• Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

Card Range To Study

through

image

Play button

image

Play button

image

Progress

1/117

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

117 Cards in this Set

  • Front
  • Back
1. Which model of radiobiology dose response is used in setting standards for radiation protection?
A. Linear nonthreshold
B. Linear threshold
C. Quadratic nonthreshold
D. Quadratic threshold
E. Exponential nonthreshold
1. Which model of radiobiology dose response is used in setting standards for radiation protection?
A. Linear nonthreshold
B. Linear threshold
C. Quadratic nonthreshold
D. Quadratic threshold
E. Exponential nonthreshold

Answer: A – on multiple old tests. This model is used b/c it overestimates risk.
2. A therapeutic dose of I-131 is administered to a female. Three hours later you discover that her urine pregnancy test was wrong and that she is in the early stages of pregnancy. What is the most appropriate next step?
A. Send to ER for stomach pumping and irrigation
B. Induce vomiting immediately
C. Administer 250mg KI then 100mg every 6 hours for 4 days
D. Administer single dose 250mg KI
E. Advise to increase fluid intake and urinate frequently
2. A therapeutic dose of I-131 is administered to a female. Three hours later you discover that her urine pregnancy test was wrong and that she is in the early stages of pregnancy. What is the most appropriate next step?
A. Send to ER for stomach pumping and irrigation
B. Induce vomiting immediately
C. Administer 250mg KI then 100mg every 6 hours for 4 days
D. Administer single dose 250mg KI
E. Advise to increase fluid intake and urinate frequently

Answer: C
2. Which of the following 30MBq-50MBq radionuclides should NOT be disposed as regular hospital waste after 1 year?
A. F-18
B. Co-56
C. Ga-67
D. Tc99m
E. Th-201
Half Life
Tc99m 6 hr
123I 13.3 hr
111In 2.8 days
67Ga 3.24 days
131I 8.05 days
89Sr 50.5 days
125I 60.2 days
57Co 270 days

th201 half life of 73 hours
. Why limit visitation and close contact with people receiving I-131 treatment?
A. Bremsstrahlung radiation
B. External gamma radiation
C. Internal Beta contamination
D. External Beta irradiation
E. Skin contamination with Beta radiation
. Why limit visitation and close contact with people receiving I-131 treatment?
A. Bremsstrahlung radiation
B. External gamma radiation
C. Internal Beta contamination
D. External Beta irradiation
E. Skin contamination with Beta radiation

Answer: B – Although there is both Beta and gamma radiation, the Beta radiation probability won't travel far enough to cause a problem.
. Which of the following is true regarding ventilation imaging in Nuclear medicine?
a. Gas should be lighter than air
b. Should be performed in negative pressure room
c. Use high Atomic number gas
d. Should be performed in a closed room with no external ventilation
e. Tech should be outside room during administration
5. Which of the following is true regarding ventilation imaging in Nuclear medicine?
a. Gas should be lighter than air
b. Should be performed in negative pressure room
c. Use high Atomic number gas
d. Should be performed in a closed room with no external ventilation
e. Tech should be outside room during administration

Answer: B
6. In US image with 250 scan lines obtained at 30 frames/second, what is the maximum depth (cm)?
a. 2
b. 5
c. 10
d. 15
e. 20
Answer: C – Wow, this is a hard question. It took about 30 minutes with an open book to figure this out. I don't think I could ever get it during a test. But, here is how you do it. There is a magical number of maximal time per line that is equal to 13 microseconds/cm. I don't know where that number comes from, but it is in the Bushberg syllabus (It was on slide 54 of the Ultrasound section in the 2008 version). This gives a magical equation that once simplified and corrected for units equals:
Frame rate = (77000cm*sec-1)/(N x D)
N is the number of scan lines and D is the depth in cm. So, if you plug in the numbers for this question:
30 sec-1 = (77000 cm*sec-1)/(250 x D)
D x 7500sec-1 = 77000 cm*sec-1
D = 77000cm/7500 = ~10 cm
6. An apparent increase in noise in digital vascular fluoroscopy is increased by:
A. Increased edge enhancement
B. Increased frame averaging
C. Increased pixel averaging
D. Increased rate of incident photons striking II
E. Use of progressive scan technology
6. An apparent increase in noise in digital vascular fluoroscopy is increased by:
A. Increased edge enhancement
B. Increased frame averaging
C. Increased pixel averaging
D. Increased rate of incident photons striking II
E. Use of progressive scan technology

Answer: A – edge enhancement in digital fluoroscopy has the same result as in CT. Edge enhancement in CT (aka bone windows) has better spatial resolution but increased noise. The difference is due to the filter used in the processing of the image. Edge enhancement uses a high pass ramp filter , which leave many of the high spatial frequencies (small objects). The high spatial frequencies contain spatial data such as the edge of the vessel or bone, which is why it is called edge enhancement. However, high spatial frequencies also contain 'all those little dots' that you see on bone windows that give the image a grainy appearance (noise). Clear as mud?
7. The kinetic energy of each detected photon from a 100 keV positron is:
a. 0.100 MeV
b. 0.256-0.500 MeV
c. 0.511 MeV
d. >1.02 MeV
7. The kinetic energy of each detected photon from a 100 keV positron is:
a. 0.100 MeV
b. 0.256-0.500 MeV
c. 0.511 MeV
d. >1.02 MeV

Answer: C – positron emission results in two 511 keV photons no matter what the energy of the positron.
8. Damage from a “dirty bomb” with 241-Am is from:
a. Ingestion
b. Skin absorption
c. Inhalation
d. Wound contamination
e. External irradiation
8. Damage from a “dirty bomb” with 241-Am is from:
a. Ingestion
b. Skin absorption
c. Inhalation
d. Wound contamination
e. External irradiation
Answer: C – Am-241 (americium) is an alpha emitter found in smoke detectors. It can be inhaled or ingested, so I'm unsure how the Mallenkrot people came to the answer C. However, they may know something I don't (high probability), so I'd go with inhalation on the test. The critical organ is the bone surface.
10. Which of the following organ systems is most sensitive to radiation induced congenital injury?
A. CNS
B. gastrointestinal
C. genitourinary
D. cardiovascular
E. musculoskeletal
10. Which of the following organ systems is most sensitive to radiation induced congenital injury?
A. CNS
B. gastrointestinal
C. genitourinary
D. cardiovascular
E. musculoskeletal

Answer: A – on multiple old tests.
11. Compared to a CXR obtained with a kVp of 80, an exam performed at a kVp of 120 and the same mAs will have:
A. increased contrast
B. increased rib conspicuity
C. increased latitude
D. greater motion unsharpness
11. Compared to a CXR obtained with a kVp of 80, an exam performed at a kVp of 120 and the same mAs will have:
A. increased contrast
B. increased rib conspicuity
C. increased latitude
D. greater motion unsharpness

Answer: C – the ribs would be less obvious. They are trying to trick us with a big word like conspicuity (those bastards). However, increasing kVp will lower contrast (due to less photoelectric effect). Lower contrast means increased latitude.
12. A characteristic of power Doppler is:
A. insensitive to motion
B. is limited by aliasing
C. provides better detail of vessel wall than color Doppler
D. can determine flow direction
E. is less subject to attenuation than color doppler.
12. A characteristic of power Doppler is:
A. insensitive to motion
B. is limited by aliasing
C. provides better detail of vessel wall than color Doppler
D. can determine flow direction
E. is less subject to attenuation than color doppler.

I think this one was poorly remembered. I have not read anything that would suggest any of these answers are correct. B and D are wrong. A is also wrong – Power Doppler is more susceptible to “flash artifacts” which occur with motion of the patient or transducer. Power Doppler is more sensitive to slow flow for whatever that is worth.
13. A region of decreased sensitivity within the center of a SPECT image of a cylinder containing a uniform concentration of activity is due to:
A. scattered radiation
B. attenuation artifact
C. beam hardening artifact
D. center of rotation artifact
E. Gibbs artifact
13. A region of decreased sensitivity within the center of a SPECT image of a cylinder containing a uniform concentration of activity is due to:
A. scattered radiation
B. attenuation artifact
C. beam hardening artifact
D. center of rotation artifact
E. Gibbs artifact

Answer: B – see notes below

Attenuation causes reduced activity in the center of the patient (15% per cm for Tc99m). Thus, it also causes increased quantum noise. This is more common in the body, chest and less in the brain. Non-uniform attenuation correction can be done with a radioactive source to measure transmission through the patient

Uniformity correction: Non uniformities such as collimator dents produce “ring artifacts” about the center of rotation (COR) in transverse images. This can be corrected by the system itself using high count floods.

COR or Axis of rotation correction: Uses point source to look for misaligned projections. This creates “doughnut artifacts,” which appears as a ring of activity instead of a point source. COR can also be fixed internally. These are usually checked weekly.

NOTE: COR cannot be checked with a flood source, so it cannot be the answer to this question. Also, the circular doughnut artifact descried above is theoretical. The resolution of the camera is not good enough to pick it up. So, poor COR simply results in decreased spatial resolution.
14. An appropriate mammographic technique for a magnification view of a 4.5 cm compressed breast with 50% glandular tissue and 50% fat is:
A. 22 kVp and 60 mAs
B. 22 kVp and 150 mAs
C. 26 kVp and 60 mAs
D. 26 kVp and 150 mAs
E. 45 kVp and 250 mAs
14. An appropriate mammographic technique for a magnification view of a 4.5 cm compressed breast with 50% glandular tissue and 50% fat is:
A. 22 kVp and 60 mAs
B. 22 kVp and 150 mAs
C. 26 kVp and 60 mAs
D. 26 kVp and 150 mAs
E. 45 kVp and 250 mAs

Answer: C – if this question looks familiar but different, that's b/c it is! The 2000 recall states the answer is 26kVp and 100mA. The learning point is mammography has 2 different mA levels. Use 30 mA for 0.1 mm focal spot (mag view) and 100 mA for 0.3 mm focal spot (regular mammogram). If you think about it, it makes total sense that the mA would be higher for larger focal spot because you need a larger cathode filament to make a bigger focal spot on the same anode. So, for a magnification view, you will use the 0.1mm focal spot (30mA). With the low mA of mag views, the exposure time is long, on the order of 2 seconds. Therefore, there are around 60mAs. Note the answers for this question are in mAs, not mA.
15. Which quality control procedure is being performed given the following scintillation camera and source arrangement?
A. intrinsic spatial resolution
B. extrinsic spatial resolution
C. intrinsic flood field
D. extrinsic flood field
15. Which quality control procedure is being performed given the following scintillation camera and source arrangement?
A. intrinsic spatial resolution
B. extrinsic spatial resolution
C. intrinsic flood field
D. extrinsic flood field

Answer: C – note the lack of collimator and the long SID.
16. Volume averaging in SPECT is most dependent upon?
A. reconstruction filter
B. type of collimator used
C. time window of the crystal
D. the width of the PHA spectrum
6. Volume averaging in SPECT is most dependent upon?
A. reconstruction filter
B. type of collimator used
C. time window of the crystal
D. the width of the PHA spectrum

Answer: B – collimator determines slice thickness in SPECT. Remember the collimators are parallel (except for the head, which are fan beam). The parallel septa demarcate the slices
Answer: B – collimator determines slice thickness in SPECT. Remember the collimators are parallel (except for the head, which are fan beam). The parallel septa demarcate the slices

17. In screen film radiography, poor screen film contact will result in:
A. decreased large area contrast
B. increased processing time
C. poorer spatial resolution
D. increased quantum mottle
17. In screen film radiography, poor screen film contact will result in:
A. decreased large area contrast
B. increased processing time
C. poorer spatial resolution
D. increased quantum mottle

Answer: C – I am told that poor contact makes the image blurry.
18. Individuals with which of the following diseases are most sensitive to radiation?
A. Ataxia telangectasia
B. Fanconi’s anemia
C. Basal cell nevus syndrome
D. Usher syndrome
E. Gardner Syndrome
18. Individuals with which of the following diseases are most sensitive to radiation?
A. Ataxia telangectasia
B. Fanconi’s anemia
C. Basal cell nevus syndrome
D. Usher syndrome
E. Gardner Syndrome

Answer: A – on multiple old tests.
19. For the given TIPS procedure, arrangement A has 3 mGy of dose. What is the dose of arrangement B (not drawn to scale)?




A. 0.1 mGray
B. 1.5 mGray
C. 3 mGray
D. 6 mGray
E. 12 mGray
19. For the given TIPS procedure, arrangement A has 3 mGy of dose. What is the dose of arrangement B (not drawn to scale)?




A. 0.1 mGray
B. 1.5 mGray
C. 3 mGray
D. 6 mGray
E. 12 mGray

Answer: D – Know how to use the inverse square law. Just be sure that the dose goes up if you move closer to the source (it's an obvious point, but you don't want to get the answer exactly backwards).
20. Keeping the exit exposure equal, increasing what parameter will decrease the entrance exposure:
A. kVp
B. mAs
C. Exposure time
D. increasing f-number of image intensifier aperture.
20. Keeping the exit exposure equal, increasing what parameter will decrease the entrance exposure:
A. kVp
B. mAs
C. Exposure time
D. increasing f-number of image intensifier aperture.

Answer: A – you should understand the general concepts. If the exit dose is the same and you increase the kVp, then the mA will be decreased. Increasing the other options will increase the dose. Note that the size of the aperture hole decreases with increasing f-number.
21. Regarding radiation induced damage to cells,
A. Rapidly dividing cells are more sensitive than slowly dividing cells
B. Double stranded breaks are more likely than single stranded breaks.
C. Oxygen level is more important for hight LET radiation compared to low LET radiation.
D. G1 is more radiosensitive than G2
21. Regarding radiation induced damage to cells,
A. Rapidly dividing cells are more sensitive than slowly dividing cells
B. Double stranded breaks are more likely than single stranded breaks.
C. Oxygen level is more important for hight LET radiation compared to low LET radiation.
D. G1 is more radiosensitive than G2

Answer: A – oxygen levels are more important for low LET radiation b/c oxygen contributes to indirect damage. High LET radiation mostly damages via direct effects. G1 is less radiosensitive than G2 b/c there is only one copy of the genome as the cell has not gone through S yet (less chance of damage).
22. In MR, the saturation of blood flowing into a slice following a 90 degree pulse will make the
A. Blood appear dark
B. Blood appear bright.
C. Dephase the blood
D. Rephase the blood
22. In MR, the saturation of blood flowing into a slice following a 90 degree pulse will make the
A. Blood appear dark
B. Blood appear bright.
C. Dephase the blood
D. Rephase the blood

Answer: A - I think, but this is a poorly worded question. Know about entrance slice enhancement (non-contrast time of flight MRA). The blood not previously in the field of view enters and is brighter than the blood there just before it. A presaturating 90 degree pulse outside the FOV will 'undo' this effect, and the blood will appear dark.
23. In MR, increasing the bandwidth of the RF receiver will
A. Increase the spatial resolution
B. Decrease the spatial resolution
C. Increase SNR
D. Decrease SNR
E. Increase imaging time
23. In MR, increasing the bandwidth of the RF receiver will
A. Increase the spatial resolution
B. Decrease the spatial resolution
C. Increase SNR
D. Decrease SNR
E. Increase imaging time

Answer: D –
Increasing bandwidth = Decreased imaging time
= Increased noise  Decreased SNR
= Decreased Chemical shift artifact.
-The SNR is increased by increasing slice thickness, decreasing matrix size, and re-
ducing RF bandwidth during signal detection.
-High magnetic-field strength also increases the SNR.
-SNR increases as the square root of the number of image acquisitions:
-
\INex•
-The tradeoff for increased SNR and the resultant improved image quality is an in-
crease in imaging time.
24. Given this intrinsic gamma camera quality control image, this defect is caused by:
A. Damaged collimator
B. Radioactivity contamination
C. Cracked Crystal
D. Proken PMT

Obviously we have no picture so just know this stuff.
24. Given this intrinsic gamma camera quality control image, this defect is caused by:
A. Damaged collimator
B. Radioactivity contamination
C. Cracked Crystal
D. Proken PMT

Obviously we have no picture so just know this stuff.
25. Geometric unsharpness in screen film radiography is worsened by
A. Increased focal spot size
B. Increase KVP
C. Decrease mAs
D. Increase screen thickness
Answer: A – tricky question. A and D both decrease spatial resolution, but geometric unsharpness is a function of focal spot.

-The penumbra is the result of x-rays arriving from slightly different locations in the
focal spot, because the focal spot is not a true point source but has a finite area.
-The resultant unsharpness is called focal spot blur or geometric unsharpness.
-Focal spot blur increases with magnification and focal spot size, as shown in Fig. 5.2.
-Reducing the focal spot size increases the sharpness and definition of object edges
by minimizing the penumbra.
26. Which arrangement in a phototimed screen/film radiography yields the lowest patient dose.
26. Which arrangement in a phototimed screen/film radiography yields the lowest patient dose.



Answer: D – The large SOD and the lack of a grid lowers dose. Also I think they meant to specify that they want skin dose. There is a difference between skin dose and “total dose.” Synonyms for total dose are integral dose and effective dose. Integral dose is total dose to the whole body and is measured in Joules (J). There is an important distinction between integral dose and “absorbed dose” which we measure in Grays (Gy) or rads. The distinction is that a Gy is a Joule per kg. Make sure you understand this concept.
27. The amount of scatter that occurs within a patient during an abdominal radiograph is decreased by
A. Increased mAs
B. Increased kVp
C. Increased distance between the patient and screen.
D. removing the grid
27. The amount of scatter that occurs within a patient during an abdominal radiograph is decreased by
A. Increased mAs
B. Increased kVp
C. Increased distance between the patient and screen.
D. removing the grid

Answer: B or C – Poorly worded question. To decrease the amount of scatter that occurs, increase kVp. Air gaps and grids decrease the amount of scatter imaged.
28. The shadowing that occurs at the edge of the cyst in ultrasound occurs from
A. image attenuation
B. refraction
C. reverberation
D. resonance
E. reflection
28. The shadowing that occurs at the edge of the cyst in ultrasound occurs from
A. image attenuation
B. refraction
C. reverberation
D. resonance
E. reflection

Answer: B
. The spread of data in a set of numbers is given by
A. range
B. mode
C. median
D. average
29. The spread of data in a set of numbers is given by
A. range
B. mode
C. median
D. average

Answer: A
31. What is the effective half life if biological half life is 6 days and physical half life is 2 days?
A. 0.5 days
B. 1.0 days
C. 1.5 days
D. 2.0 days
E. 3.0 days
31. What is the effective half life if biological half life is 6 days and physical half life is 2 days?
A. 0.5 days
B. 1.0 days
C. 1.5 days
D. 2.0 days
E. 3.0 days

Answer: C - Here is how you do it:
Teff = (Tb x Tp)/(Tb + Tp)
Teff = (6 x 2)/(6 + 2)
Teff = 12/8
Teff = 1.5 days
32. In MR, chemical shift is due to differences in
A. Proton resonance frequency
B. Proton magnetic susceptibility
C. Patient motion
D. Aliasing
E. Truncation
32. In MR, chemical shift is due to differences in
A. Proton resonance frequency
B. Proton magnetic susceptibility
C. Patient motion
D. Aliasing
E. Truncation

Answer: A – on almost every old test. Fat and water precess at different frequencies, so the artifact is seen in the frequency encoding direction. Motion artifacts occur mostly in the phase encode direction.
True coincidences occur when both photons from an annihilation event are detected by detectors in coincidence, neither photon undergoes any form of interaction prior to detection, and no other event is detected within the coincidence time-window.
A scattered coincidence is one in which at least one of the detected photons has undergone at least one Compton scattering event prior to detection. Since the direction of the photon is changed during the Compton scattering process, it is highly likely that the resulting coincidence event will be assigned to the wrong LOR. Scattered coincidences add a background to the true coincidence distribution which changes slowly with position, decreasing contrast and causing the isotope concentrations to be overestimated. They also add statistical noise to the signal. The number of scattered events detected depends on the volume and attenuation characteristics of the object being imaged, and on the geometry of the camera.

Figure 5. Types of coincidences in PET.
Random coincidences occur when two photons not arising from the same annihilation event are incident on the detectors within the coincidence time window of the system. The number of random coincidences in a given LOR is closely linked to the rate of single events measured by the detectors joined by that LOR and the rate of random coincidences increase roughly with the square of the activity in the FOV. As with scattered events, the number of random coincidences detected also depends on the volume and attenuation characteristics of the object being imaged, and on the geometry of the camera. The distribution of random coincidences is fairly uniform across the FOV, and will cause isotope concentrations to be overestimated if not corrected for. Random coincidences also add statistical noise to the data.
34. An interventional fluoroscopy procedure performed results in patient skin dose of 4 Gray; what is the effect that is most likely to occur.
A. epilation
B. telengectasia
C. blister
D. ulcer
E. dermal necrosis
34. An interventional fluoroscopy procedure performed results in patient skin dose of 4 Gray; what is the effect that is most likely to occur.
A. epilation
B. telengectasia
C. blister
D. ulcer
E. dermal necrosis

Answer: A – here is the chart of skin reactions again. You can't see this chart too many times...
36. At sea level the background radiation dose is greatest from
A. radon
B. cosmic radiation
C. nuclear waste
D. power lines
36. At sea level the background radiation dose is greatest from
A. radon
B. cosmic radiation
C. nuclear waste
D. power lines

Answer: A
37. Regarding integration of HIS and PACS, HL7 (health level 7) represents:
A. secure format to transfer medical information
B. a format compatible with DICOM.
37. Regarding integration of HIS and PACS, HL7 (health level 7) represents:
A. secure format to transfer medical information
B. a format compatible with DICOM.

Answer: Poorly remembered question. Both A and B – according to the UC Davis syllabus, “HL-7 is a standard for the transfer of alphanumeric information between clinical information systems.” HL-7 is used to securely transfer lab values, path and radiology reports etc. Here is a link to about it if you care to learn more:
38. “Person-sievert” represents what?
a) effective dose
b) collective dose
c) genetically significant dose
d) integral dose
38. “Person-sievert” represents what?
a) effective dose
b) collective dose
c) genetically significant dose
d) integral dose

Answer: B

1 rem = 0.01 sieverts (Sv)
person-sievert = (population size) x (dose (Sv) per person)
39. In a 3.5 year old child with concern for appendicitis, what is unacceptable?

kVp = 120 mA = 340

A. inappropriate filter
B. dose
C. excessive beam hardening
D. misalignment of linear attenuation coefficient
39. In a 3.5 year old child with concern for appendicitis, what is unacceptable?

kVp = 120 mA = 340

A. inappropriate filter
B. dose
C. excessive beam hardening
D. misalignment of linear attenuation coefficient

Answer: B – the mAs should be lowered in children. This image is obviously from an adult, but it was all I could find.
40. What type of effect will cause little deposition of energy if it occurs inside a patient?
A. Auger
B. Bremsstrahlung
C. Compton
D. Photoelectric effect
E. Coherent
40. What type of effect will cause little deposition of energy if it occurs inside a patient?
A. Auger
B. Bremsstrahlung
C. Compton
D. Photoelectric effect
E. Coherent

Answer: E – Coherent (Rayleigh) scattering excites the nucleus, and another photon of the same energy is emitted. No ionization occurs.
41. A woman in her second trimester is given a treatment dose of I-131. What is the effect on the fetus?
A. mental retardation
B. IUGR
C. Increased incidence of benign and malignant thyroid tumors
D. Increase in chromosome abnormalities
E. Increased risk of childhood malignancies
41. A woman in her second trimester is given a treatment dose of I-131. What is the effect on the fetus?
A. mental retardation
B. IUGR
C. Increased incidence of benign and malignant thyroid tumors
D. Increase in chromosome abnormalities
E. Increased risk of childhood malignancies

Answer: A – hypothyroidism and mental retardation are possible. Stillbirth and spontaneous abortion are also reported.
42. In mammography, a source to image distance (SID) of 600 to 750 mm is used rather than the standard 1000 mm of other modalities why?
a) To reduce the heel effect
b) To allow for a stronger beam and a shorter exposure
c) To accommodate the focal length of the grid
d) To improve geometric resolution throughout the breast
e) Decrease magnification
42. In mammography, a source to image distance (SID) of 600 to 750 mm is used rather than the standard 1000 mm of other modalities why?
a) To reduce the heel effect
b) To allow for a stronger beam and a shorter exposure
c) To accommodate the focal length of the grid
d) To improve geometric resolution throughout the breast
e) Decrease magnification

Answer: B - Okay, I changed this question and answer because they got it wrong on this years recalls. However, this question is on multiple old tests. The shorter SID allows for a more intense beam and shorter exposure times (decreases dose).
43. In a 2 minute count, a wipe test sample count was 1680. The background count was
1600. In a well counter with a sensitivity of 20 cpm/Bq what is the minimum
detectable activity (in Bq)
A. 3
B. 9
C. 18
D. 40 (or 46?)
E. 80
Answer: B - Okay, I changed this question and answer because they got it wrong on this years recalls. However, this question is on multiple old tests. The shorter SID allows for a more intense beam and shorter exposure times (decreases dose).

A. , .

Answer: A – this is a pretty hard question.

MDA = (2.71/t) + 4.65 ( Rb /t) ½
E X (A/100)

MDA – minimal detectable activity (what the question is asking for)
Rb – background count rate
t – counting interval
E – counting efficiency
A = Activity

MDA = [3√(B/t)]/ε

This is another equation dealing with same concept. Here is a link explaining this.
http://www.rss.usda.gov/publications/mdatb.htm

I just memorized the answer and hoped it would not show up again.
51. What is the effect of receiving 1 Gy radiation:
A. Lymphocyte count decrease
B. Erythema
C. Female sterility
D. Male sterility
51. What is the effect of receiving 1 Gy radiation:
A. Lymphocyte count decrease
B. Erythema
C. Female sterility
D. Male sterility

Answer: A – lymphocyte count can decrease with as little as 0.25 Gy.
52. A cause of distortion in an image intensifier:
A. External magnetic field
B. Scatter photons
C. Flat input phosphor projecting onto a curved output phosphor
D. Severe weather
E. Magic spells
Answer: A – the electrons in the vacuum of the image intensifier have charge and acceleration; their trajectory will be affected by a magnetic field. You cannot put a flouro machine in the fringe field of a MR! Option C is backwards, so be sure you read each option carefully.

You should also know about pincushion and vignetting. Both are associated with the curvature of the II input phosphor. Vinetting is the fall off in brightness in the periphery relative to that in the center, and pincushion is the relative magnification of the image near the periphery. Both are improved with decreased FOV.

Also review veiling glare, which is loss of contrast due to light scattering in the output window of the II. The light photons leaving the output phosphor can reflect at the surfaces of the output window (part of the vacuum apparatus). Here is a website about II artifacts if you want more info:

http://athene.riv.csu.edu.au/~igarbett/MIS220Page/FluoroLecPtII.pdf#search=%22pincushion%20fluoroscopy%22
53. 1% error with 95% confidence interval. How many counts are needed?
A. 40,000
B. 10,000
C. 100,000
D. 1,000,000
Answer: A – This seems really hard, but it is actually very simple once you see how to do it:
 First, realize 95% confidence interval (CI) means 2 standard deviations (67% is 1 SD and 99% is 3 SD).
 Remember N is the number of counts and the standard deviation (SD) by Poisson statistics always going to be √N (the square root of N in case the symbol it isn’t clear).
 Also, remember that the percent error (or percent SD) is equal to the √N/N.
 Now since the SD is √N, then 2SD (aka a 95% CI) is 2√N.
 Okay, set 0.01 equal to 2√N/N. It is 2√N because they wanted 2 standard deviations. Work it out and solve for N:
2√N/N = 0.01
2/√N = 0.01
√N = 2/0.01 = 200
N = 40,000
NOTE: There has been some question as to how the equation 2√N/N = %SD can be simplified to 2/√N = %SD. Well, it is a little math trick. Remember that:
(√N x √N) = N
So, you can substitute (√N x √N) for the denominator in the original equation:
2(√N)/(√N x √N) = %SD
Then you can do simple algebra “canceling out” of √N in the denominator and numerator to get the final equation:
2/√N = %SD
54. What property of calcium makes it detectable on mammography?
A. Density
B. Atomic number
C. Electron density
D. Linear attenuation coefficient
E. Magnetic spin
54. What property of calcium makes it detectable on mammography?
A. Density
B. Atomic number
C. Electron density
D. Linear attenuation coefficient
E. Magnetic spin

Answer: B – The energy of mammo is low in order to increase contrast via the photoelectric effect. PE effect is proportional to Z3 /E3 – so Z (atomic number) is why you see most things. Remember density and LAC (A & D) are important for CT. Electron density determines the chance of Compton effect; however, since almost everything (except hydrogen) has the same electron density, the actual density of the material is what matters most for CT. Magnetic spin (E) has nothing to do with mammo.
50. In PET, which tissue type has the most degradation?
A. Lung
B. Bone
C. Heart
D. Liver
50. In PET, which tissue type has the most degradation?
A. Lung
B. Bone
C. Heart
D. Liver

Answer: C – I guess degradation refers to activity... or maybe blurring from motion... either way, the heart would be the answer.
51. In an abdominal CT, what is the dose in the center relative to the skin?
A. 10%
B. 50%
C. 100%
D. 133%
51. In an abdominal CT, what is the dose in the center relative to the skin?
A. 10%
B. 50%
C. 100%
D. 133%

Answer: B – CTDIweighted = 1/3 CTDIcenter + 2/3 CTDIperiphery
(1/3)/(2/3) = ½ = 0.50
52. What is the kVp of digital subtraction angiography.
A. 100
B. 140
C. 70
D. 40
52. What is the kVp of digital subtraction angiography.
A. 100
B. 140
C. 70
D. 40

Answer: C – this is also the kVp for cardiac cath (just FYI)
55. Which of the following MR weighting sequences is not possible with gradient recalled imaging?
A. T1
B. T2
C. T2*
D. Flow
E. Spin Density
55. Which of the following MR weighting sequences is not possible with gradient recalled imaging?
A. T1
B. T2
C. T2*
D. Flow
E. Spin Density

Answer: B – a true T2 sequence uses a 180 degree pulse in creating an echo, which negates the in homogeneities of the magnetic field. Gradient echo uses alternating magnetic gradients to create an echo. The alternating magnetic fields do not cancel out the in homogeneities of the field (in fact, it exaggerates them). Therefore, a true T2 image cannot be created. GRE sequences are T2* weighted and not T2 weighted.
56. A beam distribution using a tungsten anode is displayed below. What was the kVp and resultant average keV? (Characteristic energies for tungsten = 59, 69 kVp).


kVp Avg keV
A. 30 90
B. 59 69
C. 69 59
D. 90 30
E. 90 59
56. A beam distribution using a tungsten anode is displayed below. What was the kVp and resultant average keV? (Characteristic energies for tungsten = 59, 69 kVp).


kVp Avg keV
A. 30 90
B. 59 69
C. 69 59
D. 90 30
E. 90 59

Answer: D – the average keV is approximately 1/3 to 1/2 of the max. The max keV always equals the kVp.
57. Modern fluoroscopy equipment allows for the addition of an optional copper filter. If the copper filter is used:
A. The mA is decreased to avoid tube overheating
B. The average x-ray energy will increase if the kVp is kept constant
C. The fluoroscopic receiver input dose is decreased to decrease noise.
D. The kVp is increased to increase contrast
57. Modern fluoroscopy equipment allows for the addition of an optional copper filter. If the copper filter is used:
A. The mA is decreased to avoid tube overheating
B. The average x-ray energy will increase if the kVp is kept constant
C. The fluoroscopic receiver input dose is decreased to decrease noise.
D. The kVp is increased to increase contrast

Answer: B – filtering 'hardens' the beam, meaning the HVL and the average energy are increased.
58. A possible source of gridline artifacts in mammography films:
A. Exposure time too long
B. mA set too high
C. Processor roller pressure too low
D. kVp set too high
E. Inadequate breast compression
58. A possible source of gridline artifacts in mammography films:
A. Exposure time too long
B. mA set too high
C. Processor roller pressure too low
D. kVp set too high
E. Inadequate breast compression

Answer: D – Setting the kVp too high shortens the exposure time. Shorter exposure time causes you to see the grid lines because there isn’t enough time for the grids to blur out with movement.
59. Two Screens, A &B, have the same conversion efficiency, but Screen A has twice the x-ray absorption efficiency of screen B. If you use Screen A you will get:
A. Exposure times ½ that for screen B
B Decreased geometric unsharpness
C Increased screen unsharpness (screen blur)
D Increased quantum mottle
E Increased motion unsharpness
59. Two Screens, A &B, have the same conversion efficiency, but Screen A has twice the x-ray absorption efficiency of screen B. If you use Screen A you will get:
A. Exposure times ½ that for screen B
B Decreased geometric unsharpness
C Increased screen unsharpness (screen blur)
D Increased quantum mottle
E Increased motion unsharpness

Answer: A – increasing the absorption efficiency means less radiation is used, so the exposure times can be decreased. However, since a larger percentage of the photons are absorbed at the input phosphor, the same NUMBER of photons are used to make the image. Therefore, the quantum mottle is unchanged. The screen can be thinner, so the screen unsharpness will improve. Geometric unsharpness has to do with the SID, focal spot, and OID, not the screen itself. Decreasing exposure time will decrease motion artifact.
60. In linear tomography, the plane of focus is determined by:
A. The angle over which the x-ray tube is moved
B. The distance over which the x-ray tube is moved
C. The distance from the patient to the film
D. The distance from the patient to the x-ray tube
E. The position of the fulcrum
60. In linear tomography, the plane of focus is determined by:
A. The angle over which the x-ray tube is moved
B. The distance over which the x-ray tube is moved
C. The distance from the patient to the film
D. The distance from the patient to the x-ray tube
E. The position of the fulcrum

Answer: E – fulcrum is a funny word.

Cant verify this anywhere but whatever.
61. Concerning cell survival for high LET radiation, which of these quantities is zero?
A. D0
B. Dq
C. N
D. Α
E. S
. Concerning cell survival for high LET radiation, which of these quantities is zero?
A. D0
B. Dq
C. N
D. Α
E. S

Answer: B – See below. The Dq for the high LET radiation (neutrons) is zero. Fractionating the dose leads to better survival per dose.
62. Which of the following is NOT a ultrasound quality control test?
A. Depth of Visualization
B. Distance Accuracy
C. Axial Resolution
D. Dead Zone
E. Lateral Refraction
Answer: E – never heard of this term.

The system performance of a diagnostic ultrasound unit is described by several
parameters: sensitivity and dynamic range, spatial resolution, contrast sensitivity,
range/distance accuracy, dead zone thickness, and TGC operation. For Doppler
studies, pulse repetition frequency, transducer angle estimates, and range gate sta
bility are key issues.
63. Curve C below is the attenuation for 100 kEv photons. Which curve represents the attenuation of a 100 kVp x-ray beam?
63. Curve C below is the attenuation for 100 kEv photons. Which curve represents the attenuation of a 100 kVp x-ray beam?



Answer: B – I guess it is trying to say that C is a monochromatic xray beam. Since the 100 kVp x-ray beam will be a spectrum of energies, it will have higher attenuation of the beam initially of the weaker parts of the spectrum. Then, the beam will become more and more 'monochromatic' and have a more linear attenuation.
64. Image provided. (Here is my description: There is a thin dark curvilinear line along the entire lateral aspect on an axial CT image of the body.) The artifact is most likely due to:
A. metallic artifact in the patient
B. residual contrast on the CT scanner
C. defect in 3rd generation scanner
D. defect in 4th generation scanner
64. Image provided. (Here is my description: There is a thin dark curvilinear line along the entire lateral aspect on an axial CT image of the body.) The artifact is most likely due to:
A. metallic artifact in the patient
B. residual contrast on the CT scanner
C. defect in 3rd generation scanner
D. defect in 4th generation scanner

Answer: C – Sounds like a ring artifact from a bad detector on a 3rd generation scanner. NOTE: this answer was marked D before I edited it, but that is totally wrong. The reason 4th generation scanners were developed was to eliminate ring artifact.
65. Image provided of a mobile fluoroscopy C-arm. There is a removable cylindrical spacer device on the unit. According to the FDA, the device can be removed in which of the following procedural conditions?
A. a few surgical procedures
B. large patient
D. never
C. when imaging the extremities
E. when imaging the head and neck
65. Image provided of a mobile fluoroscopy C-arm. There is a removable cylindrical spacer device on the unit. According to the FDA, the device can be removed in which of the following procedural conditions?
A. a few surgical procedures
B. large patient
D. never
C. when imaging the extremities
E. when imaging the head and neck

Answer: A – I've seen techs do this in the OR, and here is a document that states it is okay.
66. For an adult, the LD50/60 for acute low-LET exposure is:
A. 0.5 Gy
B. 1.0 Gy
C. 2.0 Gy
E. 3.0 Gy
F. 4.0 Gy
66. For an adult, the LD50/60 for acute low-LET exposure is:
A. 0.5 Gy
B. 1.0 Gy
C. 2.0 Gy
E. 3.0 Gy
F. 4.0 Gy

Answer: E – I've also seen the number 3.5 Gy.
67. This artifact can be prevented with the following:


A. increasing gate diameter
B. increasing pulsed-repetition frequency
C. increasing transducer frequency
D. larger transducer
E. increase Doppler gain
67. This artifact can be prevented with the following:


A. increasing gate diameter
B. increasing pulsed-repetition frequency
C. increasing transducer frequency
D. larger transducer
E. increase Doppler gain

Answer: B – here is a slide describing aliasing and a list of the ways you can decrease aliasing in Doppler U/S:
Increase PRF (aka - decrease SPL)
Increase angle
Decrease transducer frequency
Remember the equation for maximal velocity: Vmax = (PRF * c)/(2fi * cosθ)
The equation for the maximal Doppler shift is in the slide below.
I think the main things to memorize are the ways to lessen aliasing (allow larger velocity), which are listed above the slide.
68. Which parameters will result in the lowest skin exposure to the patient who is situated directly adjacent to the cassette? (SID = source to image distance)
A. 70 kvp, 10 mas, 40 cm SID
B. 80 kvp, 5 mas, 70 cm SID
C. 80 kvp, 20 mas, 40 cm SID
D. 80 kvp, 4 mas, 40 cm SID
E. 90 kvp, 20 mas, 70 cm SID
68. Which parameters will result in the lowest skin exposure to the patient who is situated directly adjacent to the cassette? (SID = source to image distance)
A. 70 kvp, 10 mas, 40 cm SID
B. 80 kvp, 5 mas, 70 cm SID
C. 80 kvp, 20 mas, 40 cm SID
D. 80 kvp, 4 mas, 40 cm SID
E. 90 kvp, 20 mas, 70 cm SID

Answer: B – kVp and SID are squared, mAs is direct relationship.
69. What is the positive predictive value?

A. 0.20
B. 0.33
C. 0.57
D. 0.67
E. 0.80
69. What is the positive predictive value?
Disease
Positive Negative
Test Positive 80 (TP) 60 (FP)
Negative 40 (FN) 20 (TN)

A. 0.20
B. 0.33
C. 0.57
D. 0.67
E. 0.80

Answer: C – Make sure you know this stuff, b/c it should be easy points.
PPV = TP/(TP+FP)
NPV = TN/(TN+FN)
Sensitivity = TP/(TP+FN)
Specificity = TN/(TN+FP)
In this case, PPV = 80/(80+60) = 0.57
70. Image provided. There is a monitor with wires attached to a upright cyclindrical device located approximately 2 feet below the monitor.) What is the most likely purpose of the system?
A. measure thyroid uptake
B. measure decay prior to disposal
C. measure background radiation in radiopharmacy
D. measure calibration dose
E. assay wipe test
70. Image provided. There is a monitor with wires attached to a upright cyclindrical device located approximately 2 feet below the monitor.) What is the most likely purpose of the system?
A. measure thyroid uptake
B. measure decay prior to disposal
C. measure background radiation in radiopharmacy
D. measure calibration dose
E. assay wipe test

Answer: E – This is a well counter. Dose calibrators don’t have monitors – they just have a digital readout.
71. Which statement is true in regards to fMRI?
A. oxygen is paramagnetic, and increased neuronal activity will decrease MR signal
B. deoxyhemoglobin is paramagnetic, and increased blood flow will increase MR signal
C. time of flight imaging is used in fMRI.
71. Which statement is true in regards to fMRI?
A. oxygen is paramagnetic, and increased neuronal activity will decrease MR signal
B. deoxyhemoglobin is paramagnetic, and increased blood flow will increase MR signal
C. time of flight imaging is used in fMRI.

Answer: B

Blood oxygen level-dependent BOLD and "functional MR" imaging rely on
the differential contrast generated by blood metabolism in active areas of the brain.
In flow of blood and conversion of oxyhemoglobin to deoxyhemoglobin a para
magnetic agent increases the magnetic susceptibility in the localized area and
induces signal loss by reducing T2*. This fact allows areas of high metabolic activ
ity to produce a correlated signal and is the basis of functional MRI fMRI. A
BOLD sequence produces an image of the head before the application of the stim
ulus. The patient is subjected to the stimulus and a second BOLD image is
acquired. Because the BOLD sequence produces images that are highly dependent
on blood oxygen levels, areas of high metabolic activity will be enhanced when the
presi-imulus image is subtracted, pixel by pixel, from the poststimulus image. These
fMRI experiments determine which sites in the brain arc used for processing data.
Stimuli in flvIRI experiments can be physical finger movement, sensory light
flashes or sounds, or cognitive repetition of "good" or "bad" word sequences. To
improve the SNR in the fMRT images, a stimulus is typically applied in a repetitive,
periodic sequence, and BOLD images are acquired continuously. Areas in the brain
that demonstrate time-dependent activity and correlate with the time-dependent
application of the stimulus are coded using a color scale amid are overlaid onto a
gray-scale image of the brain for anatomic reference.
72. Concerning the use of KI as a radioprotective agent:
A. reduces effects of free radicals
B. can only be used to protect the thyroid from radio-iodine
C. shortens M-phase of cell cycle
D. protects only against effects of high LET radiation
E. minimizes DNA breaks
72. Concerning the use of KI as a radioprotective agent:
A. reduces effects of free radicals
B. can only be used to protect the thyroid from radio-iodine
C. shortens M-phase of cell cycle
D. protects only against effects of high LET radiation
E. minimizes DNA breaks

Answer: B – KI is not a cure all. It only saturates the thyroid so it won't take up as much of the radioactive iodine.
73. Your group obtains a digital system for chest xrays. A female tech volunteers to have a CXR to evaluate the system performance. What is the appropriate next step?
A. have female tech do pregnancy test
B. skip pregnancy test and place lead apron around her abdomen
C. have her sign a form stating she is not pregnant
D. perform exam on male tech or radiologist
E. perform initial tests on non-human objects prior to first human patient
F. skip the pregnancy test if the tech is not married since she couldn't be pregnant without being married first.
73. Your group obtains a digital system for chest xrays. A female tech volunteers to have a CXR to evaluate the system performance. What is the appropriate next step?
A. have female tech do pregnancy test
B. skip pregnancy test and place lead apron around her abdomen
C. have her sign a form stating she is not pregnant
D. perform exam on male tech or radiologist
E. perform initial tests on non-human objects prior to first human patient
F. skip the pregnancy test if the tech is not married since she couldn't be pregnant without being married first.

Answer: E – seems like a good idea...
74. Although Tc99m decays by isomeric transition, nonpenetrating energy should be considered to the target organ from:
A. internal conversion
B. neutrinos
C. B- decay ray
D. positron
E. electron capture
74. Although Tc99m decays by isomeric transition, nonpenetrating energy should be considered to the target organ from:
A. internal conversion
B. neutrinos
C. B- decay ray
D. positron
E. electron capture

Answer: A – any isomer that decays by isomeric transition can also have internal conversion, which is sort of like an Auger electron. B- decay causes nonpenetrating radiation, but should not compete with internal conversion. Internal conversion causes the Z to decrease, and B- causes the Z to increase (sort of opposite processes).

Here is a tangent about decay processes that helped me keep things straight:
 Radionuclides produced in a CYCLOTRON have proton excess– remember this b/c the particle accelerated in the cyclotron is a proton or alpha particle (must have a + charge to be accelerated by alternating electrical current).
 Proton excess must decay by positron emission or electron capture to get rid of the extra positive charge.
 Examples of cyclotron produced radionuclides: Ga-67, I-123, In-111, C-11
 Radionuclides produced in a NUCLEAR REACTOR have neutron excess because the fuel rods release fast neutrons that interact with the target.
 Neutron excess must decay by beta (-) emission (losing a negative is the same as gaining a positive, and the atoms are proton poor).
 Examples of reactor produced radionuclides: I-131, Xe-99, Mo-99
 Metastable isomers (Tc-99m) decay by isomeric transition, which releases only energy (Z and A don't change). As above, internal conversion competes with gamma emission analogous to Auger electron (the gamma xray 'knocks out' an electron on its way out of the atom).
74. Exposure in roentgen is directly measured with:
A. photographic film
B. NaI crystal and PHA
C. TLD chip
D. ionization chamber and electrometer
E. Geiger counter
74. Exposure in roentgen is directly measured with:
A. photographic film
B. NaI crystal and PHA
C. TLD chip
D. ionization chamber and electrometer
E. Geiger counter

Answer: D – whenever you see the word 'exposure' think 'ionization chamber' (on multiple old tests). See other explanations from other test if you want more info.
75. Transmission CT and emission CT are similar in that:
A. use photon attenuation to reconstruct images
B. use filtered back projection to reconstruct images
C. can select ranges of energy to reduce scatter
D. use circular arrays of photons to acquire data
E. use same anode target material to create x rays.
75. Transmission CT and emission CT are similar in that:
A. use photon attenuation to reconstruct images
B. use filtered back projection to reconstruct images
C. can select ranges of energy to reduce scatter
D. use circular arrays of photons to acquire data
E. use same anode target material to create x rays.

Answer: B
76. Which of the following is NOT considered when deciding the lead thickness in the wall of a radiographic exam room?
A. amount of time beam aimed at wall
B. x-ray tube output and weekly use
C. fraction of time people are in an adjacent occupied area
D. distance from x-ray unit to occupied area
E. number of people within an adjacent occupied area
76. Which of the following is NOT considered when deciding the lead thickness in the wall of a radiographic exam room?
A. amount of time beam aimed at wall
B. x-ray tube output and weekly use
C. fraction of time people are in an adjacent occupied area
D. distance from x-ray unit to occupied area
E. number of people within an adjacent occupied area

Answer: E – an occupied area is an occupied are no matter how many people are in there.
77. Using spin echo MR images, what combination of TR and TE will give the LOWEST SNR to evaluate the brain?
A. TE 20 TR 2000
B. TE 20 TR 500
C. TE 80 TR 1000
D. TE 80 TR 2000
E. TE 150 TR 500
77. Using spin echo MR images, what combination of TR and TE will give the LOWEST SNR to evaluate the brain?
A. TE 20 TR 2000
B. TE 20 TR 500
C. TE 80 TR 1000
D. TE 80 TR 2000
E. TE 150 TR 500

Answer: E – In clinical imaging, T2 gives the lowest signal, and the heavier the T2 signal (longer TE) the lower signal b/c there is more time for decay. However, there is an even better answer for this question (option E), which is a combination of TE/TR never used in clinical imaging (and the reason is the low SNR!). The short TR would cause only a small amount of signal to be 'tilted' back to the transverse vector, which means the T1 signal would be small. For the same reason, the T2 signal would be very small to start, and the long TE would allow for even more signal loss. See pg 63 in MRI Made Easy for more discussion. A shortcut for this question is to realize that proton density (long TR, short TE) has the highest SNR of any sequence, and Option E is the opposite of PD.
80. What is the most appropriate power rating for the x-ray unit for DSA?
A. .01kW
B. 0.1kW
C. 1 kW
D. 10kW
E. 100kW
80. What is the most appropriate power rating for the x-ray unit for DSA?
A. .01kW
B. 0.1kW
C. 1 kW
D. 10kW
E. 100kW

Answer: D
81. With scintillation imaging, when using a parallel hole collimator, spatial resolution.
A. is highest on the surface of the collimator
B. is the same for all gamma rays
C. is greater for I131 than In111
D. is independent of the sensitivity of the collimator
81. With scintillation imaging, when using a parallel hole collimator, spatial resolution.
A. is highest on the surface of the collimator
B. is the same for all gamma rays
C. is greater for I131 than In111
D. is independent of the sensitivity of the collimator

Answer: A – collimator resolution is what limits the systems resolution.
82. The brightness of a CT pixel is determined by which tissue property of the voxel?
A. electron density
B. atomic number (Z)
C. amount of water in the voxel
D. Mass attenuation coefficient
E. Linear attenuation coefficient
82. The brightness of a CT pixel is determined by which tissue property of the voxel?
A. electron density
B. atomic number (Z)
C. amount of water in the voxel
D. Mass attenuation coefficient
E. Linear attenuation coefficient

Answer: E – on multiple old tests.
83. The spatial resolution in MRI is most improved by increasing which:
A. FOV
B. Gradient strength
C. Amplitude of the RF pulse
D. TR
E. receiver gain
83. The spatial resolution in MRI is most improved by increasing which:
A. FOV
B. Gradient strength
C. Amplitude of the RF pulse
D. TR
E. receiver gain

Answer: B – increased gradient strength will cause a thinner slice given a constant RF pulse bandwidth. Thinner slice equals better resolution.
84. The relationship between the RBE and LET of ionizing radiation is best depicted by which graph?
Answer: A – know that the higher the LET, the higher the RBE. But, the relationship is not linear. For more info, check the syllabus or this webpage:
http://www.radsource.com/gammaVx-rays.html

The term relating the effectiveness of the test radiation to the reference radiation is
called the relative biological effectiveness RBE. The RBE is defined, for identical
exposure conditions, as follows:
85. What does the shaded area in the graph represent?
A.FN (False Negative)
B. FP (False Positive)
C. TN (True Negative)
D. TP (True Positive)
E. Selectivity
Answer: B – understand this relationship b/c it should be easy points!
86. Which object is best for measuring the line spread function?
A. Star pattern
B. line pair pattern
C. pinhole aperture
D. slit line aperture
E. step wedge
86. Which object is best for measuring the line spread function?
A. Star pattern
B. line pair pattern
C. pinhole aperture
D. slit line aperture
E. step wedge

Answer: D – the slit line aperture is placed against the crystal (acts like a collimator) and a flood source is use to make a line spread function.

Whereas the MTF is best conceptualized using sine waves, in practice it is measured
differently. As mentioned above, the line spread function LSF is measured by
stimulating the imaging system to a line. For an x-ray detector, a very narrow lead
slit is used to collimate the x-ray beam down to a very thin line, For tomographic
imaging systems, a very thin sheet of contrasting material can be used to create an
LSF e.g., a 0.1-mm sheet of aluminum for CT.
87. In radiography, which best determines quantum mottle?
A. Number of photons in the incident beam
B. Number of photons exiting the patient
C. Number of photons absorbed by the screen
D. Number of light photons emitted by the screen
E. Number of light photons absorbed by the film
87. In radiography, which best determines quantum mottle?
A. Number of photons in the incident beam
B. Number of photons exiting the patient
C. Number of photons absorbed by the screen
D. Number of light photons emitted by the screen
E. Number of light photons absorbed by the film

Answer: C – quantum mottle is determined by the number of xray photons absorbed by the input phosphor. The question is basically asking where is the quantum sink? A confusing point is that QM is independent of absorption efficiency, which seems to be contrary to the answer for this question. However, the QM is determined by the NUMBER of photons absorbed, not the PERCENT! If you increase the QDE (quantum detection efficiency – aka absorption efficiency), then the the AEC (automatic exposure control) will decrease the mA, and fewer photons will be incident of the input phosphor. However, a higher percent of the photons are absorbed, and the NUMBER absorbed stays constant. Therefore QM doesn't change. Any questions??? Remember, increasing conversion efficiency INCREASES QM because the AEC will turn down the mA and fewer photons will be absorbed at the input phosphor. Sorry about the wordy explanation, but this concept gave me some trouble at first. So, the recap:
 QM is determined by the NUMBER of absorbed xray photons
 increase/decrease absorption efficiency = no change in QM
 INCREASE the conversion efficiency = INCREASE QM
 DECREASE the conversion efficiency = DECREASE QM
88. For ionizing radiation, which portion of the cell cycle is most resistant to permanent radiation induced genetic alterations?
A. G0
B. late G1
C. G2
D. late S
E. mitosis
Answer: D – The relative sensitivity is M>G2>G1>S . Lots of repair enzymes present during S phase repair the damage better. Surviving cells in a culture following radiation will be synchronized b/c most of the survivors will be in late S.
89. For a sample measured in a well counter, “S” counts were obtained. The sample is removed and “B” number of background counts were obtained (for the same amount of time as the first). What is the standard deviation of true sample counts?
A. Square root of S plus square root of B
B. Square root of S minus square root of B
C. Square root of (S minus B)
D. Square root of (S plus B)
E. Square root of (S squared plus B squared)
89. For a sample measured in a well counter, “S” counts were obtained. The sample is removed and “B” number of background counts were obtained (for the same amount of time as the first). What is the standard deviation of true sample counts?
A. Square root of S plus square root of B
B. Square root of S minus square root of B
C. Square root of (S minus B)
D. Square root of (S plus B)
E. Square root of (S squared plus B squared)

Answer: D – Kind of confusing because the syllabus says to take the square root of the sum of the SD of each of the numbers. But remember that the standard deviation is the square root of the number. So, (√N)2 = N. So D is correct.

http://radiographics.rsna.org/content/19/3/765.full.pdf+html
90. For diagnostic radiation, most DNA induced damage results from
A. Ion radicals
B. hydroxyl radicals
C. Fast electrons
D. protons
90. For diagnostic radiation, most DNA induced damage results from
A. Ion radicals
B. hydroxyl radicals
C. Fast electrons
D. protons

Answer: B – an estimated 2/3 of the damage is from indirect damage via free hydroxyl radicals.
91. What is the minification gain if the input phosphor is 9 cm and the output phosphor is 1.5 cm?
A. 2
B. 9
C. 36
D. 40
91. What is the minification gain if the input phosphor is 9 cm and the output phosphor is 1.5 cm?
A. 2
B. 9
C. 36
D. 40

Answer: C – Minification gain is di2 /do2 which is the same as (di/do) 2. So (9 /1.5)2 is 62 or 36.
92. In order to decrease the partial volume averaging in a single slice on a helical CT?
A. increase pitch
B. decrease gantry rotation time
C. increase filtration
D. decrease beam width along the z axis
92. In order to decrease the partial volume averaging in a single slice on a helical CT?
A. increase pitch
B. decrease gantry rotation time
C. increase filtration
D. decrease beam width along the z axis

Answer: D – key here is a single slice CT scanner, where slice thickness has to do with beam width (collimation).
93. In order to decrease the scatter at the image receptor, but increase the patient’s dose and scatter in the room, you would do what?
A. increase the grid ratio
B. increase compression
C. increase filtration
D. increase collimation to decrease expose field
93. In order to decrease the scatter at the image receptor, but increase the patient’s dose and scatter in the room, you would do what?
A. increase the grid ratio
B. increase compression
C. increase filtration
D. increase collimation to decrease expose field

Answer: A – funny wording. Sounds like we just want to give the patient more dose...
94. The statistics of radionuclide counting in nuclear medicine are?
A. binomial
B. Gaussian (Normal)
C. chi squared
D. Weiner
E. Poisson
94. The statistics of radionuclide counting in nuclear medicine are?
A. binomial
B. Gaussian (Normal)
C. chi squared
D. Weiner
E. Poisson
Answer: E.
96. The apparent dead time of a scintigraphic camera is greater when what changes are made?
B. using a sample with lower activity
C. increasing the window width of the PHA
D. changing to a non-paralyzable something to a paralyzable something
E. changing to a general purpose collimator from a high sensitivity
96. The apparent dead time of a scintigraphic camera is greater when what changes are made?
B. using a sample with lower activity
C. increasing the window width of the PHA
D. changing to a non-paralyzable something to a paralyzable something
E. changing to a general purpose collimator from a high sensitivity

Answer: D – more counts with the all else the same equals more dead time.
97. The noise in the subtracted image from a DSA relative to the noise in the angiogram image itself?
B. 1.4
C. 2
D. 4
E. 8
97. The noise in the subtracted image from a DSA relative to the noise in the angiogram image itself?
B. 1.4
C. 2
D. 4
E. 8

Answer: b – noise is the square root of the number of photons. The angiogram has 2x the number of photons b/c the DSA has the mask subtracted away. So, the relative noise is √2.
98. A picture of K space presented with the center blacked out? 5 MR head images given as options
A. excellent quality MR head image
B. Good MR image but with noise
C. Blurry MR head image had contrast but no edge detail
D. Image with edge detail only
E. Image with center blacked out
98. A picture of K space presented with the center blacked out? 5 MR head images given as options
A. excellent quality MR head image
B. Good MR image but with noise
C. Blurry MR head image had contrast but no edge detail
D. Image with edge detail only
E. Image with center blacked out

Answer: D – look at these in Bushberg pg 429.
99. The most radiosensitive organ is
A. stomach
B. oropharynx
C. colon
D. small intestine
99. The most radiosensitive organ is
A. stomach
B. oropharynx
C. colon
D. small intestine
Answer: D – The small bowel is the most radiosensitive organ. One tricky point is that the stomach has the highest rate of fatal radiation induced cancer.
100. The HVL in diagnostic radiology is?
A. specified in mm of copper
B. increases with mAs
C. approximately 1 cm in soft tissue
D. decreases with increasing filtration
E. must meet a minimum set by FDA
100. The HVL in diagnostic radiology is?
A. specified in mm of copper
B. increases with mAs
C. approximately 1 cm in soft tissue
D. decreases with increasing filtration
E. must meet a minimum set by FDA

Answer: E – HVL is specified in mmAl. HVL in soft tissue is 4-5cm. Increasing filtration increases the HVL (hardens the beam). There are standards for the HVL. I don't remember them exactly, but a ballpark value for mammography is the kVp divided by 100.
100. The main target of radiations-induced cell death is
A. DNA
B. RNA
C. Mitochondria
D. Lysosomes
100. The main target of radiations-induced cell death is
A. DNA
B. RNA
C. Mitochondria
D. Lysosomes

Answer: A
109. In Compton scatter, the energy of the recoil electron is the energy of the:
A. incident x-ray
B. incident x-ray minus K-shell binding energy
C. incident x-ray minus scattered photon
D. rest mass of the electron
E. scattered photon
109. In Compton scatter, the energy of the recoil electron is the energy of the:
A. incident x-ray
B. incident x-ray minus K-shell binding energy
C. incident x-ray minus scattered photon
D. rest mass of the electron
E. scattered photon

Answer: C – the electron is an outer shell electron, and is considered to be “free.” Its binding energy is so low it can be ignored. That's why B is incorrect.
110. The quantity measured by personnel radiation badges that is reported on Occupational Exposure Report:
A. dose equivalent
B. effective dose
C. absorbed dose
D. KERMA
110. The quantity measured by personnel radiation badges that is reported on Occupational Exposure Report:
A. dose equivalent
B. effective dose
C. absorbed dose
D. KERMA

Answer: A – Dose equivalent takes into effect weighting factors based on the different types of radiation, which is why the badges have different thickness filters to discern high LET from low LET radiation. The dose reports from the badges are in mrem. Plus, there is a recall question about a “pregnant intervetionalist” that gives her dose from a film badge in mSv. Effective dose is also reported in rem (Sv); however, it is a risk assessment of different organs getting cancer via the weighting factor. The total of all the weighting factors is equal to 1. Here is a chart to keep the units straight:

Description Units Notes
Exposure Ionizations (charge) per unit air Roentgen – R Use ionization chamber to measure exposure
Absorbed Dose Energy per unit mass Rad (Gy)
Dose Equivalent AD times quality factor for type of radiation Rem (Sv) What dose badges report
Effective Dose Dose equivalent times weighting factor for risk of cancer in specific organs Rem (Sv) Think – effective dose is how 'effective' the dose is at causing radiation induced cancer in specific organs. Weighting factors all add up to 1.
110. The quantity measured by personnel radiation badges that is reported on Occupational Exposure Report:
A. dose equivalent
B. effective dose
C. absorbed dose
D. KERMA
110. The quantity measured by personnel radiation badges that is reported on Occupational Exposure Report:
A. dose equivalent
B. effective dose
C. absorbed dose
D. KERMA

Answer: A – Dose equivalent takes into effect weighting factors based on the different types of radiation, which is why the badges have different thickness filters to discern high LET from low LET radiation. The dose reports from the badges are in mrem. Plus, there is a recall question about a “pregnant intervetionalist” that gives her dose from a film badge in mSv. Effective dose is also reported in rem (Sv); however, it is a risk assessment of different organs getting cancer via the weighting factor. The total of all the weighting factors is equal to 1. Here is a chart to keep the units straight:
111. A measurement of the standard deviation of CT numbers of a uniform phantom is an indirect measurement of:
A. detector MTF
B. spatial resolution
C. low contrast resolution
D. accuracy of collimator
E. ?
111. A measurement of the standard deviation of CT numbers of a uniform phantom is an indirect measurement of:
A. detector MTF
B. spatial resolution
C. low contrast resolution
D. accuracy of collimator
E. ?

Answer: C – ahh... what?
112. What ultrasound property stays constant as the pulse goes from one organ to another:
A. speed
B. frequency
C. wave length
D. amplitude
112. What ultrasound property stays constant as the pulse goes from one organ to another:
A. speed
B. frequency
C. wave length
D. amplitude

Answer: B – on multiple old tests. Frequency is a property of the transducer and does not change. The speed and wavelength change from organ to organ to keep the equation, c = fλ, balanced.
113. Most of the energy absorbed by biological tissues from ionizing radiation is transferred from:
A. ions
B. protons
C. electrons
D. characteristic x-ray
E. Compton scatter photon
113. Most of the energy absorbed by biological tissues from ionizing radiation is transferred from:
A. ions
B. protons
C. electrons
D. characteristic x-ray
E. Compton scatter photon

Answer: C – on multiple old tests.
114. In mammography, a single film, single screen emulsion detector system is usually used because of:
A. decrease noise
B. improve spatial resolution
C. improve detection of molybdenum characterist x-rays
D. lower dose
114. In mammography, a single film, single screen emulsion detector system is usually used because of:
A. decrease noise
B. improve spatial resolution
C. improve detection of molybdenum characterist x-rays
D. lower dose

Answer: B
115. The most appropriate detector for measurement of gamma-ray spectrum of a radioisotope is:
A. dose calibrator
B. NaI well counter
C. Geiger counter
D. Thermal Luminescent detector
E. Liquid scintillator detector
115. The most appropriate detector for measurement of gamma-ray spectrum of a radioisotope is:
A. dose calibrator
B. NaI well counter
C. Geiger counter
D. Thermal Luminescent detector
E. Liquid scintillator detector

Answer: B – a well counter is a spectrometer with a screen that displays photopeaks. Dose calibrator operates in the ionization region the the current-voltage chart, but is not a spectrometer b/c it is operated in current mode and information from each individual count is lost.
116. If the center of rotation of a SPECT machine is not well-aligned with the center of the projection image, the resulting clinical image will demonstrate:
A. cupping
B. degraded spatial resolution
C. enhanced star artifact
D. bright spot in center
E. some other artifact
116. If the center of rotation of a SPECT machine is not well-aligned with the center of the projection image, the resulting clinical image will demonstrate:
A. cupping
B. degraded spatial resolution
C. enhanced star artifact
D. bright spot in center
E. some other artifact

Answer: B – the classic 'doughnut' artifact (circle of bright activity instead of a point) is not actually seen in SPECT b/c the resolution is not good enough to see it. So, in real life, the effect of poor COR is a blurry image (degraded spatial resolution).
117. At distance 1 meter from the patient, in a standard abdominal radiograph, the intensity of radiation scattered at 90degrees from the patient would be what fraction of the intensity of the beam incident on the patient:
A. 0.1
B. 0.05
C. 0.01
D. 0.005
E. 0.001
117. At distance 1 meter from the patient, in a standard abdominal radiograph, the intensity of radiation scattered at 90degrees from the patient would be what fraction of the intensity of the beam incident on the patient:
A. 0.1
B. 0.05
C. 0.01
D. 0.005
E. 0.001

Answer: E – just memorize this (0.1 % or 0.001) b/c it comes up a lot on old tests.
118. This question showed representative images from PET (head)/MR (head)/ CT (abdomen)/ CXR/ ANGIO (head). Question was which image takes the most memory up on a pacs system.
A. PET
B. MR
C. CT
D. CXR
E. ANGIO
118. This question showed representative images from PET (head)/MR (head)/ CT (abdomen)/ CXR/ ANGIO (head). Question was which image takes the most memory up on a pacs system.
A. PET
B. MR
C. CT
D. CXR
E. ANGIO

Answer: D – this is true PER IMAGE. However, a CT scan has so many images, that it usually needs more space than a single CXR.
119. Which acronym stands for a series of hard drives that provides fast, short-term memory for PACS systems:
A. RAM
B. DVD
C. RAID
D. MOD
119. Which acronym stands for a series of hard drives that provides fast, short-term memory for PACS systems:
A. RAM
B. DVD
C. RAID
D. MOD

Answer: C – stands for Redundant Array of Independent Disks.
120. Which of the following will best detect low level gamma radiation:
A. Geiger counter
B. Personnel dosimetry
C. Dose counter
D. Ionization device with air at sea level
120. Which of the following will best detect low level gamma radiation:
A. Geiger counter
B. Personnel dosimetry
C. Dose counter
D. Ionization device with air at sea level

Answer: A – If you see the word 'detect' think GM counter.
121. Which best determines the binding energy of a K-shell electron:
A. Valence
B. Density
C. Number of electrons in shell
D. Atomic number
E. Mass number
121. Which best determines the binding energy of a K-shell electron:
A. Valence
B. Density
C. Number of electrons in shell
D. Atomic number
E. Mass number

Answer: D – the number of protons (positive charge) determines the binding energy of the negatively charged electrons.
122. If an abdominal image (plain film) is obtained using digital technique and “twice the normal radiation” (exact wording), what is the result
A. Too dark
B. Too light
C. Decreased latitude
D. Increased latitude
122. If an abdominal image (plain film) is obtained using digital technique and “twice the normal radiation” (exact wording), what is the result
A. Too dark
B. Too light
C. Decreased latitude
D. Increased latitude

Answer: C, I think – this question has given us a little trouble and caused a bit of discussion. On the 2006 (and #127 of this test), there is an additional option stating decreased quantum mottle, which is correct. However, this question showed up on the 2008 test with only the options above. But working with this version of the question, we need to remember the H&D curve for digital system. OD and exposure have a linear relationship. The latitude is very wide and contrast is low. However, windowing artificially changes the latitude, contrast, and grayscale to maximize the image. So, if you made no changes (no windowing), simply moving over on the x-axis of the H&D curve will give a higher OD (darker film). However, Option A is not true b/c of the wording “Too Dark.” The advantage of DR/CR is the ability to window and correct grayscale.
So, why is Option C correct? If you take the film with too much radiation, the point on the unaltered H&D graph will be high. When you window the image, the new graph must go through that point. So, your ability to window is limited. Therefore, your latitude is decreased. We've all tried to window an overexposed CXR to see a tip of a central line, and it doesn't work as well as a normal exposure. Therefore, I think the BEST answer is DECREASED LATITUDE. See the chart below:
123. In digital fluoroscopy, frame integration leads to:
A. Decreased lag time
B. Increased patient dose
C. Increased signal to noise
D. Increased contrast
E. Increased spatial resolution
123. In digital fluoroscopy, frame integration leads to:
A. Decreased lag time
B. Increased patient dose
C. Increased signal to noise
D. Increased contrast
E. Increased spatial resolution

Answer: C – frame integration “adds” multiple frames together with the goal of decreasing noise, thereby increasing signal to noise ratio.
124. If the following parameters are used in spin echo, which weighting will be obtained

A. T1
B. T2
C. PD
D. T1/T2
E. Fat suppression
124. If the following parameters are used in spin echo, which weighting will be obtained

A. T1
B. T2
C. PD
D. T1/T2
E. Fat suppression

Answer: B – review this information in MRI Made Easy if you have questions.
125. Young woman gets both head angiography and head CT. What is the lifetime risk of cataracts?
A. 0%
B. 0.1%
C. 1.0%
D. 10%
125. Young woman gets both head angiography and head CT. What is the lifetime risk of cataracts?
A. 0%
B. 0.1%
C. 1.0%
D. 10%

Answer: A – Too easy, cataracts are a deterministic thing so you either get it or you don’t. So the answer is either 0% or 100%. Well, 100% isn't one of the answer choices!
126. In planar scintillation imaging, all of the following will decrease quantum mottle EXCEPT:
A. Use high sensitivity versus general purpose collimator
B. Increase time of imaging
C. Increase administered activity
D. Use a thicker crystal
E. Use a 10% instead of a 15% photodetector window
126. In planar scintillation imaging, all of the following will decrease quantum mottle EXCEPT:
A. Use high sensitivity versus general purpose collimator
B. Increase time of imaging
C. Increase administered activity
D. Use a thicker crystal
E. Use a 10% instead of a 15% photodetector window

Answer: E – to decrease QM, you need more counts. Option E will decrease the number of counts b/c the tighter window will exclude more counts.
. An abdominal radiograph in a direct digital imaging system is acquired with 2x
normal radiation. The image will be:
A. too dark
B. too light
C. have increased latitude
D. have increased detectability (low contrast)
E. decreased noise
127. An abdominal radiograph in a direct digital imaging system is acquired with 2x
normal radiation. The image will be:
A. too dark
B. too light
C. have increased latitude
D. have increased detectability (low contrast)
E. decreased noise

Answer: E – see the discussion for #122. The answer is definitely decreased noise if that is an option.
128. In processor QC, the control film is exposed to light in what?
A. photometer
B. densitometer
C. penetratometer
D. electrometer
E. sensitometer
128. In processor QC, the control film is exposed to light in what?
A. photometer
B. densitometer
C. penetratometer
D. electrometer
E. sensitometer

Answer: E – on multiple old tests. The densitometer reads the darkness and assigns an OD (associate Densitometer and optical Density)
129. Which cell population is least radiosensitive?
A. simple transit
B. decaying
C. closed static
D. closed dividing
E. stem cells
129. Which cell population is least radiosensitive?
A. simple transit
B. decaying
C. closed static
D. closed dividing
E. stem cells

Answer: C – see chart
130. Why does the number of counts have to be greater in SPECT than in planar scintigraphic imaging?
A. resolving time is greater
B. beam hardening is greater
C. noise propagation in reconstruction is greater
D. detection efficiency is less
E. photopeak window is smaller
130. Why does the number of counts have to be greater in SPECT than in planar scintigraphic imaging?
A. resolving time is greater
B. beam hardening is greater
C. noise propagation in reconstruction is greater
D. detection efficiency is less
E. photopeak window is smaller

Answer: C