Technetium-99m is the workhorse radionuclide of general nuclear medicine. What are its principal gamma energy and physical half-life?
511 keV and approximately 110 minutes; 159 keV and approximately 13 hours; 140 keV and approximately 6 hours; 364 keV and approximately 8 days
Answer: 140 keV and approximately 6 hours. Tc-99m decays by isomeric transition, emitting a 140 keV gamma photon with a 6 hour half-life, a near-ideal match for the NaI gamma camera and for same-day imaging with acceptable patient dose. The 364 keV, 8 day combination is the tempting distractor, but those values belong to I-131; 511 keV with 110 minutes describes F-18, and 159 keV with 13 hours describes I-123.
In a Mo-99/Tc-99m generator, how is the daughter radionuclide obtained for clinical use?
Saline is passed through the column, eluting pertechnetate while Mo-99 (half-life approximately 66 hours) stays bound to the alumina; The generator is heated until Tc-99m sublimates from the column; Tc-99m is produced in the column by neutron bombardment on site; The entire alumina column is dissolved in acid each morning
Answer: Saline is passed through the column, eluting pertechnetate while Mo-99 (half-life approximately 66 hours) stays bound to the alumina. Mo-99, with a half-life of approximately 66 hours, is adsorbed onto an alumina column; its decay product Tc-99m binds much more weakly, so normal saline washes it off as sodium pertechnetate while the parent remains. Because the parent half-life is roughly ten times the daughter half-life, the system reaches transient equilibrium and can be milked daily, with the daughter activity regrowing toward maximum in approximately 24 hours. On-site neutron bombardment is a tempting-sounding distractor, but Mo-99 is produced at reactors, not in the clinic.
Before a Tc-99m eluate is used clinically, it is tested for molybdenum breakthrough. What is the regulatory limit at the time of administration?
No more than 0.15 microcurie of Tc-99m per millicurie of Mo-99; No more than 0.15 microcurie of Mo-99 per millicurie of Tc-99m; Any detectable Mo-99 requires the generator to be discarded; No more than 1 microcurie of Mo-99 total, regardless of Tc-99m activity
Answer: No more than 0.15 microcurie of Mo-99 per millicurie of Tc-99m. The NRC limit is 0.15 microcurie of Mo-99 per millicurie of Tc-99m (0.15 kBq per MBq) at the time of administration; Mo-99 contamination matters because its high-energy betas and long half-life add unnecessary patient dose. The inverted-ratio distractor is the classic exam trap, so keep the parent in the numerator. Because Mo-99 decays more slowly than Tc-99m, the ratio worsens over time, which is why the limit applies at administration.
A patient receives I-131 therapy for thyroid cancer. What are the physical half-life and principal decay mode of I-131?
Approximately 13 hours, decaying by positron emission; Approximately 8 days, decaying by beta-minus emission with a 364 keV gamma; Approximately 6 hours, decaying by isomeric transition; Approximately 60 days, decaying by electron capture
Answer: Approximately 8 days, decaying by beta-minus emission with a 364 keV gamma. I-131 has an approximately 8 day half-life and decays by beta-minus emission; the beta particles deliver the therapeutic dose to thyroid tissue while the 364 keV gamma permits imaging but also drives the radiation precautions. The 60 day electron capture distractor describes I-125, used in brachytherapy seeds, and 13 hours belongs to I-123, the diagnostic iodine isotope.
What is the approximate physical half-life of F-18, the radionuclide in FDG PET?
8 days; 20 minutes; 110 minutes; 6 hours
Answer: 110 minutes. F-18 has a half-life of approximately 110 minutes, long enough to permit regional distribution from a cyclotron plus an uptake period, yet short enough to limit patient dose. The 20 minute distractor corresponds to C-11, whose short half-life confines it to sites with an on-site cyclotron; 6 hours is Tc-99m and 8 days is I-131.
How does a PET scanner localize annihilation events without the lead collimators used in gamma cameras?
By measuring the energy of each photon precisely enough to infer its origin; By rotating a single detector head rapidly around the patient; By using a converging collimator tuned to 511 keV; By registering two 511 keV photons in opposed detectors within a nanosecond-scale coincidence window, defining a line of response
Answer: By registering two 511 keV photons in opposed detectors within a nanosecond-scale coincidence window, defining a line of response. Positron annihilation emits two 511 keV photons approximately 180 degrees apart; detecting both within a short coincidence timing window places the event on the line connecting the two detectors, so-called electronic collimation. This is why PET sensitivity is far higher than that of collimated gamma cameras. A physical collimator for 511 keV is the tempting distractor, but absorptive collimation at that energy would be prohibitively thick and inefficient and is not how ring PET works.
A liver lesion on FDG PET has an SUVmax of 6. How is the standardized uptake value (SUV) calculated?
Tissue activity concentration divided by injected activity normalized to body weight; Lesion counts divided by background liver counts; Tissue activity concentration multiplied by the patient's blood glucose; Injected activity divided by lesion volume
Answer: Tissue activity concentration divided by injected activity normalized to body weight. SUV is the decay-corrected tissue activity concentration divided by the injected activity per unit body weight, so a value of 1 represents perfectly uniform whole-body distribution. The lesion-to-liver ratio distractor is tempting because visual reads often use the liver as a comparator, but that ratio is a different metric. SUV is affected by uptake time, blood glucose, and body habitus, which is why consistent technique matters for follow-up scans.
A technologist swaps a general-purpose parallel-hole collimator for a high-resolution parallel-hole collimator. What is the expected effect on the images?
Better spatial resolution and higher count sensitivity; Magnification of the organ of interest; No change in resolution, only a change in field of view; Better spatial resolution at the cost of lower count sensitivity
Answer: Better spatial resolution at the cost of lower count sensitivity. High-resolution collimators use longer, narrower holes that accept only photons traveling nearly parallel to the hole axis, sharpening the image but rejecting more photons and lowering sensitivity, so imaging time must increase to keep counts comparable. The resolution-plus-sensitivity distractor describes a free lunch that collimator design cannot provide; the two are fundamentally traded against each other. Parallel-hole collimators neither magnify nor minify.
For a thyroid uptake and scan, the technologist selects a pinhole collimator. What imaging characteristic does the pinhole provide?
Uniform sensitivity across a large field of view; Minification of large organs such as the liver; A magnified (and inverted) image of a small superficial organ, improving effective resolution; Elimination of septal penetration at high photon energies
Answer: A magnified (and inverted) image of a small superficial organ, improving effective resolution. A pinhole collimator works like a camera obscura: photons pass through a single small aperture, projecting a magnified, inverted image when the organ is close to the pinhole, which improves effective spatial resolution for small structures like the thyroid. The minification distractor is partially true physics, since objects far from the pinhole are minified, but that is not why it is chosen clinically; sensitivity also falls steeply with distance.
A neutron-poor radionuclide can decay by either positron emission or electron capture. What condition is required for positron emission to be energetically possible?
The atom must first be ionized by an external photon; The transition energy must exceed 1.022 MeV, twice the electron rest mass; The transition energy must exceed 511 keV, one electron rest mass; The nucleus must have an excess of neutrons
Answer: The transition energy must exceed 1.022 MeV, twice the electron rest mass. Creating a positron costs one electron rest mass (511 keV), and the daughter atom also sheds an orbital electron to stay neutral, so the decay energy must exceed 2 x 511 keV = 1.022 MeV. Below that threshold, electron capture is the only available route for a neutron-poor nucleus, which is why nuclides such as Tl-201 and I-123 decay purely by electron capture. The 511 keV distractor is tempting because it is the annihilation photon energy, but it accounts for only half the required mass-energy.
Under NRC regulations, a patient treated with I-131 may be released from the hospital when which criterion is satisfied?
The patient's thyroid uptake measures below 1 percent; The total effective dose equivalent to any other individual from the patient is not likely to exceed 5 mSv (0.5 rem); The retained activity is below 1 millicurie in all cases; At least ten physical half-lives have elapsed since administration
Answer: The total effective dose equivalent to any other individual from the patient is not likely to exceed 5 mSv (0.5 rem). Release is dose-based: the patient may go home if no other individual is likely to receive more than 5 mSv (0.5 rem), a determination that can rest on administered activity, measured dose rate, or patient-specific calculations, and written instructions are required when exposures could exceed 1 mSv. The fixed activity threshold is the tempting distractor because historical practice used activity cutoffs (on the order of 30 mCi), but current NRC criteria are framed in projected dose to others, not a universal millicurie limit.
A nuclear medicine technologist stands 1 meter from an injected patient. If the technologist steps back to 2 meters, the exposure rate from the patient falls to approximately what fraction of its original value?
One eighth; It is unchanged, because the patient is a distributed source; One quarter; One half
Answer: One quarter. For a compact source, exposure rate follows the inverse square law, so doubling the distance reduces exposure to approximately one quarter, making distance the most efficient of the time-distance-shielding triad. One half is the tempting linear-thinking distractor, but intensity falls with the square of distance. A patient approximates a point source reasonably well at distances of a meter or more.
A breastfeeding mother requires a nuclear medicine study. Which agent mandates complete cessation of breastfeeding for that child, rather than a temporary interruption?
F-18 FDG for PET; Tc-99m MDP for bone scan; I-131 sodium iodide; Tc-99m pertechnetate for thyroid imaging
Answer: I-131 sodium iodide. I-131 concentrates avidly in breast milk and in the infant's thyroid, and its 8 day half-life with beta emission makes any continued feeding hazardous, so complete cessation for that child is required; ideally breastfeeding is stopped weeks before therapy to reduce maternal breast dose as well. Tc-99m pertechnetate is the tempting distractor because it also appears in milk, but its short 6 hour half-life means only an interruption of roughly 12 to 24 hours is needed; most Tc-99m agents and FDG require little or no interruption.
In PET, a 'random' (accidental) coincidence degrades image quality. What produces a random coincidence, and how does its rate behave as injected activity increases?
Detector afterglow from the previous bed position; the rate falls with activity; One annihilation photon that scattered in the patient; the rate rises linearly with activity; Two photons from a single annihilation detected outside the energy window; the rate is independent of activity; Photons from two unrelated annihilations arriving within the coincidence window; the rate rises approximately with the square of the singles rate
Answer: Photons from two unrelated annihilations arriving within the coincidence window; the rate rises approximately with the square of the singles rate. A random coincidence pairs photons from two different annihilation events that happen to arrive within the timing window, assigning a false line of response; because it requires two independent detections, its rate grows approximately as the square of the singles rate, whereas true coincidences grow only linearly. The scattered-photon distractor describes scatter coincidences, the other major contamination, which mispositions genuine pairs rather than pairing unrelated ones. Shorter timing windows, aided by fast time-of-flight detectors, reduce randoms.
A gamma camera acquiring a Tc-99m study uses a 20 percent energy window centered on the photopeak. What is the primary purpose of this energy window?
To increase the intrinsic spatial resolution of the crystal; To correct for radioactive decay during the acquisition; To prevent detector dead time losses; To reject photons that scattered in the patient, which arrive with reduced energy
Answer: To reject photons that scattered in the patient, which arrive with reduced energy. Compton scattering in the patient lowers photon energy and changes direction, so a scattered photon carries false positional information; a window of approximately 126 to 154 keV around the 140 keV photopeak discards most of these degraded events and preserves image contrast. The dead time distractor is tempting because rejecting events does reduce processed counts, but dead time management is not the window's purpose, and energy windowing does not change intrinsic resolution or decay correction.