File Name: radiography testing questions and answers .zip
Skin injuries are a result of the dose to the skin. Dose represents the energy deposited per gram of tissue, which can also be measured as Kerma kinetic energy released in matter. Kerma in air can be converted to dose in soft tissue using numerical constants. Dose-area product represents the dose times the area exposed.
Kerma depends on mA and kV, but each alone does not give the dose. The focal spot blur does not change with electronic magnification but does change is magnified with geometric magnification. Resolution increases with magnification that's the whole point , although this is degraded in geometric magnification by focal spot blur; typically, in magnification mode the device will switch to a smaller focal spot.
Dose increases to maintain the same level of exposure noise characteristics on the magnified image. Field of view is typically smaller in a magnified image of any type. In magnification mammography, in order to improve resolution a smaller focal spot is used. To prevent anode overheating, the current is decreased, and thus the exposure takes longer.
The patient has to stay still for longer. Motion blur is also magnified with the geometric magnification. Collimation reduces the area of the patient that is exposed to radiation, thereby reducing DAP. The dose energy per mass is unchanged, although in reality since scatter is decreased , the dose actually will go down. Resolution and magnification are unchanged by collimation. In order for the magnified image to have the same noise as a full-view image, more dose is needed because the pixels are smaller.
In fact, the dose will go up with the square of the magnification. This results in increasing risk of skin injury. However, the area irradiated is similarly smaller, so DAP will typically not change.
In the case of flat-panel detector electronic magnification, the DAP may actually decrease. The maximum x-ray energy produced from an x-ray tube depends on the voltage across the tube.
Filament current and voltage, not discussed in the text, generate the electrons in the first place but do not accelerate them. Tube current affects the number of electrons - and thus number of photons - produced. As silly as this question is, it is here to remind us that if x-rays did not interact with and deposit energy in the patient, there would be no image. X-rays going right through the patient produce no information.
The image is generated by differences in how many x-rays are stopped by different tissues. By the Beer-Lambert law, x-ray intensity drops off exponentially as it passes through tissue.
The rate of decrease is determined by the linear attenuation coefficient. Because of the exponential beam attenuation, almost all of the beam is attenuated in the first few centimeters of the patient. This attenuation is what is responsible for depositing dose - thus, in the first few centimeters of the patient. Remember that this represents the skin and subcutaneous tissues closest to the x-ray tube, of course.
Since larger patients attenuate the x-ray beam more, fewer photons would be available to form the image. Tissue contrast is typically worse in large patients because of increased scatter, but that cannot be easily fixed. The predominant means by which tissues differ in their attenuation - and thus contrast is generated - is with the photoelectric effect, which is highly dependent on the elemental composition of the tissue. Compton scatter depends on the electron density, which is similar in most biologic materials.
Rayleigh scattering is only a minor effect at diagnostic energies. Increasing kV decreases dose in large patients well, in any size patients because of less attenuation at higher energies. The trade-off is worse contrast. The k-edge of iodine around 30 keV strongly increases its likelihood of photoelectric interaction, thereby increasing the attenuation. Again to re-emphasize the point: The photoelectric effect predominates at low x-ray energies and is the best way to distinguish different tissues.
Tungsten does not have characteristic x-rays in the relevant spectrum they are around keV. Overall, the average x-rays end up being higher with tungsten anodes but therefore give lower dose. Images can look terrible but contain diagnostic information and vice versa. In medicine, where there are always trade-offs between benefits and harms of any procedure or test, it is critical to obtain diagnostic information rather than focusing on how nice an image looks.
Noise is principally determined by the number of photons received by the detector more photons, less noise. While scatter could be considered a type of structured or background noise, its principal effect is to worsen image contrast. Decreasing kV decreases the x-ray penetration [and also photon flux], thus giving worse noise.
Typically the automated exposure compensation will increase the mA to compensate. However, contrast is improved at low kV, especially with iodine compared to soft tissue, given its k-edge near 30 keV. This may allow one to actually use a lower dose overall, since SNR will still be higher because the signal is greater. Improving SNR can involve increased dose, but there are other interventions. For example, one can use pixel binning to trade resolution for dose. Alternatively, one can use a contrast agent to increase the signal itself.
Changes in kV, as noted above, can also improve contrast at the expense of dose. Scatter degrades tissue contrast by adding a large amount of 'background' x-rays to every pixel. Scatter also does not distingush well between different tissue types and thus does not contribute to the generation of contrast. Scatter is a big factor in thick body parts such as the abdomen. Pediatric patients non-obese ones and adult extremities are much less affected by scatter.
Because grids attenuate primary beam x-rays, they decrease signal; they also decrease the scatter background and thereby increase noise since there are many fewer photons hitting the detector now. The combination of these effects requires an increased dose to obtain the same image noise.
By reducing scatter from elsewhere in the body, collimation improves the apparent contrast. Quantum mottle is unchanged. There is decreased radiation to the patient because a smaller area is irradiated and there is less scatter.
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Test your Dental Radiograph Knowledge! Phosphor plate receptors have greater latitude than traditional film and solid-state rigid-wired or wireless sensors. Latitude refers to the ability to capture a diagnostic image with a range of exposures. Advantages of phosphor plate receptors over solid-state sensors and traditional film are their flexible construction, greater active area of exposure, and lower retake frequency. These advantages help reduce exposure-related retakes.
Have you ever had a fracture and needed to get an X-ray? What do you know about the study? May have a cumulative effect which must be considered when monitoring for maximum permissible dose. Directly by personnel equipped with special protective clothing with speciar protective clothing. There will be a tendency for each area of the film to affect the development of the areas immediately below it. Search Speak now.
Multiple-choice exams are used frequently in medical education. They are reliable and allow for rapid marking and immediate feedback for students. The three basic levels of learning are knowledge, combined comprehension and application, and problem solving. The National Board of Medical Examiners and the Medical Council of Canada have both published guides to aid in constructing effective test questions. At the University of Ottawa, medical students are taught basic radiology as a component of a broad Foundations course. Short-answer radiology questions have traditionally been incorporated into a practical Foundations exam, in which examinees are tested on anatomy and microbiology as well. Examinees rotate through 12 radiology stations.
Radiographic Testing Level 1 (RT-1) Specific Examination . 3 Radiographic Testing Level 1 (RT-1) Answers to questions .
Preparing for the Radiography Certification exam? Bushong Bushong, Stewart C. Mosby, VitalBook file. The LET refers to the linear energy transfer, the amount of energy transferred by ionizing radiation per unit length of tissue traveled.
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The American Society for Nondestructive Tes. Views 2 Downloads 1 File size 2MB. ASNT is not responsible for the authenticity or accuracy of information herein. Published opinions and statements do not necessarily reflect the opinion of ASNT. Products or services that are advertised or mentioned do not carry the endorsement or recommendation of ASNT. No part of this publication may be reproduced or transmitted in any form, by means electronic or mechanical including photocopying, recording or otherwise, without the expressed prior written permission of The American Society for Nondestructive Testing, Inc. The following contributors assisted with the review of this book including updating old questions, writing new questions and updating references: Gary Alderson Gary E.
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