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A series of multiple-choice questions covering various aspects of medical imaging, particularly x-ray technology and computed tomography (ct). It delves into concepts like x-ray tube filament current, anode attraction, power measurement, mtf values, x-ray beam filtration, photoelectric absorption, air kerma, and ct beam shaping filters. The questions are designed to assess understanding of fundamental principles and practical applications in medical imaging.
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The most likely x-ray tube filament current (mA) is: a. 0. b. 4 c. 40 d. 400 e. 4,000 - โ โ e- X-ray tube filaments are about 4 A, or 4,000 mA. Which of the following would most likely be attracted to an anode? a. Proton b. Neutron c. Electron d. Positron
e. Alpha particle - โ โ c- Electron, since it has a negative charge that is attracted to the positive anode. Which quantity is the best measure of power? a. Joule b. Tesla c. Watt d. Coulomb
e. Newton - โ โ c- Watt is a unit of power, where 1 W= 1 J/s. The MTF value (%) at the lowest spatial frequencies is most likely: a. 100 b. 75 c. 50 d. 25
d. 100
e. 1,000 - โ โ d- 100 kW is typical of the power of x-ray generators in radiography and CT. X-ray quantum mottle is best characterized by quantifying: a. x-ray beam filtration b. detector air kerma c. average photon energy d. scintillator conversion efficiency
e. image receptor thickness - โ โ b-The detector air kerma determines the number of x- ray photons used to make an image. X-ray tube output would likely in- crease the most when increasing the x-ray tube: a. voltage b. anode angle c. target Z
d. current
e. exposure time - โ โ a- The x-ray tube output is (approximately) proportional to the square of the x-ray tube voltage. A CT beam shaping filter (bow tie) is most likely made out of: a. aluminum b. copper c. molybdenum d. Teflon
e. tin - โ โ d- Teflon (i.e., tissue like) is used as the CT beam shaping filter material to minimize beam hardening artifacts. Actual vertical resolution (line pairs) achieved with a 525-line TV monitor is: a. 180 b. 262 c. 370
b. Scattered photon c. K-shell x-ray d. L-shell x-ray e. Auger electron - โ โ b- There are no scattered photons in photoelectric absorption; scatter occurs with Compton interactions. Air kerma is the kinetic energy released per unit: a. distance(m) b. area (m^2 ) c. volume (m^3 ) d. mass (kg)
e. density (kg/m^3 ) - โ โ d. Mass (energy transferred to electrons per unit mass and measured in Gy). An air kerma of 1 mGy is likely to result in an absorbed dose (mGy) to bone (no back scatter) of:
a. 1 b. 2 c. 4 d. 8 e. 16 - โ โ c- An air kerma of 1 mGy will result in an absorbed dose to bone of โผ4 mGy when there is no backscatter. An air kerma of 1 mGy will most likely to result in an absorbed dose (mGy) to soft tissue (no backscatter) of: a. 0. b. 1. c. 1. d. 2.
e. 4.0 - โ โ c- An air kerma of 1 mGy will result in an absorbed dose to tissue of โผ1. mGy when there is no
At 65 kV and with a tungsten target, the percentage (%) of K-shell x-rays in the x-ray beam is most likely: a. 0 b. 1 c. 10 d. 50
e. 99 - โ โ a- There will be no characteristic x-rays, as the electron kinetic energy (65 keV) is insufficient to eject W K-shell electrons that have a binding energy of 70 keV. At the same peak voltage, which generator likely deposits most energy into an anode? a. Constant potential b. High frequency c. Three phase (12pulse) d. Three phase (6 pulse)
e. Single phase - โ โ a- Constant potential, since it has negligible ripple and the voltage across the x-ray tube is always the maximum possible value. Changing x-ray tube current (mA) most likely changes the x-ray: a. field of view b. maximum energy c. average energy d. anode angle
e. beam intensity - โ โ e- Tube current controls the x-ray beam intensity, or the total number of x-ray photons produced. CT beam hardening artifacts are minimized by increasing the: a. tube voltage b. tube current c. scan time
e. matrix size - โ โ c- Tube currents are generally reduced in CT fluoroscopy. For a given absorber, if the Compton attenuation coefficient at 50 keV is 0.1 cm^-1 , its value at 100 keV (cm^-1 ) is most likely: a. 0. b. 0. c. 0. d. 0.
e. 0.2 - โ โ c- 0.05 cm^-1, since Compton interactions vary as 1/E, so doubling the photon energy will halve the probability of this interaction. For comparable image mottle in an abdominal radiograph, which image receptor would likely result in the highest patient dose? a. Screen-film
b. Photo stimulable phosphor c. Direct flat panel detector d. Indirect flat panel detector e. Digital photo spot - โ โ b-Photo stimulable phosphor requires more radiation as it must be thin to minimize light scatter during the readout process. For electromagnetic radiation, which increases with increasing photon energy? a. Wavelength b. Frequency c. Velocity d. Charge
e. Mass - โ โ b- Frequency, which is directly proportional to the photon energy For specification of anode heat capacities, one heat unit corresponds to energy (J) of: a. 0. b. 0.
c. 45 d. 60 e. 85 - โ โ e- Head techniques in a 1-year- old are reduced by 15%, so 85% would be used. If an average of 10,000 photons are detected per mm2, the chance (%) of detecting between 9,700 and 10,300 counts in any exposed mm is: a. 67 b. 90 c. 95 d. 99
e. 99.9 - โ โ d-Ninety-nine percent, since the standard deviation is 100, and the limits correspond to three standard deviations. If the attenuation of bone is 0.5 cm^-1, the fraction of x-ray photons transmitted through 1 cm is most likely:
a. 0. b. 0. c. e^-0. d. e^+0. e. (1 - e^-0.5) - โ โ c- e^-0.5 is the fractional transmission of photons through 1 cm, when the attenuation of bone is 0.5 cm^- If the distance from a radiation source is halved, the radiation intensity increases by a factor of: a. 2^โ b. 2^โ c. 2^ d. 2^+
e. 2^+2 - โ โ e- 2^2 (i.e., 4). Halving the distance quadruples the radiation intensity (inverse square law).
Improvement of lesion contrast (%) by the use of a grid in abdominal radiography would most likely be: a. 10 b. 25 c. 50 d. 100
e. 200 - โ โ e- Use of grids improves contrast by a large factor (e.g., 200% or more). In a standard x-ray tube, the maximum power loading (kW) on the 0.6 mm focal spot is most likely: a. 1 b. 2 c. 5 d. 10
e. 25 - โ โ e- The small focal spot can tolerate power levels of 25 kW (higher power would require the large focal spot). In abdominal imaging, the scatter to primary ratio of photons leaving the patient is most likely: a. 0. b. 0. c. 1 d. 2
e. 5 - โ โ e- There are about five scattered photons exiting the patient for every one primary photon. In bone, at what photon energy are photoelectric and Compton effects approximately equal? a. 4. b. 25
