1. Which of the figures above depicts a pair of spectra that result from a change in anode target material?
A) #1
B) #2
C) #3
D) #4
Answer:
C) #3
If the target material is changes, one would expect the characteristic x-rays to change. Option C) is the only diagram with different characteristic peaks.
Figure 6 (click here to view video). During characteristic radiation production photons are produced at energies equal to the difference in the atom’s shell binding energies. So the spectrum shows a slight increase (i.e. peaks) in the number of x-rays at these energy difference values due to the combined characteristic and Bremsstrahlung radiation production.
2.
An x-ray tube’s voltage is increased from 65 to 75 kVp. What adjustment to the mAs should be made to achieve the same amount of radiation exposure to the image detector?
A) Increase the mAs by ½
B) Increase the mAs by ¼
C) Reduce the mAs by ¼
D) Reduce the mAs by ½
Answer:
D) Reduce the mAs by ½
Rule of thumb, increase the kV by 15% doubles the exposure. 75/65=1.15 or increased by 15%
3.
What is the principal source of x-rays produced in an x-ray tube?
A) Conversion of electrons to photons
B) Removal of electrons from the K-shell of a target material
C) Stopping high speed electrons
D) Production of Auger electrons
Answer:
C) Stopping high speed electrons
Stopping should make you think of braking or Bremsstrahlung radiation, from the charts in question 1) most of the xrays are Bremsstrahlung , (with a few characteristic peaks).
4.
What attribute of tungsten makes it a better target material for a diagnostic x-ray tube as compared to using gold?
A) Efficiency of producing Bremsstrahlung radiation
B) Higher melting point
C) Higher atomic number
D) Better electrical conductance
Answer:
B) Higher melting point
Th atomic number of Gold (Au) is 79, higher than the atomic number of Tungsten (W) which is 74.
5.
For a given filament size, what is the effect as the anode angle is decreased?
A) Increases in the apparent focal spot size and decreases in the heat load capacity
B) Increases in the apparent focal spot size and increases in the heat load capacity
C) Decreases in the apparent focal spot size and decreases in the heat load capacity
D) Decreases in the apparent focal spot size and increases in the heat load capacity
Answer:
C) Decreases in the apparent focal spot size and decreases in the heat load capacity
In general, a smaller focal spot increases heat loading, or decreases heat load capacity.
6.
Which of the following would make the heel effect more pronounced?
A) Larger apparent focal spot size
B) Small anode angle
C) Utilization of rotating anode
D) Decreasing the field of view
Answer:
B) Small anode angle
Figure 14. The use of beveled anode surface results in the x-rays intensity being unevenly distributed across the x-ray field of view.
7.
Which of the following are characteristics of x-ray beam filtration?
A) The total filtration is mostly due to inherent filtration in the x-ray tube
B) Added filtration becomes unnecessary for high kVp techniques, e.g. above 125 kVp
C) Filters made from higher atomic number material attenuate more higher energy x-rays
D) None of the above
Answer:
D) None of the above
8.
Which of the following will increase the quality (penetrating ability) of a diagnostic x-ray beam?
A) Increase the kVp
B) Decrease the beam filtration
C) Increase the mAs
D) All of the above
Answer:
A) Increase the kVp
9.
Which of the following is true regarding the K-edge filtration utilized in mammography?
A) It produces a small range of photon energies in the output spectrum
B) It results from the absorption of electrons in the K-shell of an atom
C) Electron kinetic energy is greatly reduced at the edge of the K energy shell
D) The atomic number of the filter must be lower than that for the anode material
Answer:
A) It produces a small range of photon energies in the output spectrum
10.
What is the result if a 50 keV photon interacts in a photoelectric manner with a K-shell electron having a binding energy of 12 keV?
A) Characteristic x-rays and/or Auger electrons are produced
B) A scattered photon with energy less than 38 keV is produced
C) Emission of a 12 keV photoelectron
D) None of the above occur
Answer:
A) Characteristic x-rays and/or Auger electrons are produced
11.
Which of the following determines the maximum energy of Bremsstrahlung x-rays produced in an x-ray tube operated at 100 kVp?
A) Whether the generator is single or three phase.
B) The peak voltage applied between the cathode and anode
C) The type target material used
D) The product of the mA and exposure time used
Answer:
B) The peak voltage applied between the cathode and anode
12.
Which of the following determines the maximum energy of characteristic x-rays in an x-ray tube operated at 100 kVp?
A) Whether the generator is single or three phase.
B) The peak voltage applied between the cathode and anode
C) The type of the target material used
D) The product of the mA and exposure time used
Answer:
C) The type of the target material used
Really basic concept, related to the answer for Question 1
13.
What is the effect of changing the focal spot size from 0.6 mm to 1.0 mm?
A) The amount of heat the anode can absorb is increased.
B) The resolution in the image produced is improved.
C) The ability to visualize high contrast objects in the image is improved.
D) The patient dose is decreased.
Answer:
A) The amount of heat the anode can absorb is increased.
Again, increased size of focal spot increases the amount of hear that the anode can absorb.
14.
What would be the result if the amount added filtration is changed from 2 mm to 1 mm of Aluminum filtration?
A) The radiation dose to the patient would decrease and the HVL would increase
B) The radiation dose to the patient would decrease and the HVL would decrease
C) The radiation dose to the patient would increase and the HVL would increase
D) The radiation dose to the patient would increase and the HVL would decrease.
Answer:
D) The radiation dose to the patient would increase and the HVL would decrease.
Less filtration will increase the dose to the pt. Since the beam will have less low energy absorbed it will be a softer beam and therefore easier to stop. The HVL would decrease.