On a CT dose report, the CTDIvol for an abdominal scan is listed as 12 mGy. CTDIvol is best described as which of the following?
The patient's effective dose from the scan; The skin entrance dose at the scan midpoint; A standardized measure of scanner radiation output to an acrylic phantom, not the dose to the individual patient; The average absorbed dose received by that patient's abdominal organs
Answer: A standardized measure of scanner radiation output to an acrylic phantom, not the dose to the individual patient. CTDIvol is measured in standardized 16 cm or 32 cm acrylic phantoms and characterizes the scanner's output for a given protocol; it allows protocol comparison across scanners. It is not the dose to any particular patient, whose absorbed dose depends strongly on body size. Confusing CTDIvol with patient dose is the classic trap, and size-specific dose estimate (SSDE) exists precisely to bridge that gap.
How is dose-length product (DLP) calculated, and in what units is it reported?
Tube output multiplied by rotation time, reported in mAs; CTDIvol multiplied by the scan length, reported in mGy-cm; CTDIvol divided by pitch, reported in mGy; Effective dose multiplied by scan time, reported in mSv-s
Answer: CTDIvol multiplied by the scan length, reported in mGy-cm. DLP integrates scanner output over the scanned length: DLP = CTDIvol x scan length, in mGy-cm. It therefore reflects the total radiation delivered for the exam, whereas CTDIvol alone reflects only the intensity per slice position; two scans with the same CTDIvol but different coverage have different DLPs. DLP is the quantity converted to effective dose using body-region k-factors.
An effective dose of 8 mSv is estimated for a CT exam by multiplying the DLP by a body-region conversion (k) factor. What is the most appropriate use of this number?
Predicting this specific patient's lifetime cancer risk; Comparing relative stochastic detriment between different exam types in a reference population; Determining whether the patient exceeded a regulatory dose limit; Estimating the peak skin dose for deterministic injury tracking
Answer: Comparing relative stochastic detriment between different exam types in a reference population. Effective dose is defined for a sex-averaged reference phantom using tissue weighting factors, so it is a population-level comparison tool, not an individual risk estimate; applying it to a specific patient's risk is the classic misuse. Regulatory limits apply to occupational and public exposure, not to medical exams, and skin dose for deterministic effects is tracked with entirely different quantities (peak skin dose, air kerma). The k-factor method is a convenient approximation, which further limits individual precision.
A helical CT protocol is changed from pitch 0.5 to pitch 1.5 with the tube current fixed and no automatic compensation. Pitch is defined as table travel per rotation divided by the total beam collimation width. What is the effect on patient dose?
Dose decreases only if the rotation time is also shortened; Dose increases because the table moves faster; Dose is unchanged because kVp and mA are unchanged; Dose decreases because the same anatomy is covered with fewer rotations
Answer: Dose decreases because the same anatomy is covered with fewer rotations. At fixed mA, dose is inversely proportional to pitch: tripling the pitch from 0.5 to 1.5 spreads the same tube output over three times the coverage, cutting dose to roughly one third. The caveat worth knowing is that many scanners use an effective mAs concept (mAs/pitch) and automatically raise mA when pitch increases, which can erase the savings. Noise increases at higher pitch when mA is truly fixed because fewer photons contribute per image.
What is the purpose of automatic tube current modulation (ATCM) in CT?
To compensate for beam hardening by raising kVp through dense regions; To keep the tube current constant despite changing attenuation; To increase spatial resolution in high-attenuation regions; To adjust the mA to patient attenuation along and around the body, maintaining target image quality at lower overall dose
Answer: To adjust the mA to patient attenuation along and around the body, maintaining target image quality at lower overall dose. ATCM varies the tube current longitudinally (z-axis, based on the localizer) and angularly (as the tube rotates) so that thicker or denser projections receive more photons and thinner ones fewer. This holds image noise near a target level while avoiding unnecessary dose through less attenuating regions. It modulates mA, not kVp, and its goal is noise control rather than spatial resolution.
Compared with filtered back projection (FBP), what is the principal clinical advantage of iterative reconstruction in CT?
It eliminates all metal artifact; It shortens the acquisition time of the scan; It improves the scanner's maximum spatial resolution limit set by detector size; It lowers image noise at a given dose, permitting substantial dose reduction
Answer: It lowers image noise at a given dose, permitting substantial dose reduction. Iterative reconstruction suppresses noise relative to FBP at the same acquisition dose, so protocols can be run at meaningfully lower dose while preserving diagnostic quality. The tradeoff is an altered noise texture, sometimes described as plastic or waxy at aggressive settings, with possible loss of low-contrast detectability. It is a reconstruction method and does not change acquisition time or the hardware limits on spatial resolution.
A head CT shows a dark band in the posterior fossa between the petrous bones. Which artifact is most likely responsible?
Beam hardening; Ring artifact; Motion artifact; Aliasing
Answer: Beam hardening. As the polyenergetic beam crosses dense petrous bone, its low-energy photons are preferentially removed and the mean beam energy rises, causing the attenuation along those paths to be underestimated; the reconstruction renders dark bands or streaks between the dense structures. Photon starvation can also cause streaks through dense anatomy, but the classic dark band between the petrous ridges is the signature beam-hardening appearance. Scanners mitigate it with beam filtration and software corrections.
A CT image demonstrates complete concentric rings centered on the isocenter. What is the most likely cause?
An incorrect reconstruction kernel selection; A miscalibrated or defective detector element in a third-generation scanner; Patient motion during the acquisition; Beam hardening from dense contrast material
Answer: A miscalibrated or defective detector element in a third-generation scanner. In third-generation (rotate-rotate) geometry, each detector element samples a fixed radius from isocenter throughout the rotation, so a single miscalibrated or dead element traces a ring in the image. The fix is detector recalibration or service. Motion produces blurring or ghosting, and beam hardening produces bands or streaks between dense objects rather than complete rings.
While reviewing a liver CT, a radiologist wants to increase the displayed contrast between a subtle lesion and adjacent parenchyma. Which adjustment accomplishes this?
Decrease the window width; Increase the window width; Re-scan the patient at higher mAs; Increase the window level
Answer: Decrease the window width. Window width sets the range of HU values mapped across the grayscale: a narrower width spreads fewer HU over the same display range, so small attenuation differences become larger brightness differences. Window level simply centers that range and controls overall brightness. Window and level are display operations only; they change no acquired data, which is why re-scanning is unnecessary.
Which statement correctly describes the Hounsfield unit (HU) scale?
Water is defined as 0 HU and air as approximately -1000 HU; Air is defined as 0 HU and water as +1000 HU; The scale is anchored to cortical bone at 0 HU; HU values are absolute measurements independent of the scanner's effective beam energy
Answer: Water is defined as 0 HU and air as approximately -1000 HU. The HU scale is a linear transformation of the measured linear attenuation coefficient relative to water: HU = 1000 x (mu - mu_water)/mu_water, fixing water at 0 and air at about -1000. Dense cortical bone typically measures around +1000 HU or higher, and fat around -100 HU. Because attenuation depends on beam energy, HU values of a given tissue can shift slightly with kVp, so they are calibrated rather than truly absolute.
A cervical spine CT shows horizontal streaks across the shoulders where the image appears grainy. What is the best technical remedy for this photon starvation artifact?
Reconstruct with a sharper kernel; Lower the kVp through the shoulder region; Widen the display window width; Use automatic tube current modulation to deliver more mA through the widest projections
Answer: Use automatic tube current modulation to deliver more mA through the widest projections. Photon starvation occurs when very attenuating lateral projections through the shoulders transmit too few photons, producing noisy, streak-prone data. Tube current modulation raises the mA through exactly those projections, restoring photon statistics where they are needed without raising dose everywhere. Lowering kVp or sharpening the kernel worsens noise, and windowing changes display only, not the corrupted data.
A 3 mm hepatic lesion appears less conspicuous and its measured HU seems unreliable on 5 mm reconstructed images. Which phenomenon explains this?
Aliasing from undersampling; Partial volume averaging; Beam hardening; Ring artifact
Answer: Partial volume averaging. When an object is smaller than the reconstructed slice thickness, its voxels also contain adjacent tissue, and the reported HU is a weighted average of everything within the voxel. The lesion's measured attenuation is therefore shifted toward that of the surrounding parenchyma, reducing conspicuity and making characterization (for example cyst versus solid) unreliable. Reconstructing thinner slices through the lesion reduces the averaging at the cost of increased noise per image.
A small pediatric patient undergoes an abdominal CT reported with a CTDIvol referenced to the 32 cm body phantom. How does this CTDIvol relate to the dose the child actually absorbed?
It equals the patient dose, since the phantom standardizes the measurement; It overestimates the patient dose because small patients absorb fewer photons overall; It cannot be related to patient dose in any systematic way; It underestimates the patient's absorbed dose; the size-specific dose estimate (SSDE) applies a size-based correction factor
Answer: It underestimates the patient's absorbed dose; the size-specific dose estimate (SSDE) applies a size-based correction factor. A small patient attenuates less than the 32 cm reference phantom, so for the same scanner output the dose deposited at the patient's center is higher than the phantom-referenced CTDIvol suggests. SSDE multiplies CTDIvol by a conversion factor based on the patient's effective diameter; for small patients that factor exceeds 1, sometimes approaching 2 or more in infants. The intuitive but wrong answer is overestimation, reasoning from total energy rather than local dose.
A CT protocol's mAs is reduced to 25 percent of its original value with all other factors unchanged. What happens to image noise?
It approximately doubles; It increases by about 25 percent; It increases by a factor of 4; It is unchanged because noise depends on kVp
Answer: It approximately doubles. Quantum noise is governed by photon counting statistics: noise is proportional to 1 over the square root of the number of detected photons, which scales with mAs. Cutting mAs to one quarter therefore doubles the noise (1/sqrt(0.25) = 2). The factor-of-4 distractor applies the relationship in the wrong direction; the square-root law is one of the most frequently tested facts in CT dose optimization.
A CT dataset is reconstructed at 1 mm slice thickness instead of 5 mm from the same acquisition. What is the expected effect on the images?
Improved z-axis resolution with decreased noise; No change in resolution; only the display changes; Decreased z-axis resolution due to interpolation; Improved z-axis spatial resolution with increased image noise
Answer: Improved z-axis spatial resolution with increased image noise. Thinner slices reduce partial volume averaging and sharpen resolution along the z-axis, but each image is formed from roughly one fifth as many detected photons, so quantum noise rises (by about the square root of the thickness ratio). This is the fundamental slice-thickness tradeoff: resolution versus noise at fixed dose. Many practices reconstruct thin slices for review or reformats and thicker averaged slices for primary low-contrast interpretation.