct answer.

Explanation

High-energy photons can pass through thin septa, so a collimator with thick septa and long holes is required to prevent septal penetration. This comes at the cost of sensitivity.

Question 15Radiation Physics and Instrumentation

Which factor most directly determines the intrinsic spatial resolution of a scintillation camera?

  1. Spread of light in the crystal and light guide Correct answer.
  2. Diameter of the collimator holes
  3. Thickness of the patient
  4. Activity administered to the patient

Explanation

Intrinsic resolution is set by how precisely the position of a scintillation event can be localized, which depends on light spreading in the crystal and light guide before reaching the photomultipliers. Collimator geometry affects system resolution, not intrinsic resolution.

Question 16Radiation Physics and Instrumentation

What is the function of a pulse-height analyzer window centered on a photopeak?

  1. Correct for radioactive decay during acquisition
  2. Measure the activity of the source directly
  3. Accept photons that deposit full energy in the crystal Correct answer.
  4. Reject all photons that interact in the crystal

Explanation

Photons that deposit their full energy produce pulses in the photopeak, and the window accepts these while rejecting lower-energy scattered events. This improves contrast in the resulting image.

Question 17Radiation Physics and Instrumentation

In a well counter, what does the counting efficiency represent?

  1. Dead time divided by real time
  2. Detected counts divided by background counts
  3. Photopeak width divided by centroid energy
  4. Detected counts divided by emitted disintegrations Correct answer.

Explanation

Counting efficiency is the ratio of recorded counts to the number of disintegrations occurring in the sample, after correcting for background and decay. It depends on geometry, detector type, and photon energy.

Question 18Radiation Physics and Instrumentation

A sample measured for 1 minute gives 900 gross counts and the background measured for 10 minutes gives 2000 counts. What is the net count rate in counts per minute?

  1. 900
  2. 700 Correct answer.
  3. 100
  4. 800

Explanation

Background rate is 2000/10 = 200 counts per minute. Subtracting from the gross rate of 900 counts per minute gives 700 counts per minute.

Question 19Radiation Physics and Instrumentation

Which statement correctly describes the relationship between sensitivity and spatial resolution for a parallel-hole collimator?

  1. Both are maximized by shorter holes
  2. Improving resolution generally reduces sensitivity Correct answer.
  3. Improving resolution generally increases sensitivity
  4. The two are unrelated in collimator design

Explanation

A collimator that accepts a narrower range of photon angles improves resolution but transmits fewer photons, lowering sensitivity. Designers balance these competing goals for each clinical application.

Question 20Radiation Physics and Instrumentation

What is the principal reason for using a photopeak energy window rather than accepting all pulse heights?

  1. Scattered photons degrade image contrast Correct answer.
  2. Photopeak photons carry no useful information
  3. Low pulse heights increase detector dead time
  4. High pulse heights damage the crystal

Explanation

Compton-scattered photons deposit less energy and are recorded below the photopeak; accepting them adds misplaced counts that reduce contrast. The energy window preferentially keeps unscattered photopeak events.

Question 21Radiation Physics and Instrumentation

Which detector material offers the best intrinsic energy resolution among the following for gamma spectrometry?

  1. Lithium-drifted silicon at room temperature
  2. High-purity germanium Correct answer.
  3. Sodium iodide activated with thallium
  4. Plastic scintillator

Explanation

Germanium has a small band gap and produces many charge carriers per absorbed photon, giving superior statistical energy resolution. Scintillators produce fewer information carriers and broader peaks.

Question 22Radiation Physics and Instrumentation

A gamma camera records 100000 counts in 10 minutes with a background of 200 counts per minute. What is the net count rate in counts per minute?

  1. 9800 Correct answer.
  2. 10000
  3. 10200
  4. 800

Explanation

Gross rate is 100000/10 = 10000 counts per minute. Subtracting the background rate of 200 gives 9800 counts per minute.

Question 23Radiation Physics and Instrumentation

Using the nonparalysable dead-time model with τ = 5 microseconds and an observed rate of 50000 counts per second, what is the approximate true rate?

  1. 55000 counts per second
  2. 100000 counts per second
  3. 45000 counts per second
  4. 66667 counts per second Correct answer.

Explanation

For a nonparalysable system, n = m/(1 − mτ). Here mτ = 0.25, so n = 50000/0.75 ≈ 66667 counts per second.

Question 24Radiation Physics and Instrumentation

A source with a 13-hour half-life is calibrated at 08:00. What activity remains at 21:00 the same day if the initial activity is 800 MBq?

  1. 600 MBq
  2. 200 MBq
  3. 100 MBq
  4. 400 MBq Correct answer.

Explanation

The elapsed time is 13 hours, exactly one half-life, so the activity halves from 800 to 400 MBq.

Question 25Radiation Physics and Instrumentation

Why does the photoelectric effect produce a photopeak in a gamma spectrum?

  1. The photon is scattered with no energy loss
  2. The photon is absorbed by the detector housing
  3. The photon creates an electron-positron pair
  4. The photon transfers essentially all its energy to an electron Correct answer.

Explanation

In photoelectric absorption the incident photon disappears and its full energy is transferred to an atomic electron, so the detector records a pulse equal to the photon energy. This full-energy deposition forms the photopeak.

Question 26Radiation Physics and Instrumentation

In a scintillation detector, what is the role of the photocathode in the photomultiplier tube?

  1. It stores charge until the readout cycle begins
  2. It converts light photons into electrons by the photoelectric effect Correct answer.
  3. It focuses the incoming gamma rays onto the crystal
  4. It multiplies the number of electrons through secondary emission

Explanation

The photocathode is a photosensitive layer that ejects electrons when struck by scintillation light, converting the light signal into an electrical one. Amplification happens later at the dynodes, not at the cathode.

Question 27Radiation Physics and Instrumentation

Which radionuclide is most commonly used to calibrate the energy scale of a gamma camera because of its well-known 140 keV photopeak?

  1. Fluorine-18
  2. Technetium-99m Correct answer.
  3. Cobalt-57
  4. Iodine-131

Explanation

Technetium-99m emits a 140 keV gamma ray and is the workhorse of nuclear medicine, so its photopeak is the standard reference for setting camera energy windows. Co-57 is used for calibration in some contexts but its main peaks are 122 and 136 keV.

Question 28Radiation Physics and Instrumentation

Which quantity is expressed in units of becquerels?

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

Explanation

The becquerel is one disintegration per second and therefore measures activity, the rate of nuclear transformations. Absorbed dose is measured in gray and equivalent dose in sievert.

Question 29Radiation Physics and Instrumentation

A radiopharmaceutical has a physical half-life of 6 hours and a biological half-life of 3 hours. What is its effective half-life?

  1. 1.5 hours
  2. 2 hours Correct answer.
  3. 9 hours
  4. 4.5 hours

Explanation

Effective half-life follows 1/Teff = 1/Tphys + 1/Tbiol, giving 1/6 + 1/3 = 1/2, so Teff = 2 hours. The effective half-life is always shorter than either component half-life.

Question 30Radiation Physics and Instrumentation

A parallel-hole collimator is swapped for one with longer septa of the same hole diameter. What is the expected effect on camera performance?

  1. Reduced spatial resolution with increased sensitivity
  2. Improved spatial resolution with reduced sensitivity Correct answer.
  3. No change in either resolution or sensitivity
  4. Improved spatial resolution with increased sensitivity

Explanation

Longer holes narrow the acceptance angle of each channel, sharpening the projected image but rejecting more oblique photons. The trade-off is a loss of geometric efficiency, so sensitivity falls.

Question 31Radiation Physics and Instrumentation

What is the main advantage of a semiconductor detector such as high-purity germanium over a sodium iodide scintillator for gamma spectrometry?

  1. It operates at room temperature without cooling
  2. It requires no pulse-processing electronics
  3. It has far better energy resolution because fewer charge carriers are needed per event Correct answer.
  4. It has higher detection efficiency for high-energy photons

Explanation

In a semiconductor, only a few electron-hole pairs (about 3 eV each) are created per deposited energy, compared with roughly 30 eV per light photon plus conversion losses in a scintillator. The smaller statistical fluctuation yields much narrower peaks and superior energy resolution.

Question 32Radiation Physics and Instrumentation

A gamma camera image is acquired with a 20 percent energy window centred on the 140 keV photopeak. What is the approximate lower discriminator setting?

  1. 112 keV
  2. 120 keV
  3. 133 keV
  4. 126 keV Correct answer.

Explanation

A 20 percent window spans plus or minus 10 percent of 140 keV, that is 126 to 154 keV. The lower discriminator is therefore about 126 keV.

Question 33Radiation Physics and Instrumentation

A dose calibrator reading is checked with a sealed source and reads 10 percent higher than the certified activity. What is the most appropriate interpretation?

  1. The measurement geometry is irrelevant to the discrepancy
  2. The source has decayed faster than expected
  3. The instrument requires recalibration or correction of its response Correct answer.
  4. The source contains a radionuclidic impurity

Explanation

A consistent deviation from a known standard indicates an instrument calibration or response problem rather than a source problem. The appropriate action is recalibration or application of a correction factor after investigation.

Question 34Radiation Physics and Instrumentation

Which statement correctly describes the purpose of the septa in a parallel-hole collimator?

  1. They increase the energy of transmitted photons
  2. They focus gamma rays to a point on the crystal surface
  3. They convert scattered photons into useful signal
  4. They absorb photons travelling obliquely so that only near-parallel rays reach the crystal Correct answer.

Explanation

Septa are thin lead walls between holes that attenuate photons crossing them, so primarily photons aligned with the holes reach the detector. This defines the geometric acceptance and hence spatial resolution.

Question 35Radiation Physics and Instrumentation

A point source is imaged at 10 cm from a parallel-hole collimator and then at 20 cm. Assuming the same collimator, how does the measured spatial resolution change?

  1. It remains essentially unchanged because the collimator defines resolution
  2. It degrades only if the energy window is widened
  3. It improves because fewer scattered photons reach the detector
  4. It degrades because the projected hole size increases with distance Correct answer.

Explanation

For a parallel-hole collimator, resolution worsens linearly with source-to-collimator distance because each hole projects a larger footprint onto the crystal. Moving the source farther away therefore blurs the image.

Question 36Radiation Physics and Instrumentation

In a scintillation camera, what is the function of the position-logic circuit?

  1. It corrects the image for radioactive decay
  2. It determines the x-y coordinates of each event from the relative light distribution among tubes Correct answer.
  3. It rejects pulses outside the energy window
  4. It regulates the high voltage supplied to the photomultiplier tubes

Explanation

By comparing the signal amplitudes from neighbouring photomultiplier tubes, the position-logic circuit computes the centroid of the light distribution and assigns x-y coordinates to the event. Energy discrimination is a separate function.

Question 37Radiation Physics and Instrumentation

A gamma camera's intrinsic resolution is 3.5 mm and the collimator resolution at the source distance is 7.0 mm. What is the approximate system resolution?

  1. 10.5 mm
  2. 3.1 mm Correct answer.
  3. 5.3 mm
  4. 7.8 mm

Explanation

System resolution combines components in quadrature: R = sqrt(Ri^2 + Rc^2) = sqrt(3.5^2 + 7.0^2) = sqrt(12.25 + 49) = sqrt(61.25), about 7.8 mm. The collimator dominates, so the result is only slightly worse than the collimator alone.

Question 38Radiation Physics and Instrumentation

A well counter records a gross count rate of 5000 counts per second with a background of 200 counts per second. What is the net count rate and its approximate standard deviation?

  1. 4800 cps with a standard deviation of about 22 cps
  2. 5200 cps with a standard deviation of about 72 cps
  3. 4800 cps with a standard deviation of about 72 cps Correct answer.
  4. 4800 cps with a standard deviation of about 5000 cps

Explanation

The net rate is 5000 minus 200, or 4800 cps. For counting statistics, the standard deviation of the net rate is approximately sqrt(gross + background) per unit time, giving sqrt(5200) per second, about 72 cps.

Question 39Radiation Physics and Instrumentation

Why does the photofraction (peak-to-total ratio) of a NaI crystal increase as the crystal volume increases?

  1. Larger crystals reduce the photoelectric cross-section of iodine
  2. Larger crystals shorten the photomultiplier light-collection path
  3. Larger crystals increase the probability that scattered photons are reabsorbed and contribute to the photopeak Correct answer.
  4. Larger crystals eliminate Compton scattering entirely

Explanation

In a small crystal, many Compton-scattered photons escape and are lost from the spectrum. A larger crystal gives them more opportunity to undergo further interactions and deposit their full energy, adding counts to the photopeak and raising the peak-to-total ratio.

Question 40Radiation Physics and Instrumentation

A scanner using a single probe with a 2 cm diameter detector is used to survey a surface. What is the main reason for keeping the probe close to the surface during the survey?

  1. To increase the solid angle subtended and hence the detected count rate Correct answer.
  2. To narrow the energy window automatically
  3. To reduce the energy of the detected photons
  4. To decrease the dead time of the detector

Explanation

The fraction of emitted photons reaching a detector rises as the inverse square of distance, so keeping the probe close maximises the solid angle and sensitivity. This improves the ability to detect small amounts of activity.

Question 41Radiation Detection and Measurement

Which detector type operates by collecting electron-ion pairs produced when ionizing radiation passes through a gas-filled chamber under an applied voltage?

  1. Scintillation detector
  2. Semiconductor detector
  3. Thermoluminescent dosimeter
  4. Ionization chamber Correct answer.

Explanation

Ionization chambers apply a voltage across a gas volume and measure the current from electron-ion pairs created by radiation interactions. The collected charge is proportional to the energy deposited in the gas.

Question 42Radiation Detection and Measurement

In a Geiger-Mueller counter, the output pulse amplitude is essentially independent of the energy deposited by the incident radiation because the tube operates in which region?

  1. Ionization region
  2. Geiger-Mueller region Correct answer.
  3. Recombination region
  4. Proportional region

Explanation

In the Geiger-Mueller region, a single ion pair triggers a full avalanche along the entire anode, so every pulse saturates at the same amplitude regardless of initial energy. This makes GM tubes useful for counting but not for spectroscopy.

Question 43Radiation Detection and Measurement

A NaI(Tl) scintillation detector coupled to a photomultiplier tube produces a signal primarily through which sequence of processes?

  1. Heating of a crystal that releases stored charge as a current pulse
  2. Ionization of gas followed by charge multiplication at a wire anode
  3. Creation of electron-hole pairs collected directly at a p-n junction
  4. Absorption of radiation producing light, which liberates photoelectrons that are multiplied by dynodes Correct answer.

Explanation

NaI(Tl) emits visible light proportional to deposited energy; the photocathode converts photons to electrons, which the dynode chain multiplies into a measurable pulse. The pulse height reflects the energy deposited.

Question 44Radiation Detection and Measurement

A detector has a measured background count rate of 20 counts per minute. A sample produces a gross count rate of 80 counts per minute. What is the net sample count rate?

  1. 60 cpm Correct answer.
  2. 100 cpm
  3. 20 cpm
  4. 80 cpm

Explanation

Net count rate is the gross count rate minus the background count rate. Here 80 cpm minus 20 cpm equals 60 cpm.

Question 45Radiation Detection and Measurement

The dead time of a counting system causes which systematic effect on observed count rates?

  1. Observed rates fluctuate randomly around the true rate with no bias
  2. Observed rates are higher than true rates because of spurious pulses
  3. Observed rates are lower than true rates because events during recovery are lost Correct answer.
  4. Observed rates equal true rates because dead time cancels statistically

Explanation

Dead time is the interval after each pulse during which the system cannot register another event, so counts are lost and the observed rate is lower than the true rate. Corrections such as the paralysable or non-paralysable models restore the true rate.

Question 46Radiation Detection and Measurement

Which detector is best suited for measuring neutron dose because it relies on a hydrogen-rich material to moderate fast neutrons before detection?

  1. Geiger-Mueller tube
  2. Thin-window beta scintillator
  3. Rem meter with a polyethylene-moderated proportional counter Correct answer.
  4. Ionization chamber filled with argon

Explanation

Rem meters surround a thermal-neutron detector with a hydrogen-rich moderator such as polyethylene so that fast neutrons are slowed and their dose response approximates the dose equivalent curve.

Question 47Radiation Detection and Measurement

A gamma spectrum from a NaI(Tl) detector shows a full-energy peak at 662 keV with a smaller peak near 200 keV. The smaller peak most likely arises from which process?

  1. Pair production in the crystal
  2. Bremsstrahlung from beta particles
  3. Photoelectric absorption followed by X-ray emission
  4. Compton backscatter escaping the crystal Correct answer.

Explanation

A gamma that Compton-scatters in the detector and whose scattered photon escapes deposits only part of its energy, producing the Compton edge and the backscatter peak. The backscatter peak appears at an energy below the full-energy peak.

Question 48Radiation Detection and Measurement

Which quantity describes the probability per unit path length that a photon interacts with matter?

  1. Exposure rate constant
  2. Mass energy-absorption coefficient
  3. Linear attenuation coefficient Correct answer.
  4. Half-value layer

Explanation

The linear attenuation coefficient gives the interaction probability per unit path length in a material. It depends on photon energy and the material's density and composition.

Question 49Radiation Detection and Measurement

A radioactive source has a half-life of 8 days. Approximately what fraction of its initial activity remains after 24 days?

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

Explanation

Twenty-four days equals three half-lives, and each half-life halves the activity. After three halvings the remaining fraction is one eighth.

Question 50Radiation Detection and Measurement

The energy resolution of a detector is conventionally expressed as which quantity?

  1. The area under the full-energy peak divided by its width
  2. The difference between the highest and lowest energies detected
  3. The full width at half maximum of a peak divided by its centroid energy Correct answer.
  4. The ratio of peak counts to background counts

Explanation

Energy resolution is the FWHM of a photopeak expressed as a fraction or percentage of the peak centroid energy. Smaller values indicate better ability to distinguish closely spaced energies.