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an X Ray X rays are a very penetrating form of

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 6P Chapter VI

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 6P

an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. In Figure VI.1, what is the direction of the electric field in the region between the cathode and the target electrode? A. To the left B. To the right C. Toward the top of the page D. Toward the bottom of the page

Step-by-Step Solution:

Solution 6P Introduction: By convention, an electric field directs from the positive terminal of a power source to its negative terminal. This is the direction in which a unit positive charge experiences a force in an electric field. An electron moves in a direction opposite to this field. Step 1: In the given figure VI.1, target electrode is connected to positive potential and the cathode is maintained at negative potential of the battery. Therefore, the electric field will direct from the target electrode to the cathode. In the given figure, the cathode is placed to the left with respect to the target electrode inside the vacuum tube. So, the electric field will be directed to the left. Conclusion: So, option A is correct.

Step 2 of 1

Chapter VI, Problem 6P is Solved
Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

The answer to “an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. In Figure VI.1, what is the direction of the electric field in the region between the cathode and the target electrode? A. To the left B. To the right C. Toward the top of the page D. Toward the bottom of the page” is broken down into a number of easy to follow steps, and 392 words. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. The full step-by-step solution to problem: 6P from chapter: VI was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This full solution covers the following key subjects: Rays, ray, electrons, voltage, target. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Since the solution to 6P from VI chapter was answered, more than 409 students have viewed the full step-by-step answer.

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