RAPHEX IX

Question 1

mean free path

Answer 1

1/u

Question 2

convert Sv to rem

Answer 2

1 Sv = 100 rem

Question 3

The inherent filtration in a diagnostic x-ray tube results from _____ and is usually specified in terms of _____. A. metal sheets interposed between the x-ray tube and the patient; mm Cu equivalent. B. metal sheets interposed between the x-ray tube and the patient; mm Al equivalent. C. metal sheets interposed between the x-ray tube and the patient; mm W equivalent. D. components of the x-ray tube itself; mm Cu equivalent. E. components of the x-ray tube itself; mm Al equivalent.

Answer 3

e

Question 4

Typical x-ray transformers for general radiography have a power rating of about _____

Answer 4

100 kW

Question 5

Which one is CT?

Answer 5

C.
Graph C is a single x-ray energy spectrum with a maximum x-ray energy of 120 keV, which would be produced when the x-ray tube voltage is set to 120 kV. An x-ray tube voltage of 120 kV is typical for CT scanning.
-no ripple
-low energy photons filtered out

Question 6

which graph is for P-A chest radiograph?

Answer 6

A
The energy spectra shown in Graph A would be for a tube voltage setting of 100 kV and tube current settings of 300 and 600 mA. Graph A is a better choice than Graph D because posterior-anterior (P-A) chest radiographs are typically obtained with a higher tube voltage (kV) and current (mA) to yield a greater x-ray intensity and, consequently, a shorter exposure time. The spectrum in Graph D corresponds to a relatively low tube voltage of only 80 kV

Question 7

The intensity of an x-ray beam at the patient entrance surface would be increased most by _____. A. increasing tube current from 100 mA to 125 mA B. increasing tube voltage from 70 kVp to 85 kVp C. increasing exposure time from 0.6 second to 1.0 second D. decreasing the distance from the target to the patient surface from 120 cm to 60 cm E. decreasing filtration by 1 HV

Answer 7

D. decreasing the distance from the target to the patient surface from 120 cm to 60 cm
Due to the inverse square law, exposure will increase by a factor of 4 (22) when the target-to-patient distance is reduced by a factor of 2. X-ray intensity and exposure increases linearly with tube current and, therefore, the tube current increase in answer A will only increase patient exposure by a factor of 125 mA / 100 mA = 1.25. X-ray intensity increases as the square of the tube voltage and, therefore, the voltage increase in answer B will increase exposure by a factor of about (85/70)2 = 1.47. Decreasing filtration by one half-value layer (HVL) will increase x-ray intensity and exposure by a factor of 2

Question 8

The image below illustrates a feature now commonly utilized in CT scanning. ANSWER
This feature is known as _____.

Answer 8

tube-current modulation
To compensate for different degrees of attenuation by different anatomic regions of the body while maintaining patient radiation doses as low as practical, the CT scanner’s x-ray tube current is varied. It increases in regions of greater attenuation (such as the shoulders) and decreases where attenuation is not as great (as in the lungs), providing dose reduction while maintaining image quality (Z-modulation). It also may decrease in anterior-posterior (AP) projections and increase in lateral projections (angular modulation), as seen by the ripples in the tube current (mA) profile.

Question 9

Mammography units that can perform digital breast tomosynthesis are most likely to have a tube target composed of _____.

Answer 9

tungsten

Question 10

For tomosynthesis units, _____ is (are) commonly used as the x-ray tube filter. A. silver (Ag) B. aluminum (Al) C. rhodium (Rh) D. All of the above are true. E. None of the above is true.

Answer 10

All of the above are true.
As indicated in the table above, the x-ray units for digital breast tomosynthesis used for DBT (as well as full-field digital mammography) utilize an aluminum (Al), rhodium (Rh), or silver filter.

Question 11

An image intensifier with a 16-inch-diameter input screen, a 1-inch-diameter output screen, and a flux gain of 60 will have an overall brightness gain of about _____.

Answer 11

15,300
The brightness gain of an image intensifier is the product of the flux gain (60) and the minification gain (162), (16 in / 1 in)2  60 = 256  60 = 15,360  15,300. Note that the minification gain equals the square of the ratio of the linear dimensions of the input and output screens (in this case, 16 inches and 1 inch, respectively).

Question 12

Compared to that of a digital mammography detector, the linear dimension of a detector element (del) of a flat-panel fluoroscopy detector is _____. A. about half the size B. about 25% smaller C. about the same size D. about 25% larger E. about twice as large

Answer 12

E. about twice as large
The detector element (del) size (linear dimension) of a digital mammography detector is about 0.05 to 0.1 mm or smaller, while that of a flat-panel fluoroscopy detector is about 0.14 to 0.2 mm in size (i.e., about twice as large)

Question 13

If a CT scan is performed with 64 slices per rotation, a slice width of 0.625 mm, and a table increment of 6 cm per rotation, the pitch is _____.

Answer 13

1.5
For a CT scan, the pitch is the table increment divided by the product of the slice width and the number of slices. In this case, the pitch = 6 cm / (0.625 mm  64) = 60 mm / 40 mm = 1.5

Question 14

In the American College of Radiology (ACR) CT Accreditation Program, which of the following types of examinations is allowed to have the lowest contrast-to-noise ratio (CNR) and receive a passing score? A. adult head B. adult abdomen C. pediatric head D. pediatric abdomen

Answer 14

pediatric abdomen
Images of the American College of Radiology (ACR) CT accreditation phantom acquired with the pediatric abdomen technique must have a contrast-to-noise ratio (CNR) of at least 0.4. Images of this phantom acquired with the pediatric head technique must have a CNR of at least 0.7, and those acquired with adult head and abdomen settings must have a CNR of at least 1.0. (Reference: American College of Radiology CT Accreditation Program Requirements, American College of Radiology, Reston, VA, 201

Question 15

The U.S. Food & Drug Administration (FDA) regulatory limit for the entrance air kerma (Kair) rate in fluoroscopy is _____. A. 0.1 R/min B. 0.5 R/min C. D.
1 R/min 5 R/min
E. 10 R/min

Answer 15

10 R/min

Question 16

The focal-spot size of the x-ray tube in a fluoroscopy system

Answer 16

0.6 mm

Question 17

Compared to conventional (i.e., non-pulsed) fluoroscopy, pulsed fluoroscopy at a frame rate of 30 frames per second _____ motion blurring and _____ patient radiation dose. A. increases; reduces B. reduces; does not change C. does not change; reduces D. reduces; reduces

Answer 17

reduces; does not change
In pulsed fluoroscopy at 30 frames per second (fps), the x-ray tube current (i.e., mA) is only on for a short time, but it is increased to maintain the same exposure (i.e., mAs) per frame, as in continuous fluoroscopy. This reduces motion blurring (since the x-ray exposure for each frame is of shorter duration) but does not reduce or otherwise change the patient radiation exposure (since mAs per frame is uncha

Question 18

The U.S. Food & Drug Administration requires that all fluoroscopes sold in the United States after May 2006 have which of the following indicators related to radiation dose from a fluoroscopic procedure? A. effective dose B. peak skin dose C. focal spot-to-skin distance D. total air kerma at the reference point

Answer 18

total air kerma at the reference point

Question 19

The breast radiation dose in breast tomosynthesis is _____. A. about one-tenth of that from conventional mammography B. about one-third of that from conventional mammography C. about the same as that from conventional mammography D. three times that from conventional mammography E. 10 times that from conventional mammography

Answer 19

about the same as that from conventional mammography

Question 20

In graph below of multiple x-ray energy spectra, what x-ray tube parameter was changed in transitioning from the top spectrum to the bottom spectrum

A. tube current (mA) B. tube voltage (kV) C. filtration material D. target material E. acquisition time

Answer 20

B. tube voltage (kV)

Question 21

In fluoroscopy, the size of the field of view (FOV) can be changed to improve spatial resolution. In changing the FOV from the 30-cm mode to the 23-cm magnification mode, the exposure rate will _____. A. decrease by a factor equal to the square of the FOV ratios, (23 cm / 30 cm)2 B. increase by a factor equal to the square of the FOV ratios, (30 cm / 23 cm)2 C. be unchanged D. be reduced by approximately one-half

Answer 21

increase by a factor equal to the square of the FOV ratios, (30 cm / 23 cm)2

Question 22

A technique that can be used to help limit patient dose in fluoroscopy is _____. A. increasing the frame rate B. decreasing the field of view C. decreasing the frame rate D. opening the collimation E. using cine mode instead of fluoroscopy

Answer 22

C. decreasing the frame rate
Pulsed fluoroscopy may help reduce patient radiation dose by reducing the duration of the “beam-on” time. By lowering the frame rate, the dose to the patient dose is typically decreased.

Question 23

Pinhole collimators have long been used in nuclear medicine to produce magnified images of small structures, such as the thyroid. A patient is positioned near the pinhole aperture such that the distance from the patient’s neck to the pinhole aperture is less than the collimator cone length, yielding a magnified image. If the patient is moved farther away from the collimator, how do the detection sensitivity and spatial resolution change? A. The sensitivity increases and the resolution decreases. B. The sensitivity decreases the resolution increases. C. The sensitivity decreases and the resolution decreases. D. The sensitivity increases and the resolution increases. E. The sensitivity remains the same and the resolution decreases.

Answer 23

The sensitivity decreases and the resolution decreases.
As the distance of the object from the pinhole aperture is increased, the number of photons passing through the pinhole is reduced, reducing the sensitivity. Resolution decreases as the source-to-collimator distance is increased.

remember clara that patient is the source
M= SID/SOD

Question 24

As one proceeds from low-energy to medium-energy to high-energy collimation for gamma camera imaging, _____. A. the collimator septa are made longer and thicker B. the collimator apertures are made wider C. the system sensitivity decreases D. the system spatial resolution worsens E. All of the above are tr

Answer 24

E. All of the above are true.
As one images radionuclides emitting x-rays and gamma rays with higher and higher energy, the mean path length through the septa must be increased to reduce the septal penetration (and resulting mispositioning) of the increasingly energetic radiations to an acceptably low level. This is accomplished in practice by making the collimator septa both longer and thicker. To compensate, at least in part, for the reduced number of radiations reaching the crystal, the apertures are widened. Fewer photons nevertheless reach the crystal, resulting in lower system sensitivity. At the same time, a spatial resolution is degraded because of the wider apertures.

Question 25

The most likely cause of the gamma camera uniformity (or “flood”) image shown here is _____. A. a crack in the crystal B. an uncoupled PMT C. one or more software corrections being outdated, corrupted, or turned off
D. radioactive contamination E. incorrect collimation

Answer 25

C. one or more software corrections being outdated, corrupted, or turned off

A pattern of nonuniformity artifacts over the entire crystal (such as the PMT artifacts in the image shown) is very likely due to one or more software corrections being outdated, corrupted, or turned off.

Question 26

A new technologist attempts to count a sample containing 200 mCi of carbon-14, a pure beta-particle emitter, in a sodium iodide (NaI(Tl)) scintillation well counter. What will be the result? A. The observed net count rate will be accurate. B. The observed net count rate will overestimate the true count rate. C. The observed net count rate will underestimate the true count rate. D. The observed net count rate will be zero.

Answer 26

The observed net count rate will be zero.
Carbon-14 emits only “soft” (i.e., low-energy) beta-rays that are so non-penetrating as to be completely absorbed in much less than a one-mm thickness of any solid or liquid. Therefore, carbon-14 beta rays would likely be absorbed by any sample being counted, and they would certainly be completely absorbed by the housing of the crystal in a scintillation well counter. For this reason, pure soft beta-ray emitters such as hydrogen-3 and carbon-14 can only be counted by the destructive technique of liquid scintillation counting, in which the radioactive sample is completely solubilized and mixed in solution with an organic scintillator molecule. On the other hand, radionuclides such as phosphorus-32, strontium-89, and yttrium-90—which emit much harder (i.e., higher-energy, >~1 MeV) beta-rays—may be counted in a scintillation well counter by counting the bremsstrahlung (brake x-rays) that these harder beta-rays produce as they pass through matter

Question 27

In preparation for radioembolic therapy of liver cancer by infusion of yttrium-90 (90Y)-labeled microspheres via the hepatic artery, a pre-therapy imaging study was performed using technetium-99m (99mTc)-macroaggregated albumen (MAA) to estimate lung shunting. The images of and liver and lung region-of-interest (ROI) counts were as follow

liver has 485,912 counts, lung has 5,222
what is lung shunting?

Answer 27

1%
The lung shunt is calculated as the ratio of the 99mTc-MAA counts in the lung region of interest (ROI) divided by the sum of the counts in both the lung and liver ROIs multiplied by 100%. In this case, [5,222 / (5,222 + 485,912)]  100% = 1%.

Question 28

In lutetium-177 (177Lu) dotatate (LUTATHERATM) treatment of neuroendocrine tumors, the beta-rays from 177Lu in a lesion account for _____ percent of the radiation dose to the lesion. A. 0% B. <10% C. 25–50% D. >75%

Answer 28

D. >75%
Lutetium-177 (177Lu) beta particles, like beta particles and other particulate radiation generally, have a very short range (about 1 mm or less) in tissue. Therefore, they deposit all of their energy and deliver all of their radiation dose within 1 mm or less than their point of emission. Thus, -rays emitted by 177Lu within a neuroendocrine lesion will deposit virtually all of their energy and deliver virtually all of their dose to the lesion itself. The gamma-rays emitted by 177Lu, like other high-energy x-rays and gamma-rays, travel many centimeters in tissue, and they will deposit much of their energy and deliver much of their dose outside the lesion, contributing very little to the lesion dose.

Question 29

The current regulation for radionuclide therapy patients states that such patients can be released from the hospital when the projected dose to individuals around the patient is less than or equal to _____. A. 100 mrem B. 500 mrem C. D.
1 rem 5 rem
E. 10 rem

Answer 29

B. 500 mrem

Question 30

Which of the following are theranostic radionuclides used in nuclear medicine? A. fluorine-18 (18F) B. yttrium-90 (90Y) C. iodine-123 (123I) D. lutetium-177 (177Lu) E. None of the above is true.

Answer 30

D. lutetium-177 (177Lu)
As its name implies, a theranostic radionuclide can be used for diagnostic imaging and for internal radiation therapy. Such a radionuclide must, therefore, emit one or more “image-able”
gamma-rays and beta-particles or other particulate radiations in reasonably high abundance. 177Lu satisfy both of these requirements and is used for imaging and therapy. The other isotopes listed are used exclusively for PET (18F), gamma camera imaging (123I), or therapy (90Y)

Question 31

The gamma camera performance parameter known as multi-peak spatial registration can be measured using the _____ isotope. A. fluorine-18 (18F) B. technetium-99m (99mTc) C. iodine-123 (123I) D. iodine-131 (131I) E. indium-111 (111In)

Answer 31

Multi-peak (or multi-window) spatial registration is a gamma camera performance parameter that applies only to isotopes that emit multiple “imageable” gamma-rays, such as indium-111 (111In), which emits a 173-keV gamma-ray (90% abundance) and 247-keV gamma-ray (94%). All of the other isotopes listed are either a positron emitter used in PET (18F) or emit a single imageable gamma-ray (99mTc, 123I, 131I).

Question 32

As indicated, the activity concentration in this lesion is 8 kBq/cm3. The patient weighs 74 kg and was injected with 148 MBq of 124I-NaI. For the iodine-124 (124I)-sodium iodide (NaI) PET scan of this thyroid cancer patient, what is the standard uptake value (SUV) for the lung lesion (arrow)? A. B. C. D.
2 4 6 8
E. 10

Answer 32

B. 4
PET images are routinely parameterized in terms of the standard uptake value (or SUV), defined as the activity concentration in tissue (kBq/g) divided by the activity (kBq) administered per unit body mass (g). For a lesion with an activity concentration of 8.0 kBq/cm3 in a patient with a total body mass of 74 kg or 74,000 g who received an administered activity of 148 MBq or 148,000 kBq, the SUV is 8 kBq/g / (148,000 kBq / 74,000 g) = 8 kBq/g / 2 kBq/g = 4, assuming a tissue mass density of 1 g/cm3.

Question 33

Modern bone densitometry units _____. A. utilize pencil-beam or fan-beam x-rays B. employ dual-energy x-rays C. are capable of body composition measurements D. All of the above are true

Answer 33

D. All of the above are true.
Current-day bone densitometry units are capable of body composition measurements utilizing two distinct x-ray energies. The x-ray beam geometries are either a scanning pencil-beam or scanning fan-beam.

Question 34

Which of the following tissue-tissue interfaces will have the lowest ultrasound reflection coefficient? A. muscle-bone B. muscle-gas C. fat-muscle D. liver-kidney E. bone-lung

Answer 34

liver-kidne

Question 35

Compared to conventional Doppler imaging, Power Doppler imaging is more sensitive to low flow and is less dependent on Doppler angle, but it has the disadvantage of _____. A. producing more aliasing B. providing no flow direction information C. being incapable of detecting flow in abdominal organs D. yielding excessive phase information

Answer 35

B. providing no flow direction information
Power Doppler imaging in ultrasound provides information only on the presence of flow, but not the flow velocity or the flow direction

Question 36

The ultrasound artifact (arrows) shown on the image below is due to _____

Answer 36

Question 37

The display on clinical ultrasound units shows values of both the thermal index (TI) and a mechanical index (MI). Which of the following is the lowest value of TI at which adjustments to the examination (i.e., reduce the exposure time; reduce the power level) should be considered when that value is exceeded? A. 0.5 B. 1.0 C. 2.5 D. 3.0 E. 5.7

Answer 37

1.0
A value of 1 is considered a “safe value” of the thermal index (TIT) and the mechanical index (MIT) in ultrasound. If either of these respective indices are exceeded, adjustment of the examination should be considered

Question 38

A boundary between surfaces decreases the relative intensity of the ultrasound beam by 10 dB. What is the ratio of the transmitted to the incident beam intensities at the boundary? A. 0.1 B. 0.5 C. 1 D. 5

Answer 38

0.1
relative intensity (in dB) = 10log(I2/ I1).

Question 39

The speed of sound in ultrasound imaging is affected by the tissue property of _____. A. compressibility B. impedance C. transmittance D. attenuation

Answer 39

A. compressibility
The speed of sound in a particular medium is affected by the compressibility of the medium. As the compressibility increases, the speed of sound decreases. For example, the average speed of sound in air, which is highly compressible, is 330 m/s; the speed in soft tissue, which is far less compressible, is 1,540 m/s

Question 40

The heel effect can be reduced by _____. A. decreasing the anode angle B. increasing the tube voltage (kV) C. decreasing the tube current (mA) D. increasing the source-to-image distance (SID)

Answer 40

D. increasing the source-to-image distance (SID)
The heel effect describes the decrease in x-ray intensity toward the anode of the x-ray tube due to the increase in self-filtration of the x-ray beam by the anode itself. It can be reduced by increasing the source-to-image distance (SID). The decrease in beam intensity with distance (due to the inverse square law) will minimize the variation in intensity due to heel effect.

Question 41

The figure below shows an axial T2-weighted fat-suppressed image acquired during an MRI-guided liver biopsy. An arrow points to the titanium biopsy needle.
The bright signal adjacent to the biopsy needle is due to _____. A. accumulation of blood B. motion artifact C. compromised fat suppression D. coaxial biopsy needle design E. All of the above are tru

Answer 41

C. compromised fat suppression
The image shown is a T2-weighted fat-suppressed image. Fat suppression is performed in this instance with a spectrally selective radiofrequency (RF) pulse. The RF pulse is designed to suppress the lipids with a predefined range of resonant frequencies. However, in the presence of the B0 magnetic field inhomogeneity, the resonant frequency of lipids is shifted, thus making these lipids unaffected by the fat suppression RF pulse. In this image, it can be observed that subcutaneous fat remains unsuppressed in several regions due to local magnetic susceptibility differences between air and tissue. The difference in susceptibility between tissue and needle material causes local inhomogeneities of the B0 field which, in turn, leads to failed fat suppression along the biopsy needle in fa

Question 42

MRI scans can cause unwanted heating of the patient, which is measured by the specific absorption rate (SAR). The source of this heating is _____.

Answer 42

RF pulses

Question 43

Diffusion-weighted MRI is sensitive to random motion of water molecules in tissue. An appropriate application of this imaging method is _____. A. functional activity of the brain B. ischemic stroke C. intracranial calcifications D. cerebral microbleeds E. blood brain barrier disruption

Answer 43

ischemic stroke
Diffusion-weighted imaging (DWI) depicts contrast between tissues with different water mobilities, that is, the degree of water diffusion restriction. Stroke, accompanied by cerebral ischemia, leads to disruption of net water transfer from extracellular to the intracellular compartments. As a result, intracellular water accumulation leads to a reduced extracellular water volume, reducing diffusion of water in the extracellular matrix. Reduction in water diffusion or increased restriction of water diffusion is, therefore, readily detected with DWI.

Question 44

The device pictured to the right is the _____.
A. ACR MRI phantom B. NEMA SPECT phantom C. NCRP CT phantom D. IAEA PET phantom E. None of the above is true

Answer 44

ACR MRI

Question 45

In the Dixon MRI technique, _____. A. an inversion recovery pulse with a short inversion time (TI) pulse is utilized to suppress fat B. a short-duration RF pulse with center frequency set to spectrally stimulate fat protons is applied before the excitation pulse
C. water protons are selectively excited, leaving fat protons unchanged, with the use of a binomial pulse
D. multiple echoes are acquired with different echo times (TEs) to produce four output images, namely, water, fat, in-phase, and out-of-phase images

Answer 45

D. multiple echoes are acquired with different echo times (TEs) to produce four output images, namely, water, fat, in-phase, and out-of-phase images
Multiple echoes are acquired with different echo times (TEs) to produce four output images, namely, water, fat, in-phase images, and out-of-phase images

Question 46

The linearity quality control test for a dose calibrator checks that the response (i.e., activity reading) is linear as a function of sample _____ and should be performed _____. A. activity; annually B. volume; annually C. activity; quarterly D. volume; quarterly E. None of the above is true.

Answer 46

C. activity; quarterly

Question 47

For quality control tests in nuclear medicine, a long-lived “surrogate” radionuclide is often used in place of a “clinical” nuclide. Which of the following is(are) such a surrogate/clinical nuclide pair? A. fluorine-18 (18F) and germanium-68 (68Ge) B. technetium-99m (99mTc) and cobalt-57 (57Co) C. iodine-131 (131I) and barium-133 (133Ba) D. All of the above are true. E. None of the above is true

Answer 47

D. All of the above are true

Question 48

The item pictured below is the _____

Answer 48

E. detector assembly of a gamma camera

Question 49

Which one of the following three bone scan images was acquired with the gamma camera closest to the patient

Answer 49

The spatial resolution of gamma camera images degrades sharply and, therefore, the images appear blurrier, with increasing object (patient)-to-camera distance. This is because the photons emanating from any “point” of activity pass through more of the apertures in the collimator and are dispersed over a larger area of the crystal. In contrast, sensitivity remains largely unchanged as object-to-camera distance changes. Among the three images shown, image 2 is the sharpest (i.e., with the best spatial resolution) and, therefore, was the one acquired with the patient closest to the gamma camera.

Question 50

The following PET image _____.
A. has not been corrected for attenuation B. has been over-corrected for attenuation C. has been corrected for attenuation using a germanium-68 transmission source rather than a CT scanner
D. shows pathologic uptake of the PET tracer in the skin E. has not been properly normalized.

Answer 50

has not been corrected for attenuation
PET images not corrected for attenuation characteristically show “hot” (or intense) superficial activity and faint centrally located activity, as attenuation, or stopping, of radiation and resulting count loss increases with depth in the attenuating medium (in this instance, a patien

Question 51

The item pictured below is the _____

Answer 51

American College of Radiology (ACR) SPECT phantom

Question 52

Although serious adverse effects are rare, the only imaging modality that has been definitively associated with deterministic radiation effects is _____, and these effects have been on _____. A. cone-beam CT; the brain B. PET/CT; the lungs C. DEXA; skeletal muscle D. fluoroscopy; skin E. None of the above is true since modern imaging procedures have never been associated with deterministic radiation effects.

Answer 52

Fluoro, skin

Question 53

The images below are of the _____. The image on the left was reconstructed using a _____, and the one on the right using a _____.

Answer 53

American College of Radiology (ACR) CT phantom; soft-tissue kernel; bone kerne

Question 54

A typical gated blood pool study is gated at 16 frames per cardiac cycle. If a patient’s heart rate is 72 bpm and the count rate is 1.4 million cpm, for how long does the study need to be acquired to collect a total of 200,000 counts per frame? A. 2.3 minutes B. 2.7 minutes C. 5.8 minutes D. 7.0 minutes E. 9.4 minutes
idiotic question

Answer 54

2.3 minutes
A typical gated blood pool study is gated at 16 frames per cardiac cycle (i.e., beat), and the current patient’s heart rate is 72 bpm (beats per minute) = 72/60 = 1.2 bps (beats per second) and the count rate 1.4 million cpm (counts per minutes) = 1,400,000/60 = 23,333 cps (counts per second). At 1.2 bps, each cardiac cycle is 1/1.2 = 0.833 seconds in duration, and each of the 16 frames of the cardiac cycle 0.833/16 = 0.0521 second in duration. Therefore, the number of counts acquired per frame, at a count rate of 23,333 cps, is 0.0521  23,333 = 1,216 counts per frame. To acquire 200,000 counts in each frame then requires summation of 200,000/1,216 = 164 frames for each of the 16 frames over the cardiac cycle. At 0.0521 second per frame, the total acquisition would require 0.0521  164  16 = 136 seconds or 2.3 minute

Question 55

Among currently available multi-modality (i.e., hybrid) systems, the only one(s) in which images for both modalities may be acquired simultaneously is(are) _____. A. PET/CT B. SPECT/CT C. PET/MRI D. All of the above are true. E. None of the above is true.

Answer 55

PET/MRI
In PET/MRI, the PET data and MRI data may literally be acquired simultaneously, given the grossly different and non-interfering imaging signals for these two modalities, i.e., ionizing radiation and radiofrequency pulses (RF), respectively. Although not yet commercialized, the SPECT data and MRI data in a hypothetical SPECT/MRI system could be acquired simultaneously as well. In PET/CT and SPECT/CT systems, on the other hand, the PET or SPECT and CT data are acquired sequentially.

Question 56

Which of the following are the correct mathematical relationships among the linear attenuation coefficients (ice, water, and steam, respectively) and of the mass attenuation coefficients [(ice)m, (water)m, and (steam)m, respectively] of ice, liquid water, and steam?

Answer 56

ice > water > steam and (ice)m = (water)m = (steam)m

Question 57

The figure here shows an ultrasound transducer positioned at three different angular positions relative to a blood vessel during Doppler imaging. At which angle(s) is the Doppler frequency shift not detected? A. 0° and 90° B. 45° C. 0° D. 90° E. 90° and 180°

Answer 57

D. 90°
The Doppler frequency shift is proportional to the cosine of the Doppler angle. That is, the angle between the direction of the ultrasound beam and the direction of a moving object in this context (the direction of blood flowing).

Question 58

Answer 58

Question 59

Answer 59

C. Mass 1 is likely a breast cyst and Mass 2 has higher attenuation than Mass 1.
The left panel of the figure shows a linear array transducer used to scan a subject with several masses of unknown acoustic attenuation. The displayed image (right panel) shows that Mass 1 disrupts the background echo pattern in a way which causes enhancement of the echo signal. This happens when the attenuation by a mass such as this is less than that of background tissue, and Mass 1 could thus be a cyst filled with fluid. The echo signal distal to Mass 2 has lower signal intensity than that of the background tissue and, therefore, attenuation by Mass 2 is greater than that of background tissue and of Mass 1. Mass 3 creates a complete “shadow” behind it, with no echoes emanating that region. This could be caused by highly acoustically attenuating mass; Mass 3 could be air or a gall stone. (Adapted from Zagzebski, J. A. Essentials of Ultrasound Physics. Maryland Heights, MO: Mosby, 1996.

Question 60

The figure below shows the energy spectrum of barium-133 (133Ba) obtained with a thallium-doped sodium iodide (NaI(Tl)) scintillation detector. For which gamma camera test would this radionuclide be better suited than technetium-99m (99mTc)?
A. linearity B. sensitivity C. spatial resolution D. multi-peak spatial registration E. energy resolution

Answer 60

D. multi-peak spatial registration
Barium-133 (133Ba) has three distinct photons that can be used for the multi-peak spatial registration test. Technetium-99m (99mTc) has only one peak and cannot be used for this purpose.

Question 61

With regard to calculation of the fluoroscopic peak skin dose, which of the following is true? A. A correction factor for the displayed air kerma is not needed. B. A factor based on backscatter dose contribution is not needed. C. An attenuation correction for the table and the pad are not needed. D. A factor correcting for the difference in mass energy absorption between tissue and air is not needed.
ANSWER E. A correction factor for patient thickness is not needed.

Answer 61

E

Question 62

In direct digital radiography (DR), the most commonly used photoconductor is _____. A. CdWO4 B. CaWO4 C. NaI (Tl) D. a-Se E. CsI

Answer 62

a-Se

Question 63

Epilation caused by an acute radiation exposure can be seen starting at skin radiation dose levels of _____. A. 0–0.2 Gy B. 0.2–2 Gy C.
2–5 Gy
D. 5–10 Gy E. 1 Gy if given to the whole body

Answer 63

2-5 Gy

Question 64

_____ is the lifetime fatal-cancer risk per Sievert for the general population promulgated by the International Commission on Radiological Protection (ICRP). A. 8% B. 2% C. 5% D. 3% E. 1%

Answer 64

5%
The International Commission on Radiological Protection (ICRP)’s lifetime cancer risk factor for the general population is 5% (or 0.05) per Sievert (0.05% or 0.0005 per rem). This is a value averaged over both genders and over all ages

Question 65

The effective dose is appropriate to use for _____. A. epidemiological studies B. estimating risk of fetal teratogenesis due to irradiation in utero C. prospective dose assessment and comparison for planning and optimization of radiological protection measures
D. calculating radiation dose to the gonads

Answer 65

C. prospective dose assessment and comparison for planning and optimization of radiological protection measures
Effective dose reflects the risk of stochastic radiation effects, such as cancer induction and germ-cell damage. According to International Commission on Radiological Protection (ICRP) Publication No. 103, the main uses of effective dose are prospective dose assessment for planning and optimization in radiological protection and demonstration of compliance with regulatory dose limits. Effective dose is not recommended for epidemiological evaluations, nor should it be used for detailed specific retrospective investigations of individual exposure and risk.

Question 66

A busy female nuclear medicine technologist routinely wears a personal dosimeter. Her average annual dose readings over the past five years have been 2.4 mSv to the whole body. She declares to the Radiation Safety Office that she is three months pregnant. Based on current National Council on Radiation Protection and Measurement (NCRP) recommendations, which of the following limits apply to her fetus? A. 1 mSv per month B. 0.5 mSv per month C. 100 mSv D. 1 mSv E. 0 mSv (requiring the technologist to immediately stop working with radiation)

Answer 66

B. 0.5 mSv per month
Current National Council on Radiation Protection and Measurement (NCRP) (Report No. 180) guidelines limit fetal exposure to 0.5 mSv per month. The technologist’s body badge provides a preliminary estimate of expected fetal dose. Her reading of 2.4 mSv per year is equivalent to 0.2 mSv per month. This is less than half of the recommended monthly numeric protection criteria and, therefore, no modification of her work-related tasks and responsibilities is anticipated.

CNSC- 4 mSv to woman (not to fetus)
ICRP- 1 mSv to fetus

Question 67

define transport index

Answer 67

The transport index (TI) of a radioactive package is defined as the exposure rate in mR/hr (milliroentgen per hour) measured at 1 meter from the surface of the package.

Question 68

The regulatory effective dose limit for a patient undergoing a cardiac stress test procedure is _____. A. 50 mSv B. 20 mSv C. 15 mSv D.
5 mSv E. This is not limited by any regulation

Answer 68

E. This is not limited by any regulation.

Question 69

For estimating effective dose from a CT scan based on the dose-length-product (DLP), which of the following conversion factors (k-factor) is most appropriate to use for adult head scans, where E = DLP (mGy-cm)  k (mSv/mGy-cm)? A. 0.0020 B. 0.0060 C. 0.012 D. 0.020 E. 0.036

Answer 69

For the adult, the values of k (mSv/mGy-cm) are 0.0021, 0.0059, 0.014, 0.015, and 0.015 for adult head, neck, chest, abdomen, and pelvis, respectively. Note that for children, the corresponding coefficients are higher, indicating higher effective doses per unit of DLP for children than for adults.

Question 70

When designing a PET/CT suite and for a SPECT/CT suite, the modality that generally dictates the amount of lead shielding required is _____. A. CT for both PET/CT and SPECT/CT suites B. CT for a PET/CT suite and SPECT for a SPECT/CT suite C. PET for a PET/CT suite and CT for a SPECT/CT suite D. PET for a PET/CT suite and SPECT for a SPECT/CT suite

Answer 70

C. PET for a PET/CT suite and CT for a SPECT/CT s

Question 71

Which of the following products of water radiolysis is mainly responsible for the indirect action of ionizing radiation in both cell killing and induction of mutations? A. hydrogen B. hydrogen peroxide C. ionized water radicals D. hydroxyl radicals E. hydrated electrons

Answer 71

hydroxyl radicals
Hydroxyl radicals (OH) produced by the radiolysis of water molecules are highly reactive and unstable and account for the majority of the indirect chemical effects and biological damage of sparsely ionizing (i.e., low-LET) radiations such as x-rays, gamma-rays, and beta particles.

Question 72

The cell survival curves in the figure below best correspond to which of the following types of radiation from left to right?

Answer 72

E. alpha particles, gamma-rays, neutrons

Question 73

The graph in which of the following figures best describes the relationship between oxygen enhancement ratio (OER) and linear energy transfer (LET)

Answer 73

d

Question 74

In the planar radiograph to the right, the mean number of x-rays per pixel in the circular structure and the background are 1800 and 1600, respectively. The contrast-to-noise ratio (CNR) for the structure corresponding to the circular area is _____.

Answer 74

5

Question 75

What is the pixel depth (that is, the number of bits per pixel) required to display an imaging signal with 0.1% contrast resolution? A.
ANSWER
B. C. D.
D126.
2 4 6 8
E. 10

Answer 75

10
For a pixel depth to display a signal contrast of 0.1% (1/1,000) among image pixels, it should be able to display a dynamic range of 1,000 signal levels (i.e., 1,000 gray levels or colors). The largest number that can be stored in a single pixel is determined by the number of bits, n, used for each pixel and is given by 2^n. Therefore, a pixel depth of 10 bits per pixel stored per pixel yields 2^10 = 1,024 signal levels per pixel and thus can display a contrast resolution of 1/1,000 or 0.001  100% = 0.1%.

Question 76

A PET imaging system has a full-width half-maximum (FWHM) spatial resolution of 4 mm. For objects of what size (in terms of its linear dimensions) does the underestimation of activity due to the partial-volume effect become insignificant? A. >1 mm B. >2 mm C. >4 mm D. >6 mm E. >12 mm

Answer 76

> 12 mm
The partial-volume effect is general phenomenon encountered in imaging in general and tomographic imaging (e.g., CT, SPECT, and PET) in particular. For an object whose linear dimensions are “small” compared to the full-width half-maximum (FWHM) spatial resolution of the imaging system (i.e., less than approximately 2 to 3 times the FWHM), it results in an apparent signal less than the actual signal for that object. Therefore, for a PET imaging with a FWHM spatial resolution of 4 mm, an object’s signal is underestimated until the linear dimensions of the object increase beyond 3  4 mm = 12 mm

Question 77

A new-born pediatric patient is undergoing a cardiac catherization fluoroscopy exam. Which of the following imaging protocols would be optimal for the examination of such a patient? A. 1 pulses/s, 5 msec pulse width, 120 kVp with a standard filter, no anti-scatter grid B. 4 pulses/s, 5 msec pulse width, 70 kVp with a standard filter, no anti-scatter grid C. 8 pulses/s, 4 msec pulse width, 70 kVp with a standard filter, anti-scatter grid D. 15 pulses/s, 4 msec pulse width, 70 kVp with a standard filter, no anti-scatter grid E. 30 pulses/s, 8 msec pulse width, 60 kVp with a standard filter, anti-scatter grid

Answer 77

15 pulses/s, 4 msec pulse width, 70 kVp with a standard filter, no anti-scatter grid
If fluoroscopic equipment is calibrated and operated properly, a reduced pulse rate for cardiac catheterization examinations can significantly reduce patient dose with only a small degradation in image quality caused by a loss of temporal resolution. The pulse width is the time the x-ray beam is on for each image (or frame). To adequately “freeze” patient motion and reduce motion unsharpness in small pediatric patients, pulse widths should not exceed 5 msec. Anti-scatter grids should not be used on pediatric patients who are less than 10 cm thick in the area being examined. A grid would only otherwise increase radiation dose to the patient. Examination of large (i.e., adult) patients requires higher tube voltages to adequately penetrate such patients.

Question 78

ANSWER
In mammography, breast compression typically reduces the scatter-to-primary (i.e., unscattered) ratio from _____ to _____. A. 0.8–1.0; 0.4–0.5 B. 0.6–0.8; 0.4–0.5 C. 0.8–1.0; 0.1–0.2 D. 0.6–0.8; 0.1–0.2 E. Breast compression does not improve the scatter-to-primary ration at all, but simply serves to reduce motion blurring by immobilizing the breast.

Answer 78

0.8–1.0; 0.4–0.5
Breast compression minimizes the path length of x-rays through the breast and, therefore, reduces the likelihood of Compton scatter of the x-rays. As a result, breast compression typically reduces the scatter-to-primary ratio by about one-half, from 0.8–1.0 to 0.4–0.5, and thereby improves image contrast and lesion detectability.

Question 79

For the time-activity curve below presenting the fraction of an administered radiopharmaceutical retained in the body, A(t), versus time post-administration, t, which of the following mathematical functions most accurately describes the curve?

Answer 79

A(t) = A e–at + B e–bt
Commonly, the type of mathematical function that most accurately describes the organ or total-body activity of a radiopharmaceutical versus time t post-administration is a sum of exponentials. If such a function is plotted on a semi-log graph—which has a logarithmically scaled vertical axis and a linearly scaled horizontal axis—each exponential term in the sum-of-exponential function will appear as a straight-line segment. In this graph, there are two straight-line segments, so the time-activity function is a sum of two exponential terms, A(t) e–at + B e–b