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28th Annual

In-Training Examinationfor DiagnosticRadiology ResidentsRationalesSponsored by:Commission on EducationCommittee on Residency Training in Diagnostic Radiology

February 3, 2005

The American College of Radiology www.acr.org

-American College of Radiology

Section IX – Physics

230. In multiple-row detector CT, what is the relationship between CT Dose Index (CTDI) and slicethickness?

A. CTDI decreases as slice thickness decreases.

B. CTDI increases as slice thickness decreases.

C. CTDI is independent of slice thickness.

Question #230

Rationales:A. Incorrect.

B. Correct. As the slice thickness decreases, the CTDI per mAs increases. With thinner sections, the x-rayphoton flux is increased to compensate for the reduction in detector absorption efficiency

C. Incorrect.

Citations:J.T. Bushberg, et al. The Essential Physics of Medical Imaging (Second Edition). Lippincott Williams and

Wilkins, 2002, Chapter 13.

Diagnostic In-Training Exam 2005


231. What is the MOST likely effect of radiation exposure of 2 Gy (200 rad) during the preimplantationstage of pregnancy?

A. Embryonic death

B. Gross malformation

C. Growth retardation

D. Low birth weight

Question #231

Rationales:A. Correct. Prenatal death can be expected in about 80% of preimplantation embryos after 2 Gy.

B. Incorrect. This effect is possible if the radiation exposure to the fetus is less than 100 mGy (10 rad) duringthe second and third trimesters.

C. Incorrect. This effect is possible if the radiation exposure to the fetus is less than 100 mGy (10 rad) duringthe second and third trimesters.

D. Incorrect.

Citations:Eric Hall. Radiobiology for the Radiologist, 4th Edition, p 365.



232. Concerning fluoroscopy, ALL of the following increase scatter radiation EXCEPT:

A. Large patient size

B. Increasing field size

C. Narrow collimation

D. Increasing kVp

Question #232

Rationales:A. Incorrect. Scatter radiation increases with increasing patient size.

B. Incorrect. Scatter radiation increases with increasing x-ray field size.

C. Correct. Narrow collimation results in limited exposure and therefore minimizes scatter radiation.

D. Incorrect. Increasing kVp.



233. The advantages of multiple-row detector CT over single-row detector CT include ALL of thefollowing EXCEPT:

A. Increase in scan volume

B. Decrease in patient dose

C. Improved spatial resolution

D. Increased patient throughput

Question #233

Rationales:A. Incorrect. With MDCT it is possible to scan large volume per rotation compared to SDCT.

B. Correct. With MDCT, the patient dose tends to increase by 30% due to overbeaming compared with SDCT.

C. Incorrect. With smaller detector size, the advantage of MDCT is higher spatial resolution.

D. Incorrect. With shorter scan time, there is a significant increase in patient throughput.



234. What is the primary benefit of using rotating anode x-ray tubes rather than stationary anode tubes?

A. Decreases radiation exposure

B. Reduced voltage ripple

C. Ability to use higher kVp

D. Ability to use higher mAs

Question #234


B. Incorrect. The voltage ripple is dependent on the high voltage circuit, not the x-ray tube.

C. Incorrect. The maximum kVp is dependent on the high voltage circuit and the cathode-anode spacing (toprevent HV arcing), not stationary versus rotating anode.

D. Correct. Rotating anode tubes allow for very high mA for relatively short exposure times by spreading theheat over a larger surface area. This permits high-intensity short exposures.

Citations:A. Wolbarst, Physics of Radiology (1993) E, Chapter 10

J.T. Bushberg, et al., The Essential Physics of Medical Imaging (2002), Chapter 5



235. For an automatic exposure control (AEC) radiograph for an IVU, what is the effect of lowering thekVp from 85 kVp to 70 kVp?

A. Decrease exposure time

B. Decrease radiation dose

C. Increase scatter

D. Increased image contrast

Question #235


B. Incorrect.

C. Incorrect.

D. Correct. Lowering the kVp reduces the fraction of x-rays that pass through the patient. Since the AEC willincrease the mAs to give a defined exposure to the image receptor, the entrance dose to the patient willincrease. In general, contrast increases as one reduces kVp. This is particularly true with iodine contrastmedia with its 34 keV K-shell binding energy making it highly efficient at absorbing that portion of thex-ray spectra just above this energy.

Citations:A. Wolbarst, Physics of Radiology (1993), Chapter 19




236. A radioactive spill of Tc-99m results in a contamination level of 1000 cpm at 2 PM. What wouldyou expect the contamination level to be at 8 AM the following day?

A. < 10 cpm

B. 125 cpm

C. 250 cpm

D. 500 cpm

Question #236


B. Correct. Technicium-99m is the most commonly used radionuclide in nuclear medicine with a half-life of6 hours. By 8 AM the contamination will have decayed over 18 hours or 3 half-lives, reducing the activity toone eighth (0.5x0.5x0.5). Therefore the count rate should be around 125 cpm (counts per minute).

C. Incorrect.

D. Incorrect.

Citations:A. Wolbarst, Physics of Radiology (1993), Chapter 39




237. As the main magnetic field strength increases and maintains homogeneity, which one of thefollowing relaxation constants increases MOST?

A. T1

B. T2

C. T2*

D. Spin Density

Question #237

Rationales:Spin-lattice relaxation occurs when the molecular vibrational frequencies of the tissues approximate theangular precessional frequencies of the protons. With increasing precessional frequencies resulting fromincreasing magnetic field strength, there is less overlap of the precessional/vibrational frequencies, thusrequiring a longer time for the protons to release their energy back to the lattice.

A. Correct. As explained above.

B. Incorrect. As T2 is little affected by variations in the static magnetic field.

C. Incorrect. As T2 is not significantly affected, since homogeneity is maintained.

D. Incorrect. Spin-density is unchanged by changes in the magnetic field



238. Which of the following increases MR chemical shift artifact for a given slice thickness?

A. Increasing RF receiver bandwidth

B. Decreasing readout gradient strength

C. Decreasing matrix size

D. Decreasing field strength (B0)

Question #238

Rationales:Chemical shift occurs as a result of the difference in the resonant frequency of protons associated with veryhydrogenous (fat) tissues compared to protons associated with water molecules. Protons are localized by thevariation of resonance frequencies under the influence of an external magnetic gradient over a specific field ofview. This results in a net range of precessional frequencies across the field of view, which is sampled accordingto the matrix size. If the gradient strength is low, that means that a small range of frequencies will bemanifested over the FOV, and for a given matrix size, the bandwidth across a single pixel will be similarlysmall. Since the chemical shift results in frequency shifts of about 3 parts per million, for a 1 T magnet, thefrequency differences in fat versus water protons is about 1 T × 42.58 MHz / T × 3 × 10-6 = ~ 125 Hz. If thebandwidth across the pixel is small, then the signals due to fat can potentially be mapped into adjacent (oreven further away) pixels, and the anatomical presentation will manifest “chemical shift” artifacts.

A. Incorrect. A broad bandwidth receiver setting will result in similarly larger bandwidth across each pixel,reducing the effects of the chemical shift.

B. Correct. The lower readout gradient strength for a given FOV will result in a smaller range of frequenciesacross each pixel, and therefore will more likely map signals due to fat in different pixels than signals dueto water.

C. Incorrect. A smaller matrix size mapped to a given FOV will have larger pixels, and therefore will havelarger bandwidth across each pixel. The larger bandwidth per pixel will reduce the effects of chemical shift.

D. Incorrect. Lower main magnetic strength will cause the absolute chemical shift to also be lower (actualchemical shift = relative chem. Shift (ppm) × B

0 × γ.



239. What is the latent image in computed radiography (CR)?

A. Optical density patterns on a digital laser film

B. Absorbed x-ray photons in the phosphor

C. Electrons trapped in semi-stable energy wells

D. Digitally stored values in each pixel of the image

Question #239

Rationales:The latent image is the unobservable change of the detector arising from x-ray interactions in the object andthe absorption of the x-ray pattern onto the detector. CR (computed radiography) is a modality that usesphotostimulable storage phosphor (PSP) technology to acquire the x-ray information and subsequently rendera visible image after stimulating the latent image electrons that are stored in semistable traps within the PSP(typically made of barium fluoro bromide (BaFBr).

A. Incorrect. The laser film represents the final image, not the temporary unobservable latent image.

B. Incorrect. X-ray photons induce the storage of electrons, but do not comprise the latent image information

C. Correct. Electrons are trapped by the storage phosphor upon excitation by locally absorbed x-rays. Thelatent image information is subsequently rendered visible by excitation of a laser beam (diameter typically100 mm) and emission of shorter wavelength light photons upon return to the ground state.

D. Incorrect. The digitally stored values represent the final, observable image, not the unobservable image.

Citations:

J.T. Bushberg, et al. The Essential Physics of Medical Imaging, (Second Edition), Lippincott Williams andWilkins, 2002, Chapter 11.

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