RAPHEX V

Question 1

Mammography x-ray tubes typically use a lower tube voltage (kVp) than those used in general radiography in order to _____. A. reduce dose to the patient B. reduce the image exposure time C. utilize characteristic x-rays exclusively D. reduce shadowing artifacts E. improve image contrast

Answer 1

A lower x-ray tube voltage (i.e., kVp) produced an x-ray beam with lower-energy x-rays than a higher tube voltage. The proportion of x-rays interacting in tissue or other media via the photoelectric effect rather than Compton scatter increases with decreasing x-ray energy. The lower tube voltage used in mammography systems thus improves image contrast by reducing the number of Compton-scattered photons contributing to the image. This strategy (i.e., the use of lower tube voltages) is not adaptable to general radiography, however, because lower-energy x-rays, while able to penetrate the breast, would not penetrate the torso or other thicker body parts in sufficient numbers to produce a statistically reliable image at an acceptably low radiation dose to the patien

Question 2

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 2

E

Question 3

A hypothetical x-ray beam has a uniform energy spectrum, as shown below. Which of the lettered graphs best represents the beam’s energy spectrum after passing through and exiting a uniform slab of polystyrene? Assume photoelectric effect is the dominating absorption mechanism.

Answer 3

Remember uniform spectrum doesn’t mean mono-energetic! - just means same number of photons at all energies
-beam hardening occurs

Question 4

The image below illustrates a feature now commonly utilized in CT scanning. This feature is known as _____.
ANSWER
A. spectral filtration B. adaptive beam collimation C. cardiac-motion tracking D. iterative reconstruction E. tube-current modulation

Answer 4

e- look up what it looks like
To compensate for different degrees of attenuation in different anatomic regions of the body, and to maintain patient radiation doses as low as practical, the CT scanner’s x-ray tube current varies. 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 5

In an x-ray tube, the line focus principle is used to make the effective focal spot _____ the actual focal spot, which _____ the power loading of the anode. A. smaller than; increases B. smaller than; decreases C. the same as; maintains D. larger than; increases E. larger than; decrease

Answer 5

A. By angulating the anode with respect to the direction of travel of the incident electrons, the effective focal spot size will be smaller than the actual focal spot size. This improves spatial resolution by reducing penumbra blurring while dispersing the incident electron beam and associated heat deposition over a larger area of the anode

Question 6

A prostate-specific membrane antigen (PSMA)-binding small molecule labeled with lutetium-177 (physical half-life, Tp = 6.65 days) is eliminated from the body of a prostate cancer patient with a biological half-life, Tb, of 1.20 days. What is the effective half-life, Te, of lutetium-177 in this patient’s body? A. 1.02 days B. 1.20 days C. 6.65 days D. 7.98 days E. 8.85 days

Answer 6

A. A radioactive material administered to a patient is eliminated from a tissue or organ of the
patient by the simultaneous processes of radioactive decay (with physical half-life Tp = 6.65 days in this example) and metabolism, secretion, excretion, etc. (with biological half-life Tb = 1.20 days in this example). The overall elimination of the radioactivity is more rapid than if it were eliminated only by radioactive decay or only by biological processes, and the resulting effective half-life (Te) is given by the following formula

Question 7

A thyroid cancer patient is imaged after iodine-131 therapy. What is the cause of the star artifact shown below?
A. A metal object was in the field of view. B. The photopeak energy window was set to that for technetium-99m instead of that for 131I.

C. The collimator septa was penetrated by high-energy 131I gamma-rays.
D. There was crystal damage. E. There was a non-functioning photomultiplier tube (PMT)

Answer 7

C.
In addition to the imageable 364-keV gamma-ray emitted by iodine-131 in 81% of its decays, iodine-131 emits a 637-keV gamma-ray in 7.3% of its decays. This high-energy photon easily passes through the collimation designed for the 364-keV gamma-ray and, when scattered, may appear in the photopeak energy window set for the 364-keV photon, producing the observed star artifac

Question 8

Assuming the same crystal and collimator pair above, if the source is moved such that the distance between the source and the collimator is increased, what will be the effect on the resolution and sensitivity? A. The spatial resolution improves and the sensitivity decreases. B. The spatial resolution worsens and the sensitivity increases. C. The spatial resolution improves and the sensitivity increases. D. The spatial resolution worsens and the sensitivity stays the same. E. There will be no effect at all.

Answer 8

As a source is moved farther away from the collimator, the inverse-square law results in a lower number of photons reaching the detector. However, this is counterbalanced by the increased total aperture area of the collimator, through which photons can pass. Therefore, the sensitivity does not change. The spatial resolution is best with the source in contact with the collimator. As the source-to-detector distance is increased, photons strike the crystal over a larger area, resulting in degradation of spatial resolution.

Question 9

In a C-arm type of fluoroscope having a source-to-image distance (SID) less than 45 cm (a mini C-arm), the air-kerma rate (AKR) shall be measured at _____. A. the location specified in the manufacturer’s manual B. 30 cm from the focal spot C. 30 cm from the input surface of the fluoroscopic imaging assembly D. minimum SSD E. 1 cm above the input surface of the fluoroscopic imaging assemb

Answer 9

D. Based on Code of Federal Regulations (CFR) 1020.32 (d)(3)(iv), in a C-arm type of fluoroscope having an SID less than 45 cm, the AKR shall be measured at the minimum SSD

Question 10

For an air-kerma rate (AKR) exceeding 6 mGy/min, the displayed values shall not differ from the actual values by more than _____. A. the tolerance specified by the manufacturer B. 10% C. 2 mGy/min D. 5% for the range from 6 to 25 mGy/min and 10% for an AKR exceeding 25 mGy/min E. 35%

Answer 10

E.
Based on Code of Federal Regulations (CFR) 1020.32 (k)(6), the displayed AKR shall not deviate from the actual values by more than ±35% over the range of 6 mGy/min to the maximum indication of AKR.

Question 11

In fluoroscopy, “grid controlled” refers to _____. A. a rapid control of the tube output by a grid inside the tube B. elimination of scatter based on the proper selection of the scattering grid C. dynamic variation of the Bucky speed based on the AKR. D. a fixed selection of fluoro protocol parameters based on the procedure and the patient thickness
E. an evaluation of the SNR as a function of patient thickness

Answer 11

A.

Question 12

The units for dose-area product (DAP) and air-kerma rate (AKR) are _____ and _____, respectively. A. mGy; mGy/sec B. mGy-cm2; mGy/min C. R-cm2; mGy/R/sec D. mGy-cm; mGy-cm/min E. mGy/R-cm2; R/sec

Answer 12

B. mGy-cm2; mGy/min

Question 13

The benefit of using additional copper filtration in interventional fluoroscopic units is to _____. A. create a more homogenous x-ray field by attenuating slightly more in the center B. reduce overall patient dose by augmenting the spectrum with the copper K_alpha lines C. reduce patient dose by entrance skin sparing D. soften the contrast to enhance low-visibility blood vessels E. enhance the contrast in iodine contrast studies

Answer 13

C.
A copper filter (e.g., 0.11 mm thick) will significantly reduce the number of low-energy photons, which, based on their energy, are mainly absorbed in the patient’s skin. Therefore, this is used to achieve skin sparing.

Question 14

Which of the following is not a regulatory requirement for a stationary C-arm type of fluoroscope having a source-to-image distance (SID) less than 45 cm (a mini C-arm)? A. It must be labeled “For extremity use only.” B. The SSD must not be less than 10 cm. C. The fluoroscopic irradiation time has to be displayed within 6 seconds of termination of an exposure.
D. The display of the air-kerma rate (AKR) shall be clearly distinguishable from the display of the cumulative air kerma.
E. During cinefluoroscopy, the AKR shall not exceed 44 mGy/min.

Answer 14

E. There is no regulatory limit specified for AKR in cinefluoroscopy

Question 15

equation for DQE

Answer 15

(SNRout/SNRin)^2

Question 16

The brightness gain in a typical fluoroscopy system (12-inch diameter input phosphor) is approximately _____. A. B. C. D.
7.2 72 720 7,200 E. 72,000

Answer 16

D.
Brightness gain = electronic gain × minification gain. In a typical system, the electronic gain is about 50 and the minification gain for a 12-inch input phosphor to a 1-inch output phosphor is 144. Therefore, the brightness gain is 7200.

Question 17

kVp and mAs- how are they adjusted for a large patient?

Answer 17

kVp increase will spare dose to patient but cotnrast will be worse
increasing mAs and keeping kVp low will be good contrast but more dose

Question 18

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

Answer 18

B. 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

• I would argie that d is also right? Or I guess ratio of what is attenuated vs not is still same??? - to me less is attenuated due to high kV so less heel effect

Question 19

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

Answer 19

A. 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 speed of sound in air, which is highly compressible, is 330 m/s, while the speed in soft tissue, which is far less compressible, is 1540 m/s.

-Impedance affects what gets reflected/transmitted

Question 20

effective dose to adult from chest CT scan

Answer 20

~ 10 mGy

Question 21

Flow void is more likely to occur for _____. A. T1-weighted MR images with slow blood flow B. T1-weighted MR images with fast blood flow C. T2-weighted MR images with slow blood flow D. T2-weighted MR images with fast blood flow

Answer 21

D. Due to blood flow, the excited protons in blood in the tissue cross-section corresponding to the MR image may have exited that tissue section by the time of readout. Also, moving protons (as in flowing blood) accumulate additional phase shift compared to stationary protons. For T2-weighted images, the loss in signal is due to the fact that the fast-moving blood will miss the 180o re-phasing pulse, which reduces the readout signal. T1 has a shorter echo time (TE), so there will be less de-phasing.

Question 22

The American College of Radiology (ACR) reference dose for an adult head CT scan is _____. A. 15 mGy B. 25 mGy C. 35 mGy D. 40 mGy E. 75 mGy

Answer 22

E. The American College of Radiology (ACR) CT reference dose—actually, the volume CT dose index, CTDIvol—for an adult head study is 75 mGy (7.5 rad). Other ACR CT reference doses are, for example, as follows: pediatric abdomen, 15 mGy (1.5 rad); adult abdomen, 25 mGy (2.5 rad); pediatric head, 35 mGy (3.5 rad

Question 23

The ACR pass/fail criterion for an adult head CT scan is _____. A. 20 mGy B. 30 mGy C. 35 mGy D. 40 mGy E. 80 mGy

Answer 23

E. The American College of Radiology (ACR) CT pass/fail criterion for an adult head study is 80 mGy (8.0 rad). Other ACR CT pass/fail criteria are, for example, as follows: pediatric abdomen, 20 mGy (2.0 rad); adult abdomen, 30 mGy (3.0 rad); pediatric head, 40 mGy (4.0 rad).

Question 24

Multi-slice CT scanners are more advantageous than single-slice scanners because of _____. A. lower patient effective dose B. improved high-contrast spatial resolution C. decreased heat load on the machine D. increased scan time E. lower capital (i.e., purchase) cost of the scanner

Answer 24

C.
Decreased heat load on the machine is associated with multi-slice CTs compared to single-slice CT scanners. Patient effective dose is not reduced because the same patient volume is irradiated in both multi-slice and single-slice scanners. High-contrast spatial resolution is not improved. The scan time is decreased because data for more than one tissue section is collected per rotation. The purchase price of a multi-slice scanner is higher than that of a single-slice scanner.

Question 25

For which of the following CT images would one utilize the widest display window? A. head B. chest C. abdomen D. pelvis

Answer 25

B. The wide range of linear attenuation coefficients and, therefore, Hounsfield units (HUs) among the tissues in the chest—lung, soft tissue, and bone—generally requires a wide display window for optimal visualization of all such tissues

Question 26

How is the noise in a CT image affected when the patient is scanned with twice the slice thickness and one-half of the mA, assuming all other parameters are the same? A. It remains the same. B. It decreases by a factor of 2 times. C. It decreases by a factor of 4 times. D. It decreases by a factor of 8 times. E. It decreases by a factor of 2 times

Answer 26

A. The CT noise decreases by √2 by doubling the slice thickness. The CT noise increases by √2 by using only one-half of the mA. Therefore, since √2 / √2 = 1, there is no change in the CT image noise.

Question 27

accuracy of a diagnostic test

Answer 27

(TP + TN) / (TP + FP + TN + FN)

Question 28

In fluoroscopy, the greatest radiation exposure to the fluoroscopist is _____. A. the exit beam from the patient B. scatter from the patient C. leakage radiation from the x-ray tube D. scatter from the image intensifier

Answer 28

B.
At fluoroscopic x-ray energies, there is considerable side (i.e., 90-degree) scatter from the patient toward the fluoroscopist. The exit-beam dose from the patient is smaller due to attenuation by the patient. The leakage radiation is very small due to federal design requirements. Scatter from the image intensifier would also be a very small fraction of the exit beam from the patient.

Question 29

The patient scatter at the position of the fluoroscopist from a certain fluoroscopic procedure is 5.5 Gy of air kerma. The fluoroscopist is standing 50 cm from the patient, and the area of the fluoroscopy x-ray field entering the patient is 20  25 cm. If the x-ray field is enlarged to 25  30 cm, the scatter reaching the fluoroscopist would be approximately _____. A. B. C. D.
3.2 Gy 5.5 Gy 8.3 Gy 9.6 Gy
E. 11.3 Gy

Answer 29

C. The amount of scatter radiation produced is proportional to the area of the x-ray field entering the patient, so the new amount of scatter will be [(25) (30)] / [(20) (25)]  5.5 Gy = 1.5  5.5 Gy = 8.25 Gy ≈ 8.3 Gy.

Question 30

Changing the field of view of an older image intensifier tube from 30-cm mode to 15-cm mode will result in a change in spatial resolution from 1.2 line pairs (lp)/mm to _____. A. 0.3 lp/mm B. 0.6 lp/mm C. 1.2 lp/mm D. 2.4 lp/mm E. 4.8 lp/mm

Answer 30

D.
The spatial resolution of the image from the image intensifier will vary approximately as one over the diameter of the image. Therefore, changing the field of view of an image intensifier from 30 to 15 cm changes the resolution from 1.2 lp/mm to (30 cm / 15 cm)  1.2 lp/mm = 2  1.2 lp/mm = 2.4 lp/mm.

Question 31

A fluoroscopy unit with a flat-panel detector image intensifier (II) has a spatial resolution of 1.0 line pair/mm (lp/mm). Changing the field of view from 25-cm mode to 15-cm mode will result in a spatial resolution of _____. A. 0.5 lp/mm B. 1.0 lp/mm C. 1.5 lp/mm D. 2.0 lp/mm E. 3.0 lp/mm

Answer 31

B.
The spatial resolution of the image for a flat panel detector does not change with field size unless binning is employed, typically for field sizes of 30 cm and larger.

Question 32

The maximum allowed patient input air kerma rate with high-level control activated is _____. A. 88 mGy/min B. 176 mGy/min C. 350 mGy/min D. 700 mGy/min E. There is no upper limit

Answer 32

B. The maximum allowable air kerma rate (AKR) is 176 mGy/min (or 20 R/min input exposure rate.

Question 33

The typical material with which x-rays interact as the first step in producing a fluoroscopic image is _____. A. CsI B. ZnCdS C. amorphous selenium D. a photosimulable phosphor E. a CCD camera

Answer 33

A.
Interactions of x-rays with CsI (cesium iodide) is the first step in producing a fluoroscopic image. It serves as both the input phosphor on an image intensifier tube and the x-ray absorber in a flat-panel detector used for fluoroscopy.

Question 34

Switching from a 0.5-mm thick lead apron to a 0.25-mm thick lead apron will increase x-ray exposure to the wearer by a factor of about _____. A. B.
5

C. 12 D. 16 E. 32

Answer 34

B.
For the typical x-ray energy spectrum, a 0.5-mm-thick lead (or lead-equivalent) apron transmits 3% to 5% of the incident x-rays. Switching to a 0.25-mm-thick lead (or lead equivalent) apron would allow transmission of 12% to 17% of incident x-rays. This amounts to about a five-fold increase in the transmitted radiation and exposure to the wearer.

Question 35

The x-ray tube target most commonly used in current digital breast tomosynthesis units is _____. A. tungsten B. molybdenum C. rhodium D. silver

Answer 35

A. While molybdenum (Mo) and rhodium (Rh) targets were widely used x-ray tube targets for screen-film and many full-field digital mammography (FFDM) units, the higher average energy in the tungsten x-ray spectrum is more compatible with the detector absorption and the overall low-dose requirements of digital breast tomosynthesis units

Question 36

For contrast-enhanced digital mammography, two images of the breast are taken with iodine contrast material injected. One image is a conventional image with low x-ray tube voltage (kVp) and a thin filter. The second image is obtained with a kVp/filter combination of _____. A. 25 kVp/aluminum B. 30 kVp/rhodium C. 49 kVp/copper D. 70 kVp/aluminum E. 90 kVp/silver

Answer 36

C.
X-ray units designed for mammography are limited to x-ray tube voltages less than 50 kVp. However, in order to harden the beam for the second x-ray image, a higher-atomic-number (Z) copper filter works better than aluminum (Al) or rhodium (Rh), which will not harden the x-ray beam sufficiently, while silver (Ag) will excessively attenuate the x-ray beam.

Question 37

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

Answer 37

D.
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 a scanning fan-beam.

Question 38

For the test objects shown in the American College of Radiology (ACR) Accreditation Phantom below, which of the following must be imaged in order to pass the applicable ACR criteria?

Answer 38

B. The four largest fibers, the three groups with the largest specks, and the three largest masses are required to be seen in order for the facility to pass ACR accreditation

Question 39

The United States Food and Drug Administration (FDA), through the Mammography Quality Standards Act (MQSA), requires that monitors used for viewing digital mammograms must have _____ brightness and _____ spatial resolution than monitors used for viewing of conventional digital radiographs. A. the same; the same B. the same; greater C. greater; the same D. greater; better E. None of the above is true. The MQSA does not have specifications for viewing monitors

Answer 39

D. The Mammography Quality Standards Act (MQSA) specifies that the monitors used for viewing digital mammograms must have a minimum of 5 megapixels (5 MP), a calibrated, sustained maximum luminance of at least 450 candela (cd)/m2, (preferably 600 cd/m2), and a contrast ratio of at least 350:1. Conventional viewing monitors generally have fewer megapixels, lower luminance, and a lower contrast ratio.

Question 40

what is cause for less intense region in a uniformity flood image in nuc med?

Answer 40

A cold (i.e., less intense), circular, regularly shaped focal defect such as that in the figure is most likely a photomultiplier tube (PMT) defect caused, for example, by uncoupling of the PMT (i.e., physical detachment) from the back of the scintillation crystal or the PMT simply not working.

Question 41

In quality control (QC) of nuclear medicine instrumentation, a long-lived surrogate radionuclide is often used in place of a shorter-lived radionuclide used clinically. Among the following, _____ are such surrogate radionuclide-clinical radionuclide pairs. A. 68Ge and 18F B. 133Ba and 131I C. 57Co and 99mTc D. All of the above are true.

Answer 41

D.

Question 42

what is doubling dose

Answer 42

The doubling dose is the dose required to produce a mutation incidence that is twice the spontaneous level and, based on animal data, it is estimated to be 1 Sv (100 rem

Question 43

what model is endorsed for dose response?

Answer 43

linear no treshold

-linear quadratic model is sub-linear
-supra-linear moderl is not creditable

-LNT is picked over linear qudratic to be more conservative

Question 44

Which of the following is true in terms of the effects of high-dose radioiodine therapy of thyroid cancer on the subsequent reproductive function of patients? A. There is no effect. B. There is generally a temporary decrease in reproductive function (fertility) followed by complete recovery, but with recovery potentially taking years to occur.
ANSWER
C. There is generally a temporary decrease in reproductive function (fertility) followed by complete recovery, but with recovery potentially taking only weeks to occur.
D. Sterility is always induced and is always irreversible (i.e., is permanent). E. It is not known if there are any effects—the results of follow-up studies of the reproductive function post-radioiodine therapy among thyroid cancer patients are inconclusive.

Answer 44

B.
The limited patient data available on reproductive function (fertility) following high-dose radioiodine therapy are suggestive of a slow (up to several years) but complete recovery of such function. While fertility rates appear to be reduced within the first one to two years post-treatment, they appear to be similar to historical averages by approximately 10 years post-treatment.

Question 45

lifetime excess fatal cancer risk

Answer 45

.
According to the National Council on Radiation Protection and Measurements (NCRP), the life-time excess fatal-cancer risk estimate is 5% (or 0.05) per Sievert (Sv) or 0.05% (or 0.0005) per rem.

Question 46

A post-thyroidectomy thyroid cancer patient is treated with 7,400 MBq (200 mCi) of iodine-131 (131I)-iodide. Immediately following the 131I administration, the measured dose rate at a distance of 1 meter from the patient is 0.02 mSv/h (2 mrem/h). In compliance with the applicable radiation protection regulations, this patient can be released from the hospital _____. A. immediately B. one day after the 131I administration C. two days after the 131I administration D. four days after the 131I administration E. The time of release cannot be determined based on the information provided.

Answer 46

A.
Section 35.75 (Release of Individuals Containing Unsealed Byproduct Material or Implants Containing Byproduct Material) of 10 CFR Part 35 (Medical Use of Byproduct Material) permits a licensee to “authorize the release from its control any individual who has been administered unsealed byproduct material or implants containing byproduct material if the total effective dose equivalent to any other individual from exposure to the released individual is not likely to exceed 5 mSv (0.5 rem).” According to Column 2 in Table U.1 (Activities and Dose Rates for Authorizing Patient Release) in Appendix U (Model Procedure for Release of Patients or Human Research Subjects Administered Radioactive Materials), this 5-mSv (0.5-rem) dose-limit requirement will be satisfied for iodine-131 (131I) and the patient may be released if the dose rate at 1 meter is less than 0.07 mSv/h (7 mrem/h). In the current example, the measured dose rate at a distance of 1 meter from the patient is only 0.02 mSv/h (2 mrem/ h). Therefore, this patient may be released immediately following the 131I administration, that is, may be treated on an out-patient basis.

Question 47

The exposure rate measured at 1 m from the surface of a package containing radioactivity is 0.004 R/h. Therefore, the transport index (TI) of this package is _____. A. 0 (there is no transport index (TI) for this package) B. 1 C. 2 D. 4 E. 8

Answer 47

D. The transport index (TI) of a radioactive package is defined as the exposure rate in mR/hr (milliroentgen per hour) measured at 1 m from the surface of the package. For a package yielding an exposure rate of 0.004 R/h = 4 mR/h at 1 m, the TI is, by definition, 4. As always, one must pay careful attention to the unit

Question 48

The regulatory effective-dose (ED) limit for a patient undergoing a fluorine-18 (18F)-fluorodeoxyglucose PET/CT scan is _____. A. not limited by any regulation B. 200 mSv (20 rem) C. 100 mSv (10 rem) D. E.
50 mSv (5 rem) 10 mSv (1 rem

Answer 48

A.
In applying the principle of optimization of protection of the patient, the benefits and detriments are received by the same individual, the patient, and the dose to the patient is determined principally by the medical needs. Dose constraints or limits for patients are, therefore, inappropriate. The limitation of dose to the individual patient is not recommended because it may, by reducing the effectiveness of the patient’s diagnosis or treatment, do more harm than good. The emphasis is then on the justification of the medical procedures and on the optimization of protection. See International Council on Radiation Protection (ICRP) Publications 103 and 105 for additional discussions of this topic. In complex fluoroscopic procedures, there is a risk for acute skin tissue effects (deterministic effects) that should be evaluated according to guidance in National Council on Radiation Protection and Measurements (NCRP) Report No 168.

Question 49

How long should nursing mothers be advised to interrupt breast feeding following administration of iodine-131 (131I)? A. There are no breast-feeding precautions needed. B. 2 days following administration C. 4 days following administration D. 8 days following administration E. permanently for the current bab

Answer 49

E.
Guidance on precautions for lactating mothers who receive radioactive material can be found in the Nuclear Regulatory Commission (NRC) NUREG-1556, Appendix U. Although there are currently no universally recognized guidelines, nursing mothers should be advised to permanently discontinue breast-feeding for the currently nursing baby following the administration of iodine-131 (131I)-labeled radiopharmaceuticals.

Question 50

integral uniformity

Answer 50

U ≡
(Maximum counts per pixel – Minimum counts per pixel ) /(Maximum counts per pixel + Minimum counts per pixel)
X 100%

Question 51

max acceptable value for gamma camera integral uniformity

Answer 51

5%

Question 52

when to evaluate gamma camera uniformity

Answer 52

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

Question 53

For the fluorine-18 (18F)-fluorodeoxyglucose (FDG) PET below, _____ is the standard uptake value (SUV) for the lung lesion indicated by the arrow. As indicated, the activity concentration in this lesion is 26 kBq/cm3. The patient was injected with 370 MBq of 18F-FDG and weighs 86 kg.

Answer 53

E. 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 26 kBq/cm3 in a patient with a total-body mass of 86 kg or 86,000 g who received an administered activity of 370 MBq or 370,000 kBq, the SUV is 26 kBq/g / (370,000 kBq / 92,500 g) = 26 kBq/g / 4.3 kBq/g = 6, assuming a tissue mass density of 1 g/cm

Question 54

In the planar radiograph below, the mean number of x-rays per pixel in the circular structure and the background are 1,020 and 900, respectively. The contrast-to-noise ratio (CNR) for the structure corresponding to the circular area is _____.
ANSWER
A. 0.8 B. 1.0 C. 2.0 D. 4.0 E. 6.5

Answer 54

(1020-900)/30’
4

Question 55

For cone-beam CT scanners, such as those used in some SPECT/CT systems, which of the following statements is incorrect? A. Cone-beam CT scanners are less expensive than spiral CT systems. B. In SPECT/CT systems, the cone-beam CT image set can be registered and fused with the SPECT image set.
C. In SPECT/CT systems, the cone-beam CT data can be used for attenuation correction of the SPECT data.
D. Cone-beam CT images are of comparable quality to spiral CT images and can be used for clinical diagnoses.

Answer 55

D. Cone-beam CT images are somewhat poorer in overall quality than conventional spiral CT images (i.e., are not of diagnostic quality) and, therefore, cannot be used for radiologic diagnoses.

Question 56

The Mammography Quality Standards Act (MQSA) requires that the average glandular dose per image measured in a 4.2-cm-thick breast phantom simulating 50% glandularity is less than _____. A. B. C.
1 mGy 3 mGy 6 mGy
D. 10 mGy E. The MQSA does not specify any such dose limit.

Answer 56

The Mammography Quality Standards Act (MQSA) requires that the average glandular dose per image measured in a 4.2-cm-thick breast phantom simulating 50% glandularity is less than 3 mGy (300 mrem) per image. This MQSA dose limit actually applies only to the phantom, not to any patient. Failure of a mammography system to meet and MQSA requirement, including the phantom dose limit, requires immediate suspension of clinical mammography and corrective action.

Question 57

Compared to full-field digital mammography, the radiation dose to the breast from tomosynthesis is about _____. A. 80% lower B. 40% lower C. the same D. 40% higher E. 80% higher

Answer 57

Compared to full-field digital mammography (FFDM), the radiation dose to the breast from tomosynthesis is about 40% higher. A recent European study (Gennaro et al., Eur. Radiol. 28:573–81, 2018) found that the average breast tomosynthesis dose, 1.9 mGy (190 mrad), was about 38% higher than the FFDM breast dose, 1.4 mGy (140 mrad).

Question 58

duak energy CT

Answer 58

Dual-energy CT involves acquisition of projection data at two different x-ray tube voltages, generally at about 80 and 140 kVp, and results in improved image contrast among different soft tissues.

Question 59

In ultrasound, an attenuation of 4 dB corresponds to a reduction of the incident beam intensity by _____. A. 20% B. 30% C. 40% D. 50% E. 60%

Answer 59

E.
The intensity (I) of an ultrasound wave in decibels (dB) relative to some reference, or incident, intensity (Io) is given by the expression –10 log(I/Io). For 4 dB, 4 = –10 log(I/Io) and, therefore, log(I/Io) = –0.4 and (I/Io) = 10–0.4 = 0.4. The ultrasound beam is thus attenuated to 40% of its incident intensity. That is, its intensity is reduced by 60%

Question 60

what is plotted on ROC curve

Answer 60

The receiver-operator characteristic (ROC) curve is a plot of the true-positive fraction (= sensitivity) versus the false positive fraction (1-specificity)

Question 61

The maximum activity that can be accurately assayed in a scintillation well counter is approximately _____. A. 37 Bq B. 37 kBq C. 37 MBq D. 37 GBq

Answer 61

B.
A scintillation well counter, because of its high intrinsic and geometric efficiencies, is a very-high-sensitivity counting device. It is, therefore, susceptible to significant dead-time count losses even at what may be considered low activities, about 37 kBq (1 μCi) or greater. A scin-tillation well counter, like most radiation measurement devices, is a so-called “paralyzable” detector: as one attempts to assay higher and higher activities beyond about 37 kBq (1 μCi), the measured count rate will actually decrease because of dead-time count losses.

Question 62

what is this phantom

Answer 62

ACR MRI phantom

It is filled with a solution of nickel chloride and sodium chloride. The phantom was issued to measure the following MRI scanner performance parameters: geometric accuracy, high-contrast spatial resolution, slice-thickness accuracy, slice-position accuracy, image-intensity uniformity, percent signal ghosting, and low-contrast object detectability.

Question 63

The tumor appears darker than the normal prostate peripheral zone in the ADC image (right panel) because _____. A. the tumor has a lower b-value than the normal prostate tissue B. the tumor has a shorter T1 than the prostate tissue C. the random motion of water is more restricted in the tumor than in the normal tissue D. the ADC is lower in the benign peripheral zone than in the tumor

Answer 63

The ADC map reflects random water diffusion within a voxel. A smaller ADC indicates restricted diffusion

-the higher the b value, the stronger the diffusion effects

Question 64

A T2-weighted fast-spin-echo image requiring 4 minutes employed the following parameters: repetition time TR = 4,000 ms, echo time TE = 80 ms. The technologist decided to reduce the TR to 1,000 ms without changing any other parameters in order to shorten the acquisition time. Which of the following statements is true? A. There was no change in contrast since TE was not changed. B. Proton-density weighting increased because TR was shorter. C. The image represents a mix of T1 and T2 contrast because TR was reduced. D. The specific absorption rate (SAR) decreased because TR was reduced.

Answer 64

C. For T2 weighting, an image should have a relatively long repetition time (TR > 2,500 ms) so that complete or nearly complete T1 relaxation occurs between excitations. With TR equal to 1,000 ms, T1 relaxation will not be complete and the resulting image will reflect both T1 and T2 of the tissue.

Question 65

A single-slice spin-echo MR image is acquired using the following parameters: FOV = 360 mm, matrix = 256  256 voxels, number of acquisitions = 2. If on the next acquisition the matrix size is decreased to 256  128 and all other parameters are kept the same, the signal-to-noise ratio (SNR) will _____. A. increase by a factor of 2 B. decrease by a factor of 2 C. increase by a factor of the square root of 2 D. decrease by a factor of the square root of 2

Answer 65

A.
In MRI, the signal-to-noise ratio (SNR) is proportional to voxel volume. Decreasing the matrix size in one dimension by a factor of two while keeping the field of view the same increases the voxel volume by a factor of two and, therefore, increases the SNR by a factor of two

Question 66

what is MRI SNR proportional to>

Answer 66

-voxel volume
-square root of number of excitations or number of signals averaged* phase steps
-latter is also square root of time

Question 67

If the prescribed field of view (FOV) is smaller than the size of the object to be imaged, the resulting image will contain a wrap-around or aliasing artifact. Of the following techniques, _____ can be used to eliminate such aliasing. A. increasing the slice thickness B. over-sampling the data in the phase-encoding dimension C. using a navigator echo D. over-sampling the data in the frequency-encoding dimension

Answer 67

B. Acquiring a higher density of samples (i.e., over-sampling) in the phase-encoding dimension will effectively increase the field of view and reduce or eliminate aliasing.

-frequency encoding direction would work too no…. this is what is typically done…

Question 68

The gadolinium molecule in gadolinium-based contrast agents is paramagnetic. Which of the following statements is false? A. Gadolinium-based contrast agents can be used to enhance the conspicuity of lesions with leaky microvessels.

B. Gadolinium enhances contrast on T1-weighted images by shortening the T1 value of nearby water protons.
C. To maximize gadolinium contrast, a very long repetition time TR (>4,000 ms) should be used.
D. Gadolinium-based contrast agents are excreted mainly via the kidneys.

Answer 68

C.
Gadolinium contrast is maximized in T1-weighted images, which have a repetition time TR in the 2 ms to 500 ms range

Question 69

n ultrasound, _____ is the tissue boundary interface that results in the greatest intensity reflection. A. muscle-bone B. liver-kidney C. fat-muscle D. muscle-lung

Answer 69

The muscle-lung boundary has the largest difference in acoustic impedance (due to the large differences in mass density) and will, therefore, have the greatest intensity reflection.

Question 70

The length of the near field for a given transducer of diameter d

Answer 70

d^2/4 lambda

Question 71

angle of divergence of US beam

Answer 71

: sin(theta) = 1.22(lambda/d)

Question 72

reflection coefficient at US boundary
transmitted is 1-R

Answer 72

((Z2-Z1)/(Z2+Z1))^2

Question 73

If Doppler imaging is used to measure the velocity of the blood flow through the aorta in the image above, _____ would best describe the relevant parameters of the transducer used. A. light damping, high Q B. light damping, low Q C. heavy damping, high Q D. heavy damping, low Q

Answer 73

A.
Blood velocity measurements using Doppler imaging require narrow bandwidths in order to preserve the subtle changes in frequency from the echo of the blood flow. This requires a light-damping, high-Q transducer, which produces a long spatial pulse length by allowing more cycles to occur and produces a smaller range of frequencies within the pulse (as illustrated in the figure below). Q is defined as the ratio of the center frequency to the bandwidth (fo / bandwidth).

Question 74

A transducer frequency is used such that a spatial pulse length of 1 mm is produced. If a region of interest has boundaries that are separated by 0.4 mm in the axial direction, what, if anything, should be done in order to resolve the two boundaries? A. increase the frequency B. decrease the frequency C. increase the power D. decrease the power E. Nothing needs to be modified, as the boundaries will already be resolved.

Answer 74

A. A boundary needs to be separated by a dis-tance greater than one half the spatial pulse length (SPL) in order to prevent pulse over-lap, as illustrated in this figure. This defines axial resolution in ultrasound. If pulses over-lap, an object will not be resolved. In the question, the boundary is separated by 40 mm, and one half the SPL is 50 mm given that the SPL is 100 mm. Since the boundary separation is less than one half the SPL, the region will not be resolved, so one must increase the frequency (shortening spatial pulse length) in order to resolve the boundary.

Question 75

A _____ test is required by the American College of Radiology (ACR) to be performed as part of an annual survey of an ultrasound unit. A. system sensitivity B. geometric accuracy C. spatial resolution D. contrast resolution

Answer 75

A. The American College of radiology (ACR) requires the following tests be performed annually on accredited ultrasound units:

annual test requirements for US:
-physical and mechanical inspection
-image uniformity and artifact survey
-system sensitivity
-US electronic image display performance
-evaluation of QC program

Question 76

potentially harmful zone for prolonged US exposure

Answer 76

B.
While generally considered safe, the “potentially harmful” zone threshold for prolonged ultrasound exposure is around 0.1 W/cm2, as shown in the figure below

Question 77

The _____ scanning mode produces the highest spatial peak-temporal average intensity, ISPTA (in mW/cm2). A. B-mode B. M-mode C. pulsed Doppler D. color flow

Answer 77

C. Various parameters are used to measure intensities in pulse US. The various parameters are shown below, along with a table showing the various intensity parameters for the different scanning modes used in ultrasound. The table shows that power and spatial peak pulse average are the highest for Color flow imaging. However, pulsed Doppler has significantly higher spatial peak-temporal average (parameters shown below) than all the other scanning modes due to the prolonged pulsed nature of the imaging mode

Question 78

_____ is the approximate velocity of blood flow for a 5-MHz transducer that produces a Doppler shift of 3 kHz at a Doppler angle of 30 degrees. (Assume the speed of sound in the surrounding soft tissue is 1540 m/s.) A. B.
0.5 cm/s 5 cm/s
ANSWER
C. 50 cm/s D. 500 cm/s

Answer 78

C. The Doppler shift frequency is described by the following relationship, fd = [2ft v cos(theta)] / c, where ft is the ultrasound frequency transmitted by the scan head, v is the velocity of blood flow, c is the speed of sound in the medium, and theta is the Doppler angle, as illustrated in the figure below. Rearranging to solve for v yields v = 53 cm/s.

Question 79

Which of the following statements is correct? A. Mitotic death is the most common form of radiation-induced cell lethality at the 50% lethal dose (LD50).
ANSWER
B. Apoptosis involves little or no morphological changes in cells. C. Apoptosis is promoted by an up-regulation of bcl2 gene expression in cells. D. In tumors, apoptosis is observed 24 hours post-irradiation. E. Apoptosis is generally considered to be a p53-independent process.

Answer 79

A.
In general, mitotic cell death (i.e., reproductive failure) is the most common form of radiation-induced cell death. Apoptosis usually occurs within 4–6 hours post-irradiation and is often associated with cell-membrane blebbing followed by budding. Apoptosis is a p53-dependent phenomenon and is inhibited by bcl2 expression

Question 80

A radiation worker reported that he was exposed to a radiation source one month earlier and had forgotten to wear his dosimeter. Which of the following assays is the best method to estimate the radiation dose received by this individual? A. Collect peripheral blood lymphocytes and stain for γ-H2AX foci. B. Perform a comet assay for DNA damage. C. Perform karyotyping to find peripheral blood lymphocytes. D. Stain and quantify for p53 foci induction. E. Perform alkaline elution to measure single-stranded DNA lesions.

Answer 80

C. Karyotyping of peripheral blood lymphocytes is the only method that can be used to estimate the dose under these circumstances. However, the single-exposure threshold dose for detectable radiogenic effects on lymphocytes is 25 cGy (25 rad).