RAPHEX II Flashcards

Question 1

Q

does Sr-90 emit beta or gamma?

Answer

A

ONLY beta \
Ir-192, Co-60, Cs-137 emit betas but followed by gamma or xrays

Question 2

Q

approximation for Brems production efficiency

Answer

A

~ 1 % at diagnostic
-for MV, T* Z *(10^-9)

Question 3

Q

max Brems energy

Answer

A

energy of the electron

If this goes on to create characteristic xray, then max characteristic xray has energy of Emax - binding energy

Question 4

Q

the relative mass dnsity of bone is 1.7 times that of water. The relative electron density of bone is expected to be?

Answer

A

less than 1.7 times that of water because of high hydrogen content of water
-hydrogen has no neutrons and thus has high electron density

Question 5

Q

HU range for lung, fat, muscle, soft tissue and bone

Answer

A

lung- -500
fat- -50 to -100
lung- 40
soft tissue- 100
bone- 1000

Question 6

Q

the effective (or useful) dose rate of a FFF xray beam as compared to a beam with a FF will:

Answer

A

decrease with increasing field size
-FFF beams have higher dose rate in center of beam but offaxis they are only slightly higher than filtered beams

Question 7

Q

which beam modality would have greatest % increase in dose rate for a FFF beam as compared to FF beam?

a. 6 MV xrays
b. 15 MV xrays
c. 6 MeV electrons
d. 15 MeV electrons

Answer

A

15 MV xrays
Brems increases with energy
-also FF thckness increases with beam energy

Question 8

Q

what is changed for different energy electron beams on linac?

Answer

A

magnet current

Question 9

Q

advantages of dynamic wedge vs physical wedge

Answer

A

-approx same depth dose as open beam
-field height not limited
-therapists do not have to lift a heavy wedge
-less scatter dose outside field

Question 10

Q

when pairs of abutting AP/PA photon fields are used to cover a treatment area, it is beneficial to have equal collimator settings for both sets of fields because?

Answer

A

-eliminates hot and cold spots
-remember it is (match depth * SSD/L1) - (match depth* SSD/L2)

Question 11

Q

what does DICOM RT store?

Answer

A

treatment plan info

Question 12

Q

what does DICOM 3D store

Answer

A

pixel values and relevant patient info

Question 13

Q

what is in DICOM header file?

Answer

A

all patient info, study type etc. but not pixel values

Question 14

Q

An Ir-92 can always be safely installed in which tpe of room without additional shielding?

Answer

A

MV linac room
NOT PET because max energy of Ir-192 is 1.1 MeV; PET energy is 0.511 MeV

Question 15

Q

cumulative exposure limit for a lifetime

Answer

A

10 mSv * age

Question 16

Q

if a gantry mounted kV xray imager acquires an image while the MV treatment beam is on, the soft tissue contrast in that image would be… comparted to when MV beam is off

Answer

A

due to increased scatter, image will be noisier
therefore worse contrast

Question 17

Q

what does indexing the immobilization device do?

Answer

A

ensures immobilization device is on same position on couch

Question 18

Q

disadavtange of RF beacons

Answer

A

create distortions on MR images

Question 19

Q

can one implanted fiducial say anything about rotation or translation?

Answer

A

no because fiducial could have moved… also need pts to define a plane and determine rotation… but if fiducial did not move wrt anatomy, could determine translation only

Question 20

Q

if the edge of an xray field is defined by MLC with rounded leaf ends, the size of the light field used for patient setup wioth be what in comparison to actual xray field size?

Answer

A

smaller than actual xray field size because MLCs will let some radiation penetrate through them

Question 21

Q

why are spoliers used to TBI?

Answer

A

to increase dose at the surface

Question 22

Q

In SRS, if patient is aligned using only frame coordinate system, it is essential to verify all of the following
except:

-dynamic conformal arcs are used for treatment
-frame pos. has not changed between simulation and tx
-localixzation rods attached to head frame are visible in CT or MR
-treatment room lasers are accurately aligned with linac iso

Answer

A

a. dynamic conformal arcs are one of many methods

Question 23

Q

the advantages od treating RSEB using pairs of slightly angled beams instead of a single horizontal beam include all of the following except:
-taller patients can be treated
-reduced xray contamination dose to patient
-more uniform dose distribution
-shorter total tx time

Answer

A

-shorter tx time

more Bremss in center of beam (forward directed)- ie more contamination in center

-takes twice as long to treat with 2 beamcs but dose is more uniform

Question 24

Q

dose rate for TBI

Answer

A

should be < 0.2 Gy/min to minimize side effects

Question 25

Q

a patient is being simulated for SBRT. What is different about the CTSim process?

Answer

A

SBRT often uses non-coplanar beams, so need to extend sup-inf scan length

Question 26

Q

reasons for sharp dose fall-off in SBRT

Answer

A

-non-coplanar beams
-intermediate beam energy
-small collimator to skin distance

Question 27

Q

main reason for sharp fall-off in gamma knife

Answer

A

-small collimator to skin distance

Question 28

Q

which of the following statements are Ir-192 HDR sources is NOT true?:

Ir-192 is beta minus decay
Ir-192 eits xrays of several different energies
3.the dose distribution is isotropically symmetric
d. the dose vs distance from the source approximately obeys the IS law for distances < 5 cm

Answer

A

c
dose is anisotropic

also, for d> 5 cm, dose falls off faster than IS due to attenuation

Question 29

Q

how often dose a radiation shielding survey of adjacent areas need to be done for Ir-192 HDR source per NRC

Answer

A

whenever source is changed

Question 30

Q

purpose of using multiple lumans in HDR breast

Answer

A

better dose distribution

Question 31

Q

the fraction of dose delivered during a HDR treatment while source is on transit changes how with activity of the source?

Answer

A

-lower for lower activity source
-transit time is fixed so if source is higher activity then more dose is delivered

Question 32

Q

dose at 20 cm away for 12.7 cGy/s source?

Answer

A

(1/20)^2 * 12.7 cGy7/c

remember that cGy goes with cm and uGy goes with m

Question 33

Q

bakground counts = 400
total counts = 1600

percent standard dev of net counts

Answer

A

sigma = square root (400+1600)
net counts = 1200
percent = 3.7%

Question 34

Q

typical max range of proton therapy beam

Answer

A

30 cm

max energy about 250 MeV

Question 35

Q

do extremities require anti-scatter grids?

Question 36

Q

which of the following can be associated with deterministic radiation effects?
1. PET/CT
2. echo cardiography
3. dual energy CT
4. fluoro
5 SPECT/CT

Answer

A

-fluoro
-can get up to 10 Gy at skin which causes radiation burns

Question 37

Q

what does dual energy CT improve?

Answer

A

contrast resolution

Question 38

Q

tube current of fluoro vs radiography

Answer

A

fluoro is 1/100 of radiography

Question 39

Q

residence time in nuc med

Answer

A

1.44 * initial activity * effective half life
-include % uptake in initial activity; for example, 40% uptake in thyroid is 0.4 uCi/uCi administered

Question 40

Q

treshold energy for photodisintegration

Question 41

Q

a bi-exponential function when plotted on a semi-log graph looks like what?

Answer

A

2 straight line segments

Question 42

Q

which of the following statements regarding DAP is incorrect?
1 it is independent of distance from xray tube
2. it is highest dose to any portion of the patient’s skin during the procedure
3. it is the total amt of energy imparted to a patient during a procedure’
4. it is typically measured at surface of xray tube port

Answer

A

2 is wrong- this is peak skin dose

DAP is independent of distance because IS cancels out with area

Question 43

Q

change in Z, A, and N, with positron decay

Answer

A

A positron is a positively charged electron, having an elementary electric charge of +1. Simply stated, it is produced within the nucleus by the conversion of a proton to a neutron and a positron, with the emission of the positron produced from the nucleus. Positron decay thus reduces the atomic number by 1 to Z–1, increases the neutron number by 1 to N+1, and leaves the mass number A unchanged, as positrons (and ordinary negative electrons) have a negligibly small mass compared to those of neutrons and protons.

Question 44

Q

ROC curve stuff

Answer

A

Sensitivity = TP / (TP + FN)

Specificity = TN / (TN + FP)

Positive predictive value (PPV) = TP / (TP + FP)

Negative predictive value (NPV) = TN / (TN + FN)

accuracy is area under ROC curve

Question 45

Q

the radiation weighing factor is a simplified version of what?

Question 46

Q

The parameter that provides the best estimate of skin radiation dose during a fluoroscopy procedure is _
A. the displayed kerma area product B. the displayed dose at the interventional reference point C. the total fluoroscopic time used for the study D. the total number of DSA frames used for the study E. both A and B

Answer

A

B. the displayed dose at the interventional reference point The displayed air kerma (AK) is defined at a point 15 cm below isocenter at the interventional reference point (IRP). However, to accurately calculate skin dose, correction factors must be applied for the distance from the patient’s skin entrance to the IRP position, attenuation of the x-rays caused by the table pad, angulation of the x-ray tube, etc. In general, and without proper correction factors, the displayed AK will overestimate skin dose if there is gantry motion during the procedure. The same kerma-area product (KAP) can be delivered using a lower dose and a larger field of view, or a higher dose and a smaller field of view.

Question 47

Q

residual activity in nuclear medicine

Answer

A

activity that remains in flasks or syringe (not all of it is injected into patient)

Question 48

Q

excess of protons aligned parrallel vs anti-parralell 3T MRI

Question 49

Q

accuracy of diagnostic test

Answer

A

The accuracy of a diagnostic test equals (TP + TN) / (TP + FP + TN + FN). Note that the denominator is simply the total number of all subjects. In this example, the accuracy equals (420 + 340) / (420 + 45 + 340 + 40) = 0.90

Question 50

Q

what happens to Z, N in beta minus decay

Answer

A

the neutron-to-proton (N/Z) ratio decreases from parent to daughter In beta-minus decay, a neutron is converted to a proton plus a negative electron, with the emission of the electron as a beta-minus particle. As a result, the neutron number of the decaying nucleus decreases from N to N–1, and the atomic number increases from Z to Z+1, so that the neutron-to-proton ratio decreases from N/Z to N–1 / Z+1. Beta particles are emitted with a range, or spectrum, of energies from near-zero to a maximum, or end-point, energy characteristic of the particular radionuclide. Since the atomic numbers of the parent and daughter are different (Z and Z+1, respectively), they are nuclides of different elements and, therefore, have different chemical properties. A transition energy Q of at least 1.022 MeV is required for positron (beta-plus) but not negatron (beta-minus) decay. Following beta-minus decay (as well as other forms of decay), emission of characteristic atomic radiations commonly, but not always, occurs.

Question 51

Q

dose from tomosynthesis vs mammo

Answer

A

they are comparable

Question 52

Q

The axial resolution in ultrasound imaging can best be improved by using…?

Answer

A

a shorter pulse length Higher frequencies yield shorter pulse length and, thus, improved axial resolution

Question 53

Q

Which of the following reconstruction options would improve visibility of low-contrast structures in a chest CT scan?

A. use of a lung filter B. use of a bone filter C. use of a soft-tissue filter D. reconstructed of thinner transverse slices

Answer

A

use of a soft-tissue filter Lungs and bone filters are sharpening filters designed to enhance spatial resolution; these will increase image noise, however, and thus degrade low-contrast resolution. Reconstruction of thinner slices will result in fewer counts per reconstructed voxel and, thus, also increase noise and degrade low-contrast resolution. Soft-tissue filters, on the other hand, are low-pass filters that effectively suppress noise in the reconstructed images, but with some loss of spatial resolution.

Question 54

Q

Which of the following is not a contributing factor to average glandular dose in a mammogram? A. x-ray tube target material B. x-ray tube focal spot size C. x-ray tube voltage D. x-ray beam half-value layer (HVL) E. degree of breast compression

Answer

A

focal spot size

Question 55

Q

Which of the following will improve low-contrast resolution in a radiographic image? A. Change the anti-scatter grid from a grid ratio of 10:1 to one with a grid ratio of 8:1. B. Move the patient closer to the image receptor. C. Reduce x-ray tube current. D. Use a smaller field of view. E. Increase the x-ray tube voltage.

Answer

A

use smaller FOV
will reduce scatter

Question 56

Q

Which of the following modalities employs a storage phosphor to capture the x-ray signal? A. indirect digital radiography B. direct digital radiography C. computed radiography D. film-screen radiography E. xeroradiography

Question 57

Q

In Doppler ultrasound, which of the following angles is considered the maximum angle between the Doppler beam line and the direction of blood flow for which acceptably accurate measurements of blood velocity can be obtained? A. 12 degrees B. 30 degrees C. 60 degrees D. 70 degrees E. 84 degrees

Answer

A

Note that the Doppler shift is proportional to cosθ and, thus, increases as the angle θ decreases, reaching a maximum for θ = 0 degrees (flow directly towards or directly away from the transducer) and decreasing to 0 (i.e., no measurable Doppler shift) for θ = 90 degrees. An angle of θ = 60 degrees is considered the maximum angle at which the Doppler shift can be reliably measured

Question 58

Q

The American College of Radiology (ACR) SPECT phantom is able to assess three features of image quality. What are they?

Answer

A

contrast, resolution, uniformity

Question 59

Q

A patient is scheduled for an FDG PET scan, but the technologist discovers the patient received 185 MBq (5 mCi) of 131I just 24 hours earlier. The technologist calls you for guidance as to whether the FDG PET scan should proceed. A. The FDG study may be performed because the 131I emissions are singles that have a negligible impact on the coincidence detection of the 18F scan at the low (185-MBq) 131I activity.
ANSWER
B. There is a problem because 131I emits some coincidence 511-keV photons. C. There is a problem because 131I emits high-energy photons that would interfere with the coincidence detection of 18F annihilation gamma-rays.
D. The procedure should not be performed because the total radiation dose to the patient from the two diagnostic scans would be unacceptably high.
E. The FDG scan is not recommended because the FDG uptake might perturb the radioiodine distribution.

Answer

A

A. The FDG study may be performed because the 131I emissions are singles that have a negligible impact on the coincidence detection of the 18F scan at the low (185-MBq) 131I activity.

Question 60

Q

what is a laminar flow hood

Answer

A

A laminar flow hood is a cabinet with an air-flow system designed to minimize the possibility of air-borne radioactive materials escaping from the hood and contaminating the radiopharmacy and the radiopharmacists. The air-flow system directs radioactive particulates or gases into a filter trap. A laminar flow hood is not otherwise sealed, however

Question 61

Q

Which of the following performance measurements is not required as part of the acceptance testing of a dose calibrator? A. sensitivity B. geometry C. linearity D. constancy E. accuracy

Answer

A

Sensitivity (i.e., the measured signal per unit activity) is not a clinically relevant measure of dose calibrator performance.

Question 62

Q

One of the commonly used features in radiomics that describes the “peakedness” of an image attribute is illustrated graphically below. This feature is called positive if the distribution is narrower than a normal distribution and negative if broader than a normal distribution. What is the name of this radiomics metric?

Answer

A

excess kurdosis

Question 63

Q

n MRI, flow void is more likely to occur for _____. A. T1-weighted images in the presence of slow blood flow B. T1-weighted images in the presence of fast blood flow C. T2-weighted images in the presence of slow blood flow D. T2-weighted images in the presence of fast blood flow

Answer

A

One of the commonly used features in radiomics that describes the “peakedness” of an image attribute is illustrated graphically below. This feature is called positive if the distribution is narrower than a normal distribution and negative if broader than a normal distribution. What is the name of this radiomics metric?

Question 64

Q

PET outperforms SPECT in each of the following except _____. A. sensitivity B. energy resolution C. spatial resolution D. count-rate performance E. overall image quali

Answer

A

energy resolution The energy resolution of lutetium yttrium orthosilicate (LYSO), the most commonly used scintillation detector in current PET systems, is 16% to 20%. The energy resolution of thallium-doped sodium iodide, the scintillator detector used in gamma cameras, is considerably better, only 8% to 10%.

Answer 60

A

water motion is more restricted in the tumor than in the surrounding glandular tissue Diffusion-weighting causes reduced signal in regions where water diffuses more freely. The tumor has a higher signal because water does not diffuse as freely in the tumor as in the surrounding glandular tissue.

Answer 61

A

collimator septal penetration

Answer 62

A

cd/m2 and lux

Answer 63

A

diagnostic display monitors The abbreviation “GSDF” stands for “Grayscale Standard Display Function” and is a characteristic of display monitors. It is usually measured with a specific set of patterns featuring a series of gray-level steps.

Answer 64

A

examinations following orthopedic surgery CT slices produced with a mobile C-arm have poor low-contrast performance and exhibit pronounced motion blurring since each scan takes several seconds. Therefore, the most appropriate use would be for imaging following orthopedic surgery, since the objects of interest have high contrast and can be immobilized to eliminate motion

Answer 65

A

2160 mGy-cm The dose-length product (DLP) is equal to the CTDIvol times the scan length. The CTDIvol is equal to the CTDIw divided by the pitch. Therefore, DLP = (60 mGy/0.5) (18 cm) = 2,160 mGy-cm.

Answer 66

A

provide the radiologist with a 2D mammogram with no additional radiation dose to the patient Some digital breast tomosynthesis (DBT) units combine the information from the tomographic slices to produce conventional-looking mammograms for the radiologist at no additional dose to the patient.

Answer 67

A

80 kV, 200 mA, 0.4 s The patient entrance dose is proportional to the product of the square of the kV, and the mA times the exposure time. Combination A gives the largest value: 512,000

Answer 68

A

the geometric mean of the impedances of the PZT and the skin

Answer 69

A

interference among echoes from scatterer

Answer 70

A

Based on Snell’s law, sin(incident angle) / sin(transmitted angle) = c(medium 1) / c(medium 2) where c is the speed of the wave. Since c(fat) = 1,450 m/s and c(muscle) = 1,580 m/s, the angle of refraction as the wave passes from medium 1 (fat) into medium 2 (muscle) must be larger than the incident angle θ.

Answer 71

A

proportional to the acoustic power of the beam Intensity, by definition, is the rate at which energy passes through a unit area. The time average intensity is equal to the power in a wave divided by the area over which the power is dispersed, with units watts/m2. Of course, for the ultrasound system, other intensity descriptors, such as the pulse average intensity, result in significantly greater values than the time average intensity.

Answer 72

A

to test contrast ratio in contrast-enhanced digital mammography (CEDM) According to the website https://www.sunnuclear.com/shopgammex/mammo-ffdm-phantom, the new mammo FFDM phantom:
“Evaluates artifacts over the entire detector with a single image. Measures and monitors FFDM systems’ signal to noise, resolution and image quality. Simulates radiographic characteristics of compressed breast tissue. Eliminates backscatter and provides quick detection of objects from 0.14 to 1.0 mm.”
The new phantom is not designed for CED

Answer 73

A

18 cm The length of the near field for a given transducer is defined as d2 / 4 × lambda, where lambda is the wavelength of the beam.

Answer 74

A

PTCA Gd-DTPA is used for MRI, iodine-127 (the stable isotope of iodine) for CT, protein-shelled microbubbles for ultrasound, and carbon dioxide DSA. The abbreviation “PTCA” stands for “percutaneous transluminal coronary angioplasty.”

Answer 75

A

not applicable Section 21 CFR 1020.32 Fluoroscopic equipment provides the framework for regulatory requirements. Sub-section 1020.32(f) recognizes that there is a difference between regular fluoroscopy and cine fluoroscopy. 1020.32(d) establishes limits only for fluoroscopy, and there is no exposure rate limit specified for cine fluoroscopy.

Answer 76

A

sin(theta) = 1.22 * lambd/d

d is transducer diameter

Answer 77

A

(difference of impedance/sum of impedance)^2

Answer 78

A

improved lateral resolution, poorer temporal resolution Increasing the number of transmit focal zones will increase the lateral resolution over the image. Each additional focal zone that is added will require a separate excitation of the array for each transmit focal distance. However, the finite pulse transit time in tissue limits the transmit pulse repetition frequency. Because of this, and because subsequent processing is needed for each of the multiple focal zones being resolved, temporal resolution will suffer as this process will take more time to complete per frame, thus decreasing the total frame rate.

Answer 79

A

increase
Decreasing the depth setting using a given transducer will generally increase the frame rate. The reduction of image depth will decrease the amount of time it takes to acquire pulse-echo data for each beam line, so for a fixed number of “A” lines the frame rate will increase.

Answer 80

A

Nothing needs to be done; the region will be resolved. Axial resolution is equal to one-half of the spatial pulse length (SPL), as illustrated in the figure below. Therefore, any boundaries that are separated by a distance greater than one-half the SPL will prevent pulse overlap and, thus, be resolved. In this question, the boundaries are separated by 0.7 mm in the axial direction, and one-half the SPL is 0.5 mm given that the SPL is 1 mm. Since the boundary separation is greater than one-half the SPL, the boundaries will be resolved.

Answer 81

A

1.5 cm
The intensity half-value thickness (HVT) is defined as the thickness of material that will reduce the initial ultrasound beam intensity by half. Relative intensity is a log-based calculation defined as 10 log(I2/I1) where I1 and I2 are the comparative intensities of the incident and the returning echoes of the beam. Solving this equation for a relative intensity that is decreased by one-half will yield a result of a 3-dB reduction. Thus, any half-value thickness will reduce the ultrasound beam by 3 dB. This can then be used in the relationship HVT(cm) = 3 dB / [attenuation coefficient (dB/cm)] or in terms of frequency, HVT (cm) = 3 dB / [(0.5dB/cm) /MHz × f(MHz)]. Solving this for the given values yields 1.5 cm

Answer 82

A

geometric accuracy Physical and mechanical inspection, image uniformity, system sensitivity, and electronic image display performance are required quality assurance tests that must be performed annually for ACR-accredited ultrasound units. All other specified test are optional.

Answer 83

A

liver-kidney Intensity reflection at a boundary is determined by the acoustic impedance (Z) of the two boundaries at any given interface. The larger the difference in the acoustic impedance between two mediums, the larger the intensity reflection of the ultrasound beam. As indicated in the table below of pressure and reflection coefficients for various interfaces, the liver-kidney boundary has the smallest difference in acoustic impedance between the tissues and will have the least amount of intensity reflection.

Answer 84

A

The average latency period of radiation-induced leukemia is about five years. Based on the Hiroshima and Nagasaki A-bomb survivor data, the average latency period for radiation-induced leukemia is about five years. The biological effectiveness of alpha particles will not change with dose fractionation. With acute radiation exposure, the higher the dose, the shorter the prodromal syndrome. The higher the LET of a radiation, the lower the oxygen enhancement ratio (OER). Finally, radiation does not produce new, unique mutations, but merely increases the incidence of spontaneously occurring types of mutations.

Answer 85

A

Radiation-induced cancer has no defined threshold dose. Although controversial, radiation-induced cancer does not appear to have a definite threshold dose. In general, cells that rarely divide are the most radioresistant. The dose for gastrointestinal toxicity in humans is estimated to be 10–12 Gy. Among the Japanese A-bomb survivors, an increased incidence of GI tumors—including liver, stomach, colon, and esophageal tumors—has been demonstrated

Answer 86

A

49 mSv The total effective dose received by the radiation worker is 25 mGy × 1 (weighing factor, wr, for x-rays) + 0.2 mGy × 20 (weighing factor, wr, for heavy ions) + 10 mGy × 2 (weighing factor, wr, for protons) = 25 + 4 + 20 = 49 mSv. The fact that this was a whole-body irradiation means that all organs received the same absorbed dose

Answer 87

A

8 to 15 weeks
In utero irradiation at sufficient doses can cause mental retardation, with sensitivity for this effect being greatest at a gestational age of 8 to 15 weeks. Radiation induction of mental retardation is a deterministic effect, the threshold for which is not exceeded in diagnostic imaging studies.

Answer 88

A

NCRP
The NCRP (National Council on Radiation Protection and Measurements) is an authoritative advisory (non-governmental) organization whose mission is “to support radiation protection by providing independent scientific analysis, information, and recommendations that represent the consensus of leading scientists.” The FDA (Food and Drug Administration), DOT (Department of Transportation), and NRC (Nuclear Regulatory Commission) are governmental agencies that regulate various aspects of the medical use of radioactive materials; the latter two agencies also regulate various aspects of non-medical uses of such materials.

Answer 89

A

0.25 Gy Radiation can induce a variety of chromosomal abnormalities in mammalian cells, including dicentric chromosomes that can be detected in peripheral blood lymphocytes after doses as low as ~0.25 Gy (or 25 cGy).

Answer 90

A

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 table top, and for c-arm units the exposure rate is measured 30 cm in front of the detector.

Answer 91

A

5%
According to the National Council on Radiation Protection and Measurements (NCRP), the lifetime cancer risk estimate is about 5% (or 0.05) per sievert.

Answer 92

A

may maintain her current procedure volume Current National Council on Radiation Protection and Measurements (NCRP) recommendations limit fetal exposure to 0.5 mSv per month. The radiologist’s body badge provides a preliminary estimate of expected fetal dose. The badge’s reading of 4 mSv per year is equivalent to 0.33 mSv per month. This is two thirds of the guidance limit.

Answer 93

A

not limited

Answer 94

A

medical only is 3 mSv

total with everything else is 6 mSv

Answer 95

A

0.1 mGy in any week National Council on Radiation Protection and Measurements (NCRP) Report No. 147 recommends a shielding design goal (in air kerma) of 0.1 mGy per week for controlled area

I guess 20 uSv/h is not a design goal but just a condition to meet

Answer 96

A

0–10 mGy According to National Council on Radiation Protection and Measurements (NCRP) Report 174, Table 3.2, most diagnostic medical imaging examinations subject the embryo or fetus to doses of less than 10 mGy. Exceptions to this may be prolonged fluoroscopic procedures (including fluoroscopically guided interventional procedures), CT examinations (especially multiphase) of the abdomen and pelvis, and several nuclear medicine studies (e.g., myocardial-perfusion study, PET).

Answer 97

A

permanently for the current baby following administration Guidance on precautions for lactating mothers who receive radioactive material can be found in the Nuclear Regulatory Commission (NRC) NUREG-1556, Appendix U, and in the International Commission on Radiological Protection Publication 128, Table D.1. Although there are currently no universally recognized guidelines, recommended lengths of time nursing
mothers should be advised to interrupt breast feeding include the following: 24 hours for 99mTc radiopharmaceuticals; 12 to 24 hours for 18F radiopharmaceuticals or 68Ga radiopharmaceuticals; 7 days for 111In-white blood cells or 123I radiopharmaceuticals; 14–28 days for 201Tl-chloride, 67Ga-citrate, or 89Zr radiopharmaceuticals; and permanently for the currently nursing baby for 131I or 124I radiopharmaceuticals.

Answer 98

A

bone marrow cells Radiosensitivity is the relative susceptibility of cells, tissues, organs, etc. to the injurious action of radiation. In general, it has been found that cell radiosensitivity is directly related to the rate of cell division and inversely related to the degree of cell differentiation. This means that actively dividing, relatively undifferentiated cells are at greatest risk of damage by radiation. As such, bone marrow cells have the highest radiosensitivity among those listed.

wr= 0.12

Answer 99

A

an increase in the incidence of thyroid cancer The most significant radiation effect of the Chernobyl nuclear reactor accident was a significant increase in the incidence of thyroid cancer, most notably among children. This was attributed to the radiation dose to the thyroid delivered by iodine-131, which was released to the environment in large amounts and picked up by people in milk, drinking water, etc.

Answer 100

A

radium; GBq The curie is a conventional (i.e., non-System Internationale (SI)) unit of activity based on the activity in 1 gram of radium; it corresponds to 3.7 × 1010 disintegrations per second (dps). The SI unit is the becquerel (Bq), which is defined as an activity of 1 dps. Therefore, 1 Ci = 3.7 × 1010 Bq = 37 ×x 109 Bq = 37 GBq

Answer 101

A

0.035 mGy/minute The x-rays produced by the fluoroscope will scatter off of the patient in all directions. The scatter radiation at a meter is approximately 0.1% of the incident exposure rate. In the scenario, the DAP meter showed 35 mGy/min, making the scatter radiation at one meter 35 mGy/min × 0.001 = 0.035 mGy/min in the nurse’s position.

Answer 102

A

increased high-contrast spatial resolution Adding filtration to the x-ray beam eliminates lower-energy scatter that wouldn’t normally reach the detector. This will harden the x-ray beam and lower the patient’s entrance skin dose. While this does lower the dose, increasing the x-ray beam’s penetration can decrease subject contrast due to less absorption in the patient.

Answer 103

A

ncreasing the tube potential Increasing the tube potential will increase the penetration of the x-rays passing through the patient’s skin. This will, in turn, lower the total skin dose and the dose to the surrounding staff because there will be less scatter at the surface of the patient and less absorption. Likewise, with the automatic exposure rate control engaged, the mA will generally be lowered to preserve the baseline image quality, thus ultimately decreasing the dose to the patient’s skin

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