2009 Physics Recalls

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2009 Physics Recalls : 

2009 Physics Recalls

2009 Physics Recalls (XR Production) : 

2009 Physics Recalls (XR Production) 1.  If the dose at 1 m is 60, where will the dose be 30. A. 1.6 m     B. 2 m     C. 1.4 m     D. 3 m

2009 Physics Recalls (XR Production) : 

2009 Physics Recalls (XR Production) 1.  If the dose at 1 m is 60, where will the dose be 30.     A. 1.6 m     B. 2 m     C. 1.4 m     D. 3 m This question just wants you to know the inverse square law.  The dose is halved and so you need the inverse square to equal 1/2, this will occur with (1/1.4)2, and thus C is the answer. (Huda p.4)

2009 (XR Production) : 

2009 (XR Production) 2. If the mAs is held constant and the kVp is increased which of the following is correct.        A. Increased beam intensity and increased maximum beam energy     B. Same beam intensity and increased maximum beam energy     C. Increased beam intensity and same maximum beam energy     D. Same beam intensity and same maximum beam energy

2009 (XR Production) : 

2009 (XR Production) 2. If the mAs is held constant and the kVp is increased which of the following is correct.        A. Increased beam intensity and increased maximum beam energy     B. Same beam intensity and increased maximum beam energy     C. Increased beam intensity and same maximum beam energy     D. Same beam intensity and same maximum beam energy X-ray beam intensity is proportional to the x-ray tube current (mAs). mAs affects the x-ray beam intensity, but does not change the energy spectrum.  mAs has no affect on the maximum x-ray photon energy.  Increasing the x-ray tube voltage increases the maximum x-ray photon energy.    It also says that relative x-ray beam intensity is proportional to KV^2.  Answer should be A.  Huda p.8-9.

2009 (XR Interaction) : 

2009 (XR Interaction) 1. A 65 kVp x-ray has the greatest absorption in which tissue?     A. Bone     B. Air     C. Fat     D. Muscle

2009 (XR Interaction) : 

2009 (XR Interaction) 1. A 65 kVp x-ray has the greatest absorption in which tissue?     A. Bone     B. Air     C. Fat     D. Muscle The key to this question is the equation for photoelectric absorption which is that it is proportional to Z^3 and 1/E^3 (where E is photon E). At energies below 100 keV there is increase in absorption for higher Z materials.   Bushburg pg 55-56 –  Huda pg 21

2009 (XR Interaction) : 

2009 (XR Interaction) Recalled differently: At 60kvp, the main reason why bone absorbs more x-rays compared to soft tissues is? a) Compton scatter b) Higher atomic number c) Larger electron density d) Pair production

2009 (XR Interaction) : 

2009 (XR Interaction) Recalled differently: At 60kvp, the main reason why bone absorbs more x-rays compared to soft tissues is? a) Compton scatter b) Higher atomic number c) Larger electron density d) Pair production Compton is the dominant interaction even in bone above 35 kVp.

2009 (XR Interaction) : 

2009 (XR Interaction) 2. Most of the energy deposited from radiation is by:      A. Electrons     B. Protons     C. Neutrons     D. Positrons

2009 (XR Interaction) : 

2009 (XR Interaction) 2. Most of the energy deposited from radiation is by:      A. Electrons     B. Protons     C. Neutrons     D. Positrons This seems to be a pretty consistent question every year.  As XR photons enter tissue they eject electrons which then have further interaction. An energetic electron may produce hundreds or thousands of additional ion pairs.  (Huda p.19)

2009 (XR Interaction) : 

2009 (XR Interaction) 3. If the HVL of an xray beam is below the minimum limit set by regulations, what will be the effect      A. Increase dose     B. Increased compton scatter     C. Increased penetrability     D. Decreased Dose

2009 (XR Interaction) : 

2009 (XR Interaction) 3. If the HVL of an xray beam is below the minimum limit set by regulations, what will be the effect      A. Increase dose     B. Increased compton scatter     C. Increased penetrability     D. Decreased Dose If the HVL is below the limit, then the average photon energy is also below the limit.  This will cause less penetrability of the photons and increase in dose.  (Huda p. 23)

2009 (XR Interaction) : 

2009 (XR Interaction) 4. Linear attenuation coefficient is most dependent on what?     A. Atomic mass     B. Electron density

2009 (XR Interaction) : 

2009 (XR Interaction) 4. Linear attenuation coefficient is most dependent on what?     A. Atomic mass     B. Electron density Probably misrecalled as both are correct in different scenarios. Linear attenuation coefficient is the fraction of incident photons removed from the beam per unit distance.   The attenuation is based on all XR interactions (coherent scatter (Rayleigh scatter), Compton scatter, and photoelectric absorption). Atomic mass (more correctly atomic number, Z) dominates linear attenuation coefficient when photoelectric interactions predominate (lower kVp / lower photonenergies). However, electron density (which is proportional to physical density in soft tissues) predominates as the photoelectric effect interactions decrease as a relative proportion of overall attenuation.  (Huda p.22)

2009 (XR Interaction) : 

2009 (XR Interaction) (Recalled different) Linear attenuation coefficient is defined as?     A. Attenuation per unit distance     B. Attenuation per unit density

2009 (XR Interaction) : 

2009 (XR Interaction) (Recalled different) Linear attenuation coefficient is defined as?     A. Attenuation per unit distance     B. Attenuation per unit density

2009 (XR Interaction) : 

2009 (XR Interaction) After a photon interacts with an outer-shell electron, the photon gets deflected with no net energy loss. This is likely due to which type of process? a) Compton scatter b) Photoelectric effect c) Pair production d) Coherent scatter

2009 (XR Interaction) : 

2009 (XR Interaction) After a photon interacts with an outer-shell electron, the photon gets deflected with no net energy loss. This is likely due to which type of process? a) Compton scatter b) Photoelectric effect c) Pair production d) Coherent scatter Coherent scattering (R), also known as Rayleigh, Thomson, or classical scattering, occurs when the x-ray photon interacts with the whole atom so that the photon is scattered with no change in internal energy to the scattering atom, nor to the x- ray photon. Thomson scattering is never more than a minor contributor to the absorption coefficient. The scattering occurs without the loss of energy. Scattering is mainly in the forward direction.

2009 (Projection Radiography) : 

2009 (Projection Radiography) 1. What is the advantage of direct digital versus indirect?     A. There is no spread of light in the direct digital to affect image quality.     B. Absorption efficiency is higher

2009 (Projection Radiography) : 

2009 (Projection Radiography) 1. What is the advantage is direct digital versus indirect?     A. There is no spread of light in the direct digital to affect image quality.     B. Absorption efficiency is higher    This question is likely asking if you know how the various types of detectors (Photostimulabe, scintillators, and photoconductors) work.  Photoconductors directly detect XR's with Selenium being the most commonly used per Huda.  (Huda p. 41-42)

2009 (Projection Radiography) : 

2009 (Projection Radiography) Which of the following will decrease scatter radiation to the image receptor? a) Increase kVp b) decreased grid ratio c) Increase distance from the patient to the image receptor d) Increased atomic number object within the patient

2009 (Projection Radiography) : 

2009 (Projection Radiography) Which of the following will decrease scatter radiation to the image receptor? a) Increase kVp b) decreased grid ratio c) Increase distance from the patient to the image receptor d) Increased atomic number object within the patient Increasing the kVp will increase the penetration of the beam, thus more photons will pass through the patient and strike the image receptor. This will decrease the noise, but will increase amount of scatter radiation striking the receptor. Grids decrease scatter. Decreasing the grid ratio will increase scatter. Increasing the ODD introduces an air gap. HUDA page 55, “air gaps between the patient and cassette reduces scatter because the scattered photons are less likely to reach the receptor.”

2009 (Projection Radiography) : 

2009 (Projection Radiography) Which of the following will decrease scatter radiation? a) Decrease grid ratio b) Increase peak voltage from the x-ray tube c) Increase mA d) Increase Bucky factor

2009 (Projection Radiography) : 

2009 (Projection Radiography) Which of the following will decrease scatter radiation? a) Decrease grid ratio b) Increase peak voltage from the x-ray tube c) Increase mA d) Increase Bucky factor Similar concept to prior question. Grids decrease scatter. Decreasing the grid ratio will increase scatter, where as increasing the Bucky factor will decrease scatter (but will increase dose to the patient). See Huda book, page 56. Increasing the the kVp will increase scatter.

2009 (Projection Radiography) : 

2009 (Projection Radiography) Recalled Differently: In digital radiography (DR), which of the following is true regarding why there is better spatial resolution in direct-DR when compared to indirect-DR? a) Higher absorption efficiency of photons b) Less diffusion of light c) Faster speed of system d) Thicker screen

2009 (Projection Radiography) : 

2009 (Projection Radiography) Recalled Differently: In digital radiography (DR), which of the following is true regarding why there is better spatial resolution in direct-DR when compared to indirect-DR? a) Higher absorption efficiency of photons b) Less diffusion of light c) Faster speed of system d) Thicker screen On prior recalls. HUDA book page 90, 5th bullet point. “The charge collection process does not introduce diffusion as occurs with light spreading scintillators. Know that direct-DR uses photoconductors (Selenium), while indirect-DR uses scintillators (CsI). Look at figure 6.2 on this page.

2009 (Projection Radiography) : 

2009 (Projection Radiography) 2. What has the largest storage requirement (several pictures are given including  single images from MR, CT, DSA, CXR)      A. MR     B. CT     C. DSA     D. US     E. CXR

2009 (Projection Radiography) : 

2009 (Projection Radiography) 2. What has the largest storage requirement (several pictures are giving including  single images from MR, CT, DSA, CXR)      A. MR     B. CT     C. DSA     D. US     E. CXR   CXR has a matrix of around 2560 X 2048 and thus the equation of matrix X byte will give you the amount of MB needed for the image.  The others use smaller matrices.   (Huda p.40)

2009 (Projection Radiography) : 

2009 (Projection Radiography) 3. How many bits are needed to encode 512 gray-scale values?     A. 7 bits     B. 8 bits     C. 9 bits     D. 10 bits

2009 (Projection Radiography) : 

2009 (Projection Radiography) 3. How many bits are needed to encode 512 gray-scale values?     A. 7 bits     B. 8 bits     C. 9 bits     D. 10 bits   This is a straight forward question, making sure you know bit values.  The amount of gray scale values is 2^n, and so 9 bits are needed to give you 512 shades of grey, 8 bits for 256 and so on.  (Huda p.38)

2009 (Projection Radiography) : 

2009 (Projection Radiography) 4. Selenium is used in what type of digital imaging?     A. Indirect digital radiography     B. Direct digital flat panel radiography     C. Computed radiography     D. Screen-film radiograpy

2009 (Projection Radiography) : 

2009 (Projection Radiography) 4. Selenium is used in what type of digital imaging?     A. Indirect digital radiography     B. Direct digital flat panel radiography     C. Computed radiography     D. Screen-film radiograpy   You need to know the types of detectors (scintillators, photoconductors, photostimulators), how each detects radiation and what is used in each.  Photoconductors are direct XR detectors and use Selenium.

2009 (Projection Radiography) : 

2009 (Projection Radiography) “A” is the original x-ray spectrum. “B” is the x-ray spectrum with additional filtration. Which graph best represents the spectrums “A” and “B” ?

2009 (Projection Radiography) : 

2009 (Projection Radiography) “A” is the original x-ray spectrum. “B” is the x-ray spectrum with additional filtration. Which graph best represents the spectrums “A” and “B” ? Answer: C Filtration will transmit only the high energy beams, and allow it to penetrate the object. Therefore the average number of photons in the beam will decrease (smaller curve), however, the peak kV will increase since all the lower value energies are no longer being averaged in the sum (curve will shift to the right). However, the max keV will not change with filtration, this aspect of the energy spectrum is unaffected (both curves should intersect at the x-axis). Therefore, “C” best illustrates these changes.

2009 (Projection Radiography) : 

2009 (Projection Radiography) The advantages of digital radiography are all except: A.      Increased spatial resolution B.      Post processing imaging C.      Wider latittude D.     Faster retrievel of old films.

2009 (Projection Radiography) : 

2009 (Projection Radiography) The advantages of digital radiography are all except: A.      Increased spatial resolution B.      Post processing imaging C.      Wider latittude D.     Faster retrievel of old films.   Digital radiography which is essentially what is used everyone is better at many things than film screen such as post processing, ability to retrieve old films quickly, etc.  The only disadvantage is spatial resolution is not as good.  A is the answer.

2009 (Projection Radiography) : 

2009 (Projection Radiography) 1. Digital temporal filtering (Frame averaging) does what to lag and noise     A. Increase lag, increase noise     B. Increase lag, decrease noise     C. Decrease lag, decrease noise     D. Decrease lag, increase noise

2009 (Projection Radiography) : 

2009 (Projection Radiography) 1. Digital temporal filtering (Frame averaging) does what to lag and noise     A. Increase lag, increase noise     B. Increase lag, decrease noise     C. Decrease lag, decrease noise     D. Decrease lag, increase noise

2009 (Projection Radiography) : 

2009 (Projection Radiography) 2. ABC (automatic brightness control) refers to?     A. Constant light output to the video camera     B. Constant patient dose despite the patient size     C. Constant Brightness on the monitor     D. Constant light production at the input phosphor

2009 (Projection Radiography) : 

2009 (Projection Radiography) 2. ABC (automatic brightness control) refers to?     A. Constant light output to the video camera     B. Constant patient dose despite the patient size     C. Constant Brightness on the monitor ???     D. Constant light production at the input phosphor     ABC is an aspect of flouroscopy that regulated the radiation required to maintain a constant TV display.  mA and kV (tube voltage) can both be adjusted to maintain a constaint light level at the II outpur phosphor. A is likely the best answer.  C may also be correct. A different recall had “maintain optical density,” this is likely the most correct. (Huda p.60)

2009 (Projection Radiography) : 

2009 (Projection Radiography) Noise in flouroscopy is due mainly to:      A. II      B. TV Moniter      C. XR Tube

2009 (Projection Radiography) : 

2009 (Projection Radiography) Noise in flouroscopy is due mainly to:      A. II      B. TV Moniter      C. XR Tube   The Noise in fluoroscopy is primarily due to quantum mottle which is a result of low x-ray tube output. The TV moniter is more responsible for resolution and limits this.

2009 (Projection Radiography) : 

2009 (Projection Radiography) (similar Q) What causes quantum mottle in fluoroscopy? Light output at output phosphor Light output at input phosphor Incident x-ray on input phosphor

2009 (Projection Radiography) : 

2009 (Projection Radiography) (similar Q) What causes quantum mottle in fluoroscopy? Light output at output phosphor Light output at input phosphor Incident x-ray on input phosphor

2009 (Projection Radiography) : 

2009 (Projection Radiography) All of the following are used in calculating average glandular dose in mammography EXCEPT     A. kVp     B. filter material     C. beam collimation     D. Target material     E. Compression

2009 (Projection Radiography) : 

2009 (Projection Radiography) All of the following are used in calculating  mean glandular dose in mammography EXCEPT     A. kVp     B. filter material     C. beam collimation     D. Target material     E. Compression   Breast thickness and tissue composition strongly affect x-ray absorption.  Higher kVp (higher HVL) decreases dose. Target and filter material affect beam energy. Bushburg pg 223-224

2009 (Projection Radiography) : 

2009 (Projection Radiography) 5. For fluoroscopy with automatic brightness control, what will NOT increase when you decrease the radiation field by collimation:     A. kVp     B.  mA     C. noise     D.  resolution

2009 (Projection Radiography) : 

2009 (Projection Radiography) 5. For fluoroscopy with automatic brightness control, what will NOT increase when you decrease the radiation field by collimation:     A. kVp     B.  mA     C. noise     D.  resolution

2009 (Projection Radiography) : 

2009 (Projection Radiography) In DSA, image noise due to quantum mottle,     A.  Is higher in the subtracted image than in the mask image     B. Is higher in the mask image than the subtracted image

2009 (Projection Radiography) : 

2009 (Projection Radiography) In DSA, image noise due to quantum mottle,     A.  Is higher in the subtracted image than in the mask image     B. Is higher in the mask image than the subtracted image         Two images are subtracted from each other to create the DSA image, and noise is additive in this mathematical manipulation of the data.  This is not really covered in either Huda or Bushburg.  A is likely the correct answer.

2009 (Projection Radiography) : 

2009 (Projection Radiography) If you change an image intensifier FOV from 9 in to 4 in what will happen?     A. Increase entrance skin dose     B. Increased brightness output at the output phosphor     C. Decreased resolution

2009 (Projection Radiography) : 

2009 (Projection Radiography) If you change an image intensifier FOV from 9 in to 4 in what will happen?     A. Increase entrance skin dose     B. Increased brightness output at the output phosphor     C. Decreased resolution

2009 (Projection Radiography) : 

2009 (Projection Radiography) (Recalled a different way) If you change an image intensifier FOV from 9 in to 4 in what will happen?     A. X ray beam becomes smaller     B. X ray beam becomes larger     C. The patient becomes closer D. The patient becomes further ????

2009 (Projection Radiography) : 

2009 (Projection Radiography) What is the focal spot size for magnification mamo     A. 0.1mm     B. 0.3mm     C. 1mm     D. 3mm

2009 (Projection Radiography) : 

2009 (Projection Radiography) What is the focal spot size for magnification mamo     A. 0.1mm     B. 0.3mm     C. 1mm     D. 3mm   The non magnified focal spot is 0.3 mm and the magnified is 0.1 mm for mammography. Know these two numbers.   (Huda p.51)

2009 (Projection Radiography) : 

2009 (Projection Radiography) Compression of breast is helpful for all factors except:     A. Decreased mottle     B. more even optical density     C. Decreased average glandular dose     D. Increased image contrast E. Exposure time

2009 (Projection Radiography) : 

2009 (Projection Radiography) What artifact?

2009 (Projection Radiography) : 

2009 (Projection Radiography) What artifact? Poor screen film contact

2009 (Projection Radiography) : 

2009 (Projection Radiography) What theoretically results in the best contrast for angiography? Using monochromatic beam just below k edge of iodine Using monochromatic beam just above k edge of iodine Using polychromatic beam just below k edge Using polychromatic beam just above k edge

2009 (Projection Radiography) : 

2009 (Projection Radiography) What theoretically results in the best contrast for angiography? Using monochromatic beam just below k edge of iodine Using monochromatic beam just above k edge of iodine Using polychromatic beam just below k edge Using polychromatic beam just above k edge

2009 (Projection Radiography) : 

2009 (Projection Radiography) Compression of breast is helpful for all factors except:     A. Decreased mottle     B. more even optical density     C. Decreased average glandular dose     D. Increased image contrast     Breast compression increases sharpness, reduces dose and reduces scatter (hence increases contrast).  Mottle is not improved.

2009 (CT) : 

2009 (CT) In computed radiography (CR), spatial resolution is limited by a) crystal thickness b) absorption efficiency of the scintillator material c) readout by the laser in image processing d) speed of the film

2009 (CT) : 

2009 (CT) In computed radiography (CR), spatial resolution is limited by a) crystal thickness b) absorption efficiency of the scintillator material c) readout by the laser in image processing d) speed of the film On prior recalls. As stated in HUDA, page 90, CR uses photostimulable phosphors to capture x-ray exposure patterns that are subsequently “read out” by lasers. The readout process is considered the rate limiting step. Explained in detail in Bushberg.

2009 (CT) : 

2009 (CT) For an abdominal CT what percentage of the skin entrance dose will be measured at the center of the patient?     A. 10%     B. 25%     C. 80%     D. 50%

2009 (CT) : 

2009 (CT) For an abdominal CT what percentage of the skin entrance dose will be measured at the center of the patient?     A. 10%     B. 25%     C. 80%     D. 50%     Huda 3rd Edition pg 78 – In body scans, surface doses are generally higher than the central dose.  In a 32 cm body phantom (used to estimate dose for abdominal CT), the surface dose is twice that obtained at the center of the phantom. The 10% that keeps showing up on recalls has to do with X-ray.

2009 (CT) : 

2009 (CT) What is the contrast resolution of CT     A.0.5%     B. 5%     C. 0.05%     D.0.005%

2009 (CT) : 

2009 (CT) What is the contrast resolution of CT     A.0.5%     B. 5%     C. 0.05%     D.0.005%   The contrast resolution in CT is very good and is around 0.5%.  (Bushberg p.367)

2009 (CT) : 

2009 (CT) A pulmonary nodule has a Hounsfield unit of 60 HU on a 10 mm slice from a CT chest examination. What would the attenuation (HU) of the same pulmonary nodule be on a 2mm slice? a. 100 b. 60 c. 0 d. -60

2009 (CT) : 

2009 (CT) A pulmonary nodule has a Hounsfield unit of 60 HU on a 10 mm slice from a CT chest examination. What would the attenuation (HU) of the same pulmonary nodule be on a 2mm slice? a. 100 b. 60 c. 0 d. -60 This question tests if you know partial-volume effects. Partial-volume artifact decreases attenuation of the object in question. If you make slices thinner, you improve partial-volume effects, and the object will display its true attenuation. Therefore, the attenuation of the object in this question will be “higher” since partial-volume effects are being reversed. There is only one answer given that is higher than the original value of 60HU. HUDA page 130.

2009 (CT) : 

2009 (CT) Spatial resolution in CT is related to a. Pre-patient collimation b. Focal spot c. Post-patient collimation d. Target material

2009 (CT) : 

2009 (CT) Spatial resolution in CT is related to a. Pre-patient collimation b. Focal spot c. Post-patient collimation d. Target material HUDA page 94. “Larger matrix sizes would not improve CT resolution becuause of focal spot blur and detector blur.” “Spatial resolution is a function of pixel size, focal spot size, detector size, and choice of image reconstruction filter. Section thickness (collimation) effects resolution in the longitudinal plane and is important for sagittal and coronal reconstructions.”

2009 (Image Quality) : 

2009 (Image Quality) Increased scatter in screen film mammography:       A. Decreased contrast       B. Decreased resolution       C. Improved with lowering the grid ratio       D. Improved with decreasing the bucky factor

2009 (Image Quality) : 

2009 (Image Quality) Increased scatter in screen film mammography:       A. Decreased contrast       B. Decreased resolution       C. Improbed with lowering the grid ratio       D. Improved with decreasing the bucky factor   Contrast in imaging depends on two main things:  scatter and photon energy.  Increased scatter will decrease contrast as does increased photon energy (because of more compton scattering).  Resolution is effected by three main things: focal spot size, detector blur, and motion.  (Huda p. 87-89)

2009 (Image Quality) : 

2009 (Image Quality) If a population has 300 test positive for an illness but only 200 of them actually have the disease, what is the positive predictive value of the test       A. 0.1     B.  0.33     C.  0.5     D. 0.67     E. 0.8

2009 (Image Quality) : 

2009 (Image Quality) If a population has 300 test positive for an illness but only 200 of them actually have the disease, what is the positive predictive value of the test       A. 0.1     B.  0.33     C.  0.5     D. 0.67     E. 0.8   Simple positive predictive value question.  The TP is 200 and the false positives is 100.  Eqn for PPV is TP/TP + FP and thus 200/300 and 0.67 is the answer. (Huda p.98)   Answer: 0.67 PPV= TP/(TP+FP) = 200/ (200+100) = 2/3 = 0.67

2009 (Image Quality) : 

2009 (Image Quality) Geometric unsharpness is increased by what?     A. Increasing focal spot size     B. Decreasing Magnification     C. Increasing kVp of the tube

2009 (Image Quality) : 

2009 (Image Quality) Geometric unsharpness is increased by what?     A. Increasing focal spot size     B. Decreasing Magnification     C. Increasing kVp of the tube      Geometric unsharpness is defined by the equation: FSS (Magnification-1).  Thus increasing the Focal spot size or the magnification will increase geometric unsharpness.

2009 (Image Quality) : 

2009 (Image Quality) In a population with a low prevalence of lung cancer.  If a radiologist were to call all the CXR's negative without looking at them, what would be the result?     A. High Accuracy     B. High Sensitivity     C. Low specificity     D. High positive predictive value

2009 (Image Quality) : 

2009 (Image Quality) In a population with a low prevalence of lung cancer.  If a radiologist were to call all the CXR's negative without looking at them, what would be the result?     A. High Accuracy     B. High Sensitivity     C. Low specificity     D. High positive predictive value     Since lung cancer prevalence is low in this population the number of correct diagnosis (negative CXR) will be high and so accuracy is the best answer.

2009 (Image Quality) : 

2009 (Image Quality) With regards to noise, the relationship between noise and dose is a) Decreases linearly with dose b) Increases linearly with dose c) Decreases with the square root of dose d) Increases with the square root of dose

2009 (Image Quality) : 

2009 (Image Quality) With regards to noise, the relationship between noise and dose is a) Decreases linearly with dose b) Increases linearly with dose c) Decreases with the square root of dose d) Increases with the square root of dose

2009 (Image Quality) : 

2009 (Image Quality) (Recalled differently) Signal/Noise ratio is A.      Proportional to Square Root N B.      Equal to N C.      Equal to 2n D.      Inversely proportional to N E.       Inversely proportion to root N

2009 (Image Quality) : 

2009 (Image Quality) Signal/Noise ratio is A.      Proportional to Square Root N (where N is image acquisitions in MR, counts in NM) B.      N C.      2n D.      Inversely proportional to N E.       Inversely proportion to root N If it had said the “uncertainty interval increases as . . .” then E. As the count interval goes up, uncertainty decreases by a factor 1/√N, where N is the number of the new count interval.

2009 (Rad Bio) : 

2009 (Rad Bio) At what effective dose is temporary epilation is expected? a) 1 Gy b) 2 Gy c) 3 Gy d) 7 Gy

2009 (Rad Bio) : 

2009 (Rad Bio) At what effective dose is temporary epilation is expected? a) 1 Gy b) 2 Gy c) 3 Gy d) 7 Gy

2009 (Rad Bio) : 

2009 (Rad Bio) Effective dose relates to: A. Cancer only B. Cancer and heritable effects C. Deterministic effects D. Deterministic and heritable

2009 (Rad Bio) : 

2009 (Rad Bio) Effective dose is used to estimate: A. Cancer only B. Cancer and heritable effects C. Deterministic effects D. Deterministic and heritable       The question is essentially asking what effective dose is used for.  The answer is stochastic risk which includes both cancer and heretible effects.  (Huda p114-117)

2009 (Rad Bio) : 

2009 (Rad Bio) Concerning Cataracts due to radiation exposure:     A. Can be induced by 750 mGy     B. Interventianal radiologist is at risk if he has had more than 3 Gy during his life     C. Can tell the difference between radiation and no radiation induced

2009 (Rad Bio) : 

2009 (Rad Bio) Concerning Cataracts due to radiation exposure:     A. Can be induced by 750 mGy     B. Interventianal radiologist is at risk if he has had more than 3 Gy during his life     C. Can tell the difference between radiation and no radiation induced        For this question you must know that acute radiation dose of 2 Gy will induce cataracts and a chronic dose of 5 Gy will also induce cataracts. Cataracts from radiation occur in the posterior pole and thus C is the likely answer for this question. (Huda p.106-107)

2009 (Rad Bio) : 

2009 (Rad Bio) 750 mGy acute dose causes:     A. Prodromal radiation syndrome      B. Hematopoietic syndrome     C. GI syndrome     D. CNS syndrome

2009 (Rad Bio) : 

2009 (Rad Bio) 750 mGy acute dose causes:     A. Prodromal radiation syndrome      B. Hematopoietic syndrome     C. GI syndrome     D. CNS syndrome   750 mGy or 0.75 Gy is not enough to cause GI syndrome (10 Gy) or CNS syndrome (100Gy).  Hematopoeitic syndrome can occur from 0.5 to 10Gy and so could occur at this dose.  Prodromal radiation decribes the first stage of each of the subsyndromes and is technically correct (Bushberg p. 832-34).

2009 (Rad Bio) : 

2009 (Rad Bio) In a population of 1,000,000 people, 10 mSv dose will cause how many fatal cancers?     A. 4-8     B. 40-80     c. 400-800     D. 4000-8000

2009 (Rad Bio) : 

2009 (Rad Bio) In a population of 1,000,000 people, 10 mSv dose will cause how many fatal cancers?     A. 4-8     B. 40-80     c. 400-800     D. 4000-8000 This questions requires that you know it is estimated that fatal cancer's occur in 4-8%/Sv.  If you know that then just convert mSv to Sv and multiple by a million.  Answer is 400-800. (Huda p.117)

2009 (Rad Bio) : 

2009 (Rad Bio) Which radiologic produced cancer has the shortest latency period?     A. Breast     B. Sarcoma     C. Leukemia     D. Thyroid

2009 (Rad Bio) : 

2009 (Rad Bio) Which radiologic produced cancer has the shortest latency period?     A. Breast     B. Sarcoma     C. Leukemia     D. Thyroid     Just remember that solid tumor latency periods are measure in decades whereas leukemia latency averages 5-25 years.  (Huda p.108)

2009 (Rad Bio) : 

2009 (Rad Bio) What model is used to make the dose response curve used to estimate radiation exposure risk?     A. Linear threshold     B. Linear no threshold     C. Quadratic threshold     D. Quadratic no threshold

2009 (Rad Bio) : 

2009 (Rad Bio) What model is used to make the dose response curve used to estimate radiation exposure risk?     A. Linear threshold     B. Linear no threshold     C. Quadratic threshold     D. Quadratic no threshold   Huda 3rd edition pg. 131 – The linear no threshold model is used to extrapolate radiation risks from high to low doses Pg 108- For solid tumors, the excess cancer incidence was found to be a linear function of dose.  Leukemia data were best fitted by a linear quadratic function of dose

2009 (Rad Bio) : 

2009 (Rad Bio) If the skin entrance dose of a fluoroscopy exam is 3 mGy.  How does the dose change between picture 1 and picture 2 with all other factors on the fluoroscopy unit being held constant. Diagram description: the distance from the source to the patient changes from 0.7m to 0.5m.   0.5 mGy 1.4 mGy 3 mGy 5.9 mGy 9 mGy

2009 (Rad Bio) : 

2009 (Rad Bio) If the skin entrance dose of a fluoroscopy exam is 3 mGy.  How does the dose change between picture 1 and picture 2 with all other factors on the fluoroscopy unit being held constant. Diagram description: the distance from the source to the patient changes from 0.7m to 0.5m.   0.5 mGy 1.4 mGy 3 mGy 5.9 mGy 9 mGy

2009 (Rad Bio) : 

2009 (Rad Bio) What is the MQSA limit for average glandular dose per film?     A. 3 mGy     B. 1mGy     C. 2 mGy     D. 5 mGy

2009 (Rad Bio) : 

2009 (Rad Bio) What is the MQSA limit for average glandular dose per film?     A. 3 mGy     B. 1mGy     C. 2 mGy     D. 5 mGy     They don't specify with and without a grid.  So the likely answer, assuming that a grid was used, is 3 mGy. (Bushburg pg 224, Huda p.58) From prior recall, total dose per breast for screening mammogram would be 6.0 mGy (CC and MLO views, each 3.0 Gy).

2009 (Rad Bio) : 

2009 (Rad Bio) A 50 yo interventional radiologist has worked for 20 years, his lifetime dose is 600mGy.      A.  He is over the lifetime limit     B.  He is under the lifetime limit     C. He will likely develop permanent epilation     D. He is risk of having cataracts

2009 (Rad Bio) : 

2009 (Rad Bio) A 50 yo interventional radiologist has worked for 20 years, his lifetime dose is 600mGy.      A.  He is over the lifetime limit     B.  He is under the lifetime limit     C. He will likely develop permanent epilation     D. He is risk of having cataracts     If you divide 600 by 20, then his average yearly dose has been 30 mGy.  However, the 50 mSv/yr occupational exposure only applies to any year and not a lifetime. There are recommendation on lifetime exposure which are are 10 X the individuals age per the NCRP.  Thus his lifetime dose should be 500 or less.    (The ICRP recommend's an average yearly exposure of 20 mSv per year for a lifetime exposure).  (Huda p.123)

2009 (Rad Bio) : 

2009 (Rad Bio) What whole body exposure will be lethal to 50% of the population in 60 days     A. 1 Gy     B. 2 Gy     C. 3.5 Gy     D. 750 mGy

2009 (Rad Bio) : 

2009 (Rad Bio) What whole body exposure will be lethal to 50% of the population in 60 days     A. 1 Gy     B. 2 Gy     C. 3.5 Gy     D. 750 mGy

2009 (Rad Bio) : 

2009 (Rad Bio) What exam gives the highest entrance skin exposure?     A. CT abdomen     B. Chest x-ray     C. Cardiac cineradiography     D. Mammography

2009 (Rad Bio) : 

2009 (Rad Bio) What exam gives the highest entrance skin exposure?     A. CT abdomen     B. Chest x-ray     C. Cardiac cineradiography     D. Mammography          Chart from the new Huda gave entrance skin exposure from Body CT as 30 mGy and from 1 min fluoro as 20 mGy, so I think the answer would be cardiac cineradiography, since this takes a long time and accumulates a lot of exposure. From the list above CT abdomen has the largest “effective dose”

2009 (Rad Bio) : 

2009 (Rad Bio) Which contributes to the largest collective effective dose in the US? a) Chest radiographs b) Nuclear cardiology c) Interventional angiography d) Upper GI

2009 (Rad Bio) : 

2009 (Rad Bio) Which contributes to the largest collective effective dose in the US? a) Chest radiographs b) Nuclear cardiology c) Interventional angiography d) Upper GI  See Radiographics/AAPM article on radiation dose for different modalities. HUDA review course emphasized largest collective dose is from nuclear stress tests. From prior recalls, modality with highest increase in radiation exposure to the general population would be CT.

2009 (Rad Bio) : 

2009 (Rad Bio) Which factor will increase entrance skin dose? A. kVp B. mAs C. Increased CT collimation

2009 (Rad Bio) : 

2009 (Rad Bio) Which factor will increase entrance skin dose? A. kVp B. mAs C. Increased CT collimation   Increasing mAs will increase the intensity which is what Kerma is quantifying.  Thus the skin dose will increase.  (Huda p.7-9)

2009 (Rad Bio) : 

2009 (Rad Bio) What is the cause of acute early erythematous reaction with radiation     A. Histamine release     B. Increased capillary permeability     C. Basal cell layer damage

2009 (Rad Bio) : 

2009 (Rad Bio) What is the cause of acute early erythematous reaction with radiation     A. Histamine release     B. Increased capillary permeability     C. Basal cell layer damage     The erythema is cause by capillary dilation due to radiation induced release of histamine.  Need to know the doses that cause early and late erythema. (Bushberg p. 829) See radiobiology syllabus, web rad train document page number 13. “A few hours after doses greater than 2 Gy (200 rad) there is an early erythema similar to sunburn, which is caused by vasodilation, edema, and loss of plasma constituents from capillaries. Reactions resulting from stem cell death take longer to develop.”

2009 (Rad Bio) : 

2009 (Rad Bio) What is the largest collective effective dose burden? A.      GI Fluoro B.      Nuc Cards C.      Angio D.      Mammo

2009 (Rad Bio) : 

2009 (Rad Bio) What is the largest collective effective dose burden? A.      GI Fluoro B.      Nuc Cards C.      Angio D.      Mammo   This question is likely asking, from what studies is the effective dose the largest.  Mammo is tiny, leaving the other three choices as answers.  According to Huda,  nuclear medicine has the highest average effective dose and so Nuclear cardiology is likely the answer.

2009 (Rad Bio) : 

2009 (Rad Bio) Radiation to the fetus is most likely to affect which system?     A. CNS     B. GI     C. Cardio     D. GU     E. Skin

2009 (Rad Bio) : 

2009 (Rad Bio) Radiation to the fetus is most likely to affect which system?     A. CNS     B. GI     C. Cardio     D. GU     E. Skin     Most of the in utero radiation deals with CNS complications.  Near the end of pregrancy, the most concerning issue with radiation is childhood cancers.

2009 (Rad Bio) : 

2009 (Rad Bio) What is the definition of use factor in radiation protection     A. How many studies the machine will perform     B. The average kVP of the machine     C. How often the machine is in use     D. The time that the primary beam points towards a barrier

2009 (Rad Bio) : 

2009 (Rad Bio) What is the definition of use factor in radiation protection     A. How many studies the machine will perform     B. The average kVP of the machine     C. How often the machine is in use     D. The time that the primary beam points towards a barrier     There are three components that need to be known when deciding how much shielding is needed in an XR room.  They are Workload, Use Factor, and Occupancy factor.  Use factor is the time that the primary beam points toward a barrier.  answer.  (Huda p. 126)

2009 (Rad Bio) : 

2009 (Rad Bio) What is the dose limit to the fetus of a pregnant radiation worker for the duration of the gestation?     A.  0.5 mSv     B. 50 mSv     C. 5 mSv     D. 500 mSv

2009 (Rad Bio) : 

2009 (Rad Bio) What is the dose limit to the fetus of a pregnant radiation worker for the duration of the gestation?     A.  0.5 mSv     B. 50 mSv     C. 5 mSv     D. 500 mSv        Huda 3rd edition pg. 124 – In the United States, the regulatory dose limit for the fetus of a radiation worker is 0.5mSv/mo, which implies a total dose limit of 5 mSv.

2009 (Rad Bio) : 

2009 (Rad Bio) What is the breast dose from a CT?     A. 0.01 mGy     B. 0.1 mGy     C. 1 mGy     D. 10 mGy     E. 100 mGy

2009 (Rad Bio) : 

2009 (Rad Bio) What is the breast dose from a CT?     A. 0.01 mGy     B. 0.1 mGy     C. 1 mGy     D. 10 mGy     E. 100 mGy   This question is confusing because it doesn't define what type of CT it is (chest, abdomen, etc.)  Assuming they are talking about a Chest CT, then there is a chart in Bushberg that gives the Breast absorbed dose at 21 mGy and so the best answer is probably D. (See Bushberg p.798)

2009 (Rad Bio) : 

2009 (Rad Bio) What is the occupational dose to the eye that is acceptable     A. 15 mSv/yr     B. 500 mSv/yr     C. 50 mSv/yr     D. 150 mSv/yr

2009 (Rad Bio) : 

2009 (Rad Bio) What is the occupational dose to the eye that is acceptable     A. 15 mSv/yr     B. 500 mSv/yr     C. 50 mSv/yr     D. 150 mSv/yr   150 mSv/yr is the occupational dose, 15 is the public.    Need to know the eye, skin, whole body etc. dose limits for population and occupational. (Huda p.124)

2009 (Rad Bio) : 

2009 (Rad Bio) Which of the following is not an effective way to decrease CT dose burden in children A.      Decrease Pitch B.      Eliminate precontrast images C.      Decreased Technique D.      Eliminate Grid

2009 (Rad Bio) : 

2009 (Rad Bio) Which of the following is not an effective way to decrease CT dose burden in children A.      Decrease Pitch B.      Eliminate precontrast images C.      Decreased Technique D.      Eliminate Grid  There are a few ways to decreases CT doses in kids.  Eliminating the grid would allow more primary photons to reach the film and thus would decrease dose (though contrast would worsen).  Since dose is proportional to the inverse of pitch (remember that having a pitch less than 1 means there is overlap of scans  (Pitch = Length table moves/Beam width)), decreasing the pitch is going to increase your dose.

2009 (Nucs) : 

2009 (Nucs) Why is a lead cover placed around a newly eluted syringe of 99mTc when it is placed in the dose calibrator?     A. To measure moly breakthrough     B.  To measure aluminum breakthrough     C. Protect the radiation worker's hand     D. Avoid spilling

2009 (Nucs) : 

2009 (Nucs) Why is a lead cover placed around a newly eluted syringe of 99mTc when it is placed in the dose calibrator?     A. To measure moly breakthrough     B.  To measure aluminum breakthrough     C. Protect the radiation worker's hand     D. Avoid spilling    Lead is placed over the elution because it will block the gamma rays from Tc, thus allowing only the Mo gamma rays (higher energy) to be counted.  (Huda p. 153)

2009 (Nucs) : 

2009 (Nucs) Quality control test using 57Co in a dose calibrator set at 99Tm energy is used to test what?     A. Constancy     B. Linearity     C. Dead Time     D. Count accuracy

2009 (Nucs) : 

2009 (Nucs) Quality control test using 57Co in a dose calibrator set at 99Tm energy is used to test what?     A. Constancy     B. Linearity     C. Dead Time     D. Count accuracy   You need to know the quality assurance done for dose calibration: Accuracy, Constancy, Linearity.  The source usually used for constancy is Cesium per Huda and Co-57 per Bushberg. Accuracy usually is measured with Co-57. Thus I think that the best answer is likely A or D.  (Bushburg p.660 and Huda p.154)

2009 (Nucs) : 

2009 (Nucs) What is the Dose to red marrow of a 20mCi dose of Tc99m in bone scan     A. 1 mGy     B. 2 mGy     C. 3-20 mGy     D. 30 mGy

2009 (Nucs) : 

2009 (Nucs) What is the Dose to red marrow of a 20mCi dose of Tc99m in bone scan.     A. 1 mGy     B. 2 mGy     C. 3-20 mGy     D. 30 mGy

2009 (Nucs) : 

2009 (Nucs) What does the term absorbed fraction refer to in nuclear medicine? a. The fraction of activity at the source organ b. The activity localized at the target organ from the source organ c. The fraction of the target organ which is in the field

2009 (Nucs) : 

2009 (Nucs) What does the term absorbed fraction refer to in nuclear medicine? a. The fraction of activity at the source organ b. The activity localized at the target organ from the source organ c. The fraction of the target organ which is in the field See Mettler’s Essential of Nuclear Medicine, “absorbed fraction is the average absorbed dose in a target organ as a result of radioactive material deposited initially in a source organ.”

2009 (Nucs) : 

2009 (Nucs) If the effective half life is 2 days and the physical half life is 4 days, what is the biologic half life     A.  3 days     B. 2 Days     C. 4 Days     D. 1.33 Days

2009 (Nucs) : 

2009 (Nucs) If the effective half life is 2 days and the physical half life is 4 days, what is the biologic half life     A.  3 days     B. 2 Days     C. 4 Days     D. 1.33 Days     Simply know the equation 1/Te = 1/Tp + 1/Tb and calculate to get a biologic half life of 4.

2009 (Nucs) : 

2009 (Nucs) What kind of collimator is used in PET scanning     A. High Resolution     B. Pinhole     C. High Energy     D. Electronic collimation

2009 (Nucs) : 

2009 (Nucs) What kind of collimator is used in PET scanning     A. High Resolution     B. Pinhole     C. High Energy     D. Electronic collimation   Septal collimators are not needed for localization of photons in PET scanning because of coincidence detection.  I think the best answer is electronic collimation. Old Huda book page 143, however, says that high energy collimators are used with F18. The answer is thus D or C.  (see Huda, new book p. 152)

2009 (Nucs) : 

2009 (Nucs) Dose calibrators use what type to detector?     A. Ionization chamber     B. Geiger Counter     C. Thermoluminescence     D. Film Dosimetry

2009 (Nucs) : 

2009 (Nucs) Dose calibrators use what type to detector?     A. Ionization chamber     B. Geiger Counter     C. Thermoluminescence     D. Film Dosimetry     Huda p.122

2009 (Nucs) : 

2009 (Nucs) For gamma rays incident on a NaI (Tl) detector which of the following are true? A.  Tl holds the crystal lattice together B.  Photon energy increases with increasing incident energy of the gamma ray C.  Increased light is produced with increasing incident energy D.  With increased energy the light produced turns red

2009 (Nucs) : 

2009 (Nucs) For gamma rays incident on a NaI (Tl) detector which of the following are true? A.  Tl holds the crystal lattice together B.  Photon energy increases with increasing incident energy of the gamma ray C.  Increased light is produced with increasing incident energy D.  With increased energy the light produced turns red     Light output from the NAI crystal is proportional to the amount of energy absorbed by the scintillators.  Thus the light produced would be doubled if the energy of the photons were doubled.  (Huda p.148)

2009 (Nucs) : 

2009 (Nucs) What quality control test of SPECT uses a point source with images taken at different angles?     A. Center of rotation     B. Linearity at different angles     C. uniformity

2009 (Nucs) : 

2009 (Nucs) What quality control test of SPECT uses a point source with images taken at different angles?     A. Center of rotation     B. Linearity at different angles     C. uniformity     Center of rotation is a key aspect of SPECT imaging and a point source is used with images taken at different angle to see if the point source remains in the middle instead of forming a donut (due to non center of rotation).  (Bushberg p. 715)

2009 (Nucs) : 

2009 (Nucs) To go from 2D to 3D PET acquisition what do you need to do?      A.  Remove electronic collimation      B.  Retract lead collimators rings     C.  Use at least 2 photomultiplier tubes

2009 (Nucs) : 

2009 (Nucs) To go from 2D to 3D PET acquisition what do you need to do?      A.  Remove electronic collimation      B.  Retract lead collimators rings      C.  Use at least 2 photomultiplier tubes    The septal collimator rings are removed and coincidences are detected among many or all rings of detectors.  (Huda p.152)

2009 (Nucs) : 

2009 (Nucs) What type of material surrounding a syringe is used to protect radiation technologist handling 18F PET radiopharmaceuticals?     A. Lead     B. Lucite     C. Tungsten     D. Boron

2009 (Nucs) : 

2009 (Nucs) What type of material surrounding a syringe is used to protect radiation technologist handling 18F PET radiopharmaceuticals?     A. Lead     B. Lucite     C. Tungsten     D. Boron Answer is C, Tungsten. This is based on numerous internet searches finding vendors peddling these things (not lead) and a brief phone call to our nuc pharmacist who states that Lead is no longer used due to “patient safety concerns.”

2009 (Nucs) : 

2009 (Nucs) If 131I NaI is to be given during lactation what is the recommendation?     A. Discontinue permanently, 2 weeks prior to administration.     B. Discontinue permanently, allowing the breasts to fill 6hrs prior to administration.     C. Discontinue temporarily, while allowing breasts to fill 6hr prior to administration and pumping 4 times a day for 5 days     D. Discontinue temporarily, while allowing breasts to fill 6hr prior to administration and no pumping for 5 days.

2009 (Nucs) : 

2009 (Nucs) If 131I NaI is to be given during lactation what is the recommendation?     A. Discontinue permanently, 2 weeks prior to administration.     B. Discontinue permanently, allowing the breasts to fill 6hrs prior to administration.     C. Discontinue temporarily, while allowing breasts to fill 6hr prior to administration and pumping 4 times a day for 5 days     D. Discontinue temporarily, while allowing breasts to fill 6hr prior to administration and no pumping for 5 days.   Review the chart in Mettler's (back of the book) or in the Bushberg syllabus. Here are some common radiotracers: - most Tc products stop 24 hours after dose given then restart. - Ga 67 stop for 4 wks - I 131 stop 2 weeks before and don’t restart - I 123 stop for 3-5 d - Thallium stop for 2 weeks

2009 (Nucs) : 

2009 (Nucs) You are given an image with a linear photopenic defect.  If an initial intrinsic flood image with a point source was completely normal, what is causing this?     A. Damaged collimator     B. Defect photomultiplier tube     C. Damaged NaI     D. Computer malfunction

2009 (Nucs) : 

2009 (Nucs) You are given an image with a linear photopenic defect.  If an initial intrinsic flood image with a point source was completely normal, what is causing this?     A. Damaged collimator     B. Defect photomultiplier tube     C. Damaged NaI     D. Computer malfunction   This question is asking if you know the difference between intrinsic and extrinsic flood images.  Intrinsic is measuring without the collimator and extrinsic with.  Since the intrinsic image is normal, then the defect must be caused by the collimator.  (Huda p.153) but the crystal could have been cracked after / while the collimator was put on. I think that you would have to see the image in this instance to distinguish cracked crystal vs damaged collimator.

2009 (Nucs) : 

2009 (Nucs) Why is it advantageous to use Y-90 as a therapeutic agent     A. It’s a Beta emitter     B. It’s an alpha emitter

2009 (Nucs) : 

2009 (Nucs) Why is it advantageous to use Y-90 as a therapeutic agent     A. It’s a Beta emitter     B. It’s an alpha emitter   Ytrrium 90 is useful because it can be attached and localized to a source and then undergo Beta decay to deliver high doses to nearby tissue.   (Not in Huda or Bushburg)

2009 (Nucs) : 

2009 (Nucs) Type of decay where proton is converted to a neutron     A. Pair production     B. Isomeric transition     C.Beta Plus     D. Beta minus

2009 (Nucs) : 

2009 (Nucs) Type of decay where proton is converted to a neutron     A. Pair production     B. Isomeric transition     C.Beta Plus     D. Beta minus

2009 (Nucs) : 

2009 (Nucs) What advantage does a 2 headed SPECT camera have     A. Better resolution     B. Better contrast     C. Faster image acquisition

2009 (Nucs) : 

2009 (Nucs) What advantage does a 2 headed SPECT camera have     A. Better resolution     B. Better contrast     C. Faster image acquisition     Multiheaded cameras are used to increase sensitivity and reduce scan times.  (Huda p.150)

2009 (Nucs) : 

2009 (Nucs) What type of decay is best for imaging and protection?       A. Positron     B. Isomeric transition     C. Alpha decay     D. Internal Conversion

2009 (Nucs) : 

2009 (Nucs) What type of decay is best for imaging and protection?       A. Positron     B. Isomeric transition     C. Alpha decay     D. Internal Conversion   To be good for imaging, it need enough energy to get out of the body to be captured by a camera as well as not have too much radiation energy deposited in the body.  Isomeric transition does this best as it produces gamma rays which are adequately imaged with no particle emission (alpha or beta) to deposit radiation.  This is why Tc99 is used so frequently as it has isomeric transition decay as well as a short half life and a good energy peak.  (Huda p.140-42)

2009 (Nucs) : 

2009 (Nucs) FDG-18F is more of a radiation protection concern because: a)      beta emissions from positron emissions b)     More penetrating energy c)      Emits alpha particles

2009 (Nucs) : 

2009 (Nucs) FDG-18F is more of a radiation protection concern because: a)      beta emissions from positron emissions b)     More penetrating energy c)      Emits alpha particles   The penetrating ability of the high-energy 511-keV gamma rays produced from the annihilation reaction of a positron and an electron is greater than that of the 140-keV emissions from 99mTc-based compounds.

2009 (Nucs) : 

2009 (Nucs) Which of the following is true regarding pair production? a) It requires energy of 1.02MeV for it to occur b) Auger electrons are produced c) Results in annihilation of a positron to produce 2 photons of 511 eV each d) Is a common interaction seen in diagnostic radiology

2009 (Nucs) : 

2009 (Nucs) Which of the following is true regarding pair production? a) It requires energy of 1.02MeV for it to occur b) Auger electrons are produced c) Results in annihilation of a positron to produce 2 photons of 511 eV each d) Is a common interaction seen in diagnostic radiology Pair production (PP) can occur when the x-ray photon energy is greater than 1.02 MeV, but really only becomes significant at energies around 10 MeV. Pair production occurs when an electron and positron are created with the annihilation of the x-ray photon. Positrons are very short lived and disappear (positron annihilation) with the formation of two photons of 0.51 MeV energy. Pair production is of particular importance when high-energy photons pass through materials of a high atomic number.

2009 (Nucs) : 

2009 (Nucs) Which isotope decays by electron capture exclusively? Tl-201 I-131 F-18 Tc-99m

2009 (Nucs) : 

2009 (Nucs) Which isotope decays by electron capture exclusively? Tl-201 I-131 F-18 Tc-99m   The main ones that decay by electron capture are Ga67, I123, 111In, and Tl 201. (Huda p.143)

2009 (Nucs) : 

2009 (Nucs) Spatial resolution is worse in SPECT-CT when compared to planar imaging because a. Lack of use of a photomultiplier tube b. Pulse height analyzer is not utilized c. Filtered back projection is utilized d. Resolution decreases as the patient to collimator distance decreases

2009 (Nucs) : 

2009 (Nucs) Spatial resolution is worse in SPECT-CT when compared to planar imaging because a. Lack of use of a photomultiplier tube b. Pulse height analyzer is not utilized c. Filtered back projection is utilized d. Resolution decreases as the patient to collimator distance decreases Computed tomography uses filtered back projection. Read about this in Bushberg or Radiographics/AAPM article. In SPECT, series of planar images are called projections. After all projections are acquired, each slice are configured into a “sinogram.” A sinogram represents the projection of the tracer distribution in the body into a single slice on the camera at eery angle of the acquisition. Reconstruction is then applied on the projection data from the sinogram through Fourier transform. Data “filtering” is then applied to smooth out statistical noise. Different types of filters are available (low pass/high pass, etc). The data is then inversed so the data can then be taken from the frequency domain after fourier transform back into the spatial domain. Lastly, back-projection is applied so that the original data from the filtered sinogram can be applied along the same lines where the photons were originally emitted from. Through this whole process, spatial resolution is comprised as true data is manipulated and processed, picking up artifacts along the way.

2009 (Ultrasound) : 

2009 (Ultrasound) What doesn’t affect the Doppler shift?     A. size of reflector     B. ultrasound frequency     C.  ultrasound velocity     D. angle     E. reflector velocity

2009 (Ultrasound) : 

2009 (Ultrasound) What doesn’t affect the Doppler shift?     A. size of reflector (size of object)     B. ultrasound frequency     C.  ultrasound velocity     D. angle     E. reflector velocity     You need to know the doppler equation to answer this question: Doppler shift = 2 f v cos (angle) / C.  Thus the size of the reflector does not play any part.

2009 (Ultrasound) : 

2009 (Ultrasound) Where is spectral broadening the greatest.  There is a drawn sketch of a vessel with an area of stenosis.  1 and 2 are before the stenosis, 3 is in the stenosis, 4 is just after the stenosis and 5 is downstream     A. 1     B. 2     C. 3     D. 4     E. 5

2009 (Ultrasound) : 

2009 (Ultrasound) Where is spectral broadening the greatest.  There is a drawn sketch of a vessel with an area of stenosis.  1 and 2 are before the stenosis, 3 is in the stenosis, 4 is just after the stenosis and 5 is downstream     A. 1     B. 2     C. 3     D. 4     E. 5     Spectral broadening is caused by turbelence which is greatest after fast moving blood (within an area of stenosis) then enters a non stenotic area.  Thus 4 is the correct answer.

2009 (Ultrasound) : 

2009 (Ultrasound) How do phased array transducers work?     A. Sequentially steer the beam through the patient     B. Wobble the anode

2009 (Ultrasound) : 

2009 (Ultrasound) If pulse repetition frequency is 10,000/s, at what depth can you image?     A. 5 cm     B. 7.5 cm     C. 10 cm     D. 15 cm

2009 (Ultrasound) : 

2009 (Ultrasound) If pulse repetition frequency is 10,000/s, at what depth can you image?     A. 5 cm     B. 7.5 cm     C. 10 cm     D. 15 cm     The normal PRF is around 4000/second.  The higher the PRF you get the less the transducer "listens" to the returning wave and thus depth becomes limited.  A PRF of 10,000 will likely reach a depth of 7.5 cm (8 cm by Huda).  Remember that if the PRF is a normal 4000/s then the penetration depth is around 20 cm.  (Huda p.170) For a quick way of doing the calculation: 13 microseconds = 1.0 cm (depth beam travels) Next, convert PRF into seconds. Since PRF is a frequency in hertz (cycles/second), take the reciprocal to get units of seconds. 1/10,000 = .0001 seconds. Now convert to microseconds = 100 microseconds Set up proportion: 13 microseconds per 1.0 cm = 100 microseconds per “x” Solve for “x”: (13 microseconds)(x) = (100 microseconds)(1.0 cm) x = (100 microseconds)(1.0 cm)/(13 microseconds) = 7.5 cm.

2009 (Ultrasound) : 

2009 (Ultrasound) What has the greatest effect on LATERAL resolution?     A. Focusing of the US beam     B. Pulse length     C. PRF     D. Increase the amount of filter

2009 (Ultrasound) : 

2009 (Ultrasound) What has the greatest effect on LATERAL resolution?     A. Focusing of the US beam     B. Pulse length     C. PRF     D. Increase the amount of filter   Pulse length relates to axial resolution. Number of focusing cups and width of the beam relates to lateral resolution.

2009 (Ultrasound) : 

2009 (Ultrasound) Harmonic Imaging is most useful in A. Neonatal head B. Shoulder C. Abdomen D. Breast

2009 (Ultrasound) : 

2009 (Ultrasound) Harmonic Imaging is most useful in A. Neonatal head B. Shouldar C. Abdomen D. Breast   Harmonic images is exactly as it sounds.  The transducer listens for the harmonics (integral multiples of the fundamental pulse frequency).  The first harmonic is most often used. It is useful when there is fat or other tissue that causes significantly clutter (noise).

2009 (Ultrasound) : 

2009 (Ultrasound) The frequency of transmitted ultrasound beams depends on a. Crystal diameter b. Crystal thickness c. Tissue density d. Matching layer

2009 (Ultrasound) : 

2009 (Ultrasound) The frequency of transmitted ultrasound beams depends on a. Crystal diameter b. Crystal thickness c. Tissue density d. Matching layer   On prior recalls. Transducer thickness = wavelength/2

2009 (Ultrasound) : 

2009 (Ultrasound) Object is smaller than US beam width, what results     A. Nonspecular scatter     B. Reverberation     C. Mirror image     D. Noise

2009 (Ultrasound) : 

2009 (Ultrasound) Object is smaller (larger?) than US beam width, what results     A. Nonspecular scatter     B. Reverberation     C. Mirror image     D. Noise   When the image is smaller (larger object than ultrasound wavelength)? than the beam width you will get scattering.  This is why the liver has such a granular appearance.  The answer is A. (Huda p.165)

2009 (Ultrasound) : 

2009 (Ultrasound) What is it called when you do US with several different path to asses structures at several different angles     A.3D US     B. Spatial compounding     C. Harmonic imaging     D. Spectral doppler

2009 (Ultrasound) : 

2009 (Ultrasound) What is it called when you do US with several different path to asses structures at several different angles     A.3D US     B. Spatial compounding     C. Harmonic imaging     D. Spectral doppler   The answer is spatial compounding.  This is commonly used in biopsy's because with the multiple angle a more perpendicular angle to the needle can be obtained and allow you to see your needle better. (Not discussed in Huda or Bushberg)

2009 (Ultrasound) : 

2009 (Ultrasound) What causes comet tail artifacts?     A. Refraction     B. Absorption     C. Scatter     D. Reflection

2009 (Ultrasound) : 

2009 (Ultrasound) What causes comet tail artifacts?     A. Refraction     B. Absorption     C. Scatter     D. Reflection ??   The correct answer to this question should have been reverberation or ring down (not answers) (Huda p.176)

2009 (MRI) : 

2009 (MRI) What is the principle of MR angiography using phase contrast?        A. Bipolar Gradients are used      B. Bright signals from unsaturated protons are used C. Uses pre-saturation technique to encode velocity of blood prior to slice entrance D. Uses echo-planar imaging (EPI) to characterize water molecule diffusion

2009 (MRI) : 

2009 (MRI) What is the principle of MR angiography using phase contrast?   A. Bipolar Gradients are used B. Bright signals from unsaturated protons are used C. Uses pre-saturation technique to encode velocity of blood prior to slice entrance D. Uses echo-planar imaging (EPI) to characterize water molecule diffusion       There are two main techniques for MRA, Time of flight and Phase Contrast.  Phase contrast uses bipolar gradients to produce phase changes in flowing blood.  (Huda p.195) “C” is regarding time-of-flight imaging. “D” is regarding diffusion-weighted imaging.

2009 (MRI) : 

2009 (MRI) How can you increase spatial resolution while maintaining imaging time     A. Change FOV     B. Change TE     C. Increase frequency acquisition     D. Increase phase encoding steps.     E. Change slice thickness     F. Change NEX

2009 (MRI) : 

2009 (MRI) How can you increase spatial resolution while maintaining imaging time     A. Change FOV     B. Change TE     C. Increase frequency acquisition ??     D. Increase phase encoding steps.     E. Change slice thickness     F. Change NEX    For resolution improvement you want to decrease the FOV and increase the matrix.  Also, higher resolution is seen with higher SNR.  (Huda p.192)

2009 (MRI) : 

2009 (MRI) Which line on the diagram of MR sequences represents the frequency?     A. First Line     B. Second Line     C. Third Line     D. Fourth Line     E. Fifth Line

2009 (MRI) : 

2009 (MRI) Which line on the diagram of MR sequences represents the frequency?     A. First Line     B. Second Line     C. Third Line     D. Fourth Line     E. Fifth Line         Make sure you know what each of the lines represent in an MRI sequency image.   The frequency encode gradient is also called the read gradient and is applied when the echo is obtained,  it is usually the fourth line in a sequence diagram.   (Huda p.190) MRItutor.org

2009 (MRI) : 

2009 (MRI) In BOLD imaging, oxyhemoglobin compared to deoxyhemoglobin most affects:     A.  T1     B. T2     C. T2*     D. Proton Density

2009 (MRI) : 

2009 (MRI) In BOLD imaging, oxyhemoglobin compared to deoxyhemoglobin most affects:      A.  T1     B. T2     C. T2*     D. Proton Density          From http://radiopaedia.org/articles/bold_imaging, BOLD (Blood Oxygenation Level Dependent) Imaging is a functional MRI technique to delineate regional activity. It uses heavily T2* weighted sequences to detect the different concentration of oxygenated versus deoxygenated hemoglobin. It is possible to detect these differences because deoxygenated hemoglobin is paramagnetic whereas oxygenated hemoglobin is not, and therefore the former will cause signal drop out (just as all other paramagnetic compounds cause signal drop out on T2* sequences).

2009 (MRI) : 

2009 (MRI) Two T1 MR brain images are shown, both the same sequence (T2) with one having less signal. Why?       A. SNR is proportional to voxel size     B. Time of inversion has been changed     C. Chemical Shift     D. Motion artifact

2009 (MRI) : 

2009 (MRI) Two T1 MR brain images are shown, both the same sequence (T2) with one having less signal. Why?         A. SNR is proportional to voxel size     B. Time of inversion has been changed     C. Chemical Shift     D. Motion artifact     Essentially you need to know what increases SNR.  Increasing slice thickness, magnetism and surface coils will increase SNR.  Decreasing bandwith, Matrix will decrease SNR.

2009 (MRI) : 

2009 (MRI) 2 Images are shown.  The one on the right has the CSF black (It is bright on the one on the left).  What is responsible for the differences in these images.     A. Inversion Recovery     B. Spin echo     C. Gradient recalled Echo     D. FSE

2009 (MRI) : 

2009 (MRI) 2 Images are shown.  The one on the right has the CSF black (It is bright on the one on the left).  What is responsible for the differences in these images.     A. Inversion Recovery     B. Spin echo     C. Gradient recalled Echo     D. FSE   This is a straightforward question seeing if you know what sequence is used to suppress the CSF.  This is done with a FLAIR image by changing the time of inversion.  (Huda p.191)

2009 (MRI) : 

2009 (MRI) A Phantom image is shown with dark lines next to a square object.  What causes this?     A. Motion     B. Wraparound     C. Chemical Shift     D. Gibbs artifact

2009 (MRI) : 

2009 (MRI) A Phantom image is shown with dark lines next to a square object.  What causes this?     A. Motion     B. Wraparound     C. Chemical shift     D. Gibbs artifact    This is a Gibbs artifact which is caused by insufficient sampling of high frequencies near sharp boundaries.  (Bushburg p.455) Gibbs occurs in phase direction, chemical shift artifact in frequency direction

2009 (MRI) : 

2009 (MRI) How many Gauss in 0.1 T?     A. 1000     B. 100     C. 10,000     D. 0.1     E. 10

2009 (MRI) : 

2009 (MRI) How many Gauss in 0.1 T?     A. 1000     B. 100     C. 10,000     D. 0.1     E. 10   Remember that there are 10,000 gauss in a tesla. If you know that this becomes an easy problem.

2009 (MRI) : 

2009 (MRI) MRI SNR will increase by decreasing: Answer: Receiver bandwidth MRI SNR will increase by decreasing:   Answer: TE   Both were similar questions with different answer options and these were the two answers.

2009 (Miscellaneous) : 

2009 (Miscellaneous) On a ROC curve, what is the best quantitative method to measure observer-related detection? a) X and Y intercept b) Difference among two points along the curve c) Area under the curve d) Slope

2009 (Miscellaneous) : 

2009 (Miscellaneous) On a ROC curve, what is the best quantitative method to measure observer-related detection? a) X and Y intercept b) Difference among two points along the curve c) Area under the curve d) Slope

2009 (Miscellaneous) : 

2009 (Miscellaneous) The binding energy of L and K shells are 10 keV and 30keV respectively. What would be the dynamic energy of the photoelectrons ejected from the L and K shells if the incident photon has an energy of 50keV? a. 20 and 40 keV b. 30 and 20 keV c. 40 and 20 keV d. 50 and 50 keV

2009 (Miscellaneous) : 

2009 (Miscellaneous) The binding energy of L and K shells are 10 keV and 30keV respectively. What would be the dynamic energy of the photoelectrons ejected from the L and K shells if the incident photon has an energy of 50keV? a. 20 and 40 keV b. 30 and 20 keV c. 40 and 20 keV d. 50 and 50 keV

2009 (Miscellaneous) : 

2009 (Miscellaneous)

2009 (Miscellaneous) : 

2009 (Miscellaneous) Limiting spatial resolution in gamma camera? Answer: collimator