radiology physics Flashcards

Question

What is relationship between MTF and resolution ("MTF-curve")? Between screen-film, indirect, and direct detectors, which has best resolution on MTF-curve (curve relatively shifted to right, with resolution on x-axis)?

Answer 1

Screen film has best resolution, followed by direct, followed by indirect worst....resolution units can be in lp/mm or cycles/s). ....-Better resolution corresponds to the modulation transfer function (MTF) curve staying higher as you move towards the right on the x-axis (shifts to right) (MTF on y-axis, resolution on x-axis.) => screen-film shifted to relatively to right, direct mid, indirect left.

Answer 2

The correct answer is 'A' Explanation Collimation has no effect on resolution. Details Remember the 3 determinants of resolution: Detector Motion artifact Effective focal spot size All of the other answer options relate to one of these determinants of resolution: Geometric magnification results in geometric blur, a consequence of increased focal spot size. Patient respiration results in motion artifact CR cassette is the detector in portable x-ray machines. Filament size determines actual focal spot size, which determines the effective focal spot size.

Answer 3

B. Explanation: Bucky Factor of 5 requires patient dose with grid to be 5 x (patient dose without grid). Bucky Factor of 3 requires patient dose with grid to be 3 x (patient dose without grid). Going from Bucky Factor of 5 from Bucky Factor of 3 causes dose in this patient to increased by a multiplicative factor of (5 x dose without grid) / (3 x dose without grid) = 5/3. Memorize This patient dose with grid = (Bucky Factor) x (patient dose without grid) Details Assume the phototimer is in effect for any question involving radiography or mammography. As an aside, be aware that it is possible to override the phototimer and directly set exposure time in the manual mode of radiography machines, which technologists use on occasion. Both fluoroscopy (fluoro mode, not spot images) and CT don't use phototimers. Fluoro mode uses automatic brightness control, and CT uses tube current modulation, both of which you'll learn about in later sessions. A grid decreases the amount of scatter that reaches the detector, which improves image contrast. A grid also decreases the # of photons reaching the phototimer. As a result, the x-ray tube stays on longer, leading to more patient dose. The grid ratio tell us how many times more the patient dose will increase with the grid than without the grid (patient dose without grid). Bonus The question stem also includes some nuanced extraneous information. Grids with larger ratios (12:1) will also have more x-ray stopping power, and thus larger Bucky Factors. Also, higher kVp will typically result in lower Bucky Factors for a given grid because the harder x-rays will penetrate the grid more effectively and more rapidly shut down the phototimer.

Answer 4

Most Sensitive: breast, lung, colon, red marrow, etc. Intermediate: thyroid, etc. Least Sensitive: brain, skin, salivary glands

Answer 5

The correct answer is 'C'. Explanation 2x2 binning creates half as many pixel values along the row and column of the detector matrix: This is like doubling the detector pitch. Since resolution in lp/mm = 1 / (2 x detector pitch), the new resolution is half the old resolution, so the resolution has changed by a factor of 0.5. Memorize This--- Flat Panel Detector (FPD) Resolution Formula: resolution in lp/mm = 1 / (2 x detector pitch) The 2x is there because you need two detector elements (DELs) to represent a single line pair. Details The 2x2 binning mode groups data from 4 adjacent detectors to form a larger equivalent square detector with detector pitch that is twice as large as the original detector pitch. Binning in this case is required because the data bandwidth to transmit the full FPD's worth of detector element (DEL) values is too high at 30 frames per second (fps), even if all that data could be streamed reasonably at 7 fps. Binning is a poor-man's Netflix video compression method - it just averages values from neighboring pixels (DELs), so there are fewer pixel values to transmit in total for every frame. Bonus Some fluoroscopy units also use binning for their largest supported FOVs, since at the large FOV there may be an issue of too many DELs for a given rate of frames per second, which clogs the limited data streaming rate supported by the hardware.

Answer 6

D. Explanation: Detector Pitch for square detector = FOV length / matrix length = 200mm / 1000 = 0.2 mm Resolution = 1 / (2 x Detector Pitch) = 1 lp / 0.4 mm = 2.5 lp/mm Details: Detector pitch is defined as distance from the start of one DEL to start of the next DEL. In this case, detector pitch is 20 cm / 1000 DELs = 0.02 cm / DEL = 0.2 mm / DEL. Each line pair requires two DELs because a line pair includes a bright stripe and a dark stripe: The length of two DELs is 2 x 0.2mm = 0.4 mm. Resolution is given by number of line pairs (lp) resolvable per millimeter of detector. Since we can fit 1 lp in 0.4 mm, we have: 1 lp / 0.4 mm = 10/4 lp/mm = 2.5 lp/mm Bonus Detector pitch in fluoroscopy has nothing to do with pitch in CT. Be comfortable switching between metric units. Some examples (see latest version of Radiology Simplified, chapter on Radiation Safety, for more examples): 1 cm = 10 mm 1 mm = 0.1 cm 1 m = 100 cm = 1000 mm

Answer 7

C. Explanation Defined for image intensifiers only, minification gain explains one of the two ways that image intensifiers intensify images: 1) Minification Gain - brightness increase from electrons concentrated from a larger photocathode onto smaller output phosphor surface area 2) Electronic Gain - brightness increase from electrons accelerated across tube starting at the photocathode and moving to the output phosphor (anode) side. The brightness gain aggregates these two effects into one number: Brightness gain = Flux Gain x Minification Gain Bonus Image intensifier formulas are low yield, and hopefully will be obsolete soon. Focus on the big-picture concepts behind minification gain, electronic gain, and brightness gain. If they ask about II formulas, treat this as an FU question and move on.

Answer 8

The correct answer is 'E'. Explanation Magnification (either electronic or geometric) results in a smaller patch of anatomy filling up the same size monitor. The patch of anatomy you can see is the field of view (FOV), which has decreased with magnification. Details Since magnification involves displaying a smaller anatomical region over the full size of the display, it makes sense that FOV would decrease: This is true for image intensifiers (II), flat-panel detectors (FPD), geometric magnification, and electronic magnification.

Answer 9

A. Explanation Think about scattered radiation as the dominant contributor to operator absorbed dose. In the LAO position, the operator's face is directly accessible to scatter. In the RAO position, the FPD or II protects the operator's face. Moving to RAO, absorbed dose to the operator decreases. (Additionally, the x-ray source is also moved further away from operator's lower body, therefore diminished back scatter and spread to lower body): Detail Position names in fluoroscopy will first list the patient side closest to the detector. For example, LAO means that the detector is closest to the left anterior oblique side of the patient. The convention is opposite in standard projection radiography and mammography, where first listed is the patient side where x-rays enter. For example, the PA chest radiograph involves positioning the x-ray tube so that x-rays enter the patient's posterior side.

Answer 10

The correct answer is 'A' Explanation This is a recall question. Memorize This The lead apron is typically 0.5mm with 99% absorption (1% transmission to the operator) for typical kVp. This is a very rough rule of thumb, but it's widely memorized. Bonus Remember that half value layer (HVL) depends on kVp among other things? The protection you get from a lead apron depends on kVp, since PE absorption and Compton scatter both decrease with increasing x-ray photon energy. Not Exam Relevant, but Useful Vendors will sell "0.5mm lead equivalent” materials. This is an unregulated terminology, which means the % absorption is equal to 0.5mm pure lead for the tested kVp, but it's not required to match lead absorption across all kVp settings. Don't forget that can be wear-and-tear on these aprons, including cracks from folding, and shifting of materials, which could result in large patches without radiation protection. Please keep your hands out of the radiation field. You shouldn't be exposing your hand to primary beam for any reason. If you want a little more practical information on protecting yourself with aprons (from people other than the sales representative), check out this blog entry and this abstract.

Answer 11

A lead apron is typically 0.5mm with 99% absorption (1% transmission to the operator) for typical kVp

Answer 12

A. Explanation All of these artifacts are specific to image intensifiers (II). Details Flat-panel detectors (FPDs) are entirely immune to this conventional set of artifacts seen with IIs.

Answer 13

The correct answer is 'B' Explanation S-distortion results from external magnetic fields as they alter the trajectory of electrons flying across the image intensifier, disproportionately affecting the outer portion of the image. Details Pincushion is spatial distortion (worse at the periphery) resulting from the curved input phosphor surface. Vignetting is darkening of the image at the periphery resulting from longer path for electrons diverging from the focal point (where the electron beams come together inside the vacuum bottle) to the periphery of the output phosphor relative to its center. Flare/Veiling Glare occurs when intense light one region of the output phosphor obscures detail in adjacent less-bright regions because of lateral diffusion inside the glass window. Lung obscuring the cardiac contour is a classic example. Bonus The goal of an image intensifier is to intensify images that would appear too dim if you just used the input phosphor alone.

Answer 14

C. Explanation The photodiode is the light sensitive component. When visible light hits the photodiode, it results in a small current that can be measured. Details The cesium iodide (CsI) scintillator converts incoming x-rays into visible light photons that can be seen by the photodiode. The TFT layer contains transistors that transmit the digital signal from the photodiodes to the computer. There are some other electronics minutia (data line, etc.) that you don't need to know for the exam.

Answer 15

A. Explanation Collimation increases peak skin dose with fluoroscopy using image intensifiers (IIs). Collimation is generally encouraged predominantly because it improves image contrast and decreases KAP, but the overall decreased image intensifier brightness from collimation causes automatic brightness control (ABC) to increase tube output, which increases peak skin dose (PSD). Since AK and PSD relate to deterministic effects, this is a clinical consideration if you're using an image intensifier, and you should see your cumulative air kerma (CAK) values ticking upward faster during collimation in this case. Know This: In II, air kerma (AK) and peak skin dose (PSD) are increased by collimation. In both II and FPD, dose area product (same as KAP) is decreased by collimation. Details: Collimation improves image contrast, decreases noise, and increases contrast-to-noise ratio (CNR). Decreased Compton scatter results from adjacent tissues with the narrower radiation field. Better contrast happens because the image is better windowed and automatic brightness control is optimized around the relevant anatomy only. Collimation alone does not generally improve resolution (unlike electronic magnification). The exception is flat panel detectors (FPD) that use binning. In that case, collimation allows the FPD to stop binning because lower FOV decreases data rate requirements, and resolution improves to the value determined by the FPD resolution formula: FPD resolution in lp per mm = 1/(2 x pitch) Collimation increases AK and PSD even though it decreases KAP. This is less of a problem with FPDs. Unlike image intensifiers, flat-panel detectors (FPDs) do not require large increases in tube output after collimation to stay bright, because the dynamic range of FPDs is much larger than image intensifiers, and rescaling pixel values is easy using a look-up table (LUT) with a digital display. With FPDs, collimation leaves AK unchanged or slightly increased. Bonus If you're trying to google for definitions of AK, KAP, and other radiation units, save yourself some time and download the latest update to Radiology Simplified. The updated Radiation Safety chapter has the best available summary of radiation dose units for all x-ray based modalities, and the fluoroscopy-specific units are also presented in the Fluoroscopy chapter. Explanation The photodiode is the light sensitive component. When visible light hits the photodiode, it results in a small current that can be measured.

Answer 16

The correct answer is 'A' Explanation The II is now farther from the x-ray tube by a factor of (150 cm / 100 cm) = 3/2. If the x-ray tube technique didn't change, radiation at the II surface would drop by a factor of (3/2)2 = 9/4 = 2.25. Because automatic gain control (AGC) is in effect (assume this with fluoroscopy), the dose to the patient has to increase to maintain brightness, particularly with image intensifiers. If we further assume the AGC keeps kVp unchanged (with mA adjusted), the factor of by 2.25 increase in overall dose is needed to keep exposure to the detector unchanged. In practice, AGC may preferentially adjust kVp to mitigate dose increases. Memorize This 1/R2 rule means that if distance from a point of interest to the x-ray tube is increased by a factor of F, radiation dose at the point of interest is decreased by a factor of F2. AGC compensates reduction in dose at the detector by increasing x-ray tube output. Always assume AGC is in effect with fluoroscopy unless instructed otherwise in the question stem. Bonus FPDs have better dynamic range (latitude) than IIs. Also, the LUT can be adjusted to maintain image brightness. Even though AK decreases with distance from the x-ray tube (1/R2 rule), KAP doesn't change with distance. Why is that? Review the fluoroscopy session to remind yourself of the logic.

Answer 17

B . Explanation: Magnification in FPD results in no change in resolution because FPD resolution is dependent on the detector pitch. The exception (which you should know) is if binning* was used for the larger FOV, but the question stem rules out this possibility. Memorize This FPD Resolution Formula: resolution in lp/mm = 1/ (2 x detector pitch) Details * Binning means averaging values from adjacent DELs and discarding data from individual DELs: Binning is used if the electronics can't handle the large data rate, such as with large FOV where too many DELs are simultaneously reporting their values, or when frames per second (fps) is high (eg. 30 fps). For fluoroscopy machines that use binning for large FOV, moving to a smaller FOV (magnification) would allow binning to stop, and resolution would jump to a fixed higher value prescribed by detector pitch. Bonus Geometric magnification, not advised in fluoroscopy, would have an uncertain effect on resolution. Magnification in this case would actually increase the image size on the detector elements (more detector elements for a given anatomical part = better resolution), but there would also be increased focal spot blur. Magnification views in mammography actually use geometric magnification - how does mammography get around the focal spot blur issue while also reducing dose? You'll learn about that in the Mammography session.

Answer 18

The correct answer is 'A' Explanation The bulk of exposure to the operator occurs from scatter at the entry point to the patient. In position 2, scatter from the x-ray source hits your unshielded face, increasing absorbed dose: Memorize This Time, distance, and shielding to decrease dose. Of these factors, minimizing fluoro time is your most important tool for decreasing dose to yourself and the patient. Details With an operator covered by a traditional apron with no face radiation shielding, placing the x-ray tube above the table and incurring scattered radiation to the unprotected face is uniformly worse than LAO, RAO, or AP orientations where the x-ray tube is underneath the table and scatter tends to hit the apron. Bonus Fluoro machine positioning to minimize dose to patient and operator is critical for you to know. Please review the fluoroscopy positioning slides in Radiology Simplified. Keep the detector (II or FPD) close to the patient Keep the x-ray tube at a safe distance from the patient so that air kerma (AK) doesn't go so high that the patient receives radiation burns. If you're asked to choose between LAO and RAO, the RAO position minimizes dose to the operator, because the x-ray tube and resulting scatter are on the opposite side of the table from the operator.

Answer 19

A. ``` Explanation Detector element (DEL) Version 2 has a larger fill factor. As a result, shot noise decreases, so contrast-to-noise increases, although resolution is unchanged. ``` Details DEL version 2 has more photodiode surface area, so it will be able to collect more photons per pixel. The overall increased % of DEL surface area devoted to photosensitive material (Fill Factor) has increased. This allows for decreased shot noise (quantum mottle). This means that contrast-to-noise ratio will increase. Resolution is unchanged because detector pitch has not changed.

Answer 20

The correct answer is 'C' Explanation Using electronic magnification results in collimation followed by rescaling (magnification) of the collimated image onto the screen. This requires increased AK to compensate image quality. Without this compensation, the magnified image would be dimmer (II) or grainier (II and FPD). Details Magnification first results in collimation, since you're imaging a smaller area. With collimation, there are fewer x-ray photons reaching either the image intensifier (II) or flat panel detector (FPD). This wouldn't be a problem, except that this collimated image is now expanded to fill the screen. Uncompensated, this would cause a dimmer (II) or grainier (II and FPD) image. The automatic brightness control (ABC) system compensates by increasing x-ray tube output, particularly with II's and to a lesser degree with FPDs. Bonus The great dynamic range of FPDs and flexibility of look up tables (LUTs) mean that even with no changes to the x-ray tube output, electronic magnification would not make the image look dimmer. However, the uncompensated FPD magnified image would be grainier as a result of fewer photons contributing to the full screen image. The ABC compensates shot noise (graininess) by slightly increasing tube output (kVp and/or mA).

Answer 21

The correct answer is 'A' Explanation In fluoroscopy, geometric magnification typically requires increasing source-image distance (SID), the distance between the x-ray tube and detector. This causes automatic brightness control (ABC) to increase x-ray tube output, so AK increases. Since collimation is not applied, the KAP is unambiguously increased for both flat panel detectors (FPD) and image intensifiers (II). Details Geometric magnification refers generically to achieving magnified views using the equation Magnification = SID/SOD. In fluoroscopy, when you pull the detector away from the patient, you're increasing SID, which means fewer x-rays reach the detector, until automatic brightness control (ABC) reflexively increases tube output (kVp and/or mA). With FPDs, ABC actually controls shot noise. With IIs, both shot noise and brightness are issues. Basic safety considerations like these are highly testable. During IR procedures, you're urged to keep the detector as close to the patient to minimize dose. This also decreases dose to you as the operator since the detector acts like another x-ray shield. Bonus Even though it's not recommended during fluoroscopy, geometric magnification is actually how we achieve magnification in mammography. What's different about the process which causes dose to actually decrease during geometric magnification in mammography? You'll learn more about this in the Mammography session. Did you incorrectly select "unchanged" for KAP? You might be thinking about electronic magnification (both for image intensifiers and FPDs), where collimation is used to decrease area while AK increases to compensate graininess (FPD) or brightness (II).

Answer 22

The correct answer is 'B' Explanation Cancer risk is related to KAP. Since KAP relates to AK and area exposed, think about the effects of collimation on these two quantities. For image intensifiers, collimation actually increases AK substantially because of ABC. For FPDs, AK is either flat or slightly increased. In either case, area exposed decreases much more than AK increases, so overall KAP will decrease. Memorize This KAP = AK x area exposed Collimation ALWAYS increases contrast and decreases KAP, but may increase AK particularly with image intensifiers. In fluoroscopy: KAP ↔ stochastic risk (cancer) AK ↔ deterministic risk (erythema, epilation, radiation necrosis, etc.) Details Image intensifiers require large increases in AK following collimation to stay bright. FPDs have better dynamic range so increased AK isn't as big of a concern for FPDs. Bonus The conventional teaching is that collimation does not change resolution in fluoroscopy, because it does not change FOV, so answer "unchanged” if you're asked about resolution and collimation. With digital FPD, this teaching assumes there is no change in detector element binning.

Answer 23

B. Explanation Variations on the 1/R2 radiation safety question type are so easy to write that it's worth expecting it on your exam. Memorize This Exposure (AK) and absorbed dose decrease as 1 / R2, where R in this formula is distance from x-ray tube to the operator. Details This is a stock question type related to the 1/R2 rule. Answer this question in two steps: (1) Calculate the factor by which distance from the operator to the x-ray tube has changed : distance factor = (new distance)/(old distance). (2) Calculate the factor by which dose to the operator has changed using: dose factor = 1 / (distance factor x distance factor) (3) If needed, calculate the new dose to the operator: new dose to operator = dose factor x old dose to operator Caution Subtraction is not used in this calculation. The answer 1/49 was a distractor to catch anyone using subtraction to implement the 1/R2 rule. The other stock question type related to the 1/R2 rule is to ask how dose to the patient changes if the detector is moved away from the patient. The ABC mechanism increases dose to patient to compensate the 1/R2 drop in dose to the detector. Don't let yourself get confused between the two types of 1/R2 rule questions.

Answer 24

B. Explanation Tomosynthesis data is used to reconstruct slices through the breast tissue, as well as images that are equivalent to digital mammograms (eg. Hologic C-View). Tomosynthesis allows better separation of real lesions from benign overlapping tissue (decreased recall rate), as well as detection of more subtle malignancies (increased cancer detection rate). Memorize This: Typical cancer detection rates: 4 per 1000 for digital mammo 5 per 1000 for mammo+tomo Typical recall rates: 10% for digital mammo (also the ACR practice standard for your practice) 9% for mammo+tomo

Answer 25

The correct answer is 'A' Explanation This is a recall question related to the 1992 MQSA regulations. ``` Memorize This Lay summary (to the patient) and physician report (to referring physician) are due within 30 days of performing the mammogram. ``` Bonus When you're faced with a recall question, decide your answer before reviewing answer choices to protect yourself from getting confused.

Answer 26

The correct answer is 'C' Explanation This is a recall question. Memorize this Initial certification - 240 exams within any 6 months of last 2 years of residency Continuing certification - 960 exams per 2 years Details Other requirements you should know: Initial - 3 months of training in mammography Continuing - 15 category-1 CME credits per 3 years

Answer 27

A. Explanation Magnification in mammography involves three steps: Step 1: Place air gap (plastic ledge) under breast Step 2: Switch to smaller filament Step 3: Remove grid Memorize This Geometric Magnification FormulaMagnification = SID / SOD Remember that Magnification quantifies how many times larger the image on your detector is compared to the actual size of the anatomy. See the Mammography & Breast MRI chapter of Radiology Simplified for a quick review of source-image distance (SID) and source-object distance (SOD). Details Step 1: Air gap allows for geometric magnification, but also causes geometric blur. Step 2: Smaller filament decreases focal spot size, which compensates for geometric blur. Step 3: Since the air gap allows scattered radiation to diverge away and escape the detector, a grid isn't needed, saving dose.

Answer 28

A. Explanation This is a recall question related to ACR and MQSA. Details The ACR is one of a few agencies approved by the FDA to provide 1992-MQSA-mandated accreditation to mammography facilities. The ACR's process is called the Mammography Accreditation Program. For your facility to get accreditation from the ACR, it will need to submit clinical image samples of your best mammograms, and a sample image using the ACR Phantom. Clinical images are judged on several image quality criteria (hard to memorize), but the top image quality criteria to cause failed accreditation are positioning and compression.

Answer 29

The correct answer is 'B' Explanation Be sure you can describe lesion position. This breast mass is located in the upper outer quadrant. The MLO view tells us the lesion is upper because it projects superior to the nipple. We know the position is outer (lateral) rather than inner (medial) because the question stem tells us that the lesion "falls” to a more inferior position on the ML view compared with the original MLO view. Among the answer choices, the only 2 o'clock position is in the left breast upper outer quadrant. The MO together with the CC view confirms this: Caution: 2 o'clock left breast is in the upper-outer quadrant, but 2 o'clock right breast is in the upper-inner quadrant. This is because the clock, centered on each nipple and viewed facing the patient, is numbered the same way for both breasts (see Mammography chapter of Radiology Simplified). Interpreting lesion position based on clock face requires specifying left versus right breast. Memorize This When going TO the ML view from the MLO view, "Muffins rise, and Lead falls.” In other words, a medial breast lesion will project higher on the ML image than the MLO image, and a lateral breast lesion will project lower on the ML image than the MLO image. The CC view also determines medial versus lateral position. The outer (lateral) breast on CC view is the upper portion of the CC image for both left and right breasts. This is easy to forget if you haven't been on mammography rotation in a while.

Answer 30

B. Explanation This is a recall question related to effective focal spot sizes. Memorize This X-ray-based Modality: Effective Focal Spot Size: Conventional Radiography 1.2 mm CT 1 mm Fluoroscopy variable 0.3 mm to 1.2 mm Mammography (standard) 0.3 mm Mammography (magnification) 0.1 mm Details Mammography machines have a dual filaments in the focusing cup to generate the different effective focal spot sizes described above. Geometric magnification (SID/SOD) using an air gap also incurs geometric blurring. To compensate, machines use the smaller filament that provides a 0.1 mm effective focal spot size.

Answer 31

The correct answer is 'A' Explanation Not enough information. The mean glandular dose limit depends on whether a grid is used or not. Memorize This MQSA-mandated mean glandular dose limit, single view: 3 mGy with grid 1 mGy without grid Details The mean glandular dose is measured in mGy (energy per mass of tissue) using an ionization chamber and a fudge factor that depends on % glandular tissue, breast thickness, and x-ray energy spectrum.

Answer 32

D. Explanation This is a recall question related to breast MRI protocol. Memorize This The ACR recommends breast MRI be performed at 7-14 days. Details Normal increased breast parenchymal enhancement is seen during the secretory phase, which increases false positive and false negative rates.

Answer 33

A. Explanation This is a recall question related to breast MRI protocol. Memorize This ACR-recommended dose of gadolinium is 0.1 mmol/kg followed by 10 mL saline flush. Details Breast MRI is typically performed on a 1.5T or stronger magnet with dedicated bilateral breast coils.

Answer 34

B. Explanation This is a recall question related to MQSA. Memorize this MQSA stands for the Mammography Quality and Standards Act Congress passed the MQSA in 1992. FDA enforces MQSA through approved certification agencies (like the ACR) VA does its own oversight

Answer 35

A. Because mammographic magnification uses geometric magnification, the breast is brought closer to the x-ray source (decreased SOD) which increases entrance skin dose (ESD) and hence average glandular dose. On the other hand, there is radiation saved by removing the grid during magnification views, and the beam is collimated to a smaller FOV. These competing effects suggest that dose to the breast during magnification view is either increased or unchanged relative to standard view, but not decreased. This is based on our detailed review of Monte Carlo simulation and empirical research papers and between 1998 and present, as well as current texts; a definitive study with modern equipment does not exist as of 3/30/17. References: from source here from 1998 using film-screen; work out of Europe from 2005 with screen-film estimates mag views have about 2x more dose than screening views. A recent article in Radiology from 2010 perpetuates this sentiment without data. More recent texts suggest mag views generate dose that approaches standard views. No published research suggests mag views would decrease mean glandular dose.

Answer 36

A. Explanation Magnification is not related to compression. Compression separates overlapping tissue. Compression decreases breast motion, which increases resolution. Fewer x-rays are absorbed or scattered in the thinner compressed breast, so the AEC shuts off faster (shorter exposure time), decreasing patient dose relative to a non-compressed breast. Memorize This Per MQSA, the mammography unit should be capable of maximum breast compression force between 111 - 200 Newtons (25-45 lbs). Most residents memorize this range or at least the lower end (111 Newtons; 25 lbs.), but many residents don't realize this is a specification for the machine, and this does not mean you have to use this much compression force on every patient. Bonus Review the three determinants of resolution in the Projection Radiography chapter of Radiology Simplified. NEXT

Answer 37

Per MQSA, the mammography unit should be capable of maximum breast compression force between 111 - 200 Newtons (25-45 lbs). Most residents memorize this range or at least the lower end (111 Newtons; 25 lbs.)

Answer 38

D. Explanation This is a fatty breast (<25% fibroglandular tissue). Mo target generates the lowest characteristic peak (just below the 20 keV k-edge), and Mo filter has the lowest cutoff point (k-edge) at 20 keV. This gives the softest (lowest average energy) photons of all the options. Memorize This Target / Filter* Breast Density Mo/Mo Fatty Mo/Rh Scattered Fibroglandular or Heterogeneously Dense Rh/Rh, W/Ag Dense *pay attention to the order - target listed first, filter listed second Details The characteristic energies of Mo and Rh targets determine the typical photon energies generated by the x-ray tube. Higher characteristic energy → higher the average x-ray photon energy. The k-edge of the filter material determines the upper limit on typical x-ray energies that pass through the filter into the patient. Higher filter k-edge → higher average x-ray photon energy. Bonus Check your basic understanding now: Rh target with Mo filter isn't used. Why? ----bc the K-edge of Mo is lower than the characteristic radiation of of Rh

Answer 39

A. Explanation Tomosynthesis with synthetic 2-view mammography requires approximately the same dose as 2-view screening digital mammography. Details Exact doses vary between manufacturers. Synthetic mammograms (eg. Hologic's C-view) look like regular 2-view (CC + MLO) digital mammograms. They are computed from the tomosynthesis data, so no additional radiation is required.

Answer 40

A. Explanation Tungsten (W) target with Silver (Ag) filter releases x-rays predominantly using braking radiation. This is because the Ag k-edge (25 keV) is much lower than the several characteristic peaks of W (k-edge 70 keV). Details Most standard digital mammography systems use Mo/Mo (fatty), Mo/Rh (scattered fibroglandular, heterogeneously dense), or Rh/Rh (dense). The energy spectrum from these x-ray tubes have a large component of characteristic radiation. Some new tomosynthesis units use W targets with Ag, Rh or other filters, predominantly emitting braking radiation.

Answer 41

The correct answer is 'C' Explanation This is a recall question related to mammography facility certification implemented by the ACR on behalf of the FDA, in accordance with the 1992 MQSA. Memorize This - ACR phantom contains 6-5-5 (fibers -speck groups-masses) - Passing score requires visualizing at least 4-3-3 - The ACR phantom is equivalent to a 4.2cm breast with 50% glandularity. - MQSA stands for Mammography Quality Standards Act, passed by Congress in 1992 - ACR provides accreditation of mammography centers on behalf of the FDA in accordance with the 1992 MQSA.

Answer 42

C. Explanation The pediatric protocol will likely require an increase in mAs to accompany the decrease in kVp. Empirical studies show that decreasing kVp in CT also decreases patient dose. This also makes it easier for photons to get absorbed (PE) in the patient, resulting in grainier images (shot noise). Commonly, the mAs is increased to partially compensate shot noise, while ensuring the overall dose has been reduced. Bonus It is still true that increasing kVp in radiography (including mammo and fluoro) will generally save dose. If you think that opposite recommendations on kVp for radiography and CT is counterintuitive, you're in good company. .....the explanation for this paradox is that radiography generally utilizes a phototimer (=> i.e. auto increase mAS for loss of photon penetration at detector with dropping kVP, in order to maintain desired overall image brightness); however there is no phototimer used in CT (=> no autodose increase/mAs increase with dropping of kvp; however => will have increase in shot noise with decreased kvP...=> even with CT will try to increase mAs to minimie shot noise problem, but overall will effect dose to lesser degree as with phototimer?)

Answer 43

The correct answer is 'A' Explanation CT Abd/Pelvis results in ~25 mGy to the fetus. The major risk at third trimester is ~1% increased risk of childhood cancer. It's good to think about alternative options in this scenario (like ultrasound), and to evaluate risk-benefit for each option. Memorize This < 14 days post-conception: 100 mGy threshold for fetal death. "All-or-nothing” survival: no elevated deterministic or stochastic risk. Later in gestation: 50 mGy fetal deterministic effects threshold (teratogenicity & mental retardation esp. 2-15 wks.) 25 mGy in-utero exposure (CT abd/pelvis to mother) increases risk of childhood cancer by ~1% Detail CT abd/pelvis in pregnant women is not a negligible risk to the fetus. Some institutions recommend ultrasound as the first line assessment of appendicitis in pregnant women. Fetal thyroid dysfunction following maternal IV contrast has not been studied extensively, but is presumed to be a negligible risk, especially in the context of routine fetal thyroid screening for all newborns. Gadolinium IV contrast is contraindicated in pregnancy. Limited use of MRI without contrast is an option in pregnancy to r/o appendicitis. CT in pregnancy is a medicolegal risk, so follow the protocols specified by your institution which may include informed consent. Don't be surprised if recommendations change over the years - it's difficult to perform research on fetal radiation risks, or cancer-related radiation risk in general.

Answer 44

CT Abd/Pelvis results in ~25 mGy to the fetus.

Answer 45

A) < 14 days post-conception, 100 mGy threshold for fetal death. "All-or-nothing” survival: no elevated deterministic or stochastic risk. B) - 50 mGy fetal deterministic effects threshold (teratogenicity & mental retardation esp. 2-15 wks.) - 25 mGy in-utero exposure (CT abd/pelvis to mother) increases risk of childhood cancer by ~1%

Answer 46

The correct answer is 'C' Explanation Breast dose is the primary consideration, particularly because pregnant women are also young patients. Pregnancy additionally causes a proliferation of glandular tissue (TDLUs, ducts). The PE CT deposits a hefty dose of radiation to the breast (10-70 mGy per breast, but don't worry about memorizing this number range). Memorize This 50 mGy is threshold for induction of deterministic effects in the fetus. MQSA-specified mammography dose limit to breast per image is 3 mGy with grid and 1 mGy without grid Details Fetal dose from PE is considered negligible, similar to the fetal background radiation exposure (~1 mGy). Fetal thyroid dysfunction following maternal IV contrast has not been studied extensively, but is presumed to be a negligible risk, especially in the context of routine fetal thyroid screening. Although reasonable and preferred in many clinical cases, CT in pregnancy is still a medicolegal risk, so follow the protocols specified by your institution which may include informed consent. Don't be surprised if recommendations change over the years - it's difficult to perform research on fetal radiation risks, or cancer-related radiation risk in general. Bonus Here is a source article on PE CT in pregnancy. More recent articles have continued to reinforce the preference for PE CT over V/Q in pregnancy, although V/Q delivers similar dose to the fetus and much less dose to the maternal breasts. Here is one authoritative source on prenatal radiation risks in general.

Answer 47

The correct answer is 'A' Explanation The larger tube current means that more photons participate in the HU value calculated for each voxel, reducing shot noise. Recall that: Noise Standard Deviation ∝ 1/sqrt(# photons) As a result, Noise Standard Deviation ∝ 1/sqrt(tube current) The tube current went from 100 mA to 144 mA, which is 1.44x increase. The square-root of 1.44 is 1.2 (recall 12 x 12 = 144). The new noise standard deviation is 1/1.2 times the old standard deviation. Note that 1/1.2 = 1/(12/10) = 1/(6/5) = 5/6. Bonus Always use common sense to check the answer you've selected. Should noise increase or decrease when you increase mA? You should be able to eliminate all of the answers that suggest noise has increased.

Answer 48

A. Explanation The larger voxel size means that more photons participate in the HU value calculated for each voxel, reducing shot noise. Recall that: Noise Standard Deviation ∝1/sqrt(# photons) As a result, Noise Standard Deviation ∝1/sqrt(voxel size) Because the slice thickness has doubled, the total voxel volume (slice thickness x in-plane width x in-plane height) has also doubled. Details The noise discussed here is shot noise (quantum mottle), which is the grainy image appearance that you get when there aren't enough photons that hit the detector (photon starvation). When examiners and review sources talk about image noise, they're actually referring to the standard deviation of the image noise. Bonus People commonly use "reconstruction” to go from the raw CT data to the axial images. In contrast "reformat” refers to post-processing performed on the axial images. Correct Usage: "Can you reformat these axial images into sagittal and coronal for me?”

Answer 49

A. Explanation Recall the definition of Hounsfield unit: HU = 1000 x (μvoxel - μwater) / μwater Since μvoxel = 2μwater , we have: HU = 1000 x (2μwater - μwater) / μwater = 1000 Memorize this HU = 1000 x (μvoxel - μwater) / μwater where μ is the linear attenuation coefficient. ex.) HU of pure vacuum is -1000, because μvacuum = 0 Details Materials with larger linear attenuation coefficient are exponentially better at stopping x-rays. That's why the 2x increase in linear attenuation coefficient resulted in massive HU increase in this question. Don't worry about the calculating exponential formulas related to this topic, because they get painful quickly. Materials with high μ also have small HVL distance, because materials that are better at stopping x-rays need less thickness to stop half of incoming x-ray photons.

Answer 50

B. Explanation Axial acquisition is more subject to stair-step artifact, although it provides less z-axis volume averaging than helical acquisition. The other answers are distractors. Memorize This Stair-step artifact is decreased by: overlapping slices using thinner slices Details Helical acquisition is named after the helical shape carved out by the x-ray beam from continuous table movement. Because it doesn't provide a full set of projections for every axial slice, helical acquisition needs to interpolate data between table positions. The strength of helical acquisition is that slice thickness and overlap can be set in post-processing. Both axial and helical acquisition can make thin slices. Helical acquisition makes it easy to have overlapping slices using interpolation after data acquisition, where traditional axial scanning would require setting pitch <1, which incurs extra dose. Isotropic resolution in CT creates voxels that are equal and high resolution in all three dimensions, which allows for high quality reformats in arbitrary slice planes.

Answer 51

The correct answer is 'A' Explanation When people talk about pitch (with no other modifier), they mean collimator (beam) pitch. Remember the definitions of DLP and CTDIvol: DLP = CTDIvol x scan length CTDIvol = CTDIw / pitch Going from pitch of 1.5 to pitch of 0.5, the pitch is multiplied by 1/3, so CTDIvol will get multiplied by 1/(1/3) = 3. Details Beam (Collimator) Pitch is the distance traveled by the table (gantry) during one x-ray tube rotation divided by the width of the x-ray beam. Lower pitch causes more energy to be deposited per kilogram of tissue (increased dose).

Answer 52

A. Explanation More than one of these factors is adversely affected. Specifically, both contrast-to-noise ratio and resolution (both longitudinal and in-plane) are decreased when pitch is increased. Because the table is moving faster, there are less photons released by the x-ray tube per slice of tissue, increasing shot noise and decreasing contrast-to-noise ratio. Both longitudinal and in-plane resolution are decreased because there are fewer projections to calculate each slice of tissue. The good news is that increasing pitch also decreases scan time and decreases dose. Memorize This CTDIvol relates only to beam pitch. Details When questioners ask generically about pitch without other specification, they are referring to beam pitch (also called collimator pitch). Beam pitch (unitless) is table movement (cm) per rotation, normalized to beam width (cm).

Answer 53

E. ``` Explanation Beam pitch (unitless) is table movement (cm) per rotation of the x-ray tube, divided by beam width (cm). Since the question stem tells us that the x-ray tube takes the same amount of time to complete a single rotation, the larger beam pitch also means that the table travels farther per second. Since the beam pitch is 3x what it used to be (going from 0.5 to 1.5), covering the same scan length will take one-third less time. ``` Detail Intuitively, beam (collimator) pitch is the distance the table travels during a single rotation of the x-ray tube, divided by beam width. This question could be solved by writing down equations, but an intuitive understanding of beam pitch should help you answer this question without doing any serious math. Bonus One benefit of increased pitch is decreased motion artifact, resulting from shorter scan time for the same scan length.

Answer 54

A. This answer is correct. Explanation This is a recall question. Memorize This Exam Type Optimal kVp CT angiography 100 DSA 70 Barium Enema 90-110 Details The general rule is that you want the average energy to be slightly larger than the k-edge of your contrast agent (Iodine - 33 keV, Barium - 37 keV) to get the best contrast; CTA uses iodine. This rule works only approximately, and it's best to memorize the values above.

Answer 55

The correct answer is 'B' Explanation This is a recall question related to CT technology. 3rd generation is the most common type of CT scanner in radiology departments. Don't Memorize This You could memorize the details of hardware configuration for each generation of CT scanner, but memorizing this information is of limited practical consequence.

Answer 56

The correct answer is 'A' Explanation SDCT is more radiation dose efficient than MDCT. Memorize This Geometric efficiency is better for SDCT than MDCT. Within MDCT types, geometric efficiency goes up with # slices. Details Radiation dose efficiency quantifies the fraction of radiation passing through the patient that actually contributes to the image. Radiation dose efficiency includes absorption efficiency and geometric efficiency: Geometric efficiency is the fraction of x-rays passing through the patient that enter the detectors. The z-axis spillover of x-rays is called the penumbra. Geometric efficiency is determined by geometry - the x-ray beam shape relative to the detector array shape. Absorption efficiency is the fraction of x-rays entering the detector that are absorbed by the detector, approximately the same between SDCT and the various MDCT answer choices. Absorption efficiency is a property of the detector and how well the detector stops/absorbs x-rays. SDCT has better geometric efficiency than MDCT for two reasons: MDCT uses septa between the rows to reduce scatter, and the septa create dead space on the detector panel where x-rays aren't registered. MDCT needs radiation spill-over along the z-axis (penumbra) to ensure uniform x-ray exposure across detector rows. SDCT is able to ensure no spill-over, because with only one row, there is no concern for uneven exposure across rows. MDCT scanners with larger # slices have better geometric efficiency. This is because more rows requires a wider x-ray beam width (collimation), which means that the z-axis penumbra of wasted radiation will be a smaller percentage of the total radiation that passes through the patient to illuminate the detectors.

Answer 57

The correct answer is 'A' Explanation In-plane resolution relates to several factors, including: ``` # of measurements per rotation reconstruction kernel (eg. bone vs. tissue) geometry (tube-detector distance, detector density within axial slice, effective focal spot) ``` In-plane resolution is not a specific advantage of MDCT over SDCT. You might reasonably argue that motion artifact is a contributor to in-plane resolution, and thus in plane resolution should be better in MDCT, but this is the weakest of the available answers. Details Here are the major advantages of MDCT: MDCT reduces scan time, because the multiple rows allow data to be gathered from longer z-axis segments of tissue with each rotation. Roughly speaking, scan time is proportional to 1/# detector rows. In other words, multiplying # detector rows by M cuts scan time by 1/M. Shorter scan time allows you to complete acquisition within one breath hold (reduced motion artifact), and lets you reduce the contrast bolus duration (more efficient contrast media usage). MDCT allows for thinner slices (increased z-axis resolution), partly because the z-axis thickness of each MDCT row is smaller than the older single detector-row CT (SDCT) technology. Additionally, MDCT acquires multiple rows of data per tube rotation. This makes it practical to acquire lots of thin slices on MDCT, which would have otherwise caused x-ray tube overheating and massively long scan time with SDCT. The better z-axis resolution (thinner slices) also allows for isotropic voxels (equal x, y, and z dimensions) which helps reduce stair-step artifact and supports excellent reformats in arbitrary planes.

Answer 58

The correct answer is 'C' Explanation The bowtie filter shapes the x-ray beam intensity and removes low energy photons. Details The bowtie shape lowers x-ray intensity more at the periphery (where the patient is thinnest) than at the center (where the patient is thickest). This evens out the noise level throughout the image because of a more uniform photon count over the detector, and overall decreases patient dose. The bowtie filter also serves as the x-ray tube filter that removes low energy photons that would otherwise deposit in the patient without contributing to image contrast, also decreasing patient dose. This is the standard role of a filter that you learned for projection radiography.

Answer 59

The correct answer is 'A' Explanation Angular tube current modulation causes sinusoidal ripples corresponding to differences in attenuation based on tube angle. For example, the lateral projection will require more mA than the frontal projection in patients that are wider in the transverse dimension than the anteroposterior direction. The tube voltage (kVp) is constant (not modulated) during tube current modulation. Detail Tube current modulation (Siemens CareDose, GE Auto-mA) guarantees constant image quality by decreasing tube current (mA) for beam paths with less attenuation. Modulation occurs with rotation (angular modulation) and along the z-axis (longitudinal modulation). Modulation is planned using the scout image, and/or using data gathered during the scan. Bonus Care should be taken with very obese patients where tube current modulation can increase dose versus standard protocols, if mA is allowed to escalate unreasonably in response to photon starvation.

Answer 60

D. Explanation With obese patients, x-rays travel a longer path through tissue on their way to the CT detector array. As a result, a larger fraction of photons get absorbed, and fewer photons are left to form the image, a process called photon starvation. The end result is grainer images, representing increased shot noise, also called quantum mottle.

Answer 61

A. Explanation Beam width is 100mm = 10 cm. Beam pitch = 2 with beam width 10 cm implies the table moves 20 cm per rotation: (2 beam pitch) x (10 cm beam width) = 20 cm table movement per rotation. We need the table to travel 60 cm, which requires 3 tube rotations at this pitch and beam width: rotations = (total z-axis distance to travel) / (z-axis distance traveled per rotation) = 60 cm / 20 cm per rotation = 3 rotations. Memorize This ``` Beam pitch (unitless) is table movement (cm) per rotation, divided by beam width (cm). When questions ask about pitch, they're talking about beam (collimator) pitch. ``` Detail This question tests your understanding of beam (collimator) pitch. Other than the formula for beam pitch, you shouldn't try to memorize any other formula that's included in the above answer to this question. Instead, make sure you're comfortable changing between the various metric units (mm to cm) and handling units in general, eg. cm / (cm/rotation) = rotation.

Answer 62

B. Explanation This is a recall question. Memorize This Sound travels at 1540 meters/s in soft tissue Speed of sound is faster in denser material: bone > muscle > soft tissue > fat Bonus Pay attention to the units here, 1540 meters per second.

Answer 63

The correct answer is 'A' Explanation Decreased PRF causes increased axial FOV and lower frame rate. Increased TF increases axial and lateral resolution, but it also increases signal attenuation in deeper structures. The FOV is unrelated to TF. Memorize This Speed of sound in soft tissue is 1540 meters/s Sound attenuation in soft tissue (dB) = 0.5 dB / cm of distance traveled / MHz of transducer frequency. Axial resolution is SPL / 2 PRF = 1/PRP Details Decreased pulse repetition frequency (PRF) is equivalent to longer time between two pulses (larger pulse repetition period, abbreviated PRP). This provides more time to gather echoes returning from deeper tissue, which translates into greater axial FOV. Increased TF increases axial resolution because spatial pulse length (SPL) decreases with decreasing wavelength. The improvement in lateral resolution with increased TF happens in the far zone. Since attenuation in soft tissue is increased with increasing TF, the deeper tissues also return less echo, appearing darker.

Answer 64

The correct answer is 'A' Explanation Axial field of view (maximum depth that is included in the image) is determined by pulse repetition period (PRP) and by pulse repetition frequency (PRF). Memorize This PRP = 1 / PRF PRP is measured in seconds PRF is measured in hertz (pulses per second) Note that PRP = 1 / PRF is mathematically the same formula as PRF = 1 / PRP. Details PRP is the amount of time between two ultrasound pulses. This is the time that the machine has available to listen for echoes before the next pulse begins. As a result, PRP determines the maximum depth a pulse can dive before it has to return to the transducer. This maximum depth is the axial field of view (FOV). Caution Don't confuse transducer frequency (TF) and pulse repetition frequency (PRF) because the word "frequency” is in both terms. Think of a telephone ringing. The pitch of the ring is the transducer frequency. The number of rings per second is the pulse repetition frequency. Don't confuse the axial FOV with the axial resolution.

Answer 65

The correct answer is 'A' 'B' Bottom Line Be sure you understand how key parameters affect image quality for every modality. Memorize This Minimum resolvable distance between two points = SPL/2 You may see this written as "Axial resolution = SPL/2". This is imprecise. Better the resolution corresponds to a smaller minimum resolvable distance between two points, i.e. smaller SPL/2. TF determines the SPL, because a single pulse is composed of ~2-3 pulses. The spatial width of a single pulse is the wavelength λ = c/TF. So the higher the TF, the smaller the SPL. Details Recall that PRP = 1/PRF. The PRP determines the axial field of view.

Answer 66

The correct answer is 'A' 'C' Bottom line The wider the thickness of the crystal in the Z-axis, the more spread out the ultrasound beam, and worse the elevational resolution. It's also the case that the ultrasound beam is able to stay more narrow with higher TF. So crystal thickness and TF are both correct answers. Details A convenient way to conceptualize resolution in ultrasound is to ask how narrowly spaced two point phantoms could be while still being able to resolve that those point phantoms are separate on imaging. Note If you answered either crystal thickness or TF, but not both, your answer may be graded as wrong. The correct answer is both.

Answer 67

D. Explanation Smaller angles of insonation are better, preferably below 60 degrees. Memorize This Traditional exam teaching recommends 30-60 degrees as the optimal angle* (for practical purposes, with the more sharp angle better for technical accuracy purposes). Doppler shift at 90 degrees is zero, so you can't measure blood flow when the transducer is at perfect right angles to the blood flow. Power Doppler is relatively insensitive to angle of insonation, unlike spectral and color Doppler. *Data actually shows that smaller insonation angles are more accurate (0 angle would be best). Anatomy can sometimes make small angles hard to achieve. Details The angle of insonation is the acute angle between the ultrasound beam and the vessel: At insonation angles larger than 60 degrees, small changes in the angle lead to large errors in the predicted velocity. This comes from the cos(theta) in the Doppler shift equation, which changes rapidly when theta > 60. Bonus The term "spectral” in Spectral Doppler refers to the multiple speeds of blood particles within a small region of interest. Spectral broadening, where the spread of velocities in a small area grows wider, occurs both in pathology (like post-stenotic turbulent flow), and normal anatomy (turbulent flow at bifurcations, small vessels like pediatric vertebral arteries, etc).

Answer 68

The correct answer is 'A' Explanation Interfaces 1 and 2 create stronger reflections than Interface 3, because the difference in acoustic impedance (Z) is larger for materials at Interfaces 1 and 2 than 3. The ordering of the materials in interfaces 1 and 2 doesn't matter, so the resulting brightness is equal between the two options. Memorize This Materials with higher density also have larger acoustic impedance. Details An interface is just the place where two different materials meet, like bone and muscle. Two things happen to the incident ultrasound beam at the interface: Transmission Reflection The fraction of the incident ultrasound energy that is reflected depends only on the difference between acoustic impedances of the two materials that form the interface, not their specific order, because of this formula that you do not need to memorize: Reflected Fraction of Incident Energy = (Z1 - Z2)2/(Z1+Z2)2 where Z1 and Z2 are the impedances of the material forming the interface. The formula is ugly enough that expecting you to calculate stuff with the formula is fully unreasonable, so don't waste extra space memorizing it.

Answer 69

A. Explanation Power Doppler displays signal related to speed rather than direction of blood flow. Increased sensitivity to flash artifact is a disadvantage of power Doppler. Flash artifact is Doppler signal in non-flowing tissue resulting from transducer or soft tissue motion. Details Power Doppler has three major advantages over color Doppler: Increased sensitivity to blood flow Independent of Doppler angle Not susceptible to aliasing Power Doppler has 2 major disadvantages relative to color Doppler: Increased sensitivity to flash artifact Slower frame rate

Answer 70

Details The sector transducer has a small, flat footprint - that's why it's good for jamming in between ribs to image the liver or sub-xiphoid to look at the heart. There are also microconvex transducers that are simply small convex transducers (no phased array capability), but you'll be able to guess they're microconvex because they will have the same convex curve that the regular-sized convex transducer has. Bonus The term transducer just means that the ultrasound probe converts from one type of energy to another. To produce the ultrasound pulse, the probe goes from electrical energy to mechanical energy. To record the echo, the probe reverses this process.

Answer 71

A. Explanation Pulse repetition frequency is the best answer here. Although PRF does affect aliasing and accuracy of the velocity estimate, it doesn't directly enter into the Doppler equation. Memorize This Velocity is estimated by combining the measured Doppler frequency shift with these parameters: Transducer frequency (TF) Speed of sound (c) Angle of insonation (cos θ) Velocity = c x (Doppler Shift / TF) x 0.5 cos(θ) Details Doppler frequency shift = TF - frequency of the echo By convention, velocity that moves towards the probe is positive. Bonus If the Doppler formula keeps leaking out of your memory, don't stress out about it. You'll still do better than if you hadn't seen this material at all.

Answer 72

A. Explanation Frame rate depends on these factors: ``` Pulse repetition frequency (PRF) # scan lines ``` Memorize This Increasing PRF also increases framerate, but axial FOV decreases. PRP = 1 / PRF, where PRP is the pulse repetition period Detail Don't worry about memorizing the following formula, but in case you were wondering: time to gather one image (frame) in B-mode ultrasound = pulse repetition period (PRP) x # scan lines

Answer 73

The correct answer is 'A' 'B' 'D' Explanation Lateral resolution depends on these factors: - Beam width - Transducer frequency (TF). Higher TF (lower wavelength) => narrower beam width in far field. The TF is also tied to SPL. - Density of scan lines Side Note If you answered SPL for this question, reserve SPL for when you're asked about axial resolution. Although TF improves all three types of resolution and decreases SPL, the traditionally-asked relationship between SPL and axial resolution is more likely to be emphasized. Memorize This Focal Zone is the axial (depth) position with the narrowest ultrasound beam width, and best lateral resolution. Bonus Near field is just the region we use for imaging, close to the transducer where the beam width is reasonable. Far field, the region deep to the near field, is not used for imaging because the beam width rapidly diverges. There are deeper technical details including equations related to near and far field that you can disregard.

Answer 74

Explanation M-mode is used to demonstrate the fetal heart rate. B-mode is the standard grayscale ultrasound image. A-mode is a graph of echo amplitude versus time (or depth), used in ophthalmology. Spectral Doppler plots the range of velocities measured at a single point in space versus time. Color Doppler plots velocity as a colorized image. When this color map is superimposed on B-mode, the composite image is called Duplex. Power Doppler plots speed-related signal as a colorized image. Details M-mode, most commonly used in cardiac imaging including first trimester fetal heart rate visualization, displays time along the x-axis and position along the y-axis. Bonus Spectral Doppler should not be used in first trimester, because heating (thermal index) could damage the fetus. A stands for amplitude, B stands for Brightness, and M stands for Motion. Irrelevant for the exam, but useful to impress referring clinicians.

Answer 75

B. Explanation The sound wave attenuates over the full 4 cm round trip between the probe and the lesion which is 2 cm deep to the skin. The attenuation of sound formula depends on the full travel length and the transducer frequency (TF), but not the PRF. Memorize This Attenuation of sound (dB) = (0.5 dB/cm/MHz) x (full travel path in cm) x (TF in MHz) Details This type of question is so easy to write and grade that understanding how to solve it is essential, although you aren't likely to break out this formula in the reading room. Plugging into the formula, we have: Attenuation of sound (dB) = (0.5 dB/cm/MHz) x (4 cm) x (10 MHz) = 20 dB Caution Forgetting to use the full length of travel when appropriate is a common mistake with residents. That's why I've purposefully included 10 dB as a distractor.

Answer 76

The correct answer is 'A' Explanation Aliasing is signal distortion resulting from low sampling rate. In Doppler ultrasound, pulse repetition frequency (PRF) needs to be set high enough relative to the Doppler shift in order to avoid aliasing. Nyquist Rate is the minimum sampling rate (PRF) needed to avoid aliasing. There are two major ways to decrease aliasing: Increase PRF. You may need to switch to a more superficial region of interest, since PRF also determines axial field of view. Decrease TF. This decreases the Doppler shift, which eases the Nyquist requirement for high PRF. Memorize This Doppler Shift is proportional to TF and cos(theta) Spectral and Color Doppler suffer aliasing. Power Doppler does not. Details The sampling rate is the number of times per second (Hz) that we record the value of a signal. For example, sampling rates for audio in MP3 players is 44.1 kHz. Aliasing is signal distortion caused by insufficient sampling rate. You'll also see aliasing in MRI when you learn about wraparound artifact. Pulsed-wave versus continuous wave Doppler. Radiologists use pulsed-wave Doppler to simultaneously measure velocity within a vessel and the precise location where this flow is occurring. Instead of keeping the ultrasound continuously on at a fixed frequency (continuous wave Doppler), the ultrasound machine releases short pulses at the PRF. Spectral Doppler, color Doppler, and power Doppler are types of pulsed-wave Doppler. The hand-held device you use in IR to listen to a non-palpable dorsalis pedis arterial pulse uses continuous wave Doppler. In pulsed-wave Doppler, PRF controls both axial field of view (FOV) and aliasing. Deeper vascular structures require lower PRF to include them in the FOV. This makes it particularly hard to image fast flowing arteries in deep structures without incurring aliasing. Note that pulse repetition period (PRP) is just 1 / PRF, so increasing PRF is equivalent to decreasing PRP. Bonus If you're looking for a way to keep PRF and TF organized in your mind, think of the ultrasound probe as an old-fashioned telephone ringing. The transducer frequency (TF) is the pitch of each ring. The pulse repetition frequency (PRF) is the number of telephone rings (pulses) per second. Using special techniques (compressed sensing, finite rate of innovation theory), it's sometimes possible to sample far below the Nyquist rate and avoid aliasing entirely. Don't worry about this for the exam.

Answer 77

The Nyquist limit represents the maximum Doppler shift frequency that can be correctly measured without resulting in aliasing in color or pulsed wave ultrasound. Physics: The Nyquist limit always equals Pulse Repetition Frequency (PRF)/2. The US machine can display the Nyquist limit either as the maximum measurable blood flow velocity, or in kHz, the latter representing the maximum measurable Doppler shift. If the blood flow velocity exceeds this limit the device will incorrectly register the direction and velocity of the flow, resulting in color or spectral Doppler aliasing artifact. Practical points Choosing the appropriate PRF for the interrogated vessel is key to avoiding errors caused by exceeding the sampling limit. For high-velocity flow the PRF should be increased accordingly. As a rule of thumb the lowest PRF without aliasing should be chosen. In spectral Doppler this can be ensured not only by changing the PRF, but also by correctly setting the baseline of the Doppler curve.

Answer 78

C. Explanation Depth to lesion is calculated by the range equation: Lesion depth = (c) x (t/2) where c is the speed of sound in the imaged tissue and t is the roundtrip time for the echo from the lesion. Since the sound wave takes half the roundtrip time (t/2) to get to the lesion, we use (t/2) to calculate distance to the lesion. With a fatty liver, cassumed > ctrue because sound is slower in fat than soft tissue; sound travels faster in denser tissue. As a result, depthassumed > depthactual. Memorize This Differences between assumed and true speed is basis for speed displacement artifact. Details The classic example of speed displacement artifact is focal fat in the liver which causes the liver-diaphragm border to appear farther from the transducer, because the machine assumes a faster speed than the focal fat accommodates.

Answer 79

D. Explanation The sector probe has the specific advantage of a small footprint and large field of view (FOV). Details The sector probe is also called the phased array probe, because it coordinates timing of the individual transducers, using constructive and destructive interference to sweep the ultrasound beam across a wide field of view despite its small footprint.

Answer 80

A. This is a reasoning question. You'll need to memorize and manipulate the half-life formula: 1/effective-half-life = 1/physical-half-life + 1/bio-half-life For these numbers, effective half life = 2. Be sure you understand each step of this calculation, including how to add fractions, and how we solved for effective half life in the last step. Details In nuclear medicine, half life is the amount of time (minutes or days) required to see the activity of a radiopharmaceutical drop by half. The total rate of emissions coming from your patient decays (effective half life) because the radiopharmaceutical decays (physical half life) and because the drug is excreted (biological half life). Bonus This type of question is so easy to write and grade that you should absolutely know how to solve it.

Answer 81

A. Explanation This reasoning question checks whether you know the difference between decay constant (λ, lambda) and half life (t1/2). Recall that Tc-99m has a half life (t1/2) of 6 hrs. Using the formula: lambda = 0.693/half-life Memorize This" Half life of Tc-99m is 6 hrs. lambda = 0.693/half-life (approx: lambda =0.7/half-life) (equivalent: half-life = 0.693/lambda) Bonus: Can you appreciate the mathematical similarity between (λ) and HVL? What is the HVL-related quantity that is analogous to (λ)? ---- Answer: linear attenuation coffecient! (notated by mu, rather than lambda!), formula is same!: HVL = 0.693/mu)

Answer 82

D. Explanation The decay from Tc-99m to Tc-99 is called isomeric transition because mAss number (A) and atomic number (Z) haven't changed - the parent and daughter radioisotopes are isomers. Details Both Tc-99m and Tc-99 have the same number of protons and neutrons: they are nuclear isomers. The difference between two isomers relates to their energy states, not their actual ingredients. The "m” next to Tc-99m stands for metastable, which just means that the Tc-99m hangs around for a while - it doesn't instantly decay into Tc-99. Bonus Memorizing all of the decay schemes is excessively painful. Save yourself time and focus on isomeric transition because it describes Tc99m → Tc99, the workhorse of nuclear imaging departments. What is the half life of Tc99m? Is this a physical, biological, or equivalent half life? ---A: Both physical and bio half-live of tc-99m are 6h! (=> effective half-life of Tc99m alone are = 1/6 + 1/6 = 1/effective-half-lfe => eff-half-life = 3 hrs.)

Answer 83

Memorize This A) Keep the gamma camera close to the patient to maximize resolution B) Use thicker and longer septa to improve resolution. Downside to these approaches is lower sensitivity since you're rejecting more photons Details There are different collimator types (see Nuclear Imaging chapter of Radiology Simplified): parallel, diverging, converging, pinhole. Parallel hole collimators come in four flavors: ``` Low Energy All-Purpose (LEAP) - wider septa spacing, short septa Low Energy High-Resolution (LEHR) - narrow septa spacing, taller septa Medium Energy (Ga-67, In-111) - thicker septa High Energy (I-131, F-18 FDG) - thickest septa ``` Bonus Don't get too frustrated memorizing the difference between converging and diverging collimators. They're poorly named and difficult to keep straight.

Answer 84

Parallel hole collimators come in four flavors: ``` Low Energy All-Purpose (LEAP) - wider septa spacing, short septa Low Energy High-Resolution (LEHR) - narrow septa spacing, taller septa Medium Energy (Ga-67, In-111) - thicker septa High Energy (I-131, F-18 FDG) - thickest septa ```

Answer 85

The correct answer is 'D' and 'E' Explanation The well counter uses NaI crystal as a scintillator. Remember that a scintillator converts x-ray or gamma photons to visible light photons. Memorize This Photon Detector Instruments Method NaI Well Counter, Gamma Camera (Planar Imaging + SPECT) CsI Indirect Flat Panel Detectors (FPD) for X-Rays (Fluoroscopy) BGO, LSO, GSO PET Gas-filled detector G-M Pancake Probe, Dose Calibrator, Cutie Pie Survey Meter Bonus NaI is used in Nuclear Imaging. ("N" for Nucs; vs CSI used in projection radiography indirect flat panel detectors).

Answer 86

A. This answer is correct. Explanation Septal penetration occurs when a photon from the radiotracer penetrates the lead septa of the collimator on the Gamma camera: Higher energy photons more easily penetrate these septa. Details When the Gamma camera registers a photon that has penetrated collimator septa, the photon gets registered at the wrong position, degrading spatial resolution. Since higher energy photons can penetrate collimator septa more easily, this problem is worse for higher energy photons (I-131 with 365 keV photons) or concentrated spatial distribution of radiotracer (Tc-99m sulfur colloid at the injection site for lymphoscintigraphy)

Answer 87

B. Explanation This is a recall question related to quality control (QC) for nuclear imaging instrumentation. Memorize This ``` The dose calibrator QC schedule follows the CLAG acronym: Constancy - Daily Linearity - Quarterly Accuracy - Annually Geometry - On repair ``` To memorize the maintenance times, try "Dairy Queen Always Open" (Daily, Quarterly, Annually, On-repair) Read details on what these QC tests actually involve in the Nuclear Imaging chapter of Radiology Simplified. Bonus Pay attention to QC for dose calibrators and Gamma cameras. Save time and bypass details of Nuclear Imaging QC for other instruments.

Answer 88

A. Explanation SPECT uses radiotracers which emit a single photon per decay. On the other hand, sodium fluoride (F18-NaF) bone scans use F-18, which is a positron emitter (also used in F18-FDG), which means that it emits a positron when it decays; the positron undergoes annihilation which results in two 511 keV gamma photons being released simultaneously in opposite directions. This requires a PET scanner to interpret. Details Planar Imaging refers to a single projection taken with the Gamma camera, frequently on the same equipment used for SPECT. The whole-body bone scan is commonly performed as planar imaging. SPECT stands for Single Photon Emission Computed Tomography. SPECT takes images at multiple projections using the Gamma camera, and performs iterative reconstruction (similar to CT) to get crossectional and 3D images. SPECT-CT combines SPECT with a CT scanner for better attenuation correction and anatomic localization. Tc-99m sestamibi parathyroid scan is performed with SPECT-CT for anatomic localization. PET stands for positron emission tomography. The PET machine performs coincidence detection to form images and uses radioisotopes like F-18 that are positron emitters. PET is optimized to detect and localize the 511 keV gamma photons that are released simultaneously in opposite directions when the positron meets an electron and undergoes annihilation. PET is also combined with CT for attenuation correction and anatomic localization (PET-CT). FDG-PET is the workhorse of cancer followup imaging. Bonus Quality control (QC) for SPECT is similar to what you've memorized for Gamma cameras in the Nuclear Imaging chapter of Radiology Simplified. Don't worry about additional details.

Answer 89

C. Explanation This is a recall question. Memorize This The term isotope describes two elements that have the same atomic number: atomic number = # of protons Detail Let's take a closer look at the shorthand notation we use when we say I-131, I-123, or Tc-99. The letter represents the element from the periodic table: I stands for iodine, and Tc stands for technetium. The number is the mass number. I-123 and I-131 have different mass numbers: mass number = # protons + # neutrons Every element has a unique atomic number. Since I-131 and I-123 represent different forms of the same element (iodine), so we know that they have the same atomic number. They are isotopes because they have the same number of protons. (Isobar = atoms of different chemical elements that have the same number of nucleons (total protons+neutrons), same mass number, A; i.e. Z1+N1 = Z2 + N2, A1=A2.)

Answer 90

A. Explanation This is a recall question. The "99” in this shorthand notation refers to the mass number. Since the mass number is equal between Mo-99 and Tc-99m, they are called isobars. Memorize This The term isobar describes two elements that have the same mass number: mass number = # protons + # neutrons

Answer 91

The correct answer is 'A' Explanation Tc-99m < 100 mCi qualifies as minor spill. This does not require leaving the area immediately. You should be able to clean up the spill yourself, using protective gear that will also subsequently be discarded as radioactive material. You'll need to tell the RSO, but the FDA is not involved. Nuclear Regulatory Commission (NRC) reporting may be required. The FDA is not involved. Memorize This Know the major and minor spill criteria for the essential radioisotopes. Longer half-life radioisotopes tend to have lower thresholds for major spill because they stay radioactive for longer. See the Nuclear Safety chapter in Radiology Simplified. Details Minor and major spills are handled differently, as described in Radiology Simplified. It can be hard to memorize the details. You should know that cleaning the site yourself is not part of the major spill protocol. Bonus The Nuclear Safety chapter of Radiology Simplified summarizes the essential points that you should know going into the Core Exam.

Answer 92

The correct answer is 'A' Explanation The depicted scenario is called transient equilibrium, because parent and daughter activity are approximately equal (equilibrium), but continually decreasing (transient). Memorize This Transient equilibrium: t1/2 parent > t1/2 daughter (i.e., daughter half life is slightly less than parent half life) Secular equilibrium: t1/2 parent >> t1/2 daughter (i.e., daughter half life is much less than parent half life) Details In both transient and secular equilibrium scenarios, both the parent and daughter are continually undergoing decay. The parent decays into the daughter. The daughter decays into something else. When t1/2 parent >> t1/2 daughter, the parent activity essentially remains constant while the daughter accumulates. The daughter activity flattens out when then rate of daughter production equals the rate of daughter decay. Bonus Mo-99/Tc-99m generator is a key example of transient equilibrium.

Answer 93

A. Explanation F18-FDG is the workhorse of PET. This radiopharmaceutical is chemically very similar to glucose, and it's tagged with radioactive fluorine (F18), which is a positron emitter (make sure you know what this term means). Because it acts like glucose, FDG is pushed into muscle cells when the patient receives insulin (also with intense exercise prior to the exam). This means that less radiotracer is available for uptake in tumor cells. Additionally, there's more background uptake from muscle. Overall, this decreases the contrast between tumor and non-tumor. Memorize This Standardized Uptake Value (SUV) = (Mean ROI activity / activity administered) / (body weight) Mean ROI activity (mCi, mL) is the value you get from the ROI tool covering the area of interest. Activity administered (mCi) is the total activity of radiotracer injected in the patient. Body weight is measured in grams. The units in the SUV formula are very hard to memorize, so don't worry about memorizing them. Details SUV is used both in SPECT and PET. Many factors affect SUV, making it difficult to reliably compare SUV across patients, or even within the same patient across time. For example, high serum glucose levels will cause the FDG face more competition in getting absorbed by tumor cells.The net effect is that high serum glucose decreases SUV measured in tumor cells. Elevated SUV uptake alone is nonspecific for inflammation versus tumor. The brain shows avid FDG uptake in normal scans. This is your clue that you're looking at an FDG-PET study.

Answer 94

SUV relates to ROI activity, Activitity admnistered, and body weight, as follows Memorize This: Standardized Uptake Value (SUV) = (Mean ROI activity / activity administered) / (body weight) Mean ROI activity (mCi, mL) is the value you get from the ROI tool covering the area of interest. Activity administered (mCi) is the total activity of radiotracer injected in the patient. Body weight is measured in grams. The units in the SUV formula are very hard to memorize, so don't worry about memorizing them. Details SUV is used both in SPECT and PET. Many factors affect SUV, making it difficult to reliably compare SUV across patients, or even within the same patient across time. For example, high serum glucose levels will cause the FDG face more competition in getting absorbed by tumor cells.The net effect is that high serum glucose decreases SUV measured in tumor cells. Elevated SUV uptake alone is nonspecific for inflammation versus tumor. The brain shows avid FDG uptake in normal scans. This is your clue that you're looking at an FDG-PET study.

Answer 95

A. Explanation Xe-133 is a gamma emitter with photon energy 81 keV. Tc-99m is a gamma emitter with photon energy 140 keV. Downscatter from Tc-99m MAA will contaminate the Xe-133 energy window, unless Xe-133 is administered first. Memorize This Xe-133 gamma photon energy is 81 keV Tc-99m gamma photon energy is 140 keV Details For any radiopharmaceutical, the primary photon energies that we memorize are called photopeaks because we see them as peaks in the energy spectrum. Compton scatter from inside the patient causes many gamma photons to lose energy and change their trajectory. This is called downscatter, because the energy spectrum gets smeared "downward” to include lower energy ranges. Including downscattered photons in the image decreases contrast-to-noise ratio (CNR). Gamma cameras detect the energy of incoming photons, so they use energy windows to select the photopeaks and reject downscattered photons and preserve CNR. Since the energy window for Xe-133 is centered on its 81 keV photopeak and Tc-99m photopeak is 140 keV, many downscattered photons from Tc-99m would fall into the Xe-133 window, unless Xe-133 ventilation was administered and measured before the patient was ever exposed to Tc-99m MAA. This is why we administer Xe-133 ventilation before Tc-99m MAA. Bonus How is it possible to perform a V/Q scan with Tc-99m DTPA ventilation and Tc-99m MAA perfusion? (A: is this bc the bio-half-life is much shorter with DTPA (1hr) presumably bc it is exhaled? => when it decreases enough can image ok with MAA, which is also given in higher dose?)

Answer 96

Memorize This Xe-133 gamma photon energy is 81 keV Tc-99m gamma photon energy is 140 keV

Answer 97

The correct answer is 'B' Explanation This is a reasoning question. You're asked to recall and apply NRC regulation. Memorize This By default, the fetus is permitted 5 mSv over the duration of the entire pregnancy, including the time before the woman declared the pregnancy. If the fetus has already received more than 5 mSv by the time the pregnancy is declared, the fetus is allowed another 0.5 mSv. If the pregnancy is not declared by the woman, there are no dose limits on the fetus. The woman can later undeclare her pregnancy to continue working. Bonus Nuclear safety is a critical part of the Core Exam, because it incorporates the RISE Exam which was used historically by the ABR to grant Authorized User eligibility. Make sure that you know everything in the Nuclear Safety chapter of Radiology Simplified before walking into exam day. Most states are "agreement states” which means they manage nuclear safety instead of the NRC. If you're in an agreement state, you'll need to file separate paperwork after passing the Core Exam to become an Authorized User (AU) in that state.

Answer 98

The correct answer is 'A' Explanation This is a recall question. Memorize This If you're asked about how long nuclear-related records should be kept, select three years. I call this the three-year rule. Details Disposing radioactive material in the way described in this question is only allowed for radioisotopes with: physical half life less than 120 days indistinguishable from background radiation before disposal all labels removed Bonus The three-year rule isn't 100% correct, but it's not worth memorizing the exceptions. Physical half-life of I-131 is 8 days. See the Nuclear Imaging chapter of Radiology Simplified for the chart of radioisotope half-lives that you should memorize.

Answer 99

3-years ("the three-year rule" for nucs) The three-year rule isn't 100% correct, but it's not worth memorizing the exceptions.

Answer 100

- physical half life less than 120 days - residual activity indistinguishable from background radiation before disposal - all labels removed

Answer 101

The correct answer is 'A' Explanation This is an important recall-style question related to Nuclear Safety. Memorize This Maximum NRC-allowed contamination: 0.15 μCi Mo-99 / mCi Tc-99m eluate 10 μg Aluminum / mL of Tc-99m eluate Caution: Pay close attention to units here, especially micro (μ) vs. milli (m), and Curie (Ci) vs. grams (g) vs. liters (L). Bonus Aluminum contamination is detected using a paper strip test. Tc-99m is the workhorse of the Nuclear Medicine department, so expect to be tested on it.

Answer 102

Maximum NRC-allowed contamination: 0.15 μCi Mo-99 / mCi Tc-99m eluate 10 μg Aluminum / mL of Tc-99m eluate

Answer 103

Explanation This is an important recall question related to Nuclear Safety. Memorize This Maximum NRC-allowed contamination: 0.15 μCi Mo-99 / mCi Tc-99m eluate 10 μg Aluminum / mL of Tc-99m eluate Caution: Pay close attention to units here, especially micro (μ) vs. milli (m), and Curie (Ci) vs. grams (g) vs. liters (L). Bonus NRC stands for Nuclear Regulatory Commission. Most of the nuclear safety questions relate to standards enforced by the NRC. Tc-99m is the workhorse of the Nuclear Medicine department, so expect to be tested on it.

Answer 104

A. Explanation MRCP is heavily T2 weighted. Details Magnetic resonance cholangiopancreatography (MRCP) is a generic term for high resolution imaging of the biliary ducts. Various pulse sequence techniques have been used over the years, and all MRCP sequences share these two qualities: heavily T2 weighted high resolution Bonus Don't worry about digging into the detailed history of MRCP techniques. Do be prepared to identify basic diseases of the bile ducts on MRCP.

Answer 105

A. Explanation This is a recall-type question. Memorize This Gadolinium causes T1 and T2 shortening 0.1 mmol/kg of Gadolinium is the patient-weight-dependent dose for breast MRI Bonus Walk into exam day having memorized the numbers in the MRI chapter of Radiology Simplified. Use the first couple minutes of the exam to "core dump” many of your memorized facts onto the sheet that is provided by the test center.

Answer 106

C. Explanation The precession frequency of the magnetic moment vector in a static magnetic field of strength B is given by the Larmor Equation: Precession frequency (MHz) = γ x B where γ is a constant called the gyromagnetic ratio, and B is the strength of the local magnetic field. Memorize This ``` The gyromagnetic ratio (γ) is 42 MHz / Tesla Precession frequency (MHz) = (42 MHz / Tesla) x (Magnetic field strength in Tesla) One Megahertz (MHz) = 1,000,000 Hertz (Hz) ``` Details Precession is the wobble of the nuclear magnetic dipole moment vector that results in the presence of an magnetic field. For most MRI applications, we use the magnetic moment vector of the hydrogen (H) nucleus, which is just a single (unpaired) proton, but other elements also have nuclear magnetic moment vectors that can be used for MRI provided they have with unpaired (odd numbers of) protons or unpaired neutrons. Bonus The transmit coil in the MRI emits a burst of radio waves called a radiofrequency (RF) pulse. In order to tip a magnetic moment vector of a hydrogen proton in the patient's tissue, this pulse has to match the precession frequency of that magnetic moment vector. It makes sense to expect basic physics questions like this one related to the building-block concepts of MRI as well as more clinically-relevant concepts like the effect of scan parameters on SNR, tissue contrast, resolution, scan time, and artifacts. The MRI literature can feel overwhelming, but most of it isn't relevant to the intent and scope of the Core Exam. Use the MRI chapter of Radiology Simplified to anchor your study.

Answer 107

The precession frequency of the magnetic moment vector in a static magnetic field of strength B is given by the Larmor Equation: Precession frequency (MHz) = γ x B where γ is a constant called the gyromagnetic ratio (gyromagnetic ratio (γ) is 42 MHz / Tesla), and B is the strength of the local magnetic field. Memorize This ``` The gyromagnetic ratio (γ) is 42 MHz / Tesla Precession frequency (MHz) = (42 MHz / Tesla) x (Magnetic field strength in Tesla) One Megahertz (MHz) = 1,000,000 Hertz (Hz) ```

Answer 108

B. T1 relaxation time is short for fat and long for water. Memorize This Fat has short T1 and short T2 relaxation times Water has long T1 and long T2 relaxation times Gadolinium shortens both T1 and T2 relaxation times Details How do you memorize T1 and T2 relaxation times for fat and water? Remember that fat is bright on T1-weighted images. Memorize by convention* that hyperintense voxels on T1-weighted images have short T1. It follows that fat has short T1. + Try this logic for T2-weighted images, where the convention* is that hyperintense voxels on T2-weighted images have long T2. + Try out this logic for water instead of fat. * This "convention" is essentially to display larger echoes as brighter pixels. For the T1-weighted sequence, shorter T1 relaxation substances have larger echoes. This is because of two contributing factors in the T1-weighted spin-echo sequence: short TR and short TE: Because TR is short, only substances with short T1 will have recovered longitudinal magnetization substantially before the 90 degree pulse that rotates the magnetization vector into the x-y plane for the length of that vector to be measured as an echo. Because TE is short, the rotated vector won't have decayed much as a result of T2 relaxation, so the echo size won't reflect differences in T2 between voxels. For the T2-weighted sequence, longer T2 relaxation substances give larger echoes. This because of two contributing factors in the T2-weighted spin-echo sequence: long TR and long TE: Because TR is long, all voxels will have recovered longitudinal magnetization, regardless of their T1 relaxation times before the 90 degree pulse that rotates the magnetization vector into the x-y plane. The size of this vector (and hence the size of the echo) will have nothing to do with differences in T1 times. Because TE is long, the rotated vector will have decayed differently between voxels with different T2 relaxation times. This time, the echo size will reflect differences in T2 between voxels. Bonus Be careful that you understand the essentials of MRI without getting bogged down in advanced material. The MRI chapter of Radiology Simplified was written specifically to encompass the relevant level of detail.

Answer 109

Cardiac MRI is broadly divided into two families of sequences based on the signal intensity of blood on images. Know the names of some examples from each family: 1) dark-blood sequences - FSE (Fast Spin Echo) - HASTE (half-Fourier single-shot turbo spin-echo) 2) bright-blood sequences - SSFP (Siemens True FISP) 3) You'll ALSO want to know about these bread-and-butter sequences: - Delayed enhancement sequences use gadolinium contrast, imaged 15 minutes after injection, together with an inversion recovery (IR) pulse that selectively nulls (zeroes) signal from normal myocardium so that abnormal myocardium shows brightly. - Phase contrast imaging calculates blood velocity information (speed and direction) using a phase encoding gradient along the long axis of the vessel. Slices are typically oriented to cut through the short axis of the vessel. To avoid aliasing in phase contrast imaging, set the velocity encoding parameter (Venc) slightly larger than the maximum velocity. Cine Imaging with ECG gated acquisition is broadly similar to cardiac CT in that it can be performed prospectively or retrospectively. Bonus For cardiac MRI, be sure to review all of the cases in the cardiac clinical chapter of Radiology Simplified, which includes cine clips. Make sure you can identify the various basic orientations and anatomical components of cardiac MR images from example cases. Don't worry about additional technical details of the various cardiac MRI pulse sequences. For example, did you know the Venc is the blood velocity that produces a phase shift of 180 degrees for a phase contrast imaging protocol? Awesome information, but beyond the scope of initial certification.

Answer 110

Memorize This Scan Time = (NEX * #slices * TR * Ny) / Turbo Factor NEX = number of repetitions TR = time between 90 degree pulses Ny = # phase encoding steps turbo factor = 1 for regular spin-echo, >1 for turbo spin-echo sequences that acquire more than one echo per TR Details The # of phase encoding steps directly relates to scan time, and equals the # of voxels along the phase encoding axis. By choosing the shortest FOV axis to represent the phase encoding axis, we can maintain a reasonable # of phase encoding steps while having enough voxels relative to the length of anatomy being covered along the phase encoding axis.

Answer 111

D. Explanation This is a recall type question. Details T1 relaxation occurs from spin-lattice interaction T2 relaxation occurs from spin-spin interaction* * T2 has the word "spin” twice ("2") in the type of interaction. Bonus The underlying physics of what spin-spin and spin-lattice interaction actually mean are wonderful, but beyond the scope of the Core Exam.

Answer 112

The correct answer is 'A' . Explanation Dixon fat suppression permits post-contrast imaging and is robust to magnetic field inhomogeneity. Details Chemical fat suppression uses a 90 degree RF pulse (presaturation pulse) at the resonant frequency of fat protons to selectively eliminate longitudinal magnetization of fat before imaging begins. This technique can be used with almost any pulse sequence, so T1, T2, and T1-post contrast images are all possible. Since the resonant frequency is sensitive to local magnetic field changes, chemical fat suppression is ruined by magnetic field inhomogeneity, like from metal implants or with large field-of-view (FOV) where magnetic field can vary. STIR uses inversion recovery (see Radiology Simplified for details). It is robust to magnetic field inhomogeneity. However, STIR only produces T2-weighted images, and inversion-recovery-based fat suppression also nulls gadolinium, so STIR cannot be used for contrast-enhanced imaging. Dixon fat suppression combines in-phase and out-of-phase image data to compute images entirely without water or entirely without fat. Dixon fat suppression can be used to visualize contrast enhancement. The three-point Dixon is able to correct for magnetic field inhomogeneity. We use this routinely for fat-suppressed post-contrast imaging of the neck. Bonus You should be comfortable with the technical details of chemical fat suppression, STIR, and Dixon fat suppression. Review the MRI section of Radiology Simplified for details.

Answer 113

B. Explanation Think of this as a recall type question. Memorize This T1 weighting: short TR, short TE T2 weighting: long TR, long TE PD weighting: long TR, short TE This is critical knowledge. Details The basic spin-echo sequence checks how much transverse magnetization (Bxy) is left at the echo time (TE). By choosing TR and TE, we can select whether the transverse magnetization better reflects the rates of T1 relaxation (recovery of longitudinal magnetization Bz) or rates of T2 relaxation (loss of Bxy). It helps to visualize graphs of T1 and T2 relaxation versus time. The best separation between fat and water happens early on the T1 graph and late on the T2 graph. This is why we choose a short TR for T1 weighting and long TE for T2 weighting. By choosing a short TE for T1 weighting, we can ensure the image is insensitive to voxel differences in T2 times, because T2 relaxation hasn't had a chance to happen. By choosing a long TR for T2 weighting, we can ensure the image is insensitive to voxel differences in T1 times, because T1 relaxation has already occured for all voxels regardless of T1 time. Bonus Check out the MRI chapter in Radiology Simplified for the time ranges that qualify as "short” or "long” for TR and TE. Try to memorize this, but don't stress out if the numbers aren't sticking.

Answer 114

The correct answer is 'C' Explanation Doubling the FOV along the frequency encoding axis does not change the number of frequency encoding steps. Memorize This Dimensions of the k-space matrix equal the dimensions of the actual diagnostic image in voxels, controlled separately along frequency and phase encoding axis. Spacing between samples in k-space controls the FOV, also controlled separately along frequency and phase encoding axis. Voxel length along each axis = (FOV along that axis) / (# of voxels along that axis) Details Increasing FOV along either frequency or phase encoding axis in the clinical image causes k-space values to be sampled more narrowly along that axis in k-space. The total number of frequency encoding steps is not changed, so the number of voxels is unchanged. Larger FOV with unchanged # of voxels results in larger voxels, a.k.a. decreased spatial resolution. Bonus Focus your energy on understanding the basic spin-echo sequence and the turbo spin-echo variant.

Answer 115

A. Explanation Parallel imaging does not increase SNR. The remaining statements are true, and you should be familiar with them as essential points about parallel imaging. Details Parallel imaging uses multiple coils to decrease scan time by the acceleration factor. Individual coils are intentionally set to have small FOV by skipping lines in k-space. This allows for faster scans, but result in lower SNR, as well as aliasing that must be corrected during reconstruction by pooling data across coils. Higher acceleration factor results in lower SNR because more k-space data is skipped per coil, and increases parallel imaging artifact because the aliasing problem gets harder to resolve. Bonus The two widely available parallel imaging methods on scanners are called SENSE and GRAPPA. Don't worry about the technical details of how these methods perform calibration to learn individual coil sensitivities and reconstruct images.

Answer 116

C. Explanation It's ok to treat this as a recall-type question. The technical details of gradient-echo sequences are fun, but prioritize your time to master details of the spin-echo family of sequences. Memorize This Increase the flip angle on the SPGR sequence to improve T1 contrast. Details The SPGR pulse sequence is part of the gradient-echo (also called gradient recalled echo) family of pulse sequences. Pulse sequences in the gradient-echo family postpones and recall the free induction decay using a magnetic field gradient, which reflects T2* weighting for long TE. The T2* weighted sequence that we call GRE on neuroradiology service uses this capability to detect micro-hemorrhage. In contrast, pulse sequences in the spin-echo family of sequences use the 180 degree rephasing pulse to generate an echo which reflects T2 weighting for long TE. Gradient-echo pulse sequences were designed for fast scanning. They achieve this using partial flip angles which tip the longitudinal magnetization by less than 90 degrees towards the transverse plane. Because longitudinal magnetization can recover faster with a partial flip angle, the TR can be decreased, reducing scan time. The penalty is decreased T1 weighting. Bonus SPGR sequences are also special because they are acquired as 3d-volumetric acquisitions. Instead of exciting a slice of tissue with each TR, 3d-volumetric acquisition excites a slab (cube) of tissue. The z-axis of the signal is determined using a phase-encoding axis in that direction, which is applied in addition to the in-plane frequency and phase encoding axes. This allows us to reconstruct thin contiguous slices* with small isotropic voxels (equal length in every dimension). Small isotropic voxels provide elegant multi-planar reformatting and surface reconstruction, so SPGR is great for the surgical planning software that neurosurgeons use, despite the moderate sacrifice in T1 contrast. * Trying for contiguous slices with standard 2D acquisition results in cross-talk between slices due to the slice-selection RF pulse.

Answer 117

Doubling bandwidth (BW) changes SNR by a factor of 1/sqrt(2). Memorize This SNR ~ voxel volume * sqrt(Nx*Ny*NEX/BW) -Nx and Ny are the number of frequency and phase encoding steps, and -BW refers to the receiver bandwidth (rBW). If a question stem mentions "bandwidth", assume receiver bandwidth (rBW) unless explicitly told otherwise. Details This detail section is your chance to understand why increasing BW causes decreased type 1 chemical shift artifact. The receiver bandwidth (rBW) is the range of frequencies that we listen to in each echo, determined by the rate at which we digitize the echo. Listening to a wider range of frequencies ends up introducing more noise relative to signal, which decreases SNR. This is why increasing rBW decreases SNR. The frequencies present in the echo relate to position in space along the frequency encoding axis, as determined by the magnetic field gradient (Gx) along the frequency encoding axis. As a result, there is an intimate relationship between bandwidth, Gx, and field of view along the frequency encoding axis (FOVx), that you don't need to memorize: FOVx = rBW /(gamma*Gx) Again, don't memorize this formula, but just as an FYI, this is how your scanner allows you to control rBW without changing FOVx and without changing resolution, by adjusting automatically adjusting Gx. As a result, for the exam, you should think of rBW as # of Hertz per voxel along the frequency encoding axis. By increasing rBW, you're keeping water and fat in the same voxel, even though protons in fat have a slightly shifted Larmor frequency. This is why increasing BW decreases chemical shift. Bonus Receiver bandwidth (rBW) is unrelated to transmit bandwidth (tBW), even though they both use the word bandwidth. Memorize the SNR proportionalities above, because reasoning these values out can get tricky during exam time.

Answer 118

A. Explanation Spatial resolution increases (gets better). Memorize This ``` Spatial resolution (voxel size) along each axis = FOV along that axis / # voxels along that axis Dimensions of the k-space matrix equal the dimensions of the actual diagnostic image in voxels. # of phase encoding steps = # rows in k-space matrix = # voxels along phase encoding axis of image # of frequency encoding steps = # columns in k-space matrix = # voxels along frequency encoding axis of image ``` Details In basic spin-echo MRI, each TR provides one echo centered on the TE time. The shape of this echo fills out a single row of k-space. Although k-space is continuous, we digitally sample the echo so that it fits into a matrix (chart) of values. The 2-Dimensional Fourier Transform (2DFT) converts the k-space matrix into a digital image with the same voxel dimensions as the k-space matrix. The field of view (FOV) is adjusted separately by the technologist for the phase and frequency encoding axes of the image. What's actually happening inside the scanner software when FOV is increased along the phase encoding axis? The scanner plans gradients to make the rows of k-space matrix data more narrowly spaced along the phase-encoding dimension of k-space. Similarly, if FOV is increased along the frequency encoding axis, columns of the k-space matrix are sampled more narrowly along the frequency encoding axis. Bonus For the Core Exam level of mastery, focus your energy on understanding the basic spin-echo sequence and the turbo spin-echo variant in detail. FOV can be increased along the frequency encoding axis either by (1) increasing the frequency encoding gradient, or (2) by increasing the receiver bandwidth (a.k.a. sampling the echo more rapidly), but the default scanner convention is to keep the receiver bandwidth fixed if FOV is being changed. If you don't understand why this is, it's because we haven't covered the relationship between receiver bandwidth, frequency encoding gradient, and FOV in detail. We'll cover this material at the in person review.

Answer 119

Phase wrap is an aliasing artifact that occurs in the phase encoding direction in practice. - To fix phase wrap, increase the FOV to include all of the anatomy along the phase encoding. - Increased FOV is achieved by decreasing spacing between k-space values along the corresponding axis (frequency or phase encoding axis). - Increased # of voxels along a given axis is achieved by increasing the total # of k-space values along that axis. - If you keep FOV constant and increase # of voxels, this is called increased resolution. Example Increasing FOV with a fixed k-space data matrix size will decrease resolution (increase voxel size). This is because the new diagnostic image has the same unchanged voxel dimensions as the k-space matrix but spread over a larger FOV. Details The solution to phase wrap is to increase the FOV along the phase encoding axis, but what does the scanner actually do when you request larger FOV in the phase encoding axis direction? In response, the scanner actually decreases spacing in k-space between phase encoding steps in order to increase FOV along the phase encoding axis. If you don't also change the total # of phase encoding steps, your matrix size will remain unchanged. As a result, the same number of voxels will need to cover a larger FOV, increasing voxel size and decreasing spatial resolution. Bonus Why do we seldom see wrap artifact along the frequency encoding axis, although it's theoretically possible? This is because decreasing spacing in k-space between frequency encoding steps in order to increase FOV along the frequency encoding axis without sacrificing resolution doesn't require extra scan time. The technical details behind this are interesting but beyond the scope of the Core Exam.

Answer 120

B. Explanation Chemical shift artifact type 1 is most easily seen in voxels that contain both fat and water, such as the interface between the kidney/perirenal fat, and the thecal sac/epidural fat. What's happening here? Hydrogen protons in fat have slightly different precession frequencies than hydrogen protons in water. Since position along the frequency encoding axis depends on precession frequency, the fat-related signal in a voxel gets shifted into a higher frequency position along the frequency encoding axis, just a few pixels away. The effect is most apparent along the frequency encoding axis in water-rich structures surrounded by fat, creating a dark rim on the lower frequency side (orange arrows) and a bright rim on the higher frequency side (blue arrows). Memorize This Increasing receiver bandwidth on the MRI console decreases type 1 chemical shift artifact. The penalty for increasing receiver bandwidth on the MRI console is decreased SNR. When people ask you about "bandwidth”, they are talking about receiver bandwidth. Chemical shift type 1 artifact occurs only along the frequency encoding axis. You can do well by memorizing these facts without really understanding why they are true. Details Receiver bandwidth (rBW) is the total range of frequencies along the frequency encoding axis that you choose to include in computing the image. The range of frequencies produced by the tissue along this axis is determined by the strength of the magnetic field gradient Gx along the frequency encoding axis (x). For a fixed resolution, the receiver bandwidth is directly proportional to the # of Hz included per voxel along the frequency encoding axis. Type 1 chemical shift artifact occurs because protons in water and fat have slightly differrent Larmor frequencies, so they map onto slightly shifted parts of the frequency encoding axis. The receiver bandwidth controls the # of Hz per voxel along the frequency encoding axis. By increasing rBW, you decrease chemical shift artifact. This is because you're assigning a wider range of frequencies to the same voxel, causing the water and fat signals to be contained within the same voxel, despite the differences in Larmor frequency between fat and water protons. By default, scanners let you change receiver bandwidth without affecting voxel resolution or FOV. If you were to increase receiver bandwidth without changing Gx, you would actually increase the field of view along the frequency encoding axis (FOVx). However, your scanner is configured to adjust Gx in order to keep the FOV constant when you adjust the receiver bandwidth parameter. Since the size of your k-space matrix is not changed by the bandwidth, the resolution (voxel size) can be held constant while you change receiver bandwidth. This is why it's ok to think of rBW as controlling # of Hz per voxel along the frequency encoding axis. Bonus Don't confuse transmit and receiver bandwidth. These are different terms, despite both containing the word "bandwidth.” Transmit bandwidth is a different concept entirely related to the 90 degree radiofrequency (RF) excitation pulse used to excite the slice being imaged. High transmit bandwidth RF pulses have larger SAR (more tissue heating) and excite a larger volume. Check out the MRI chapter in Radiology Simplified for essential coverage of chemical shift type 2.

Answer 121

B. Explanation The dimensions of the k-space matrix equals the number of voxels in the corresponding phase and frequency encoding axes of the diagnostic image. Details MRI data is gathered in k-space, which records the amount of each spatial frequency present in an image. The 2-dimensional Fourier Transform (2DFT) converts the k-space data into an image for us to interpret. The math behind the 2DFT preserves the matrix dimensions for the k-space and image space, as depicted below:

Answer 122

The correct answer is 'C' Explanation Decreasing voxel volume proportionately decreases SNR. Voxel volume is now 1/8 of the original voxel volume, so SNR changes by a factor of 1/8 (bc decrease by 2^3) Memorize This: SNR ~ voxel volume * sqrt(Nx*Ny*NEX/BW) Nx and Ny are the number of frequency and phase encoding steps, and BW refers to the receiver bandwidth (rBW). If a question stem mentions "bandwidth", assume rBW unless explicitly told otherwise. Bonus Memorize the SNR proportionalities above, because reasoning these values out can get tricky during exam time.

Answer 123

The correct answer is 'A' Explanation Larger static magnetic fields (B0) cause increased susceptibility artifact. Practically speaking, you'll see more metal artifact or blooming from micro-hemorrhage on 3T versus 1.5T. For example, when protocoling spines with hardware, 1.5T is often preferred to 3T. Memorize This Increasing B0 has various effects. Here are some of the most important to know: increases SNR Increases Type 1 chemical shift artifact assuming fixed receiver bandwidth (because resonant frequency difference between fat and water widens) increases susceptibility artifact Longer T1 relaxation time results because stronger B0 weakens the spin-lattice interaction that causes T1 relaxation, so it takes longer. Longer TR times for the same desired tissue contrast results directly from the longer T1 relaxation time. Details With larger static magnetic fields (B0), microscopic differences in susceptibility result in larger local magnetic field differences. This causes faster transverse plane magnetic dipole moment dephasing, resulting in faster free induction decay (shorter T2* time).

Answer 124

B. This is a recall question, and a classic statistic in hospital management.

Answer 125

The correct answer is 'A'

Answer 126

D. Explanation Key performance indicators (KPI) are financial and non-financial metrics used to evaluate the success of an organization. Bonus You may actually hear an investor on Shark Tank ask the entrepreneur, "What are your key performance indicators?” to help the investor understand how the entrepreneur defines success.

Answer 127

The correct answer is 'A' Explanation Don't be faked out. The "Do” in the Plan-Do-Study-Act (PDSA) cycle just means go ahead and collect data. Memorize This Plan - Choose data to be obtained Do - go ahead and collect data Study - analyze data Act - Devise and implement plan for improvement based on analysis

Answer 128

The correct answer is 'A' | ?,not Fishbone, d?

Answer 129

E. Bonus The fishbone diagram is also called an Ishikawa diagram.

Answer 130

A. Explanation Did this question stem seem too brief to narrow down on one answer? Memorize this phrase and associate it exclusively with ROC curves: The ROC curve is used to analyze the performance of a diagnostic system.

Answer 131

The correct answer is 'A' Explanation All of these are recommendations of the Choosing Wisely Campaign except for the back pain recommendation. Details Another ACR imaging recommendation from the Choosing Wisely Campaign is to avoid CXR for admission or pre-op in ambulatory patients with unremarkable medical history and physical exam. Bonus Review the criteria for adnexal cyst followup.

Answer 132

The correct answer is 'B'

Answer 133

A. Explanation Weber effect is the name for the phenomenon where adverse event reporting tends to increase in the first two years after introduction of a new agent or use for a new indication, peaks at the end of the second year, and then declines. Details The important point here is that reporting might decrease because of reporter fatigue or other causes, and not necessarily because adverse events are occurring less frequently.

Answer 134

C. Explanation Hawthorne effect is the name for the phenomenon where being observed changes behavior. Details For example, if you announce that you're measuring scan-to-report time, you should expect that radiologists in your department will work faster to minimize this time in response to being observed.

Answer 135

D. Explanation Here are the Six Quality Aims from the Institute of Medicine (IOM). Memorize each of these six aims and be capable of reciting them. ``` Effective Efficient Equitable Safe Timely Patient-Centered ``` Bonus If you're given a multiple choice question related to the Quality Aims, don't be distracted by options that sound like one of these aims but aren't actually on this list.

Answer 136

A. Value in healthcare is closely related to efficiency: efficiency = results achieved / resources expended

Answer 137

~3days (same for all the "iums")

Answer 138

- Tc-99m.......6hr - I-123.............13 hr - I-131..............8 day - Indium, Gallium, Thallium.........~3days each (all the "iums") - Xe-133.........5 days - F-18........... 110min

Answer 139

365 KeV ("like the days of the year")

Answer 140

100, 200, 300, 400 KeV (roughly)

Answer 141

80 KeV (this is really x-ray level, rather than true gamma)

Answer 142

Short TE: <30ms Long TE >150ms Short TR: <500ms Long TR: >15000ms (Note TR is always longer the TE comparitively, bc TR comes after TE in the sequences! At least in "basic" spin echo that is.)

Answer 143

This various paths taken of spatial encoding gradients for filling-in of K-space. - HASTE = "Half-Fourier Acquired Single-Shot Turbo Spine Echo". It reduces number of k-space lines acquired, exploiting k-space symmetry. Faster acquisition, lower SNR, reduced SAR ("specific absorption rate"). (ex uses: T2 scout images, fetal, abdominal imaging). - BLADE (or PROPELLOR) = fills in diagonal pattern. Redundant sampling of k-space center used for motion compensation at cost of slower acquisition. (ex uses: brain in disoriented patient). - SPIRAL = fills k-space from center out. More-efficient than standard Cartesian trajectory, good for ultra-short-TE sequences needed for dynamic MRI. (ex uses: cardiac).

Answer 144

4 W/Kg for 15min (which ensures less the 1degree C rise in core temperature).

Answer 145

- Magnevist-----gadopentetate-----linear - Mutlihance-----gadbenate----------linear - Omniscan-------gadodiamide------linear - Gadavist---------GADOBUTROL----MACROCYCLIC (Macrolcyclics are safer, less chance of free Gd being reliease from chelate=>less NSF risk....I guess this is why we tend to use Gadavist?...so know that generic name most closely; the others listed are linear). Omniscan has highest risk of NSF. ("GadoDIE-amide")

Answer 146

-Mnemonic "GIIT" Gallium, (I-123), Indium, Thallium -All have approx same half-lives: ~3days.

Answer 147

Mechanisms: Thallium (mimic K+-uptake at K/Na Atp-ase pump => reflects "alive tissue"). Gallium (iron-mimic, acute-phase reactant => reflect inflammatory activity) Photopeaks: - Thalium--> low energy ("x-ray level"), 70-80 KeV )2 peaks) - Gallium --> 4 peaks, ~100,200,300,400 Kev This vs That: - Lungs in Aids: - Kaposi Sarcoma (Th-Hot, Ga-Cold; bc its "alive" (tumor) without generating inflamm reaction (low-grade tumor))....VS......PJP pneumonia (Th-cold, Ga-Hot; bc its infection, not alive but inflam). - Brain in Aids. ring-enhancing lesion: - Toxo (Th-cold, Ga-Hot; bc its infection=> inflam, but not alive)...VS.....Lymphoma (Th-hot, Ga-Cold; bc its tumor => alive, with no sig inflam) - Radiation necrosis (Th-cold, Ga-Hot; bc its not alive, but inflam).....VS tumor recurrence (Th-hot, Ga-Cold; bc its alive, but no sig inflam).

Answer 148

Energy = Plank's constant * (speed of light / wavelength) => Energy inversely proportional to wavelength => Energy proportional to frequency. => Doubling the photon's frequency -> doubling the energy...

Answer 149

Y-axis = "True Positive Rate" = Sensitivity X-axis = "False Positive Rate" = (1- true positive rate)= 1 -Sensitivity. Note, "specificity" is not really part of this diagram (Specificity actually = True Negative Rate)

Answer 150

In poisson distribution: SD= sq(of the mean). => if average is 100, then the SD =10.

Answer 151

Typical imaging depth for a 3mHz US probe is..... 20cm Typical depth for a 10mHz probe is.... 6cm

Answer 152

Larmor frequency for H = 42MHz/T => in 3T 42MHz*3 = 126 MHz. (****Larmor frequency of H at 1T = 42MHz)

Answer 153

- Head: 3 W/kg | - Body: 4 W/kg

Answer 154

-Semi-annually (every 6 months)

Answer 155

Ga-67 for spinal osteo! (Mnemonic: Sixty-7 for Spine) Vs Tagged-WBC for GI/abdomen! (Bc Ga-67 normally has some GI activity => can obscure abn foci; while tagged-WVC does not go to bowel normally!).

Answer 156

``` A) Short TE: <30ms Long TE >150ms Short TR: <500ms Long TR: >15000ms (Mnemonic: Short is TE 30, TR 500; with Long is x3 short for each...) ``` B) 1) T1: Short TR (2000-3000ms) and Short TE (25-30 ms) 2) T2 : Long TR (>2000ms) and Long TE (>70 ms) 3) PDIL Long TR (2000-3000ms) and Long TE (>70 ms)

Answer 157

less the 5 minutes.

Answer 158

Gradient Echo (most commonly SSFP), which allows for fast image aquisition with high spatial resolution.

Answer 159

A and D. Both peds and IR commonly use additional copper filteration. Copper filtration is used to reduce the radioation dose in pediatric imaging and to reduce the risk of skin burns in IR.

Answer 160

- Plain film: 100kW | - Mammo: 3kW

Answer 161

``` Power = Current * Voltage (P = IV) ``` Mnemonic = "A peripheral IV (P=IV) give an electrical circuit power"

Answer 162

Matching layer part of probe that is interface between the transducer element/PZT and the tissue; its prupose is reduce acoustic reflection that would otherwise occur at probe/skin interface, by prodiving a layer with an impedance value that is partially between the PZT and skin/soft-tissue.

Answer 163

Fast Spin Echo

Answer 164

Remember, T1 is "a measure" of T1 longitudinal relation time (related to, but not identical to it). Recovery of longitudinal magnetization occurs exponentially time at a predictable rate, T1 is defined as the time when 63% of the maximal longitudinal magnetization has recovered...... by T1*4 it has NEARLY all recovered (at T1*2, just 87%; at T1*3 95%). Remember, T2 is independent (not related) to longitudinal magnetization relaxation time...

Answer 165

10min post Gd administration

Answer 166

Beta-catenin subtype. Which also does not contain fat. Which also has somewhat higher risk of malignant transformation (Mnemonic:" only non-alpha beta-men take hormones and get hepatic adenomas. These same hormone-taking beta-men are hate having any fat, they'd rather get cancer!")

Answer 167

D. Both! with dose increase from electronic mag > geometric mag.

Answer 168

TE decreases! Bc a broader bandwidth has a higher amplitude readout gradient which allows for faster sampling of the MR signal.

Answer 169

Btwn the THYRO-PHARYNGEAL (above) and the CRICO-PHARYNGEAL (below), which together make up the inferior constrictor of the pharynx.

Answer 170

Complication of otomastoiditis, triad -> +petrous apicitis (involving the petrous apex) + Abducens nerve palsy (secondary to involvement of the nerve as it passes adjacently though Dorello's canal) + facial pain distribution of V2/V3 (due to inflammation extending into Meckel's cave). It is treated with IV abx

Answer 171

25% (1/4.... mnemonic: "4 words in Slipped Capital Femoral Epiphysis" => 1/4) - Klein's line (line along lateral aspect of femoral neck should intersect femoral head.... if it doesn't, the fem head has slipped) - Treated with pinning.

Answer 172

Typically low T1, high T2 signal (similar to fluid signal due to the high fluid content of myxoid tissue, which involves a high level of mucinous extra-cellular matrix; => myxoid tumors might be a mimic of cystic lesions at times)

Answer 173

1.0 mGy per hour measured at 1m from the source (mnemonic: "1 and 1" leak allowed)

Answer 174

1Gy = 100 rad

Answer 175

4-5mCi for infection 10mCi for tumor (higher dose than for just infection) (BONUS: Half-like of Ga-67 is 78hr (~80hr, which also like other "ium" Thallium which 9s 73hr)

Answer 176

To monitor O2 levels in the event of an emergent quench.... bc when the helium gets released in a quench it will discplace the O2 in the room and could lead to aspgyxiation. => O2 levels should always be monitored, specifically in the event of a quench.

Answer 177

0.5 rem (5mSv). Note this is 1/10th the threshold than for an adult radiation-worker level threshold allowed, which is 5 rem (50mSv)/year. It is the same as what a fetus max or under-18 worker thresholds!

Answer 178

Beryllium.

Answer 179

Increase. Bc Near Field inversely proportional to wavelength (=> smaller wavelength -> bigger near field; increasing the frequency -> decreasing the wavelength => increasing the near zone)

Answer 180

4N (Near Field = (element size)^2/wavelength ...explanation of this concept, i.e. 5 elements in an array are often pulsed simultaneous, effectively creating an element is 5x wider,=>lengthening the near zone by a factor of 25)

Answer 181

Semi-annually (q Six Months).....( mnemonic: S in Sealed for Semi-annually/Six-months)

Answer 182

Increase both quantity and quality of the x-ray beam. => Additionally, Reducing tube ripple will increase the average xray photon energy (bc increases the quality), which in turn will decrease tissue contrast. (whatever the hell "xray ripple" is). => Ripple is inversely proportional to avg x-ray energy and directly proportional to tissue contrast. (Increasing ripple --> decrease xray energy, increased contrast)

Answer 183

Mottle is inversely related to the SQUARE ROOT of the number of photons incident on the detector. => i.e. increase the number of photons incident by 4 (AKA increase the tube mAs by 4) ->decrease mottle by 2 (square root of 4).

Answer 184

Accuracy = (TP + TN)/all-patients

Answer 185

in Sulfur Colloid: Liver > Spleen (Mnemonic: its called a "liver-spleen scan" => liver is more bc it comes first in the name).... if you see spleen>liver in Sulfur-colloid scan, may reflect "colloid shift" in hte setting of liver dysfunction. In In111-WBC: Spleen > Liver (the other one is the other one; plus spleen is bloody organ, therefore make sense blood cells will go there more).

Answer 186

Typical rads: 3 mGy (1-10 mgy) Typical flouro: 30 mgy/min (20-50 mgy/min) Typical IR: 300-500 mgy/min **Remember easily as: Radiograhy "ones", Flouro "tens", IR "hundreds"

Answer 187

Enterance skin dose actually tends to be 50% more than enterance air kerma! (=> if the avg enterance air kerma is measured as 3 mGy, then the apprx skin dose is ~5mGy).

Answer 188

At least 5 years.

Answer 189

ANYTIME you have MULTIFOCAL tumor involvement in the SAME LUNG it is either going to be T3 or T4 disease...... ....to detemine whether it is T3 or T4, ask: is the additional tumor nodule in the same lobe as the dominant lesion or in a seperate lobe? (Same lobe is T3; different lobe same lung is T4) ....Not the seperate tumor in contralateral lung is considered M1a.

Answer 190

-RBC scan: 0.1 mL/min........vs............angio 1 mL/min

Answer 191

----------- 0.5 explanationL the system's OVERALL modulation transfer function is THE PRODUCT of all the indivual modulation transfer functions (in this case =0.7*0.8*0.9 = 0.5)

Answer 192

Lymphocele (typically presents 1-2months post transplant) ...FYI can also result in ipsilateral lower extremity edema from femoral vein compression.

Answer 193

Want RI's <0.7. If >0.7, suggests a problem...

Answer 194

Malakoplakia: Urinary bladder/Ureter nodules related to chronic UTI's....they are NOT PREMALIGNANT => no surgery indicated, get better with abx. -often seen in immunocomp females. vs. Leukoplakia: =Squamus metaplasia of urinary bladder, also causing mural filling defect. ...IS PREMALIGNANT->squamus cell CA (as opposed to malakoplakia which is not premalignant);