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
