Question 1

Why is rectification of AC to DC important to the operation of the x-ray tube?

Accepted Answer

Rectification of AC to DC is important to protect the cathode side of the tube from electron bombardment. This is because without rectification, the electron flow would be bidirectional and could damage the cathode. Rectification ensures that the electron flow is unidirectional, allowing for safe operation of the x-ray tube.

Question 2

Where does controlled thermionic emission physically take place?

Accepted Answer

Controlled thermionic emission takes place at the cathode filament, where electrons are emitted due to thermal energy.

Question 3

Which of the following is a variable aperture beam-limiting device?

Accepted Answer

A collimator is a variable aperture beam-limiting device that can adjust the size of the beam by changing the size of the opening or aperture. Aperture diaphragms, cones, and cylinders are not variable and have a fixed size or shape.

Question 4

What is the function of the high voltage circuit within the x-ray tube?

Accepted Answer

The high voltage circuit is responsible for accelerating and driving electrons from the cathode to the anode, which creates the x-ray production.

Question 5

Because electron potential energy is greater the farther away a bound electron is from the nucleus, what results from an outer shell electron falling into an inner shell vacancy?

Accepted Answer

When an outer shell electron falls into an inner shell vacancy, the electron loses potential energy as it moves closer to the nucleus. This lost energy is released in the form of electromagnetic radiation, such as a photon. Therefore, choice D ("Electromagnetic energy is released") is the correct answer. The other choices are not applicable to this process - the size and weight of the nucleus do not change, the binding energy of the outer shell electron does not decrease (in fact, it may increase due to the lower electron-electron repulsion), and the nucleus does not emit a proton.

Question 6

Select the advantage automatic exposure control (AEC) has over other methods of determining x-ray exposure time.

Accepted Answer

Automatic Exposure Control (AEC) is designed to adjust the exposure time based on the density and composition of the part of the body being imaged, ensuring consistent image quality while accommodating for variations in patient size and density.

Question 7

Which of the following combinations would produce the smallest effective focal spot?

Accepted Answer

Using a small filament with a small anode angle reduces the effective focal spot because it produces a smaller electron beam. The anode angle is less important in determining the focal spot size.

Question 8

How much energy is released when a K-characteristic x-ray is emitted from a given element?

Accepted Answer

When a K-characteristic x-ray is emitted from a given element, it is due to the transition of an electron from an outer shell (such as L or M) to the K-shell. The energy released is equal to the difference in binding energy between the two shells, which is the amount of energy required to overcome the electrostatic attraction between the electron and the nucleus. This energy is converted into the energy of the emitted x-ray photon. Therefore, the energy released when a K-characteristic x-ray is emitted is the difference in binding energy between a K-shell electron and an L-shell electron.

Question 9

The penetration of the x-ray beam is increased under which condition?

Accepted Answer

The penetration of x-ray beam depends on the energy of the photons, which is directly proportional to the kVp. As the kVp is increased, the energy of the photons increases, resulting in increased penetration of the x-ray beam. Therefore, option B is correct. Option A is incorrect because decreasing the kVp results in decreased penetration. Options C and D are incorrect because the mA affects the quantity, not the quality of the x-ray beam. Option E is incorrect because increasing the time only affects the quantity of the x-ray beam, not the quality.

Question 10

Choose the electrical requirements necessary to operate a 300/125 x-ray machine that is hard-wired into a chiropractic office in the United States.

Accepted Answer

X-ray machines typically require higher voltage and amperage than standard electrical outlets can provide. The 220-volt requirement is common for medical equipment, and the 100-amp requirement is within the range of what a typical commercial building can provide. The machine likely operates on alternating current (AC) as opposed to direct current (DC).

Question 11

Which component part is designed specifically to prevent leakage radiation?

Accepted Answer

The tube housing is designed specifically to prevent leakage radiation by absorbing any x-rays that do not contribute to the image formation, ensuring that only the x-rays directed towards the patient through the window of the housing are used for imaging.

Question 12

A lateral cervical spine view is taken at a 40-inch FFD using 70 kVp, 100 mAs, and .2 second. If the FFD were changed to 60 inches, which of the following techniques would produce the same exposure to the film as the original?

Accepted Answer

To maintain the same exposure when increasing the focal film distance (FFD) from 40 inches to 60 inches, the square law for radiography can be used. This law states that the intensity of the radiation is inversely proportional to the square of the distance from the source. When the distance is increased, the exposure time or mA must be adjusted to compensate for the decrease in intensity. The original mAs (100 mAs) needs to be increased to maintain the same exposure level. The formula to calculate the new mAs when changing distance is: new mAs = original mAs × (new distance² / original distance²). Plugging in the values: new mAs = 100 mAs × (60² / 40²) = 225 mAs. Option B, which offers 300 mA for .15 second, results in 45 mAs, which does not directly match the calculated increase. However, given the options and the need to adjust for the increased distance, option B is the closest in terms of adjusting the technique to maintain exposure, considering a mistake in my initial calculation. Correcting the calculation: new mAs = 100 mAs × (60² / 40²) = 225 mAs. The correct approach involves finding the combination of mA and time that closely matches this new mAs requirement while keeping the kVp constant. Upon reevaluation, none of the options directly match the calculated mAs, indicating a need to correct my explanation based on the provided options and the principles of radiographic exposure. The correct answer should be chosen based on the closest match to the needed adjustment for maintaining exposure, considering the principles of distance and mAs relationship. However, my initial explanation incorrectly calculated the adjustment needed. The correct calculation should guide the choice, but an error in my explanation led to confusion.

Question 13

According to federal standards, a 90-kVp beam requires which of the following measurements to be the equivalent of aluminum of total filtration to minimize soft radiation exposure to patients?

Accepted Answer

Federal standards require a 90-kVp beam to have a minimum of 2.5 mm aluminum equivalent total filtration to effectively reduce patient exposure to soft radiation.

Question 14

Select the relationship of wavelength to frequency in an electromagnetic waveform.

Accepted Answer

Wavelength and frequency are inversely proportional to each other in an electromagnetic waveform. This means that as the wavelength increases, the frequency decreases, and vice versa. This relationship is described by the equation v = λf, where v is the velocity of the wave, λ is the wavelength, and f is the frequency.

Question 15

Which type of generator will produce the greatest effective tube current when identical exposure factors are used?

Accepted Answer

100 KHz high frequency generators are the most efficient in producing a consistent and high-quality x-ray beam with the least amount of ripple, resulting in the greatest effective tube current for a given set of exposure factors.

Question 16

A scattered x-ray is an x-ray that __________.

Accepted Answer

Scattered x-rays are those that have changed direction after interacting with matter, typically resulting in an increase in wavelength and a corresponding decrease in frequency due to the loss of energy during the scattering process.

Question 17

Select the entire energy range that represents diagnostic x-rays.

Accepted Answer

Diagnostic x-rays typically have an energy range from about 40 keV to 150 keV. This range is broad enough to include the energies used for various diagnostic imaging techniques, such as radiography and computed tomography (CT), allowing for effective penetration and contrast in the images produced.

Question 18

Which one of the following controls the initiation of electron flow from cathode to anode?

Accepted Answer

The exposure switch controls the initiation of electron flow from cathode to anode. When the exposure switch is pressed, it completes a circuit that allows the flow of electrons from the cathode to the anode, resulting in the production of X-rays. The other options listed do not directly control this process.

Question 19

Which of the following control console selectors determines focal spot selection?

Accepted Answer

The mA selector on the control console determines the focal spot size by selecting the filament current, which in turn affects the size of the electron beam and the focal spot on the anode.

Question 20

Which statement is true of the anode heel effect?

Accepted Answer

The anode heel effect is less evident at longer focal film distances because the divergence of the x-ray beam reduces the intensity variation across the field.

Quiz 1: Plain Radiographic Imaging