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ABNM Nuclear Medicine practice questions

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Question 1Physics, Instrumentation, and Radiation Safety

Which unit expresses radioactivity in the International System of Units (SI)?

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

Explanation

The becquerel is the SI unit of activity and equals one nuclear transformation per second. The gray measures absorbed dose, while the sievert expresses equivalent or effective dose.

Question 2Physics, Instrumentation, and Radiation Safety

A radiopharmaceutical has a physical half-life of 6 hours. What fraction of the initial activity remains after 18 hours?

  1. 1/8 Correct answer.
  2. 1/4
  3. 1/16
  4. 1/2

Explanation

Eighteen hours corresponds to three half-lives, since 18 divided by 6 equals 3. Activity falls by a factor of two with each half-life, so the remaining fraction is (1/2)^3, which is 1/8.

Question 3Physics, Instrumentation, and Radiation Safety

In a gamma camera, what is the primary role of the thallium-activated sodium iodide crystal?

  1. To convert light flashes into electronic pulses
  2. To collimate incoming photons by septal absorption
  3. To amplify the light signal before digitization
  4. To convert incident gamma photons into visible light flashes Correct answer.

Explanation

NaI(Tl) is a scintillator: absorbed gamma energy excites the crystal, which then emits visible light photons. The light is subsequently converted and amplified by the photomultiplier tubes.

Question 4Physics, Instrumentation, and Radiation Safety

What is the main purpose of the collimator on a gamma camera?

  1. To increase the energy of detected photons
  2. To restrict the detector's field of view to a defined direction Correct answer.
  3. To reduce the absorbed dose to the patient
  4. To improve the uniformity of the crystal response

Explanation

The collimator is a lead or tungsten plate with holes that absorbs photons not traveling along the hole axis, projecting the source distribution onto the crystal. It defines the geometric field of view and strongly influences spatial resolution.

Question 5Physics, Instrumentation, and Radiation Safety

The physical half-life of technetium-99m is approximately:

  1. 13 hours
  2. 20 minutes
  3. 6 hours Correct answer.
  4. 2.7 days

Explanation

Tc-99m decays by isomeric transition with a physical half-life of about 6 hours. This permits same-day imaging while limiting how long a prepared dose remains usable.

Question 6Physics, Instrumentation, and Radiation Safety

The ALARA principle in radiation protection requires that exposures be:

  1. Eliminated entirely whenever any exposure is involved
  2. Reduced to a level that is as low as reasonably achievable Correct answer.
  3. Shared equally among all staff in a department
  4. Kept below the annual regulatory limit at all times

Explanation

ALARA requires that all exposures be kept as low as reasonably achievable, with economic and social factors taken into account. It applies even when doses are already below regulatory limits.

Question 7Physics, Instrumentation, and Radiation Safety

Under U.S. NRC regulations, the annual occupational total effective dose equivalent limit for a radiation worker is:

  1. 50 mSv (5 rem) Correct answer.
  2. 5 mSv (0.5 rem)
  3. 20 mSv (2 rem)
  4. 500 mSv (50 rem)

Explanation

NRC regulations set the annual total effective dose equivalent limit for radiation workers at 50 mSv (5 rem). Separate, lower limits apply to members of the public.

Question 8Physics, Instrumentation, and Radiation Safety

On a gamma camera energy spectrum, the photopeak corresponds to:

  1. Photons that have undergone Compton scatter in the patient
  2. Photons that escape the crystal without interacting
  3. Electronic noise below the lower discriminator level
  4. Complete absorption of the photon energy in the crystal Correct answer.

Explanation

The photopeak arises when a gamma photon deposits essentially all of its energy in the crystal, usually through photoelectric absorption. Its position identifies the photon energy, while scattered photons appear at lower energies.

Question 9Physics, Instrumentation, and Radiation Safety

What is the function of the photomultiplier tubes in a gamma camera?

  1. Converting gamma photons directly into electrical current
  2. Providing mechanical support for the collimator
  3. Converting scintillation light into amplified electronic signals Correct answer.
  4. Filtering scattered photons by their energy

Explanation

Photomultiplier tubes convert scintillation light into electrons at the photocathode and multiply them through a dynode chain. The resulting pulse amplitude is proportional to the light, and hence energy, deposited in the crystal.

Question 10Physics, Instrumentation, and Radiation Safety

Which change would most improve the energy resolution of a NaI(Tl) gamma camera?

  1. Improving light collection efficiency at the photocathode Correct answer.
  2. Increasing the thickness of the crystal
  3. Widening the pulse height analyzer window
  4. Increasing the diameter of the collimator holes

Explanation

Energy resolution depends on the statistical variance in the number of photoelectrons produced per absorbed gamma. Better light collection and higher light yield increase that number and narrow the photopeak.

Question 11Physics, Instrumentation, and Radiation Safety

For a parallel-hole collimator, increasing the hole length while keeping hole diameter and septal thickness constant will:

  1. Improve spatial resolution and increase sensitivity
  2. Improve spatial resolution and decrease sensitivity Correct answer.
  3. Degrade spatial resolution and increase sensitivity
  4. Degrade both spatial resolution and sensitivity

Explanation

Resolution improves as the ratio of effective hole length to hole diameter increases, because the accepted range of incident angles narrows. The same narrowing reduces the solid angle of acceptance, so sensitivity falls.

Question 12Physics, Instrumentation, and Radiation Safety

A 20% energy window is centered on the 140 keV photopeak of Tc-99m. Which events are primarily rejected by this setting?

  1. Photons with energies above 168 keV only
  2. Compton-scattered photons with energies outside the window Correct answer.
  3. Photons depositing exactly 140 keV in the crystal
  4. Photons with energies below 112 keV only

Explanation

The analyzer accepts only pulses whose amplitudes fall within the window, which for Tc-99m spans roughly 126 to 154 keV. Scattered photons depositing energy outside that range are largely rejected, improving contrast.

Question 13Physics, Instrumentation, and Radiation Safety

A gamma camera with paralyzable dead time is imaging a source of very high activity. As activity increases further, the observed count rate will:

  1. Continue to increase in direct proportion
  2. Remain unchanged because dead time is fixed
  3. Plateau at a constant value equal to 1/tau
  4. Decrease after reaching a maximum Correct answer.

Explanation

In a paralyzable system each detected event extends the dead time, so at high rates the observed count rate passes through a maximum and then falls as activity rises further. The nonparalyzable model instead approaches a plateau of 1/tau.

Question 14Physics, Instrumentation, and Radiation Safety

In PET imaging, the 511 keV annihilation photons are detected in coincidence because they are emitted:

  1. In the same direction with a small angular spread
  2. Sequentially with a fixed time delay
  3. At 90 degrees to each other from the nucleus
  4. Back-to-back at approximately 180 degrees Correct answer.

Explanation

Annihilation of a positron with an electron produces two 511 keV photons emitted nearly back-to-back. Detecting both within a short coincidence window defines a line of response without physical collimation.

Question 15Physics, Instrumentation, and Radiation Safety

The physical half-life of fluorine-18 is approximately:

  1. 2 minutes
  2. 20 minutes
  3. 110 minutes Correct answer.
  4. 12 hours

Explanation

F-18 has a physical half-life of about 110 minutes, long enough for radiosynthesis and transport but short enough to limit patient dose. It decays predominantly by positron emission to oxygen-18.

Question 16Physics, Instrumentation, and Radiation Safety

In PET/CT, the CT transmission data are used to:

  1. Shorten the positron range in tissue
  2. Correct for attenuation along each line of response Correct answer.
  3. Increase the injected activity that can be administered
  4. Reduce random coincidence events

Explanation

CT images provide a map of tissue density that is converted into attenuation coefficients at 511 keV. This map is used to correct each line of response, preventing deep structures from appearing artificially low in activity.

Question 17Physics, Instrumentation, and Radiation Safety

A Tc-99m generator can be eluted again with useful activity after several hours because:

  1. Mo-99 decaying in the column continuously produces new Tc-99m Correct answer.
  2. The generator column is replaced after each elution
  3. Mo-99 and Tc-99m have identical half-lives
  4. Tc-99m is added to the column between elutions

Explanation

Mo-99 on the column decays to Tc-99m, so activity regrows after elution and approaches transient equilibrium over roughly a day. This regrowth allows the generator to be eluted repeatedly.

Question 18Physics, Instrumentation, and Radiation Safety

Scattered photons that fall within the photopeak window of a gamma camera image mainly cause:

  1. Shortened acquisition time without penalty
  2. Increased energy resolution
  3. Reduced image contrast Correct answer.
  4. Improved spatial resolution

Explanation

Scatter accepted by the window carries incorrect position information, raising background counts and lowering image contrast and quantitative accuracy. Energy discrimination with an appropriate window is the main defense.

Question 19Physics, Instrumentation, and Radiation Safety

A lead shield has a half-value layer of 0.3 mm for a given gamma energy. What thickness is required to reduce the intensity to 1/8 of its original value?

  1. 0.6 mm
  2. 2.4 mm
  3. 0.9 mm Correct answer.
  4. 0.3 mm

Explanation

Reaching one-eighth of the initial intensity requires three half-value layers, because 2^3 equals 8. With a half-value layer of 0.3 mm, the required thickness is 3 times 0.3 mm, or 0.9 mm.

Question 20Physics, Instrumentation, and Radiation Safety

A vial of liquid Tc-99m is dropped and breaks on the floor. The first appropriate action is to:

  1. Restrict access and notify the radiation safety officer Correct answer.
  2. Mop the spill with a wet cloth and discard the cloth
  3. Vacuum the area to remove the liquid
  4. Leave the room and return after 6 hours

Explanation

The initial response is to warn others, restrict the area, and notify radiation safety personnel while preventing the spread of contamination. Cleaning and thorough surveying follow, and the event is documented.

Question 21Physics, Instrumentation, and Radiation Safety

The spatial resolution of a parallel-hole collimator at source distance b is R = d(le + b)/le, where d is hole diameter and le is effective hole length. If d = 1.5 mm and le = 30 mm, what is R at b = 100 mm?

  1. 6.5 mm Correct answer.
  2. 13.0 mm
  3. 7.5 mm
  4. 5.0 mm

Explanation

Substituting the values gives R = 1.5 mm times (30 + 100)/30, which equals 1.5 times 4.33, or about 6.5 mm. Resolution degrades linearly as the source moves away from the collimator face.

Question 22Physics, Instrumentation, and Radiation Safety

For two detector elements with singles rates S1 and S2 and coincidence resolving time tau, the random coincidence rate is approximately:

  1. S1S2/tau
  2. 2tauS1S2 Correct answer.
  3. tauS1S2/2
  4. tau(S1 + S2)

Explanation

Random coincidences occur when two unrelated photons arrive within the resolving time, giving a rate of about 2 times tau times the product of the singles rates. Shortening the resolving time or lowering the singles rates reduces randoms.

Question 23Physics, Instrumentation, and Radiation Safety

Time-of-flight PET primarily improves image quality by:

  1. Localizing events along the line of response Correct answer.
  2. Eliminating the need for attenuation correction
  3. Increasing the detector crystal light output
  4. Reducing the positron range of the radionuclide

Explanation

Time-of-flight detectors measure the arrival-time difference of the two annihilation photons, constraining the event to a short segment of the line of response. This improves the signal-to-noise ratio, especially in large patients.

Question 24Physics, Instrumentation, and Radiation Safety

A small sphere containing high activity is surrounded by background activity. The measured activity concentration in the sphere is lower than the true value mainly because of:

  1. Random coincidence events
  2. Dead time losses in the detector
  3. Partial volume effect and spill-out of counts Correct answer.
  4. Attenuation of photons by the sphere

Explanation

Small structures suffer from the partial volume effect: counts from the object are diluted and spread into neighboring voxels, while background counts spill in. The measured concentration underestimates true activity for objects smaller than roughly two to three times the reconstructed resolution.

Question 25Physics, Instrumentation, and Radiation Safety

In the MIRD formalism for internal dosimetry, the S-value represents:

  1. Absorbed dose per unit cumulated activity in a target Correct answer.
  2. The effective half-life of the radionuclide in the target organ
  3. The total number of disintegrations in a source organ
  4. The fraction of emitted energy absorbed in the whole body

Explanation

The S-value gives the absorbed dose to a target region per unit cumulated activity in a source region, incorporating geometry, attenuation, and radionuclide emission data. Dose is obtained by multiplying S by the cumulated activity and summing over all source regions.

Question 26Physics, Instrumentation, and Radiation Safety

A vial contains 800 MBq of Tc-99m (physical half-life 6 h). What activity remains in the vial 12 h later?

  1. 200 MBq Correct answer.
  2. 100 MBq
  3. 50 MBq
  4. 400 MBq

Explanation

Activity falls by one-half with each elapsed half-life, so the number of half-lives is 12 h / 6 h = 2. Two halvings reduce 800 MBq first to 400 MBq and then to 200 MBq. No correction for biological excretion applies to an in vitro vial.

Question 27Physics, Instrumentation, and Radiation Safety

The physical half-life of iodine-131 is closest to which value?

  1. 13.2 hours
  2. 60 days
  3. 8.0 days Correct answer.
  4. 2.7 days

Explanation

I-131 decays with a physical half-life of about 8.02 days, which is long enough to permit delayed imaging and outpatient therapy dosing. Its principal gamma emission is 364 keV, which is why it requires a high-energy collimator for imaging.

Question 28Physics, Instrumentation, and Radiation Safety

Which dose quantity is obtained by weighting equivalent doses in each organ by tissue weighting factors and summing over the whole body?

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

Explanation

Effective dose combines equivalent dose in each tissue with that tissue's weighting factor, giving a single whole-body risk index for stochastic effects. Equivalent dose already includes radiation weighting but not tissue weighting. Effective dose is expressed in sieverts.

Question 29Physics, Instrumentation, and Radiation Safety

Which scintillator is used in most modern PET/CT detectors because it combines a short decay time with high light output?

  1. BGO
  2. NaI(Tl)
  3. CsI(Tl)
  4. LYSO Correct answer.

Explanation

LYSO (and closely related LSO) provides high stopping power for 511 keV photons, high light yield, and a fast decay time that supports narrow coincidence timing windows and time-of-flight capability. These properties give better count-rate performance and timing resolution than older materials.

Question 30Physics, Instrumentation, and Radiation Safety

A gamma camera has an intrinsic spatial resolution of 3.5 mm and a collimator resolution of 4.0 mm at a given source distance. What is the approximate system spatial resolution?

  1. 3.5 mm
  2. 5.3 mm Correct answer.
  3. 4.0 mm
  4. 7.5 mm

Explanation

Intrinsic and collimator blurring combine in quadrature: R = square root of (R_intrinsic squared + R_collimator squared). Using 3.5 mm and 4.0 mm gives the square root of (12.25 + 16.0), or about 5.3 mm. The system resolution is therefore always worse than either component alone.

Question 31Physics, Instrumentation, and Radiation Safety

A pinhole collimator has its aperture 20 cm from the detector face, and a small source is placed 5 cm from the aperture. What is the resulting magnification of the image?

  1. 0.25 Correct answer.
  2. 1.25
  3. 4
  4. 5

Explanation

For a pinhole collimator the magnification equals the aperture-to-detector distance divided by the aperture-to-source distance, so 20 cm / 5 cm = 4. Moving the source closer to the aperture increases magnification and improves effective resolution but reduces the field of view.

Question 32Physics, Instrumentation, and Radiation Safety

What is the accepted limit for Mo-99 breakthrough in eluted Tc-99m at the time of administration?

  1. 0.15 Bq per MBq
  2. 0.15 kBq per MBq Correct answer.
  3. 15 kBq per MBq
  4. 1.5 kBq per MBq

Explanation

Regulations limit Mo-99 contamination to 0.15 kBq of Mo-99 per MBq of Tc-99m (historically 0.15 microcuries per millicurie) at the time of use. Breakthrough is measured by assaying the eluate in a dose calibrator or by a molybdenum assay method that discriminates the 740 keV and 780 keV Mo-99 photons from the 140 keV Tc-99m photopeak.

Question 33Physics, Instrumentation, and Radiation Safety

A dose calibrator is calibrated using 1 mL of Tc-99m in a syringe at a fixed position. The identical activity is then placed in the same position but in a 10 mL volume. What is the most likely effect on the reading?

  1. It will read lower because of dead-time losses
  2. It will read exactly the same activity
  3. It will read higher because of the larger volume
  4. It will read lower because of self-attenuation Correct answer.

Explanation

Ionization-chamber response depends on source geometry and on attenuation of photons within the sample, so increasing the volume for the same activity lowers the measured current. Activity should therefore be assayed in a consistent, geometrically reproducible container and position. Dead-time losses are relevant at very high activities, not to this modest change in volume.

Question 34Physics, Instrumentation, and Radiation Safety

A chi-square test performed on a daily flood image is used to do which of the following?

  1. Calculate the collimator's sensitivity
  2. Measure the system's intrinsic spatial resolution
  3. Detect nonuniformity beyond statistical noise Correct answer.
  4. Verify the energy window setting

Explanation

The chi-square analysis compares regional count densities with the variation predicted by Poisson counting statistics, so an abnormally high value signals nonuniformity such as a photomultiplier tube or tuning problem. It is a quantitative supplement to visual inspection of the flood field. Resolution, energy peaking, and sensitivity are assessed with other dedicated tests.

Question 35Physics, Instrumentation, and Radiation Safety

A patient is to receive phosphorus-32 colloid therapy. Which material is most appropriate for shielding the syringe and for lining the waste container?

  1. Tungsten
  2. Concrete
  3. Lead foil
  4. Acrylic plastic Correct answer.

Explanation

P-32 is a pure beta emitter, and beta particles interacting with high-atomic-number materials such as lead generate bremsstrahlung that is more penetrating than the betas themselves. Low-Z materials like acrylic absorb betas while producing minimal bremsstrahlung. The acrylic is then supplemented with a thin outer lead shield if needed.

Question 36Physics, Instrumentation, and Radiation Safety

A SPECT system is operated with a misaligned center of rotation. Which artifact is most characteristic on the reconstructed images?

  1. Ring artifact Correct answer.
  2. Motion blur
  3. Photopenic defect
  4. Pincushion distortion

Explanation

Center-of-rotation error offsets projections in opposite directions at opposing angles, so a point source traces a circle during backprojection and produces a ring-shaped artifact centered on the axis of rotation. It is detected with a point-source COR acquisition and corrected by recalibration or software offset. Motion blur and photopenic defects have other causes.

Question 37Physics, Instrumentation, and Radiation Safety

The exposure rate 1 m from an unshielded point source is 40 microsieverts per hour. Ignoring attenuation and scatter, what is the exposure rate at 4 m?

  1. 5 microsieverts per hour
  2. 10 microsieverts per hour
  3. 2.5 microsieverts per hour Correct answer.
  4. 20 microsieverts per hour

Explanation

For a point source, exposure rate falls with the inverse square of distance. Increasing distance from 1 m to 4 m multiplies the dose rate by (1/4) squared, or 1/16, giving 2.5 microsieverts per hour. Doubling the distance quarters the rate.

Question 38Physics, Instrumentation, and Radiation Safety

A collimator designed for 140 keV Tc-99m imaging is used to image the 364 keV photons of I-131. What is the most likely consequence?

  1. Complete absorption by the septa
  2. No change, since collimators are energy-independent
  3. Improved spatial resolution at higher energy
  4. Septal penetration with loss of contrast Correct answer.

Explanation

Septal thickness is chosen so that a specified fraction of photons is absorbed in the septa at the design energy, and attenuation falls steeply as photon energy rises. At 364 keV a Tc-99m collimator allows a substantial fraction of photons to cross the septa, producing count skimming and degraded contrast and resolution. A high-energy collimator with thicker septa is required.

Question 39Physics, Instrumentation, and Radiation Safety

A narrow beam of 140 keV photons passes through 10 cm of soft tissue with a linear attenuation coefficient of 0.15 per cm. Approximately what fraction of the incident photons is transmitted?

  1. 0.05
  2. 0.15
  3. 0.86
  4. 0.22 Correct answer.

Explanation

Narrow-beam transmission follows I/I0 = exp(-mu times x), so exp(-0.15 times 10) = exp(-1.5), which is about 0.22. This means roughly 78 percent of the photons are removed by interaction over 10 cm of tissue at this energy. Broad-beam geometry would give a somewhat higher apparent transmission because scattered photons are not removed.

Question 40Physics, Instrumentation, and Radiation Safety

A Tc-99m generator has just been eluted to exhaustion. Approximately what fraction of its maximum Tc-99m activity is available if it is eluted again 22 h later (Mo-99 half-life 66 h)?

  1. About 50% Correct answer.
  2. About 75%
  3. About 33%
  4. About 21%

Explanation

Ingrowth of new Tc-99m after elution follows first-order buildup driven by the Mo-99 parent, approximately 1 minus exp(-0.693 times t / 66 h). With t = 22 h the exponent is about 0.23, giving roughly 0.20 to 0.21 of the maximum activity. This is why a typical daily elution performed at 24 h yields about 25 percent of maximum.

Question 41Radiopharmaceuticals and Radiochemistry

What is the physical half-life of technetium-99m?

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

Explanation

Technetium-99m decays by isomeric transition to technetium-99 with a physical half-life of about 6 hours. The 140-keV gamma emission and short half-life allow efficient same-day imaging with a relatively low radiation burden.

Question 42Radiopharmaceuticals and Radiochemistry

The Mo-99/Tc-99m generator column most commonly uses which support material?

  1. Silica gel
  2. Cation-exchange resin
  3. Alumina Correct answer.
  4. Activated charcoal

Explanation

The standard Mo-99/Tc-99m generator uses an alumina column to which molybdate binds. Saline elution removes pertechnetate while leaving molybdate on the column.

Question 43Radiopharmaceuticals and Radiochemistry

Which positron-emitting radionuclide has a 110-minute half-life and is used to label fluorodeoxyglucose?

  1. Fluorine-18 Correct answer.
  2. Gallium-68
  3. Nitrogen-13
  4. Oxygen-15

Explanation

Fluorine-18 has a 109.8-minute half-life and is the positron emitter used in FDG. Its longer half-life permits transport from a cyclotron to imaging sites.

Question 44Radiopharmaceuticals and Radiochemistry

Which radiopharmaceutical is routinely used for skeletal scintigraphy?

  1. Tc-99m MAG3
  2. Tc-99m DTPA
  3. Tc-99m MDP Correct answer.
  4. Tc-99m sulfur colloid

Explanation

Tc-99m MDP localizes to bone by chemisorption onto hydroxyapatite and is the standard skeletal scintigraphy agent. Renal agents are used for kidney imaging, and sulfur colloid is taken up by reticuloendothelial cells.

Question 45Radiopharmaceuticals and Radiochemistry

Which radioiodine is used therapeutically for Graves disease and differentiated thyroid cancer?

  1. I-124
  2. I-131 Correct answer.
  3. I-123
  4. I-125

Explanation

Iodine-131 emits beta-minus particles that deliver therapeutic dose to thyroid tissue and also emits gamma photons for imaging. Iodine-123 is used primarily for diagnostic imaging.

Question 46Radiopharmaceuticals and Radiochemistry

A germanium-68/gallium-68 generator produces Ga-68 from the decay of which parent radionuclide?

  1. Germanium-68 Correct answer.
  2. Tungsten-188
  3. Molybdenum-99
  4. Strontium-82

Explanation

A Ge-68/Ga-68 generator relies on the decay of germanium-68 to gallium-68, which is eluted for PET labeling. The other parent-daughter pairs listed correspond to different generator systems.

Question 47Radiopharmaceuticals and Radiochemistry

Which radiopharmaceutical is used for pulmonary perfusion imaging?

  1. Tc-99m Technegas
  2. Tc-99m DTPA aerosol
  3. Tc-99m MAA Correct answer.
  4. Xenon-133 gas

Explanation

Tc-99m macroaggregated albumin particles are trapped in pulmonary capillaries after intravenous injection, allowing perfusion imaging. The other agents are used for ventilation studies.

Question 48Radiopharmaceuticals and Radiochemistry

The principal gamma-ray energy used for imaging Tc-99m is approximately:

  1. 140 keV Correct answer.
  2. 511 keV
  3. 364 keV
  4. 159 keV

Explanation

Tc-99m emits a principal gamma photon of about 140 keV, which is well suited to Anger camera imaging. The other energies correspond to commonly used PET or other SPECT radionuclides.

Question 49Radiopharmaceuticals and Radiochemistry

Radiochemical purity of a Tc-99m radiopharmaceutical is defined as the:

  1. fraction of activity that remains in the vial after injection
  2. fraction of total radioactivity attributable to the desired radionuclide
  3. fraction of total radioactivity present in the desired chemical form Correct answer.
  4. fraction of activity bound to plasma proteins after administration

Explanation

Radiochemical purity is the proportion of radioactivity in the intended chemical species, such as pertechnetate or a labeled complex. Radionuclidic purity instead describes the fraction of activity from the desired radionuclide.

Question 50Radiopharmaceuticals and Radiochemistry

In a Tc-99m labeling kit, stannous chloride primarily serves to:

  1. reduce pertechnetate to a lower oxidation state Correct answer.
  2. chelate free technetium permanently
  3. oxidize pertechnetate to a higher valence state
  4. buffer the solution at alkaline pH

Explanation

Stannous chloride reduces Tc-99m pertechnetate from the +7 oxidation state to lower, reactive states that bind the kit ligand. Without sufficient reducing agent, labeled yield falls and free pertechnetate remains.