radiology physics Flashcards
Photoelectric and Compton scatter are equally likely at these energies:
25 keV in soft tissue
40 keV in bone
At lower energies, PE absorption dominates.
At higher energies, Compton scatter dominates.
average x-ray beam energy (keV) value is estimated how?
= about 1/2 the maximum tube voltage (kVp);
note, this formula works for general projection radiography (CXR, KUB, etc.) which uses predominantly braking radiation, and not most mammography which uses predominantly characteristic radiation (i.e. doesnt apply to Mo/Mo, Rh/Rh, Mo/Rh mammography, bc energy distributions in those type of mammo tubes are dominated by characteristic radiation (rather than braking radiation in all other projection radiography, and that is where that formula matters)
What is the 15% rule in x-ray technique?
The 15% rule (memorize this) is that increasing tube voltage (kV or kVp) by 15% would require mA to be cut in half in order to maintain the same total # of x-rays produced per second.
(=> in other words, increasing kVp by 15% -> 2x mA)
a) What is the “focal spot” in x-ray?
b) Whats the relevance of focal spot size for resolution? You’ll learn more about this in the Projection
c) How does decreasing focal spot size affect technique parameters (i.e. kvp, mA, exposure time) that may need to be adjusted?
a) focal spot is the actual area on the anode target metal exposed to electrons in the x-ray tube.
b) smaller focal spot size increases resolution, countering the effects of geometric blur.
c) Since we’re exposing a smaller area of metal, it heats up more easily (resulting in melting). To counter melting, we need to turn down the tube current (mA) which controls the # of electrons hitting the metal per second. If we decrease mA, we need to increase exposure time to keep the total # of photons coming from the tube constant.
What percent of energy that goes into an x-ray tube comes out as x-rays?
1% of energy that goes into an x-ray tube comes out as x-rays. The rest (99%) goes into heat, which is dissipated by rotating the metal anode target and bathing the x-ray tube in oil.
What are the k-shell binding energies (units of kiloelectron-volts , abbreviated keV) of the following materials:
Anode-target/Filter materials: Mo , Rh , Ag , W
Contrast agents: I , Ba
(units of kiloelectron-volts , abbreviated keV)
Mo - 20,
Rh - 23,
Ag - 25,
W - 70,
I - 33,
Ba - 37
**Remember that the highest characteristic energies are slightly less than the k-shell binding energy.
Note, bonus:
The higher the atomic number (Z, the # of protons), the higher the k-edge. This makes sense - the k-shell binding energy is higher if there are more protons to pull on the k-shell electron. Thus, I know the Z number of Rh is higher than Mo just because I know the corresponding k-shell binding energies
Formula for likelihood of
(a) photo-electric interaction with material.
(b) Compton scatter interaction with material
and relation to Z
(a) PE proportional to Z^3/E^3
(dependent on Z and physical density)
(b) Compton proportional to 1/E
(dependent on electron density and physical density, not really that related to Z; => adjusting Z alone does not really affect Compton scatter contribution)
(Z= atomic number, E=photon energy)
mnemonic “Zebra sits on the Elephant” (Z on top of E)
=>that’s why PE dominates at lower energies, while compton dominates at higher energies (the liklihood of a PE falls of by 1/E^3, but while for compton just by 1/E; note of course PE cannnot happen at all without involving at least the minimum E for k-edge)
*Memorize:
What is the best kVp for subject contrast with these various modalities:
catheter angiograms -
CT angiograms -
Barium GI fluoro studies -
catheter angiograms - 70 kVp
CT angiograms - 100 kVp
Barium GI fluoro studies - 90-110 kVp
(note: …In general Subject contrast is best when average x-ray energy is slightly more than the k-edge of the contrast agent; i.e. in case of iodine, since the average photon energy from an x-ray tube is about half the kVp, if we set kVp to 70 kV, then the average x-ray photon energy is 35 keV, which is slightly higher than the k-edge of iodine (33 keV)!…. however, because this simplified reasoning isn’t bulletproof, it’s useful to memorize the above best kvp values.)
You are performing a diagnostic cerebral angiogram in the IR suite and you switch from imaging the groin for puncture to the skull, where the fluoroscopy machine automatically increases average beam energy to penetrate bone. What happens to the HVL of your lead apron?
Choose only ONE best answer.
A
HVL increases.
B
HVL decreases.
C
HVL is unchanged.
D
HVL increases initially, then settles back.
E
HVL decreases initially, then settles back.
The correct answer is ‘A’ .
HVL depends on the material and the x-ray beam energy.
Denser materials have smaller HVLs because they stop x-rays more efficiently.
Higher energy x-rays cause the HVL to increase for any material, because those x-rays are more penetrating (less likely to engage in Compton Scatter or photoelectric absorption).
You’re thinking about getting a new shielding apron (2mm HVL for typical x-rays) in one of two thicknesses. You can get either the Lizard model (6mm thick) or the Hippo model (12mm thick). How many times more attenuating is the Hippo model versus the Lizard model?
Choose only ONE best answer.
A
1/2
B
2
C
6
D
8
E
64
D. 8
Explanation:
You’ll need to be comfortable with basic calculations using half value layer (HVL).
The x-ray intensity drops by half for every HVL of thickness.
First, convert the thicknesses in mm to #’s of HVLs, where one HVL is 2mm for this particular scenario:
Lizard Model: 6 mm is 3 HVL of material
Hippo Model: 12 mm is 6 HVL of material
Second, use the HVL formula, which relates the input x-ray intensity (I0) to the output x-ray intensity (I1) in terms of the distance travelled by the x-ray beam (d):
I1 = I0 x (1/2)(# of HVL)
Lizard Model: I1 = I0 x (1/2)3 = (1/8) I0
Hippo Model: I1 = I0 x (1/2)6 = (1/64) I0
The Hippo Model is 8x more attenuating than the Lizard model.
Bonus
Practice this type of question ahead of time. If your head is spinning at the sight of math on exam day, don’t stress out. Just move on to the next question.
A grid’s height is 1.2 mm, its grid bar (septum) width is .6 mm, and its interspace width is 0.2 mm. What is the grid ratio?
Choose only ONE best answer.
A
6:1
B
1/6:1
C
1/5:1
D
2:1
E
1.2:1
A.
Explanation
Grid ratio is 1.2 mm grid height divided by 0.2 mm interspace width. 1.2/0.2 = 12/2 = 6. The width of septa themselves is irrelevant to grid ratio. By convention, this grid ratio is reported as 6:1.
Memorize This
Grid ratio = height / interspace width
You’ll see the answer reported using the colon (:) symbol as (height/interspace width) : 1.
When memorizing this formula, remember “height is high”, “D is in the denominator”.
Typical grids have ratios like 6:1 or 12:1. It makes sense that they are taller than their interspace width, so height should be in the numerator.
Increasing anode angle results in which of these?
Choose only ONE best answer.
A
Increased effective focal spot size, decreased anode heel effect
B
increased effective focal spot size, increased anode heel effect
C
decreased effective focal spot size, increased anode heel effect
D
decreased effective focal spot size, decreased anode heel effect
E
unchanged effective focal spot size, unchanged anode heel effect
A.
Explanation
Increased anode angle causing larger effective focal spot size is purely a result of geometry, called the “line focus principle.”
The large anode angle also makes the metal target more parallel with the detector, so the detector gets irradiated more evenly, which makes for decreased anode heel effect.
Details
Mnemonic to remember which angle is the anode angle: The Large anode angle looks like the letter L (L for Large). The Very Small anode angle looks the letter V.
Bonus
Anode heel effect is less of an issue with digital radiography than film, because post-processing algorithms can partially compensate the spatial inhomogeneity in x-ray intensity.
The CR reader extracts the x-ray exposure pattern using different colors of light as follows:
Choose only ONE best answer.
A
apply red light, read blue light, erase with white light
B
apply blue light, read red light, erase with white light
C
apply blue light, read red light, erase with green light
D
apply white light, read red light, erase with blue light
E
apply red light, read white light, erase with blue light
The correct answer is ‘A’
Explanation:
The CR cassette stores x-ray exposure events with electrons that move from ground to metastable state. The CR reader stimulates the cassette with red light. Metastable electrons return to ground state, releasing blue light, marking the spots where x-rays previously landed. The cassette is completely erased with intense white light.
Memorize This
-Computed Radiography (CR) cassettes store the x-ray image using photostimulable phosphors (PSPs).
-The CR cassette reader uses red, blue, and white light in these different ways: Red-light stim-> Triggering Blue-light release (“hit it with red to release it; it will burn blue to let it be red”); Erase it with bright white.
(The order of things is RED, BLUE, WHITE; RBW “Ruth Bader Winsburg”)
What focal spot size is most commonly used in standard projection radiography for chest, abdomen and extremity imaging?
Choose only ONE best answer.
A 1.2 mm B 0.3 mm C 0.1 mm D 1.0 mm E 2.0 mm
A.
This is a recall question. Standard projection radiography uses the 1.2 mm focal spot size.
Memorize this:
X-ray-based Modality / Effective Focal Spot Size
Standard Radiography / 1.2 mm CT / 1.0 mm Fluoroscopy / variable from 0.3 mm to 1.2 mm Mammography (standard) / 0.3 mm Mammography (magnification) / 0.1 mm
Details
Review the difference between actual and effective focal spot size.
Actual focal spot size is determined by filament length. Smaller actual focal spot sizes require more limited tube current to avoid anode target melting because heat is concentrated onto a smaller area of the metal.
Effective focal spot size is proportional to actual focal spot size (filament length) and anode angle.
Effective focal spot size is one of the three determinants of resolution.
Memorize:
What are the effective focal spot sizes used in the various x-ray modalities:
Standard Radiography CT Fluoroscopy Mammography (standard) Mammography (magnification)
X-ray-based Modality / Effective Focal Spot Size
Standard Radiography / 1.2 mm CT / 1.0 mm Fluoroscopy / variable from 0.3 mm to 1.2 mm Mammography (standard) / 0.3 mm Mammography (magnification) / 0.1 mm
What is the standard resolution limit of digital projection radiography for chest, abdomen, and extremity imaging?
Choose only ONE best answer.
A 7 lp/mm B 1 lp/mm C 2 lp/mm D 5 lp/mm E 11 lp/mm
D. 5 lp/mm
Explanation:
Every x-ray based modality has different resolution limits, and line pairs per millimeter (lp/mm) is one measure of resolution. CR-based digital projection radiography has an approximate resolution limit of 5 lp/mm.
Memorize this
These numbers are rough, so you’ll see a wider range of values reported across the literature. Have a general idea for how each modality ranks by resolution limit.
Modality Resolution Limit (lp/mm) CT 1 Fluoroscopy 1-4 DSA 2 CR-based digital radiography 5 film radiography 7 digital mammography 7 film mammography 11 & 13 (perpendicular vs. parallel to focal spot, respectively- MQSA standards)
What are typically quotes Resolution Limits for following modalities (lp/mm)
- CT
- Fluoroscopy
- DSA
- CR-based digital radiography
- film radiography
- digital mammography
- film mammography
Modality Resolution Limit (lp/mm):
CT 1 Fluoroscopy 1-4 DSA 2 CR-based digital radiography 5 film radiography 7 digital mammography 7 film mammography 11 & 13 (perpendicular vs. parallel to focal spot, respectively- MQSA standards)
Which one of these detector types is most commonly used for chest, abdomen, and extremity digital radiography?
Choose only ONE best answer.
A Photostimulable phosphor B Intensifying screen C Cesium Iodide scintillator D Sodium Iodide scintillator E Photodiodes with thin-film transistors
The correct answer is ‘A’
Explanation: Computed radiography (CR) cassettes are the detector types most commonly used in digital projection radiography. The underlying technology is a photostimulable phosophor (PSP) that stores x-ray exposure events, which can be read later using a CR reader.
Details
Intensifying screen is the “screen” part of film-screen, the original detectors in projection radiography. The screen converts x-rays into visible light and brightens the image before it hits the film, decreasing the patient dose needed to form an image. That all you need to know about film-screen.
Cesium Iodide (CsI) scintillators are the “indirect” part of indirect flat panel detectors (FPD) in Fluoroscopy. Scintillators (in the most general sense of the word) convert x-rays into visible light. With indirect digital flat panel detectors (FPD), CsI first converts x-rays into visible light so that it can be detected by photodiodes (light-sensitive detectors) connected to thin-film transistors (TFT) that digitally record the photodiode signal. The charge-coupled device (CCD) is an alternate light-sensitive digital circuit that may be used in conjunction with the CsI scintillator layer instead of the photodiode-TFT combination.
Amorphous selenium is used in direct digital FPD detectors for Mammography. This technology forms a digital image from x-rays without the CsI scintillator or photodiodes. There is still a TFT layer to read out the digital signal.
Sodium Iodide (NaI) scintillators are used in Nuclear Medicine. The NaI converts gamma rays into visible light for gamma cameras (planar and SPECT imaging) and well counters.
Bonus
Keep your scintillators straight. NaI is used in Nuclear Medicine.
On an x-ray characteristic curve ( which describes the relationship between x-ray exposure and the resulting image), Which parts of this curve correspond with image contrast and latitude?
- Contrast is proportional to the slope of the characteristic curve. The slope is called gamma.
- Latitude is the x-axis range over which the curve hasn’t plateaued, also called the dynamic range.
Details
The characteristic curve describes the relationship between x-ray exposure and the resulting image.
In digital imaging, windowing/leveling at the viewing station simulates adjustments in contrast and latitude in the image.
Bonus
The major advantage of digital radiography is virtually infinite latitude versus the latitude of any type of film. (bc film has an S-shaped curve, so has plateaus at either end and therefore has a narrow region of latitude centrally; but digital is a straight-linear curve, so essentially has no points where it plateaus.) (you lose contrast/slope at points of the curve that are flat/plateau/”outside the region of latitude”)
You move to a higher ratio grid. How does your image quality change?
Choose only ONE best answer.
A Unchanged shot noise, increased contrast B Increased shot noise, increased contrast C Decreased shot noise, increased contrast D Decreased shot noise, decreased contrast E Unchanged shot noise, unchanged contrast
The correct answer is ‘A’
Explanation
The grid decreases scatter (the higher grid ratio means less scatter gets through). Scatter deteriorates contrast. The phototimer ensures the # of photons forming the image, and hence shot noise (quantum mottle), is unchanged (we assume of coarse a phototimer is being used). Overall, the contrast-to-noise ratio increased.
Details
Contrast can be quantified by placing a radiopaque disc at the center of the field. You can then compare the pixel intensity at the center to the periphery. Scatter causes the photon count from the radiolucent periphery to average into the image area for the radiopaque disc, leading to an overall grayer image for the disc. Contrast has decreased from scatter- there’s less pixel intensity difference between disc and periphery.
Bonus
- Reading through answers can be confusing when the choices are small variations of each other. Minimize confusion in this scenario by deciding the correct answer without reading every distractor option in detail.
- On the exam, assume the phototimer is in effect with questions related to projection radiography, unless a question stem explicitly tells you otherwise.
A chest radiograph irradiates the breast in a pregnant woman being assessed for TB. Which unit of radiation dose accounts for the fact that breast tissue is being irradiated, elevating stochastic (cancer) risk?
Choose only ONE best answer.
A Absorbed dose B Effective dose C Air Kerma D Equivalent dose E Kerma-Area Product
The correct answer is ‘B’.
Explanation
Effective dose is the best option here for relating to cancer risk.
Details
Absorbed, equivalent, and effective dose are all related to each other by weighting factors:
exposure → absorbed → effective
Exposure (gray, abbreviated Gy = Joules/kilogram) - how much x-ray energy (Joules) was released per kilogram of air.
(-Fluoro uses air kerma (AK - units of mGy) and kerma-area product (KAP- units of mGy∙cm2 ) extensively - those are important units of exposure, and we’ll see more of that in fluoroscopy. In fluoroscopy, you’ll learn that AK relates to deterministic effects (epilation, radiation necrosis, etc.) and KAP relates to stochastic effects (cancer). Since this question doesn’t relate to fluoroscopy, it’s best to avoid choosing AK or KAP. Moreover, neither AK or KAP account for tissue type.)
- Absorbed dose (gray, abbreviated Gy = Joules/kilogram) - how much x-ray energy gets into the tissue per kilogram of tissue.
- Equivalent dose (Sieverts, abbreviated Sv) - accounts for different amounts of DNA damage caused by different types of radiation. Both x-rays and gamma rays have weighting factor = 1, so the numerical value of equivalent dose in Sv = absorbed dose in Gy for diagnostic radiology. Although this unit accounts a little bit for propensity to cancer, it does not account for tissue type, and thus is not the best answer.
Effective dose (Sv) - accounts for tissue irradiated with regards to stochastic (cancer) risk. Each organ has its own weighting factor based on it’s level of sensitivity to radiation, including breast. Effective dose tells you the effect of the radiation on cancer risk, accounting for tissue type. This is the best answer.
Bonus
In x-ray, the weighting factor for type of radiation is 1 => absorbed dose (mGy) = equivalent dose (mSv); effective dose will vary based on the specific tissue in question and its tissue weighting factor.
Don’t Memorize This
Forget about radiation dose units like rad, rem, and roentgen - these aren’t SI, so they shouldn’t be asked on the exam. Plus, they’re a headache to keep straight.
You are evaluating possible investment in a CR cassette manufacturing startup that’s decided to go from a thin photostimulable phosphor (PSP) to a new thicker PSP. What do you expect will happen to the modulation transfer function (MTF) and detective quantum efficiency (DQE) curves when the company moves to the thicker PSP?
Choose only ONE best answer.
A MTF down, DQE up B MTF up, DQE down C MTF up, DQE up D MTF down, DQE down E MTF unchanged, DQE up
A. is correct.
Explanation
Thicker PSP means x-rays are more effectively captured and recorded, increasing DQE (reflecting better image signal-to-noise ratio). Unfortunately, thicker PSP also causes more light spreading during the reading process (when red light is shown onto the cassette, triggering release of blue light from within which then spreads out as it propogates through the thicker PSP to the photodioide), which causes the MTF curve to go down (loss of resolution).
Memorize this
Increased DQE allows for lower patient dose to achieve the same required image quality. Want better image quality with lower patient dose? Seek higher DQE.
Details
DQE is the ratio of output signal-to-noise ratio (SNR) to input SNR. The term SNR is equivalent to contrast-to-noise (CNR) in imaging. You can think of “input SNR” as the CNR for a perfect imaging system. The “output SNR” is the CNR from your actual equipment with all of its imperfections.
DQE varies with kVP and with spatial frequency.
Direct digital has the best DQE, followed by indirect digital, and film-screen with the worst DQE across all spatial frequencies.
You are evaluating possible investment in a CR cassette manufacturing startup that’s decided to go from a thin photostimulable phosphor (PSP) to a new thicker PSP. What do you expect will happen to the modulation transfer function (MTF) and detective quantum efficiency (DQE) curves when the company moves to the thicker PSP?
Choose only ONE best answer.
A MTF down, DQE up B MTF up, DQE down C MTF up, DQE up D MTF down, DQE down E MTF unchanged, DQE up
A. is correct.
Explanation
Thicker PSP means x-rays are more effectively captured and recorded, increasing DQE (reflecting better image signal-to-noise ratio). Unfortunately, thicker PSP also causes more light spreading during the reading process (when red light is shown onto the cassette, triggering release of blue light from within which then spreads out as it propogates through the thicker PSP to the photodioide), which causes the MTF curve to go down (loss of resolution).
Memorize this
Increased DQE allows for lower patient dose to achieve the same required image quality. Want better image quality with lower patient dose? Seek higher DQE.
Details
DQE is the ratio of output signal-to-noise ratio (SNR) to input SNR. The term SNR is equivalent to contrast-to-noise (CNR) in imaging. You can think of “input SNR” as the CNR for a perfect imaging system. The “output SNR” is the CNR from your actual equipment with all of its imperfections.
DQE varies with kVP and with spatial frequency.
(Direct digital has the best DQE, followed by indirect digital, and film-screen with the worst DQE across all spatial frequencies).
You are evaluating possible investment in a CR cassette manufacturing startup that’s decided to go from a thin photostimulable phosphor (PSP) to a new thicker PSP. What do you expect will happen to the modulation transfer function (MTF) and detective quantum efficiency (DQE) curves when the company moves to the thicker PSP?
Choose only ONE best answer.
A MTF down, DQE up B MTF up, DQE down C MTF up, DQE up D MTF down, DQE down E MTF unchanged, DQE up
A. is correct.
Explanation
Thicker PSP means x-rays are more effectively captured and recorded, increasing DQE (reflecting better image signal-to-noise ratio). Unfortunately, thicker PSP also causes more light spreading during the reading process (when red light is shown onto the cassette, triggering release of blue light from within which then spreads out as it propogates through the thicker PSP to the photodioide), which causes the MTF curve to go down (loss of resolution).
Memorize this
Increased DQE allows for lower patient dose to achieve the same required image quality. Want better image quality with lower patient dose? Seek higher DQE.
Details
DQE is the ratio of output signal-to-noise ratio (SNR) to input SNR. The term SNR is equivalent to contrast-to-noise (CNR) in imaging. You can think of “input SNR” as the CNR for a perfect imaging system. The “output SNR” is the CNR from your actual equipment with all of its imperfections.
DQE varies with kVP and with spatial frequency.
(Direct digital has the best DQE, followed by indirect digital, and film-screen with the worst DQE across all spatial frequencies)
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)?
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.
You’re struggling to search for a retained sponge in an uncomplicated appendectomy case on a portable radiograph from the OR. Which one of these factors does not affect spatial resolution?
Choose only ONE best answer.
A collimation B geometric magnification C CR cassette D patient respiration E filament size
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.
You are choosing between two grids for chest radiography. The 6:1 grid has a Bucky Factor of 3 at 100 kVp. The 12:1 grid has a Bucky Factor of 5 at 100 kVp. By what factor is dose increased going from the 6:1 grid to the 12:1 grid?
Choose only ONE best answer.
A 1/2 B 5/3 C 1 D 2 E 3/5
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.
Based on tissue weighting factor (wT) which is used to calculate effective dose from equivalent dose, how the common tissues of concern grouped by degree of radiosensitivity?
Most Sensitive: breast, lung, colon, red marrow, etc.
Intermediate: thyroid, etc.
Least Sensitive: brain, skin, salivary glands
A commercial FPD (Flat Panel Detector) that normally runs at 7 fps with no binning requires a 2x2 binning mode in order to sustain a 30 fps frame rate. By what factor has the resolution (in lp/mm) changed in going to 30 fps from 7 fps?
Choose only ONE best answer.
A
2
B
1
C
0.5
D
30/7
E
0.8
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.
Given an square FOV of 20 cm x 20 cm and FPD detector matrix size of 1000 x 1000, what is the resolution?
Choose only ONE best answer.
A
2 lp/mm
B
0.4 lp/mm
C
0.2 lp/cm
D
2.5 lp/mm
E
5 lp/mm
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
The image intensifier is designed to make the image brighter than if you used just the input phosphor alone:
Image intensifier output is brightened in part simply by the fact that the electron beam is concentrated from a wider input photocathode radius onto a narrower output phosphor radius. This phenomenon is called:
Choose only ONE best answer.
A
Flux gain
B
Brightness gain
C
Minification gain
D
Electronic magnification
E
Pincushion artifact
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.
Magnification is performed. What happens to FOV?
Choose only ONE best answer.
A
depends on geometric versus electronic magnification
B increased C unchanged D depends on II versus FPD E decreased
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.
The operator stands in the usual position to the patient’s right during a percutaneous biliary tube placement. The C-arm is switched from LAO to RAO position. What happens to operator absorbed dose?
Choose only ONE best answer.
A decreases B increases C unchanged D variable E not enough information
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.
What is the approximate dose reduction to the operator and thickness of a typical lead apron?
Choose only ONE best answer.
A
99%, 0.5mm
B
80%, 0.25mm
C
95%, 1.0mm
D
99%, 2.0mm
E
50%, 0.25mm
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.
What is the typical lead apron thickness? And how much x-ray radiation is typically absorbed by it?
A lead apron is typically 0.5mm with 99% absorption (1% transmission to the operator) for typical kVp
Which of these artifacts are seen in FPD (Flat Panel Detector) systems?
Choose only ONE best answer.
A
None of the above
B
Pincushion distortion
C
S-distortion
D
Vignetting
E
Flare/Veiling Glare
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.
Which of these artifacts does not relate to the intrinsic geometry or material construction of the image intensifier?
Choose only ONE best answer.
A
Pincushion distortion
B
S-distortion
C
Vignetting
D
Flare/Veiling Glare
E
None
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.
Which one of these is the light-sensitive portion of the indirect flat-panel detector (FPD)?
Choose only ONE best answer.
A
Thin Film Transistor (TFT) layer
B Cesium iodide (CsI) scintillator
C
Photodiode
D
Data line
E
None of these
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.
Which one of these is not true about the effect of collimation with image intensifiers?
Choose only ONE best answer.
A
decreased peak skin dose
B
increased contrast
C
decreased dose area product
D
increased contrast-to-noise ratio
E
decreased noise
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.
You briefly fluoro to identify the femoral head before a groin puncture, where an image intensifier (II) is 100 cm from the x-ray source. You then raise the II away so that it’s 150 cm away from the x-ray source in order to access. You fluoro again before removing the needle to verify position. By what factor does the entrance skin dose to the patient change with the II in the new position? Assume kVp is fixed.
Choose only ONE best answer.
A
2.25
B
0.25
C
1.5
D
0.67
E
1
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.
You electronically magnify to inspect the biliary anastomosis in an OLT patient using FPD fluoroscopy without binning. What happens to the resolution?
Choose only ONE best answer.
A
increases
B
unchanged
C decreases D variable E not enough information
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.
Your attending insists that you perform hand injection of contrast during fluoroscopic cone-beam CT cerebral angiography (DynaCT, XperCT). You’re standing on the right side of the patient. She says you’ll be fine since you’re wearing an apron. What is your absorbed dose when the C-arm is at Position 2 ( PA, source above table/detector underneath) relative to Position 1 (LAO, source below table/detector above-to left of patient) during the rotation?
Choose only ONE best answer.
A increased B decreased C unchanged D variable E not enough information
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.
You’re the chief medical officer of a new indirect FPD startup that wants to know your opinion about the clinical impact of moving to DEL version 2 (relatively smaller TFT), from the existing DEL version 1 (relatively larger TFT). Assume data, gate, and bias lines are perfectly electrically isolated, and that the total DEL size is unchanged.
Choose only ONE best answer.
A Increase contrast-to-noise ratio B Increase shot noise C Decrease scatter D Increase resolution E Decrease fill factor
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.
You’re trying to get a better view of a renal artery aneurysm for coiling, so you use electronic magnification. What does this do for patient dose (AK)?
Choose only ONE best answer.
A
unchanged
B
decreases
C
increases
D
depends on FPD versus II
E
insufficient information
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).
You’re trying to get a better view of a renal artery aneurysm for coiling, so you use geometric magnification. What does this do for KAP?
Choose only ONE best answer.
A increases B decreases C unchanged D depends on FPD versus II E insufficient information
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).
You’ve just aggressively collimated during fluoroscopy to get a better view of your microcatheter in the renal artery. What is the best choice for what happens to the cancer risk?
Choose only ONE best answer.
A
increases
B
decreases
C unchanged D variable E Depends on II versus FPD
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.
You’ve just moved from the patient’s side (1 meter from the x-ray tube) to the corner of the room (8 meters from the x-ray tube) during an auto-injected DSA run. By what factor has dose changed going from the patient’s side to the corner of the room?
Choose only ONE best answer.
A
64
B
1/64
C
8
D
1/49
E
1/2
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.
How does digital breast tomosynthesis compare with mammography in terms of screening performance?
Choose only ONE best answer.
A
decreased recall rate, unchanged cancer detection rate
B
decreased recall rate, increased cancer detection rate
C
increased recall rate, unchanged cancer detection rate
D
decreased recall rate, decreased cancer detection rate
E
unchanged recall rate, unchanged cancer detection rate
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
Lay summaries of the mammogram must be provided to the patient within how many days of the mammogram?
Choose only ONE best answer.
A 30 B 15 C 10 D 5 E 60
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.
MQSA requires that you perform how many mammograms under supervision of an interpreting physician for initial certification and continuing certification?
Choose only ONE best answer.
A
400 initial, 1000 per 2 years continuing
B 150 initial, 300 per 2 years continuing C 240 initial, 960 per 2 years continuing D 250 initial, 800 per 2 years continuing E 300 initial, 600 per 2 years continuing
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
On a diagnostic mammogram, you perform magnified views of calcifications using the air gap, but the magnified views look blurry. Which component is faulty?
Choose only ONE best answer.
A filament switching B grid C filter D detector E Computer assisted diagnosis (CAD)
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.
The ACR Mammography Accreditation Program for mammography facilities requires that the posterior nipple line (PNL) length on CC should be within how many centimeters of the PNL length on MLO?
Choose only ONE best answer.
A 1 cm B 2 cm C 3 cm D 4 cm E 5 cm
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.
The irregular mass in the left breast of this MLO view (lesion located superiorly on this MLO view) has a more inferior craniocaudal position on the ML view. What is the most appropriate clock face for this mass?
Choose only ONE best answer.
A
4 o’clock
B 2 o’clock C 6 o’clock D 8 o’clock E 10 o’clock
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.
What effective focal spot size is used for magnification view in digital mammography?
Choose only ONE best answer.
A 0.3 mm B 0.1 mm C 1.2 mm D 1 mm E variable 0.3 mm to 1.2 mm
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.
What is the MQSA-mandated limit for average glandular dose from a single view of one breast?
Choose only ONE best answer.
A not enough information B 3 mGy C 2 mGy D 1 mGy E 0.5 mGy
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.
What is the recommended period of the menstrual cycle during which to perform breast MRI?
Choose only ONE best answer.
A 0-6 days B 15-22 days C 23-30 days D 7-14 days E any time
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.
What is the standard dose of Gadolinium contrast for breast MRI?
Choose only ONE best answer.
A 0.1 mmol/kg B 1 mmol/kg C 10 mmol/kg D 100 mmol/kg E 0.5 mmol/kg
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.
Which agencies created and enforce MQSA?
Choose only ONE best answer.
A FDA created, states enforce B Congress created, FDA enforces C FDA created, FDA enforces D ABR created, ABR enforces E FDA created, ABR enforces
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
Which is NOT a benefit of magnification in mammography?
Choose only ONE best answer.
A Decreased patient dose B Decreased scatter C Increased CNR D Increased spatial resolution E Better characterization of calcifications
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.
Which is NOT a benefit of the compression in mammography?
Choose only ONE best answer.
A Increased magnification B Decreased scatter C Separation of overlapping tissue D Decreased patient dose E Increased spatial resolution
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
How much compression (force) should a mammo compression unit be capable of?
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.)
Which target/filter combination is best for screening mammography of this breast (mammogram shows a predominantly fatty breast)?
Choose only ONE best answer.
A Mo/Rh B Rh/Rh C W/Ag D Mo/Mo E None of these
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
Your patient is concerned about radiation from tomosynthesis. What is the dose from digital breast tomosynthesis + synthetic 2D images relative to 2-view screening digital mammography?
Choose only ONE best answer.
A about the same B 2x C 0.5x D 4x E 3x
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.
Your sales rep tells you that most digital mammography predominantly uses characteristic radiation. Which one of these target/filter combinations predominantly use braking radiation?
Choose only ONE best answer.
A W/Ag B Ag/W C Mo/Rh D Rh/Mo E Rh/Rh
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.
You’re reviewing sample mammography images from your practice group to send to the ACR. Which one of these constitutes successful visualization of the ACR phantom?
Choose only ONE best answer.
A 4 fibers, 2 speck groups, 2 masses B 5 fibers, 2 speck groups, 3 masses C 4 fibers, 3 speck groups, 3 masses D 6 fibers, 5 speck groups, 2 masses E 6 fibers 2 speck groups, 3 masses
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.
A new pediatric protocol recommends decreasing kVp to reduce dose for CT abdomen/pelvis. What adjustment do you expect to see the protocol require for mAs?
Choose only ONE best answer.
A decrease B unchanged C increase D variable E not enough information
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?)
A pregnant woman at third trimester reports to the ER with symptoms of appendicitis at a rural hospital with no access to MRI. The ER would like you to consent the patient for single-phase CT abd/pelvis with IV and oral contrast. What is the major risk?
Choose only ONE best answer.
A Stochastic risk to fetus B Deterministic risk to fetus C Stochastic risk to mother D Deterministic risk to mother E Fetal thyroid dysfunction
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.
How much estimated dose to fetus is there in a standard CT abd/pelvis exam on a pregnant patient?
CT Abd/Pelvis results in ~25 mGy to the fetus.
What is the dose-threshold and result of irradiating a fetus at following gestational ages:
A) < 14 days post-conception:
B) Later in gestation:
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%
A pregnant woman is short of breath and the ER wants you to consent the patient for CT Pulmonary Embolus protocol. What is the major risk to this patient?
Choose only ONE best answer.
A Fetal demise B Neonatal thyroid malfunction C Maternal breast dose D Radiation burns E No risk relative to general population
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.
Current-exposure time product (mAs) was increased from 100 mAs to 144 mAs. No other parameters have changed. By what factor does the image noise change?
Choose only ONE best answer.
A
5/6
B
1/sqrt(2)
C
1/2
D
1.2
E
2
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.
Slice thickness during post-acquisition image reformats was increased from 0.6mm to 1.2mm. No other parameters have changed. By what factor does the image noise change?
Choose only ONE best answer.
A 1/sqrt(2) B 0.50 C 2 D 4 E 1
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?”
The linear attenuation coefficient of a phantom is 2x that of water. What Hounsfield unit should it register on the CT scanner?
Choose only ONE best answer.
A 1000 B 0 C -1000 D -400 E 3
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.
The senior neuroradiologist prefers axial acquisition for CT Head because helical acquisition causes z-axis volume averaging. Which one of these is a downside of axial acquisition?
Choose only ONE best answer.
A
Increased pitch
B Stair-step artifact C Decreased SNR D Detector artifact E Beam hardening artifact
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.
The technologist adjusts the pitch from 1.5 to 0.5. All other scan parameters are held constant. By what factor does DLP change?
Choose only ONE best answer.
A 3 B 0.33 C 2 D 1 E not enough information
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).
The technologist increases pitch on a helical acquisition CT scan for an ER patient who is unable to hold still on the table. Which one of these factors is adversely affected?
Choose only ONE best answer.
A More than one of these factors B Contrast-to-noise ratio C Scan time D Resolution E Dose
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).
The technologist repeats a CT Chest over the same scan length, after changing the pitch from 0.5 to 1.5. Tube rotation speed is held constant. By what factor does scan time change?
Choose only ONE best answer.
A 3 B 1/2 C 2 D 1 E 1/3
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.
What is the optimal kVp setting for CT angiography?
Choose only ONE best answer.
A 100 kVp B 70 kVp C 120 kVp D 140 kVp E 200 kVp
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.
Which generation CT scanner is in common use?
Choose only ONE best answer.
A 2 B 3 C 1 D 4 E All are equally represented
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.
Which one of these CT scanners is capable of the greatest radiation dose efficiency?
Choose only ONE best answer.
A SDCT B 512 slice MDCT C 256 slice MDCT D 128 slice MDCT E 64 slice MDCT
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.
Which one of these is not an advantage of multidetector CT over single detector CT?
Choose only ONE best answer.
A better in-plane resolution B More efficient contrast media usage C capability for isotropic resolution D single-breath-hold acquisition E thinner slices
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.
Which one of these is the role of a bowtie filter?
Choose only ONE best answer.
A maintain uniform distribution of photon flux on detector B remove low energy photons C more than one of these answers D decrease noise E Increase spatial resolution
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.
You employ tube current modulation to save dose while preserving image quality. What explains the small ripples in the tube current, while rotating within a single plane/anatomic region of interest?
Choose only ONE best answer.
A angular modulation B longitudinal modulation C kVp modulation D noise E motion
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.
Your patient is obese. Which one of these is the major image quality issue will your protocol need to address?
Choose only ONE best answer.
A Streak artifact B Subject contrast C Image contrast D Shot noise E None of these
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.
Your patient requires 60 cm craniocaudal coverage in a CT Abdomen/Pelvis. The technologist sets pitch to 2, and the software computes a beam thickness of 100 mm. How many full rotations does the x-ray tube complete?
Choose only ONE best answer.
A 3 B 60 C 2 D 15 E 30
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.
Sound travels at what speed in soft tissue?
Choose only ONE best answer.
A
1540 cm/s
B
1540 m/s
C 1440 m/s D 1450 cm/s E 1430 cm/s
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.
The pulse repetition frequency (PRF) has decreased and the transducer frequency (TF) has increased, while all other parameters are unchanged. What is the most likely outcome?
Choose only ONE best answer.
A increased FOV B decreased FOV C unchanged FOV D increased frame rate E unable to determine
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.
Which of these quantities determines axial field of view (FOV)?
Choose only ONE best answer.
A More than one answer B PRP C PRF D SPL E TF
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.
Which of these quantities determines axial resolution?
Choose ALL answers that apply.
A Transducer frequency (TF)
B Spatial pulse length (SPL) C Pulse repetition period (PRP) D Pulse repetition frequency (PRF) E More than one of these (and if so, which?)
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.
Which factors matter for elevational resolution?
Choose ALL answers that apply.
A Crystal thickness B SPL C TF D Depth-gain compensation E PRF F More than one of these (and if so, which?)
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.
Which one of these choices is the best angle of insonation for spectral and color Doppler ultrasound?
Choose only ONE best answer.
A
85
B
60
C
80
D
30
E
90
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).
Which one of these interfaces is expected to reflect the largest percentage of the incident ultrasound beam energy?
1: Bone/Fluid
2: Fluid/Bone
3: Fluid/Soft-Tissue
Choose only ONE best answer.
A Interface 1 or 2 B Interface 1 C Interface 2 D Interface 3 E All equal
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.
Which one of these is not an advantage of power Doppler?
Choose only ONE best answer.
A Decreased sensitivity to flash artifact B Increased sensitivity to blood flow C Independent of Doppler angle D Not susceptible to aliasing E More than one of these is a valid answer
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
What is “special” about the “sector tranducer” type in ultrasound?
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.
Which one of these parameters does not enter directly into the calculation of velocity in Doppler ultrasound?
Choose only ONE best answer.
A Pulse repetition frequency (PRF) B Doppler frequency shift C speed of sound in patient (c) D Transducer Frequency (TF) E Angle of insonation (theta)
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.
Which one of these quantities determines frame rate?
Choose only ONE best answer.
A PRF B TF C SPL D beam width E wavelength
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
Which one of these quantities determines lateral resolution?
Choose ALL answers that apply.
A
TF
B Width of the ultrasound beam C SPL D Axial position relative to focal zone E More than one of the above (and if so, which?)
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.
Which one of these ultrasound modalities is typically used to demonstrate fetal heart rate in first trimester ultrasound?
Choose only ONE best answer.
A M-mode B B-mode C A-mode D Spectral Doppler E Color Doppler
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.
You examine a breast lesion that is 2 cm from the skin surface. The transducer frequency is 10 MHz, and the pulse repetition frequency (PRF) is 5 kHz (0.005 MHz). How much has the sound wave emitted by the probe attenuated by the time it reflects off the lesion and returns to the transducer?
Choose only ONE best answer.
A 0.005 dB B 20 dB C 0.0025 dB D 10 dB E 0.001 dB
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.
You observe aliasing on your spectral Doppler imaging of a stenotic renal transplant arterial pedicle. Which parameter do you increase to fix this problem?
Choose only ONE best answer.
A PRF B TF C PRP D beam width E FOV
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.
What is the Nyquist limit?
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.
You’re measuring the depth of a liver lesion from skin surface with ultrasound. The ultrasound machine assumes the speed of sound in the liver is 1540 m/s, but this liver is extremely fatty. What is the depth of the liver lesion on the ultrasound image relative to the actual depth?
Choose only ONE best answer.
A Image depth is shallower than actual B Image depth equals actual C Image depth is deeper than actual D Image depth variable relative to actual E Insufficient information
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.
You’re trying to define the relation of the diaphragm to a pleural effusion through the ribs in preparation for thoracentesis. What transducer type would give you the largest field of view without causing rib shadows?
Choose only ONE best answer.
A linear B curvilinear C none of these answers D sector E not enough information
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.
A new radiopharmaceutical has a physical half life of 6 hours, and a biological half life of 3 hours. What is the effective half life in hours?
Choose only ONE best answer.
A 2 B 3 C 6 D 9 E 1
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.
What is the decay constant (λ) of Tc-99m?
Choose only ONE best answer.
A 0.1155 hr-1 B 0.2150 hr-1 C 6 hr D 8 hr E 0.167 hr
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)
Which decay scheme occurs when Tc-99m turns into Tc-99?
Choose only ONE best answer.
A
Beta particle emission
B positron emission C Electron capture D Isomeric transition
E
None of these
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.)
For max resolution in nucs gamma camera imaging:
A) Gamma camera close or far?
B) thicker or thinner, longer or shorter septa? What is downside to maximizing resolution this way?
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.
What are the four flavors of parallel hole collimators in nucs?
For the more specialized types, which radiopharmaceutical are they specially used for?
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
Which nuclear imaging instrument uses Sodium Iodide (NaI) scintillator crystals?
Choose ALL answers that apply.
A G-M Pancake Probe B Dose Calibrator C Indirect FPD D Well Counter E Gamma Camera F More than one of above (and if so, which)
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).
Which one of these image quality problems gets worse with increasing radiotracer photon energy?
Choose only ONE best answer.
A Septal penetration B Scatter C Off photopeak D Center of Rotation Artifact E Non-uniformity
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)
Which one of these is not a QC test for the dose calibrator?
Choose only ONE best answer.
A
Constancy
B
Uniformity
C Linearity D Accuracy E Geometry
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.
Which one of these radiotracers is never used with SPECT-CT?
Choose only ONE best answer.
A F18-NaF bone scan B Tc-99m sestamibi cardiac imaging C I-131 thyroid cancer followup D In-111 WBC scan E Thallium-201 cardiac imaging
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.
Which one of these terms best describes the relationship between I-131 and I-123?
Choose only ONE best answer.
A
isomer
B
isotone
C
isotope
D
isobar
E
none of these
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.)
Which one of these terms best describes the relationship between Mo-99 and Tc-99m in the technetium-99m generator?
Choose only ONE best answer.
A isobar B isotone C isomer D isotope E none of these
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
You accidentally spill three vials of Tc-99m sulfur colloid on the ground, amounting to 90 mCi of activity. Which one of these activities is an appropriate step?
Choose only ONE best answer.
A Clean up the spill yourself B Leave the area immediately C Shield the source if possible D Notify the FDA E None of these
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.
You are experimenting with a generator for a new radioisotope. Which statement is accurate about half lives of the parent and daughter based on this graph?
Choose only ONE best answer.
A
t1/2 parent > t1/2 daughter
B
t1/2 parent»_space; t1/2 daughter
C
t1/2 parent < t1/2 daughter
D
t1/2 parent = t1/2 daughter
E
t1/2 parent «_space;t1/2 daughter
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»_space; 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»_space; 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.
Your patient takes insulin in the waiting room a few minutes before they receive your F18-FDG injection for their cancer follow-up study. Which answer best describes the resulting image?
Choose only ONE best answer.
A Intense uptake in the muscles B No effect on image C Intense uptake in the prostate D FDG completely excreted E Anaphylaxis due to interaction of insulin with F18-FDG
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.
How is SUV calculated? (what is the formula?)
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.
Your technologist asks you to protocol a V/Q scan using Xenon-133 ventilation and Tc-99m MAA perfusion studies. Which is the best protocol?
Choose only ONE best answer.
A Xe-133 followed by Tc-99m MAA B Tc-99m MAA followed by Xe-133 C Both Xe-133 and Tc-99m MAA simultaneously D Xe-133 and Tc-99m MAA are incompatible. Switch to Tc-99m DTPA for ventilation. E Xe-133 cannot be used for perfusion.
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?)
What is the photopeak of Tc-99m? Xe-133?
Memorize This
Xe-133 gamma photon energy is 81 keV
Tc-99m gamma photon energy is 140 keV
Your technologist is pregnant, and has already accumulated 2 mSv exposure to the fetus by the time she formally declares her pregnancy. How much additional exposure is permitted to the fetus for the remainder of the pregnancy?
Choose only ONE best answer.
A
50 mSv
B 3 mSv C 48 mSv D 5 mSv E 5 Sv
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.
You’re planning to discard some excess I-131 after letting it decay in storage. How many years are you required to keep a record of the disposal?
Choose only ONE best answer.
A 3 B 5 C 1 D 2 E No record needed
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.
How long must “nuclear-related records” be kept?
3-years (“the three-year rule” for nucs)
The three-year rule isn’t 100% correct, but it’s not worth memorizing the exceptions.
What criteria must be satisfied in order to be able to “dispose of radioactive material in the regular trash”
- physical half life less than 120 days
- residual activity indistinguishable from background radiation before disposal
- all labels removed
What is physical half-life of I-131
8 days
You’ve just prepared your batch of Tc-99m for the week from a Mo-99 / Tc-99m generator. What is the maximum NRC-accepted level of Aluminum breakthrough in the eluate?
Choose only ONE best answer.
A 10 μg Al / mL Tc-99m B 1.5 μg Al / mL Tc-99m C 15 μg Al / mL Tc-99m D 1.5 μg Al / L Tc-99m E 1.5 mg Al / mL Tc-99m
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.
What maximum level of contamination of M0-99 and Al in a prepared eluate of Tc-99m?
Maximum NRC-allowed contamination:
0.15 μCi Mo-99 / mCi Tc-99m eluate
10 μg Aluminum / mL of Tc-99m eluate
You’ve just prepared your batch of Tc-99m for the week from a Mo-99 / Tc-99m generator. What is the maximum NRC-accepted level of Mo-99 breakthrough in the eluate?
Choose only ONE best answer.
A 0.15 μCi Mo-99 / mCi Tc-99m B 1.5 μCi Mo-99 / mCi Tc-99m C 15 μCi Mo-99 / mCi Tc-99m D 1.5 mCi Mo-99 / mCi Tc-99m E 1.5 mCi Mo-99 / mCi Tc-99m
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.
How many half-lives does it take for Tc-99m to reach transient equilibrium with Mo-999 in the Mo-Tc generator?
4.
The MRCP sequence is used to characterize intra- and extra-hepatic biliary ductal structure. What type of weighting is used in MRCP?
Choose only ONE best answer.
A Heavily T2 weighted B Heavily T1 weighted C T2* weighted D Proton density weighted E Gadolinium-enhanced T1 weighted
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.
What is the appropriate dose of gadolinium for contrast-enhanced MRI of the breast?
Choose only ONE best answer.
A 0.1 mmol/kg B 0.1 mmol C 1 mmol/kg D 1 mmol E 100 μmol/mL
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.
What is the precession frequency of the hydrogen nuclear magnetic dipole moment vector in a 0.5T magnet when no gradients are applied?
Choose only ONE best answer.
A 42 MHz B 84 MHz C 21 MHz D 14 MHz E 40 MHz
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.
What is the Larmor frequency equation?
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)
What is the T1 relaxation time of fat relative to water?
Choose only ONE best answer.
A longer B shorter C equal D variable E not enough information
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.
Describe the “families” of cardiac MRI sequences.
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.
What is the MRI scan-time formula?
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.
Which of these physical interactions determines the T2 time constant?
Choose only ONE best answer.
A spin-lattice B lattice-lattice C all of these D spin-spin E none of these
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.
Which one of these fat suppression methods is compatible with post-contrast images and robust to magnetic field inhomogeneity?
Choose only ONE best answer.
A Dixon fat suppression B chemical fat suppression C Short tau inversion recovery (STIR) D None of these are robust E All of these are equally good
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.
Which one of these settings will give you T1 weighting on your spin-echo sequence?
Choose only ONE best answer.
A Long TR, long TE B Short TR, short TE C Short TR, long TE D Long TR, short TE E None of these
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.
You double the FOV along the frequency encoding axis. All other parameters are unchanged. By what factor does the number of frequency encoding steps change?
Choose only ONE best answer.
A 2 B 0.5 C 1 D sqrt(2) E 4
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.
Your practice group’s MRI director is constantly advocating for parallel imaging MRI at the partner meetings. Which one of these is not true about parallel imaging?
Choose only ONE best answer.
A
Increases SNR
B
Decreases scan time by the acceleration factor
C
Larger acceleration factors result in more noise and artifact
D
Parallel imaging artifact looks like residual aliasing from incomplete correction by the reconstruction process.
E
Parallel imaging is best for high-SNR applications like 3T MRI and contrast-enhanced MRA
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.
Your referring neurosurgeon is not happy with the T1 weighting on the SPGR (BRAVO) pre-surgical volumetric images. Which one of these parameter adjustments will improve the T1 weighting?
Choose only ONE best answer.
A Increase NEX B decrease flip angle C increase flip angle D Increase TE E Partial k-space acquisition
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.
Your technologist doubles the bandwidth. All other parameters are unchanged. By what factor does the signal-to-noise ratio get multiplied?
Choose only ONE best answer.
A 1/sqrt(2) B 2 C 1/4 D 4 E sqrt(2)
Doubling bandwidth (BW) changes SNR by a factor of 1/sqrt(2).
Memorize This
SNR ~ voxel volume * sqrt(NxNyNEX/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.
Your technologist increases the number of phase encoding steps. All other parameters are unchanged. What happens to spatial resolution?
Choose only ONE best answer.
A increases B decreases C unchanged D variable E not enough information
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.
What is phase-wrao artifact in MRI and how do you fix it?
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.
Your technologist notices Type-1 chemical-shift artifact and suggests that you increase receiver bandwidth (while keeping FOV constant) to decrease its severity. What is the drawback of this proposed adjustment?
Choose only ONE best answer.
A Phase wrap artifact B Increased noise C Aliasing D Decreased resolution E Increased zipper artifact
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.
You’ve chosen a 128 x 256 dimensional k-space matrix (# phase encoding steps x # frequency encoding steps). The phase encoding axis is anteroposterior (AP) and the frequency encoding axis is transverse (TR). What is the size of your MRI image expressed in (# of voxels AP x # voxels TR)?
Choose only ONE best answer.
A 128 x 128 B 128 x 256 C 256 x 256 D 256 x 128 E not enough information
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:
You’ve chosen to decrease voxel dimensions from 2mm x 2mm x 2mm to 1mm x 1mm x 1mm to get better spatial resolution on an MRI of a suspected glomus tympanicum. What is the effect of your manipulation on SNR?
Choose only ONE best answer.
A 8 B 1/4 C 1/8 D 4 E 1/2
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(NxNyNEX/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.
You’ve replaced your 1.5T magnet with a 3.0 T magnet, where the strength of the magnet refers to the static (B0) magnetic field. Which one of these will not result from increasing B0.
Choose only ONE best answer.
A Decreased susceptibility artifact B Increased SNR C Longer T1 relaxation time D increased chemical shift E Longer TR settings for the same desired contrast weighting
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).
How many deaths annually can be attributed to medical errors, per the 1998 Institute of Medicine (IOM) Report?
Choose only ONE best answer.
A
22,000 – 44,000
B
44,000 - 98,000
C 98,000 – 128,000 D 20,000 – 30,000 E 3,000 – 5,000
B.
This is a recall question, and a classic statistic in hospital management.
How would you classify an error caused by inadequate monitoring or follow-up of treatment?
Choose only ONE best answer.
A Preventive Error B Diagnostic Error C Treatment Error D Communication Error E Other Error
The correct answer is ‘A’
KPI stands for which of these?
Choose only ONE best answer.
A
Key predictive information
B Kaiser performance standard C Key predictive indicators D Key performance indicators
E
Kwality performance index
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.
PDSA stands for Plan, Do, Study, Act. What does the “Do” part mean?
Choose only ONE best answer.
A Collect data B Choose data to be obtained C Analyze data D Devise and implement plan for improvement based on analysis E None of these
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
The Six Sigma methodology aims to reduce the rate of defects to less than:
Choose only ONE best answer.
A 3.4 per million B 1 per million C 4 per million D 5 per million E 10 per million
A.
Which type of chart is used to illustrate the steps and decision points that make up a work process?
Choose only ONE best answer.
A
ROC curve
B Control chart C Fishbone diagram D Flow Chart E Pareto Chart
D.
Which type of chart, also called a Shewart chart, is used to analyze performance of a system as a function of time?
Choose only ONE best answer.
A Control chart B Fishbone diagram C ROC curve D Pareto Chart E Flow Chart
A
This type of diagram is used to diagram the steps in a process in order to identify decision or stress points, areas of vulnerability, and opportunities for improvements.
Choose only ONE best answer.
A Flow Chart B Pareto Chart C ROC curve D Fishbone diagram E Control chart
The correct answer is ‘A’
?,not Fishbone, d?
Which type of diagram is used to used to visually display a rank ordering of quality, safety or risk-factor issues by importance or impact?
Choose only ONE best answer.
A Pareto Chart B Flow Chart C ROC curve D Fishbone diagram E Control chart
A.
Which type of graph is used in Root Cause Analysis (RCA) to identify all contributing causes to an identified problem?
Choose only ONE best answer.
A Control chart B ROC curve C Pareto Chart D Flow Chart E Fishbone diagram
E.
Bonus
The fishbone diagram is also called an Ishikawa diagram.
This type of graph is used to analyze the performance of a diagnostic system.
Choose only ONE best answer.
A ROC curve B Pareto Chart C Flow Chart D Fishbone diagram E Control chart
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.
Which of these is not one of the five specific recommendations of the Choosing Wisely Campaign of the ACR?
Choose only ONE best answer.
A Don’t do imaging for lower back pain. B Don’t image uncomplicated headache. C Don’t image low pretest PE D Don’t CT for r/o appendicitis in children unless ultrasound has been considered. E Don’t recommend f/up imaging for clinically inconsequential adnexal cysts
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.
Which one of these campaigns targeted at the imaging community focuses on adult radiation protection?
Choose only ONE best answer.
A
Image Gently
B
Image Wisely
C Choosing Wisely D ALARA E ACR Appropriateness Criteria
The correct answer is ‘B’
Which one of these describes the phenomenon where adverse event reporting peaks about two years after a new agent or use?
Choose only ONE best answer.
A Weber effect B Hawthorne effect C Moire pattern D Lean process E Kanban
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.
Which one of these describes the phenomenon where being observed changes behavior?
Choose only ONE best answer.
A
Moire pattern
B Weber effect C Hawthorne effect D Proton density weighted E Kanban
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.
Which one of these is not a National Patient Safety Goal defined by the Joint Commission?
Choose only ONE best answer.
A
Keep radiation as low as reasonably achievable (ALARA)
B
Conduct a pre-procedure verification process
C
Mark the procedure site
D
Perform a time-out before the procedure
E
Label all medications, medication containers, and other solutions on and off the sterile field in perioperative and other procedural settings
A.
Which one of these is not one of the Institute of Medicine’s Six Quality Aims for healthcare?
Choose only ONE best answer.
A Effective B Safe C Timely D Educational E Efficient
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.
Which one of these terms best describes a continuous process of making things better, including retrospective and prospective review?
Choose only ONE best answer.
A Quality Improvement B Quality Assurance C Quality Control D Effectiveness Research E Six Sigma
A.
Which one of these terms best relates to the term “value” in healthcare?
Choose only ONE best answer.
A Quality / Cost B Quality C Cost D Revenue E Profit
A.
Value in healthcare is closely related to efficiency:
efficiency = results achieved / resources expended
Half-life of Tc-99m?
6hr
Half-life of I-123?
13hr
Half-life of I-131
8days
Half-life of Indium, Gallium, Thallium?
~3days (same for all the “iums”)
Half-life of Xe-133?
5 days
Half-life of F-18?
110min
Half-lifes of:
- Tc-99m
- I-123
- I-131
- Indium, Gallium, Thallium
- Xe-133
- F-18
- 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
what is main photon energy of Tc-99m?
140 keV
what is main photon energy of I-123
159 KeV
what is main photon energy of I-131
365 KeV (“like the days of the year”)
what is main photon energy of Ga-67
100, 200, 300, 400 KeV (roughly)
what is main photon energy of all PET agents (including F-18, Rubidium…etc)? (i.e. the gamma photon resulting from positron emission and annihilation)
511 KeV
what is main photon energy of Thallium?
80 KeV (this is really x-ray level, rather than true gamma)
What counts as “short TR”, “long TR”, “short TE”, “long TE” on Spin Echo MRI sequences?
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.)
What do these Terms/Sequences refer to in MRI?
- HASTE
- BLADE/PROPELLOR
- SPIRAL
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).
What is the FDA-limit on SAR (Specific Absorption Rate) in MRI to whole-body?
4 W/Kg for 15min (which ensures less the 1degree C rise in core temperature).
Gadolineum-based contrast agents, what are their generic names and structures?
- Magnevist
- Mutlihance
- Omniscan
- Gadavist
- 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”)
Which radionuclides decay by “electron capture”? What their half-lives?
-Mnemonic “GIIT”
Gallium, (I-123), Indium, Thallium
-All have approx same half-lives: ~3days.
Thallium vs Gallium:
Mechanisms?
Photopeaks?
This vs That conditions?
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).
Relationship between photon wavelength and its energy (i.e. what does doubling the frequency do to the energy):
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…
In an ROC curve, what is the y-axis and x-axis?
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)
What is the Standard Deviation in a Poisson distribution?….
In poisson distribution:
SD= sq(of the mean). => if average is 100, then the SD =10.
Typical imaging depth for a 3mHz US probe is…..
Typical depth for a 10mHz probe is….
Typical imaging depth for a 3mHz US probe is….. 20cm
Typical depth for a 10mHz probe is…. 6cm
Half life of Xe-133…
…5 days
What is the Larmor frequency of protons in a 3T magnet?
Larmor frequency for H = 42MHz/T => in 3T 42MHz*3 = 126 MHz.
(**Larmor frequency of H at 1T = 42MHz)
What is the maximum SAR (Specific Absorbtion Rate) for:
- head imaging over 10 minute?
- whole-body imaging over 15 minutes?
- Head: 3 W/kg
- Body: 4 W/kg
For MQSA, how frequently must performing compression quality control testing be done in mammo?
-Semi-annually (every 6 months)
For evaluating osteomyelitis on nuclear imaging study, what’s better tagged-WBC (Indium-111-WBC) or Ga-67 scan?
Vs
Which of these same agents is better for assessing GI/abdominal infection/inflammation?
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!).
A)
what are the typical time ranges for “short” vs “long” TR and TE, respectively, on Spin Echo?
B)
What are the typical TR and TE times for:
1) T1-weighted ?
2) T2-weighted?
3) Proton Density Imaging?
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)
The normal time to peak for cortical uptake of Tc-99m-DTPA and Tc-99m-MAG3 is less than__________ minutes.
less the 5 minutes.
Cine cardiac MRI imaging is most reliant on what sequence type?
- Spine Echo
- Gradient Echo
- Inversion Recovery
Gradient Echo (most commonly SSFP), which allows for fast image aquisition with high spatial resolution.
In which clinical situations is copper filtration of x-ray tube output most likely to be used?
(Select all the apply)
a. Pediatric digital radiography
b. Barium flouroscopy
c. Adult chest radiography
d. Interventional radiology
e. CT imaging
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.
After delivery of a package of radioacive material to a nuc med facility, within how much time from the reciept of delvery must the recieving facility inpect the package (i.e. wipe test, one meter test)?
3 hours
How much power(kW) is generally used to obtain an image with plan film radiography? with mammo?
- Plain film: 100kW
- Mammo: 3kW
In an electrical circuit, what the definition of power (what is the relationship of Power, current and voltage)?
Power = Current * Voltage (P = IV)
Mnemonic = “A peripheral IV (P=IV) give an electrical circuit power”
What is the “matching layer” in a US-probe and its purpose?
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.
Pulsation artifact is most likely to be seen in which MRI sequence type?
Fast Spin Echo
Complete recovery of longitudinal magnetization is achieved after a time period equal to... A. T1 B. T2 C. 2 * T1 D. 2 * T2 E. 4 * T1 F. 4 * T2
E. 4* T1
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 T14 it has NEARLY all recovered (at T12, just 87%; at T1*3 95%).
Remember, T2 is independent (not related) to longitudinal magnetization relaxation time…
What is the typical timing to acquire delayed post-contract MR images for detecting delayed subendocardial enhancement in acute MI?
10min post Gd administration
Which hepatic adenoma subtype is unique in that it is only seen in MALES who TAKE HORMONES?
Additionally, which hepatic adenoma subtype usually does NOT CONTAIN FAT?
Additionally, which sub-type has a somewhat higher risk for malignant transformation?
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!”)
Which type of magnification results in increased patient dose?
A. Geometric mag
B. Electronic mag.
C. Both the same.
D. Both, but one more than the other and if so, which?
E. Neither
D. Both! with dose increase from electronic mag > geometric mag.
In MR imaging, if the receiver bandwidth is increased, what is the effect on TE? Increase, Decrease, or Stay Same? And, why or why not?
TE decreases! Bc a broader bandwidth has a higher amplitude readout gradient which allows for faster sampling of the MR signal.
Between which two muscles does a Zenker’s diverticulum posteriorly project from the cervical esophagus?
Btwn the THYRO-PHARYNGEAL (above) and the CRICO-PHARYNGEAL (below), which together make up the inferior constrictor of the pharynx.
What is Gradenigo Syndrome?
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
How of often is Slipped Capital Femoral Epiphysis bilateral?
And, What’s the name of the line used to assess for SCFE on x-ray?
And, how is SCFE treated?
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.
What are the typical MRI characteristics of “myxoid” tissue?
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)
What is the typical Bucky factor for mammo grids?
2-3
What is threshold for leakage radiation from x-ray tube housing?
1.0 mGy per hour measured at 1m from the source (mnemonic: “1 and 1” leak allowed)
What is the conversion between Gray and Rad units?
1Gy = 100 rad
What is typical dose for Ga-67 imaging, for infection vs tumor imaging?
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)
What is the standard dose for MDP bone scan?
20-25mCi
Why must MRI scanners have a gas level monitor?
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.
What is the approximate entrance air kerma for an abdominal radiograph?
A. 0.03 mGy
B. 0.3 mGy
C. 3 mGy
D. 30 mGy
C. 3 mGy
Before releasing a patient who recieved therapeutic I-131, a facility must ensure that the potential dose to another indivual who is maximally expose to the patient (i.e. a household member) will not exceed __________ rem (____ mSv) in a year.
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!
X-ray tube windows in mammo are made of _____________
Beryllium.
If US beam frequency increases, how will the near zone (fresnel zone) be affected? Incease, decrease, or not change?
Increase. Bc Near Field inversely proportional to wavelength (=> smaller wavelength -> bigger near field; increasing the frequency -> decreasing the wavelength => increasing the near zone)
Assuming all other factors kept equal, how will doubling the element size affect size of the near field (N) in US imaging?
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)
NRC requires “sealed sources” be leak-tested how frequently?
Semi-annually (q Six Months)…..( mnemonic: S in Sealed for Semi-annually/Six-months)
What does decreasing “x-ray ripple” do?
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)
How is mottle related to the the number of photons incident on a detector?
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).
What is the formaula for calculating “accuracy” of a diagnostic test?
Accuracy = (TP + TN)/all-patients
This v That.
in Sulfur Colloid vs In111-WBC scan, what is the difference in normal liver vs spleen uptake?
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).
What are the typical enterance air kermas for typical radiography, vs typical flouro, vs typical IR/Cardiac interventions?
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”
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”
what is the relationship between enterance air kerma and entrance skin dose?
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).
How long must a mammo facility keep a patients mammos/records on file?
At least 5 years.
Lung CA ‘T’ grading, T3 vs T4….
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.
Sensitivity of nucs tagged RBC scan vs angio, for detecting occult GI-bleed.
-RBC scan: 0.1 mL/min……..vs…………angio 1 mL/min
An imaging system is found to have unacceptable poor spatial resolution, and modulation transfer functions for the focal spot , motion, and detector blur are found to be 0.7, 0.8, and 0.9 respectively…. which of the following best describes the imaging systems’ OVERALL modulation transfer function?
- 5
- 6
- 7
- 8
- 9
0.5
explanationL the system’s OVERALL modulation transfer function is THE PRODUCT of all the indivual modulation transfer functions (in this case =0.70.80.9 = 0.5)
What is the FDA-mandated maximum pt skin entrance exposure rate for STANDARD flouroscopy?
88mGy/min
What is the most common post-renal transplant fluid collection to cause transplant hydronephrosis?
Lymphocele
(typically presents 1-2months post transplant)
…FYI can also result in ipsilateral lower extremity edema from femoral vein compression.
What is threshold for normal RI post-renal transplant?
Want RI’s <0.7. If >0.7, suggests a problem…
What is Malakoplakia? How is it treated?
vs.
Leukoplacia?
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);
