RAPHEX XIII

Question 1

convert Sv to rem

Answer 1

1 Sv = 100 rem

Question 2

All of the following statements regarding x-ray production are true except for which statement? A. X-rays are produced by two different mechanisms: bremsstrahlung and characteristic x-ray emission.

B. Characteristic x-rays have discrete energies. C. Bremsstrahlung x-rays have a continuous spectrum of energies. D. The efficiency of x-ray production is about 1% for 100-kVp beams. E. None—all of these statements are true.

Answer 2

None—all of these statements are true

Question 3

Typical x-ray transformers for general radiography have a power rating of about _____. A. 1 W B. 100 W C. 1 kW D. 100 kW

Answer 3

D. 100 kW
According to the electrical power formula, electrical power P (in watts, W) equals the product of the current I (in amperes, A) and voltage V (in volts, V), P = I × V. For a typical x-ray tube for general radiography, the tube voltage is of the order 100 kV = 100,000 V, and the tube current is about 1,000 mA = 1 A, yielding a power P of about 100,000 V × 1 A = 100,000 W = 100 kW

Question 4

The SI unit of air kerma is the _____. A. roentgen (R) B. sievert (Sv) C. rad D. curie (Ci) E. gray (Gy)

Answer 4

E. gray (Gy)
The SI unit of air kerma is the gray (Gy) (or joule per kilogram of air, J/kg), the same as the unit of absorbed dose. It is formally defined as the sum of the initial kinetic energies of all the charged particles liberated by uncharged ionizing radiation (i.e., indirectly ionizing radiation, such as photons and neutrons) in a volume of air, divided by the mass of the volume of air. However, kerma can be different from absorbed dose, depending on the radiation energies involved. This is because ionization energy is not taken into account. While kerma approximately equals absorbed dose at low energies, kerma is higher than absorbed dose at higher energies, because some energy escapes from the absorbing volume in the form of bremsstrahlung x-rays or high-energy electrons, and is not counted as absorbed dos

Question 5

_____ is the predominant interaction of 120-kVp x-rays, commonly used in CT scanners, in soft tissues. A. Rayleigh scatter B. Photoelectric interaction C. Pair-production D. Compton scatter

Answer 5

D. Compton scatter
For an x-ray spectrum typically produced in a CT scanner at 120 kVp, the average beam average energy is high enough for Compton scatter to predominate in soft tissues. At the same time, the average energy is well below the threshold energy, 1.022 MeV, for pair production.

Question 6

In plain radiography, _____ reduces patient dose. A. using a grid B. filtering the beam C. lowering the tube voltage (i.e., kVp) D. decreasing the source-to-object distance

Answer 6

B. filtering the beam
Filtering removes lower-energy, less-penetrating photons from the x-ray beam. These photons, since they are more likely to be absorbed in the patient, increase patient dose. By preferentially removing these x-rays from the beam, filtering reduces the dose. All other choices listed increase the patient dose.

Question 7

The radiographic modality with the highest beam quality is _____. A. diagnostic radiology B. computed tomography C. fluoroscopy D. mammography

Answer 7

B. computed tomography
Beam quality, which essentially refers to the average x-ray energy, increases with increasing x-ray tube voltage (i.e., by kVp) and filtration. Since CT scanning is performed with a relatively high tube voltage of 100 kVp or greater and with a highly filtered x-ray beam, CT has the highest beam quality among radiographic modalities.

Question 8

The automatic exposure control system of a CT scanner determines the tube current for a particular scan based on a planning view (scout) image acquired with the tube stationary under the patient’s bed. If the patient centerline is positioned below scanner isocenter, which of the following will be reduced? A. image noise B. low-contrast visibility C. spatial resolution D. patient dose

Answer 8

A. image noise
The closer the patient is to the x-ray tube, the more magnification will occur, with the patient appearing to be larger (i.e., have a greater body thickness) than he/she actually is. This will result in increasing the exposure (i.e., milliampere-second, mA), which reduces the statistical noise in the resulting images, but increasing the radiation dose to the patient’s spatial resolution is not affected by the exposure (i.e., mAs

dose is higher if patient is near source
relative noise is lower because more mA

Question 9

The device that produced the image shown here was _____. A. a CT scanner B. an MRI scanner C. a C-arm image-intensifier (II) fluoroscopy unit D. a mobile DR E. a bone densitomet

Answer 9

C. a C-arm image-intensifier (II) fluoroscopy unit
The image is an artifact from stray magnetic fields distorting the image in the image intensifier (II) of a C-arm fluoroscopy unit.

Question 10

In the American College of Radiology (ACR) CT Accreditation Program, which of the following ranges of CT numbers in Hounsfield Units (HUs) would be the allowed (i.e., passing) range for air insert? A. 970 to 1005 B. 850 to 970 C. –7 to 7 D. –850 to –970 E. –970 to –100

Answer 10

E. –970 to –1005

Question 11

The test tool shown in the image below is used to evaluate _____.
A. spatial resolution B. focal spot size C. contrast D. modulation transfer function (MTF) E. CT slice thickne

Answer 11

B. focal spot size The tool shown is the star pattern used to measure x-ray tube focal spot size

Question 12

If the flux gain and overall brightness gain of an image intensifier are 50 and 5000, respectively, the image intensifier’s minification gain is _____. A. B.
100 500

C. 2,500 D. 10,000 E. None of the above is true

Answer 12

A. 100
The flux gain of an image intensifier (II) is the number of light photons emitted at the output phosphor for each photon emitted at the emitted phosphor. The minification gain is the increase in image brightness resulting from the reduction in image size from the input to the output phosphor. The overall brightness gain is the product of the flux gain and the minification gain. The minification gain is, therefore, the overall brightness gain divided by the flux gain, in this case, 5000 divided by 50 or 100.

Question 13

Fluoroscopy is typically performed with an x-ray tube current of about _____. A. 0.3 mA B. 3 mA C. 30 mA D. 300 mA

Answer 13

3 mA

Question 14

Compared to continuous fluoroscopy, pulsed fluoroscopy _____. A. uses a higher x-ray tube current B. may reduce the patient dose C. results in sharper images D. All of the above are true.

Answer 14

D. All of the above are true.
In conventional fluoroscopy, the x-ray tube current is very low (about 3 mA) but is on, and x-rays are produced continuously. In pulsed fluoroscopy, the tube is on intermittently (3-msec frames) but at a much higher current (about 30 mA). Pulsed fluoroscopy reduces the patient dose by 30% to 40% when acquiring frames at a rate of fewer than 30 frames/sec. And because the frames are so short (about 3 msec/frame for pulsed fluoroscopy versus 30 msec/frame for conventional fluoroscopy), blurring due to patient motion is reduced and the images are, therefore, sharper

Question 15

An interventional fluoroscope is set up to perform fluoroscopy at 15 frames/sec, cinefluorography at 10 frames/sec, and digital subtraction angiography (DSA) at 3 frames/sec. The correct order of patient entrance dose rates from lowest to highest entrance dose rate, when tested with a 20-cm-thick water phantom, is _____. A. cinefluorography, DSA, fluoroscopy B. cinefluorography, fluoroscopy, DSA C. DSA, cinefluorography, fluoroscopy D. DSA, fluoroscopy, cinefluorography E. fluoroscopy, cinefluorography, DSA

Answer 15

E. fluoroscopy; cinefluorography; DSA
For consistency of image quality, fluoroscopy systems are generally configured such that the same image receptor dose is achieved for 1 DSA frame, 10 cinefluorography frames, or 100 fluoroscopy frames. Patient entrance dose approximately tracks image receptor dose. For a one-minute (or 60-second) run, therefore, the “relative dose” for each modality is as follows:
Fluoroscopy “relative dose” of 1 per frame × 15 frames/sec × 60 seconds → “relative dose” of 900
Cinefluorography “relative dose” of 10 per frame × 10 frames/sec × 60 seconds → “relative dose” of 6000
DSA “relative dose” of 100 per frame × 3 frames/sec × 60 seconds → “relative dose” of 18,000

Question 16

What is the relationship between the modulation transfer function (MTF) and the line spread function (LSF) of an imaging system? A. The MTF is the derivative of the LSF. B. The MTF is the log transform of the LSF. C. The MTF is the integral of the LSF. D. The MTF is the Fourier transform of the LSF. E. The MTF and LSF are unrelated.

Answer 16

D. The MTF is the Fourier transform of the LSF.

Question 17

The standard deviation and percent standard deviation of the number of counts per pixel are _____ and _____, respectively. A. 25 counts; 20% B. 50 counts; 14% C. 100 counts; 10% D. 200 counts; 7

Answer 17

C. 100 counts; 10%

I think they meant to ask for number of counts

Question 18

For an image intensifier-based fluoroscope, switching from 9-inch to 6-inch magnification mode results in a _____. A. dose decrease by a factor of (9/6)2 B. dose decrease by a factor of (9/6) C. dose increase by a factor of (9/6)2. D. dose increase by a factor of (9/6)

Answer 18

C. dose increase by a factor of (9/6)2
For image intensifier-based fluoroscope, the dose increases by a factor determined by the ratio of magnification squared.

Question 19

For an image intensifier-based fluoroscope, switching from 9-inch to 6-inch magnification mode results in _____. A. poorer resolution (line pairs/mm) by a factor of (9/6)2 B. poorer resolution (line pairs/mm) by a factor of (9/6) C. improved resolution (line pairs/mm) increase by a factor of (9/6)2 D. improved resolution (line pairs/mm) increase by a factor of (9/6)

Answer 19

D. improved resolution (line pairs/mm) increase by a factor of (9/6)
For an image intensifier-based fluoroscope, the resolution increases by a factor determined by the ratio of magnification.

Question 20

Pure beta particle-emitting radionuclides are best shielded using _____. A. plastic B. lead C. steel D. titanium E. zinc

Answer 20

A. plastic
Beta particles are considered non-penetrating radiation and are best shielded by materials with a low atomic number (Z) such as plastic, a relatively thin layer of which will completely attenuate any incident beta particles with negligible bremsstrahlung production. Higher-Z materials such as metals would produce more bremsstrahlung. High-energy beta particles, such as those emitted by 90Y, can produce more bremsstrahlung, and the low-Z plastic shield can be supplemented by lead shielding to attenuate any bremsstrahlung produced in the plastic. Typically, the amount of bremsstrahlung thus produced will be very small in any case

Question 21

The device pictured to the right is a/an _____, and it is used to assess the _____ of a gamma camera. A. ACR phantom; sensitivity B. Mobius phantom; uniformity C. transmission phantom; dead time D. four-quadrant bar phantom; spatial resolution

Answer 21

D. four-quadrant bar phantom; spatial resolution
A technetium-99m (99mTc) transmission image of a four-quadrant bar phantom depicted is widely used to assess gamma camera spatial resolution

Question 22

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 follows

lung ROI counts = 4168
liver ROI counts = 485912

Therefore, the lung shunting is _____. A. 0.5% B. 0.9% C. 1.6% D. 2.5%

Answer 22

B. 0.9%
The lung shunt is calculated as the ratio of the 99mTc-MAA counts in the lung region of interest (ROI) counts divided by the sum of the counts in both the lung and liver ROIs multiplied by 100%. In this case, [4168 / (4168 + 485,912)] × 100% = 0.85% ≈ 0.9

Question 23

In iodine-131 (131I)-iodide treatment of metastatic thyroid cancer, the β-particles from 131I in a lesion account for _____ percent of the radiation dose to the lesion. A. 0 B. <10 C. 25–50 D. >75

Answer 23

D. >75
Iodine-131 (131I) beta particles, like beta particles and other particulate radiation generally, have a very short range (about 1 mm or less) in tissue and deposit all of their energy, and deliver all of their radiation dose, within 1 mm or less of their point of emission. Thus, beta-particles emitted by 131I within a thyroid cancer lesion will deposit virtually all of their energy
in and deliver virtually all of their dose to the lesion itself. The 364-keV gammas emitted by 131I, like other high-energy x-and gamma-rays, travel many centimeters in tissue and will deposit much of their energy and deliver much of their dose outside the lesion, contributing very little to the lesion dose.

Question 24

What is the intensity half-value thickness for a 6-MHz ultrasound beam, assuming the soft-tissue attenuation coefficient is 0.5 dB/cm/MHz? A. 0.5 cm B. 0.75 cm C. 1 cm D. 1.5 cm E. 2.0 cm

Answer 24

C. 1 cm
The intensity half-value thickness (HVT) is defined as the thickness of material that will reduce the initial ultrasound beam intensity by one-half. Relative intensity equals 10 × log(I2/I1) where I1 and I2 are the intensities of the incident and reflected intensities, respectively, of the beam. Solving this equation for a relative intensity that is decreased by one-half will yield a 3-dB reduction. Thus, a half-value thickness (HVT) will reduce the ultrasound beam intensity by 3 dB. In this case, we have 0.5 dB/cm/MHz at 6 MHz, which means an attenuation coefficient of 3 dB/cm. Thus, 1 cm material will reduce the intensity by 3 dB or 50%

Question 25

The gamma camera performance test known as multi-peak spatial registration is performed using _____ isotopes. A. fluorine-18 (18F) B. gallium-67 (67Ga) C. zirconium-89 (89Zr) D. technetium-99m (99mTc) E. iodine-123 (123I)

Answer 25

B. gallium-67 (67Ga)
Multi-peak (or multi-window) spatial registration is a gamma camera performance test that should be done with isotopes that emit multiple “image-able” gamma-rays, such as gallium-67 (67Ga), which emits a 93-keV gamma-ray (40% abundance), a 185-keV gamma-ray (24%), and a 300-keV gamma-ray (22%). All of the other isotopes listed are either a positron emitter used in PET (18F, 89Zr)) or emit a single image-able gamma-ray (99mTc, 123I). Gamma cameras essentially acquire a separate image for such gamma-ray and then superimpose (i.e., sum) them to yield the final image. Multi-peak spatial registration expresses how accurately these multiple images are spatially registered. Typical values of this parameter are about 1 mm, meaning that the multiple gamma-ray images are registered with each other to within 1 mm.

Question 26

A lung cancer patient weighing 74 kg undergoing a fluorine-18 (18F)-fluorodeoxyglucose (FDG) PET scan is injected with 370 MBq of 18F-FDG. What is the standard uptake value (SUV) for the lung lesion (arrow) having an activity concentration of 16 kBq/cm3? A. 1.0 B. 2.4 C. 3.2 D. 4.0 E. 4.8

Answer 26

C. 3.2
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 16 kBq/cm3 in a patient with a total-body mass of 74 kg or 74,000 g who received an administered activity of 370 MBq or 370,000 kBq, the SUV is 16 kBq/g / (370,000 kBq / 74,000 g) = 16 kBq/g / 5 kBq/g = 3.2, assuming (as usual) a tissue mass density of 1 g/cm3.

Question 27

For modern bone densitometry units, which of the following statements is not true? A. Such units utilize pencil-beam or fan-beam photons. B. Modern bone densitometry utilizes gadolinium-153 (153Gd) gamma-rays for the photon beam.
C. Such units are capable of body composition measurements. D. The effective dose (ED) to the patient for a whole-body bone densitometry study is considerably lower than that for a chest/abdomen/pelvis CT scan.

Answer 27

B. Modern bone densitometry utilizes gadolinium-153 (153Gd) gamma-rays for the photon beam.
Current-day bone densitometry units are capable of body composition measurements utilizing dual-energy x-rays rather than gadolinium-153 (153Gd) gamma-rays. Radioactive 153Gd sources were used at one time in bone densitometry units but were replaced by x-ray tubes a number of years ago. The x-ray beam geometries are either a scanning pencil-beam or scanning fan-beam. Bone densitometry/body composition analysis using dual-energy x-ray absorptiometry is a dosimetrically favorable procedure, delivering considerably lower radiation doses than most other radiological studies, with effective doses (EDs) of well under 100 mrem or 0.1 rem. In contrast, a typical chestabdomen/pelvis CT delivers an ED of ~1.5 rem to an adult patient.

Question 28

A blood sample counted for 2 minutes yields 1800 counts. A 2-minute background measurement is made yielding 200 counts. What is the standard deviation σ of the net (i.e., background-subtracted) count rate (in counts per minute, cpm)? A. 100 cpm B. C. D. E.
88 cpm 44 cpm 22 cpm 12 cpm

Answer 28

D. 22 cpm
The standard deviation σ of a total of N counts is . For equal counting intervals, the net counts Nn equals the difference of the gross, or total, counts Ng and the background counts Nbg, that is, Nn = Ng – Nbg. By the formula for propagation of error, the standard deviation of the net (or background-subtracted) counts σN equals
N  where σg =
gbg 22
σbg = 200 42 4 14 14
22 ..
1800 = 42.4 and
= 14.1. Therefore, the standard deviation σN of the net count counts equals = = 45 counts, and the standard deviation of the net count rate equals
45 counts / 2 minutes = 22 cpm.

propagation of errors!

Question 29

The American College of Radiology (ACR) phantom, or “Jaczszak,” pictured here is used to evaluate which performance parameters of SPECT systems? A. count rate, spatial resolution, contrast B. spatial resolution, count rate, sensitivity
C. uniformity, contrast, spatial resolution D. sensitivity, uniformity, center-of-rotation alignment

Answer 29

C. uniformity; contrast; spatial resolution
The Jaczszak phantom is widely used to evaluate the overall system performance of SPECT systems. The phantom includes a plain section for the evaluation of the tomographic system uniformity, a section containing non-radioactive, or “cold,” spheres ranging from 9.5 to 31.8 mm in diameter for evaluation of contrast, and a section containing solid (non-radioactive) rods ranging from 4.8 to 12.7 mm in diameter for evaluation of spatial resolution.

Question 30

A 10-ml syringe containing 2-mCi of technetium-99m (99mTc) in 1 ml is assayed in a dose calibrator. An additional 1 ml of plain saline is added to the syringe, and the syringe is assayed again. This process is repeated, adding 1-ml aliquots of plain saline to the syringe and re-assaying the syringe after the addition of 1 ml aliquot until the total volume in the syringe is 10 ml. These measurements are used in the evaluation of the _____ dose calibrator test. A. geometry B. linearity C. constancy D. accuracy

Answer 30

A. geometry
Dose calibrators are required to be tested for geometry dependence at installation, after major repair, or after being relocated. This is done by adding successive aliquots of plain saline or water to a 1-ml radioactive solution in a syringe or vial and assaying its activity before and after each addition of plain saline or water. Over a total volume range of 1 to 10 ml, the activity readings should not differ by more than 10%.

Question 31

A 1-microCi iodine-129 (129I) standard source is measured in a scintillation well counter and yields a gross, or total, count rate of 5000 counts per minute (cpm). A background count rate of 100 cpm is then measured. _____ is the efficiency of the well counter for 129I. A. 20% B. 10% C. 1% D. 0.2%

Answer 31

D. 0.2% The efficiency ε of a counting system is defined as follows:
epsilon=(Cradionuclide-Cbackground)/Aradionuclide * 100%
where Cradionuclide is the gross count rate (in cpm) of the standard source, CBackground is the background count rate (in cpm), and Aradionuclide is the activity of the standard source (in disintegrations per minute, dpm). In this example, Aradionuclide = 1 microCi = 37,000 Bq = 37,000 decays/sec × 60 sec/min = 2,220,000 decays per minute (dpm). Therefore, ε = 0.0022 ×
100% = 0.22%. Importantly, this value of the efficiency applies only to this specific isotope, 129I, counted with the particular counter energy window used for this measurement.

Question 32

equation for doppler US

Answer 32

fdoppler = (2* f*vcos(theta))/c
c is speed of sound in tissue

Question 33

typical T1 values at 3 T for fat, CSF, white matter, grey matter

Answer 33

typical T1 values at 3 T are approximately 300 ms for fat, 3000 ms for CSF, 1000 ms for white matter, and 1200 ms for grey matter.

Question 34

The figure below shows an axial T2-weighted fat-suppressed image acquired during an MRI-guided liver biopsy. The 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 true.

Answer 34

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 fat

Question 35

what type of MRI trajectory is less sensitive to motion artifacts caused by respiration in abdominal MR imaging. A. cartesian B. EPI C. radial D. spiral E. None of the above is true

Answer 35

C. radial
The major benefit of radial k-space sampling is its relative insensitivity to motion artifacts. The center of k-space in radial sampling is oversampled and continuously updated. Therefore, motion on one or a few radial spokes is less likely to cause severe motion artifacts than other k-space sampling strategies.

Question 36

MRI aliasing or wrap-around is a frequently encountered MRI artifact when the field-of-view (FOV) is not sufficiently large to cover the body part being imaged. Below is a diagram for normal (a) and aliased (b) axial brain MRI images. Which of the following is the correct phase encoding direction of the MRI scan in (b)?

Answer 36

C. anterior to posterior or posterior to anterior
Aliasing or wrap-around artifact is characterized by folding over of the image signals in the opposite side when field-of-view (FOV) is not sufficient to cover the body part being imaged, such as the shaded area in the anterior and posterior of (b). In modern MRI scanners, aliasing in the frequency direction is eliminated by oversampling of the data during MR signal digitization. However, oversampling in the phase encoding direction needs to acquire more k-space lines with increased scan time and, therefore, aliasing usually occurs in the phase encoding direction. For this question, the phase encoding direction is from anterior to posterior or from posterior to anterior

Question 37

Specific absorption rate (SAR) is the radio frequency (RF) power absorbed by the patient. Both the International Electrotechnical Commission and the United States Food and Drug Administration have specific safety guidance on the amount of energy that can be absorbed during an MRI scan. Which of the following is not a factor that can affect SAR? A. flip angle B. body weight C. field strength of the magnet (e.g., 1.5 T, 3 T) D. gradient strength E. duty cycle

Answer 37

D. gradient strength
(A) Specific absorption rate (SAR) is proportional to the square of the RF pulse flip angle. (B) SAR increases significantly with body size, so accurate body weight is needed for every MRI scan to estimate SAR. (C) SAR is proportional to the square of field strength B0, so the SAR of the same sequence in the same patient at 3 T is about four times of the SAR at 1.5 T. (D) SAR is not directly affected by gradient strength. (E) Duty cycle is the percentage of time in a pulse sequence when the RF pulse is being transmitted, so a higher duty cycle results in higher SAR.

Question 38

Receiver bandwidth (BW) is an important MRI parameter that can affect signal-to-noise ratio (SNR), minimum echo time (TE), and artifacts in the image. Below is the acquisition of MRI signal in the readout direction. Assuming the matrix size (total number of samples) in the readout direction is 256 and the time between samples, or the dwell time (td), is 30 microseconds, which of the following is the correct bandwidth in Hz/pixel?

Answer 38

B. 130
Receiver bandwidth (BW) = 1/Ts (Hz/pixel), where Ts is the whole sample period. Given the interval between two adjacent samples (dwell time td) = 30 microseconds and the total number of samples = 256 in the readout, we can calculate Ts = N × td = 256 × 30 × 10–6 = 0.00768 seconds. One can then calculate BW = 1/Ts = 1 / 0.00768 = 130 Hz/pixe

Question 39

Paramagnetic materials have a _____ susceptibility and diamagnetic materials a _____ susceptibility. A. positive; negative B. negative; positive C. zero; negative D. negative; zero

Answer 39

A. positive; negative
Magnetic susceptibility is the extent to which a material becomes magnetized when placed in an external magnetic field. Paramagnetic atoms such as gadolinium placed in a magnetic field produce an increase in the local magnetic field, that is, have a positive susceptibility. Diamagnetic atoms result in a decrease in the local magnetic field, that is, have a negative susceptibility.

Question 40

If the images from an MRI scan look noisy, which of the following will not improve the signal-to-noise ratio (SNR) of the scan? A. reduce the receiver bandwidth (BW) B. increase the slice thickness C. increase the parallel imaging acceleration factor D. increase the field of view (FOV)

Answer 40

C. increase the parallel imaging acceleration factor
Reducing the receiver bandwidth (BW) will increase signal-to-noise ratio (SNR). Noise is distributed evenly across the entire frequency spectrum. Therefore, reducing BW limits the amount of noise mixed in the MRI signal and thus increases the SNR of the image. Increasing the slice thickness will increase SNR because it increases the voxel size and thus the amount of MRI signal received by the individual voxel. Increasing the number of signal average will increase SNR by a factor of square root of the averages at the cost of increased scan time. Increasing the parallel imaging acceleration factor will decrease SNR. In parallel imaging, increasing acceleration factor means decreased MRI data sampling (fewer k-space lines) and increased noise amplification during parallel image reconstruction.

***Increasing the field of view (FOV) will increase SNR because it increases the voxel size if the matrix size keeps unchanged, so the amount of MRI signal received by individual voxels increases.

Question 41

does increasing FOV in MRI affect noise?

Answer 41

it will if matrix is unchanged, because then voxel size will increase

Question 42

what is echo train length

Answer 42

The number of echoes acquired in a given TR interval is known as the echo train length (ETL) or turbo factor.

The time period between the excitation pulse and this echo is the effective echo time. (TSE TF) The turbo factor is the number of echoes acquired after each excitation. This is a measure of the scan time acceleration, e.g. at turbo factor 3 the scan time is 3 times faster as a SE sequence with comparable parameters.

Question 43

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

Answer 43

C. the response (i.e., activity reading) is linear as a function of sample activity; quarterly
The linearity quality-control test for a dose calibrator checks that the response (i.e., activity reading) is linear and should be performed quarterly. Linearity is most often checked by the so-called “decay method.” One begins with an activity higher than that used clinically (often as high as ~1 Ci), independently calibrated technetium-99m source and assays its activity at 6-to 12-h intervals over three consecutive days. Over that time, equivalent to 12 half-lives of technetium-99m, the activity has to decay to less than 0.027 mCi. The measured activities are then plotted versus time on a semi-logarithmic graph, and the best-fit straight line drawn through the data points thus plotted (either “by eye” or using a least-squares curve-fitting algorithm). For each data point, the difference between the measured activity and the activity on the best-fit straight line at that point should be less than 10%. An alternative approach to checking linearity is the so-called “shield method,” in which lead sleeves of increasing thickness are placed in the dose calibrator with a technetium-99m source. By interposing increasing “decay-equivalent” thicknesses (as specified of by the manufacturer of the set of lead sleeves) between the source and the dose calibrator’s sensitive volume, a decay-equivalent activity is measured for each sleeve. While the shield method is much faster than the decay method for checking linearity (taking minutes instead of days), an initial decay-based calibration of the set of sleeves is recommended to accurately determine the actual decay equivalence of each shield

Question 44

Which of the following statements regarding hybrid PET/MRI scanners is not true? A. The PET and MRI scans are acquired simultaneously on such hybrid scanners. B. Photomultiplier tubes (PMTs) are used as part of the PET scintillation detectors of hybrid PET/MRI scanners.
ANSWER
C. The variable MR gradients may induce eddy currents in conductive materials of the PET detectors and distort the applied gradient fields.
D. In hybrid PET/MRI scanners, a two-point Dixon VIBE MR sequence is often used in deriving PET attenuation-correction data.
E. None of the above

Answer 44

B. Photomultiplier tubes (PMTs) are used as part of the PET scintillation detectors of hybrid PET/MRI scanners.
Conventional PET detectors were based on PMTs, which do not operate properly in the presence of a magnetic field due to distortion of the trajectory of electrons as they travel among the dynodes with a PMT. In current commercially available PET/MRI hybrid scanners, the PMTs have been eliminated and replaced by solid-state silicon photomultipliers (SiPMs). SiPMs are relatively insensitive to magnetic fields

Question 45

The optimum approach to balancing the mAs and radiation dose for a pediatric CT scan is to _____. A. use patient thickness-based technique charts B. use patient weight-based technique charts C. use patient age-based technique charts D. use the same mAs for pediatric as for adult patients E. simply use an mAs of 100 and activate the scanner’s automatic exposure control (AEC).

Answer 45

A. use patient thickness-based technique charts
To optimally balance signal-to-noise ratio (SNR) and radiation dose among patients, all CT techniques should be tailored to account for differential x-ray attenuation among patients. Attenuation is most closely related to patient thickness. Although automatic exposure control (AEC) is helpful in maintaining a consistent SNR among different patient thicknesses, the operator must understand how the AEC operates for his/her particular scanner—a single default starting mAs for all pediatric sizes may not be appropriate

Question 46

Although serious adverse effects are rare, the only imaging modality which 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 46

D. fluoroscopy; skin
In cardiovascular fluoroscopy and interventional imaging generally, skin absorbed doses can be up to several Gy (several hundred rad). The acute threshold absorbed dose for skin effects of ~2 Gy (200 rad) may, therefore, be exceeded. In rare instances, skin doses may be considerably higher than 2 Gy (200 rad) and result in necrotic damage so severe as to necessitate surgical intervention, particularly as fluoroscopically guided procedures have become more complex and beam-on times have increased. Largely for this reason the U.S. Food and Drug Administration (USFDA) in 1994 issued a health advisory entitled, “Avoidance of Serious X-ray Skin Injuries to Patients during Fluoroscopically Guided Procedures” and established regulatory limits for the entrance dose rate, specifically for the air kerma rate (Kair or AKR). With certain exceptions, the AKR shall not exceed 88 mGy/minute (10 R/min) at the measurement point. (The foregoing regulatory limit applies specifically to fluoroscopic equipment manufactured on or after May 19, 1995.

Question 47

In myocardial perfusion imaging with thallium-201 (201Tl)-thallous chloride, imaging is typically performed within 20 minutes of injection, but it can also be performed as late as ~24 hours post-injection. This is because _____. A. 201Tl is retained long-term in the mitochondria of myocardial cells B. 201Tl clears very slowly from the myocardium C. 201Tl readily crosses cell membranes and is continuously redistributed in the body D. 201Tl has a short biological half-life in the body

Answer 47

C. 201Tl readily crosses cell membranes and is continuously redistributed in the body
Thallium-201 (201Tl) is an analog of potassium and, like potassium, readily crosses cell membranes (i.e., into and out of cells). Imaging of the myocardium immediately (i.e., within ~20 minutes) post-injection will, therefore, reflect regional myocardial perfusion and thus delineate areas of hypoperfusion resulting from coronary artery disease, particularly in response to exercise-or pharmacologically induced stress. Because 201Tl can exit and then re-enter cells (i.e., re-distribute) and has a fairly long physical half-life of 73 hours, it can be imaged as late as ~24 hours post-injection.

Question 48

A patient is to undergo a gated cardiac blood-pool study (i.e., multi-gated acquisition, or MUGA) with the heart cycle (R-R interval) binned into 16 frames. The patient is in normal sinus rhythm and has a heart rate of 75 beats per minute (bpm). What would the time per frame (in milliseconds, ms) be for this study? A. 50 ms B. 22 ms C.
0.29 ms
D. 1.5 ms E.
3.4 ms

Answer 48

A. 50 ms
For a MUGA study binned into 16 frames per cardiac cycle (i.e., beat) and a patient’s heart rate of 75 bpm (beats per minute) = 75/60 = 1.25 bps (beats per second), each cardiac cycle is 1/1.25 = 0.80 second in duration, and each of the 16 frames of the cardiac cycle is, therefore, 0.80/16 = 0.050 second = 50 ms in duration.

Question 49

When analyzing a gated blood pool (i.e., multi-gated acquisition, or MUGA) study, incorrectly placing the background region over part of the spleen will have what effect on the resulting calculated value of the left ventricular ejection fraction (LVEF)? A. The calculated LVEF will be higher than patient’s true LVEF. B. The calculated LVEF will be lower than patient’s true LVEF. C. The calculated LVEF will not be affected. D. The effect will be different, depending on whether the patient’s cardiac function is normal or abnorma

Answer 49

A. The calculated LVEF will be higher than patient’s true LVEF

spleen gives hot background- underestimate deniminator so LVEF is overestimated

Question 50

For many years, molybdenum (Mo) has been the most common target material used in the x-ray tube of mammography systems. Newer mammography systems are increasingly using tungsten (W). Which of the following statement(s) is/are correct regarding W-target versus Mo-target mammography systems? A. W-target-based systems produce more bremsstrahlung x-rays than Mo target-based systems.
ANSWER
B. W targets are more heat tolerant and have a higher melting point than Mo targets. C. The x-ray beam produced by W-based mammography systems, typically operated at an x-ray tube voltage of ~30 kVp, does not include characteristic x-rays.
D. All of the above are true.

Answer 50

D. All of the above are true.
Tungsten (W) has a higher atomic number, Z = 74, than molybdenum (Mo), Z = 42, and thus will produce considerably more bremsstrahlung x-rays than Mo. Tungsten also has a much higher melting point than Mo, 6200 °C versus ~2600 °C. Characteristic W x-rays are higher in energy than characteristic Mo x-rays, ~50 keV versus ~20 keV, and thus would not be included in the x-ray beams produced with a tube voltage of 30 kVp typically used in mammography.

Question 51

The reference-point air kerma recorded during fluoroscopically guided interventional procedures refers to the air kerma at _____. A. 30 cm in front of the image intensifier B. 15 cm from isocenter toward x-ray tube C. 1 cm above the table D. At the surface of x-ray tube port E. 10 cm from patient entrance surface

Answer 51

B. 15 cm from isocenter toward x-ray tube
The International Electrotechnical Commission (IEC) defines the interventional reference point to be 15 cm from the isocenter of gantry toward the x-ray tube. This is supposed to be a location representative of the patient’s skin. The FDA has separate reference points for measuring dose when testing a fluoroscopic unit. For units with the x-ray tube under the table (GI units), the exposure rate is measured at 1 cm above the tabletop, and for c-arm units the exposure rate is measured 30 cm in front of the detector

Question 52

Which of the following statements regarding Dose-Area Product (DAP) is incorrect? A. It is independent of distance from the x-ray tube. B. It is the highest dose to any portion of the patient’s skin during the procedure. C. It is the total amount of energy imparted to a patient during a procedure. D. It is typically measured at the surface of the x-ray tube port.

Answer 52

B. It is the highest dose to any portion of the patient’s skin during the procedure.
The highest dose to any portion of the patient’s skin is characterized by a different quantity called the Peak Skin Dose (PSD), which includes backscatter. The Dose-Area Product (DAP) is typically measured using a meter at the x-ray port, does not include backscatter, and is independent of distance because the inverse square law from the point source cancels out with the area at any point from between the focal spot and the detector.

Question 53

Which of the following tissue boundary interfaces has the lowest ratio of reflected-to-incident ultrasound beam intensity? A. liver-kidney B. muscle-bone C. fat-muscle D. muscle-lung E. bone-lung

Answer 53

liver-kidney

Question 54

The figure below is the count profile across a gamma camera image of a technetium-99m (99mTc)-filled capillary tube. Such a source is often used to determine the spatial resolution of a gamma camera. The spatial resolution derived from this count profile is approximately _____.

Answer 54

C. 8 mm
The spatial resolution of a gamma camera is generally expressed as the full-width half-maximum (FWHM) of the camera’s count profile across a line source, that is, of the line spread function (LSF). In the LSF shown in the figure, the maximum height is about 1200 counts, which means that the half-maximum is about 600 counts. There are two points close to the 600-counts line, one at about 16 mm (to the left of the maximum) and the other at about 24 mm (to the right of the maximum). The FWHM corresponds to 24 mm – 16 mm = 8 mm.

Question 55

what is integral uniformity

Answer 55

(max counts- min counts)/(max counts + min counts) * 100%

acceptable value is within 5 %

Question 56

how often must gamma camera uniformity be evaluated?

Answer 56

A. daily
Gamma camera uniformity must be evaluated daily, at least for technetium-99m (99mTc) or its surrogate isotope cobalt-57 (57Co). Uniformity for other radioisotopes can be evaluated less frequently (e.g., monthly).

Question 57

When designing PET/CT suites and SPECT/CT suites, 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 57

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

Question 58

If a CT display window is set at a width of 800 with the center at 500, the anatomy that will be appear as white is _____. A. fatty tissue B. soft tissue C. bone D. lung

Answer 58

C. bone
With the CT image display set to a center value of 500 HU and a width of 800 HU, the range is 100 to 900 HU. CT numbers lower than 100 HU and greater than 900 HU are outside this window. Tissues such as bone (at least some boney tissues) with CT numbers greater than 900 HU appear white. The CT numbers of lung, fatty tissue, soft tissue, and bone are typically –800 to –900 HU, –50 to –100 HU, 20 to 100 HU, and 800 to 1000 HU, respectively. Tissues such as lung with CT numbers less than 100 HU will appear black.

Question 59

Regarding CT scanner bow tie filters, which of the following statement(s) is/are correct? A. The shape of the bow tie filter used depends on the body part to be scanned. B. The attenuation of the x-ray beam by such a filter increases with increasing distance from the beam center to the beam periphery.
C. Bow tie filters help ensure that all of the detectors receive similar x-ray exposures. D. All of the above are true.

Answer 59

D. All of the above are true.
With the fanbeam geometry used in modern CT scanners, in the absence of a bow-tie filter the x-ray detectors along the central axis of the beam would receive a lower exposure than those toward the periphery of the beam because the central-axis x-rays pass through a greater thickness of the patient. Interposing a bow-tie filter in the fanbeam as it exits the x-ray tube will result in greater attenuation of the periphery of the x-ray beam and lesser attenuation along the central portion of the beam, thereby helping to equalize the x-ray exposures among the detectors.

Question 60

The radiation dose to lung from a chest radiograph is _____ of that from a chest CT scan. A. <10% B. about 20% C. about 40% D. about 60% E. about 80%

Answer 60

A. <10%
A chest CT scan of an adult patient delivers absorbed dose of 10 to 20 mSv to the lung. A single chest radiograph delivers an absorbed dose of ~0.1 mSv to the lung or less than 10% of that from a chest CT scan

Question 61

Mammography units that can perform digital breast tomosynthesis (DBT) have an x-ray tube with an angular range of motion around the breast up to _____. Α. ±5° Β. ±20° C. ±45° D. ±90° Ε. ±–120°

Answer 61

B. ±20° ±°
Among DBT units, the angular range of motion of the x-ray tube varies from ±7.5° to ±25°, as shown in the table below.

Question 62

Which of the following medical physicist tests or evaluations should be performed by a qualified medical physicist on a newly installed digital mammography unit prior to its clinical use? A. phantom image quality B. AEC system performance C. collimation assessment D. spatial resolution E. evaluation of technologist’s QC program

Answer 62

A. phantom image quality

Question 63

When the x-ray beam of a fluoroscopy system is directed laterally at the patient, the operator will receive the lowest radiation exposure _____. A. standing at the x-ray tube side of the patient B. standing at the image intensifier side of the patient C. The exposure will be the same at either position. D. The exposure will vary depending on where the operator is positioned, but it may be lower or higher at either position, depending on various other factors.

Answer 63

B. standing at the image intensifier side of the patient
Because backscatter will not contribute to the operator’s exposure when the he/she is standing at the image-intensifier side of the patient, exposure will be minimized at this position.

Question 64

When a fluoroscopy system is operated in the “high-level” mode, the patient entrance air kerma (entrance exposure) rate must not exceed _____. A. 43 mGy/min (5 R/min) B. 88 mGy/min (10 R/min) C. 176 mGy/min (20 R/min) D. There is no regulatory upper limit on entrance air kerma rate

Answer 64

C. 176 mGy/min (20 R/min)
By FDA regulation, the maximum permissible patient entrance air kerma (entrance exposure) rate in the high-level fluoroscopy mode is 176 mGy/min (20 R/min) and in normal mode 88 mGy/min (10 R/min)

Question 65

The value of the cumulative air kerma displayed on the console of a fluoroscopy system may significantly over-or underestimate the actual peak skin dose because _____. A. air kerma does not account for radiation backscattered to the skin B. the x-ray beam may not always be incident on a single area of the skin C. the skin may be closer to the x-ray tube than the reference point D. displayed air kerma does not take into account attenuation of table or pad E. All of the above are true.

Answer 65

E. All of the above are true. All of the factors listed may result in an erroneous value of the cumulative air kerma

Question 66

If a CT scan covers an axial length of 20 cm of the patient with 320 × 0.625-mm slices, the quantum noise in each image will be about _____ times noisier than if the scan had been performed to yield 40 × 5-mm slices. A. 2.8 B. 3.6 C. 8.0 D. 12.5 E. 16.0

Answer 66

A. 2.8
Quantum noise varies as the square root of the number of photons forming the image. Since the 0.625-mm thick slices are 0.625 mm / 5 mm = 1/8 as thick and will, therefore, have 1/8 of the number of photons per slice as the 5-mm thick slices, the 0.625-mm thick images will be 8 = 2.8 times noisier

Question 67

The CT artifact shown in the figure below could be avoided by _____

A increasing the mAs B. increasing the kV by about 10 C. having the patient raise his/her arms above their head D. switching from step-and-shoot to helical mode E. switching to a smoothing reconstruction algorithm

Answer 67

C. having the patient raise his/her arms above their head
The artifact is due to inadequate coverage of the entire patient in the transverse plane by the scan field, with the x-ray beam for certain views passing through the arms. Raising the patient’s arms will prevent them from attenuating the x-ray beam for those view

Question 68

The type of ultrasound image artifact seen in the figure (right panel) below is _____.

A. reverberation B. shadowing C. speed displacement D. side-lobes E. mirror image

Answer 68

E. mirror image
Mirror-image artifact arises from multiple beam reflections between a mass and a strongly reflecting structure, such as the diaphragm. Multiple echoes result in a mirror image of a mass beyond such a structure (in this case, the diaphragm). Reverberation artifacts occur between two strong reflectors, producing equally spaced signals of diminishing amplitude in the image. Shadowing occurs distal to objects of high ultrasound attenuation, resulting in hypointense signal. Speed-of-sound variation in the tissues can cause mis-mapping of anatomy. Side-lobes energy emissions in transducer arrays can cause anatomy outside of the main beam to be mapped into the main beam

Question 69

If the Thermal Index (TI) or the Mechanical Index (MI)—which are both displayed on the ultrasound unit’s screen—exceed 1.0, which of the following would NOT be an appropriate adjustment? A. shorten the examination time B. change Doppler to standard B mode operation C. reduce power output D. reduce area of color flow measurements E. increase the pulse repetition frequency (prf

Answer 69

E. increase the pulse repetition frequency (prf)
Increasing the pulse repetition frequency (prf) causes more energy per unit time to be deposited in the patient, thereby increasing (rather than decreasing) the Thermal and Mechanical Indices (TI and MI, respectively).

Question 70

Harmonic ultrasound imaging with an initial frequency setting of 4 MHz will image ultrasound reflections of _____. A. 2 MHz B. 4 MHz C. 6 MHz D. 8 MHz E. 16 MH

Answer 70

D. 8 MHz
Harmonic ultrasound imaging will set the system to respond to the first harmonic of the incident frequency, which is 2 times the incident frequency, 2 × 4 MHz = 8 MHz in this case

Question 71

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 ratio at all, but simply serves to reduce motion blurring by immobilizing the breast

Answer 71

A. 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 72

A PET scanner has a full-width half-maximum (FWHM) spatial resolution of 5 mm. For lesions of with a minimum linear dimension _____, the under-estimation of the lesion activity due to the partial-volume effect becomes insignificant. A. >15 mm B. >7.5 mm C. >5 mm D. >2.5 mm E. >1.0 mm

Answer 72

A. >15 mm
The partial-volume effect is a 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 × 5 mm = 15 m

Question 73

The image shown to the right is a whole-body gamma-camera scan of 131I-iodide in a patient with metastatic thyroid cancer. It exhibits a pronounced “starburst” image artifact in the neck area. What is the cause of this artifact? A. low-pass filter artifact B. high-pass filter artifact C. collimator septal penetration D. damaged collimator aperture E. motion artifact

Answer 73

C. collimator septal penetration
Collimator septal penetration is more pronounced with higher-energy photon emitters such as 131I, but is rarely seen with radionuclides such as 99mTc which emit lower-energy photons. Septal penetration produces a characteristic starburst artifact, a consequence of the paths of least attenuation through collimator septa and the hexagonal shape of the collimator apertures. The starburst artifact often seen with 131I is mainly due to septal penetration of several gamma-rays with energies greater than ~500 keV rather than the 364-keV gamma-rays actually used in imaging 131I. These higher-energy gamma rays may be down-scattered into the 364-keV photopeak energy window.

Question 74

For an occupationally exposed radiation worker, such as a nuclear medicine physician or a radiologist, the annual maximum permissible dose (i.e., effective dose equivalent) is _____. A. 0.5 rem/year B. 1 rem/year C. 5 rem/year D. 10 rem/year

Answer 74

C. 5 rem/year

50 mSv/year

Question 75

All of the following populations have shown evidence of increased risk of induced cancer attributable to radiation exposure, except _____. A. patients who have undergone PET/CT scans B. the A-bomb survivors C. patients who have undergone chest fluoroscopy for tuberculosis D. the radium dial painters E. patients who have undergone external-beam radiation therapy

Answer 75

A. patients who have undergone PET/CT scans
The National Academy of Science BEIR (Biological Effects of Ionizing Radiation) VII Report reviewed available radiation epidemiological evidence, as have various UNSCEAR (United Nations Scientific Committee on Effects of Atomic Radiation) reports. Answers B through E are high-dose studies providing evidence for increased cancer risk attributable to radiation. PET/CT scans have not been causally implicated in any such study to date, likely due to the low overall doses. Preston et al. studied the A-bomb survivors and noted that statistically significant effects have not been seen in sub-populations exposed at less than about 15 rem. The AAPM (American Association of Physicists in Medicine) and the HPS (Health Physics Society) have stressed that evaluating risk at doses less than about 5 to 15 rem is not appropriate because of the large uncertainty in any such risk estimation in that dose range.

Question 76

Based on studies conducted in animal models, the dose of radiation required to double the mutation rate in humans is roughly _____. A. 10 mSv B. 50 mSv C. 10 mSv D. 1 Sv E. 5 Sv

Answer 76

D. 1 Sv
The doubling dose of radiation-induced germ-cell (i.e., heritable) mutations in man, largely based on studies in animal models such as mice, is estimated as1 Sv (= 100 rem).

Question 77

A busy female interventional radiologist routinely wears both a personal dosimeter on her collar above the lead apron and a second dosimeter on her torso beneath the apron. Her average annual dose readings over the past five years have been 60 mSv to the collar badge and 2.4 mSv to the body (torso) badge. She declares that she is 3-months pregnant. Based on current National Council on Radiation Protection and Measurement (NCRP) recommendations, she _____. A. should reduce her procedure volume by 25% B. should reduce her procedure volume by 50% C. should reduce her procedure volume by 75% D. may maintain her current procedure volume E. should immediately stop working with radiation

Question 78

Which of the following is the correct relationship among the effective doses (EDs) of a chest radiograph, a mammogram, and an FDG PET/CT scan? A. chest radiograph > mammogram > FDG PET/CT B. chest radiograph < mammogram < FDG PET/CT C. chest radiograph = mammogram = FDG PET/CT D. The effective dose (ED) cannot be determined for any of these three imaging procedures.

Answer 78

B. chest radiograph < mammogram < FDG PET/CT
The effective dose (ED), a single-value metric of overall stochastic risk (i.e., risk of cancer and heritable genetic damage), is the weighted sum of the dose equivalents to the various tissues of the body; it can be calculated for any type of radiation exposure. Radiation doses (i.e, effective doses) to patients from diagnostic imaging procedures can vary widely, from as low as ~0.1 mSv for a chest radiograph to ~3 mSv for a mammogram to 5 to 20 mSv for such common procedures as a chest/abdomen/pelvis CT scan and FDG PET/CT scan

Question 79

The penetrability of an ultrasound beam _____ the frequency. A. decreases with the square of B. increases with the square of C. is directly proportional to D. is inversely proportional to E. is unrelated to

Answer 79

D. is inversely proportional to
Attenuation in ultrasound is proportional to the frequency. Lower-frequency probes emit an ultrasound beam with greater tissue penetrability, while higher-frequency probes emit less-penetrating beams.