University Physics with Modern Physics (14th Edition)
14th Edition
ISBN: 9780321973610
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
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Chapter 38, Problem 38.45PP
To determine
Which option will generate higher-energy photons in this linear accelerator.
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Higher-energy photons might be desirable for the treatment of certain tumors. Which of these actions would generate higher-energy photons in this linear accelerator? (a) Increasing the number of electrons that hit the tungsten target; (b) accelerating the electrons through a higher potential difference; (c) both (a) and (b); (d) none of these.
Provide the answers in 90 minutes, and count as 2 questions if needed.
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the electron?
- does no depend;
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Chapter 38 Solutions
University Physics with Modern Physics (14th Edition)
Ch. 38.1 - Silicon films become better electrical conductors...Ch. 38.2 - Prob. 38.2TYUCh. 38.3 - Prob. 38.3TYUCh. 38.4 - Prob. 38.4TYUCh. 38 - Prob. 38.1DQCh. 38 - Prob. 38.2DQCh. 38 - Prob. 38.3DQCh. 38 - Prob. 38.4DQCh. 38 - Prob. 38.5DQCh. 38 - Prob. 38.6DQ
Ch. 38 - Prob. 38.7DQCh. 38 - Prob. 38.8DQCh. 38 - Prob. 38.9DQCh. 38 - Prob. 38.10DQCh. 38 - Prob. 38.11DQCh. 38 - Prob. 38.12DQCh. 38 - Prob. 38.13DQCh. 38 - Prob. 38.14DQCh. 38 - Prob. 38.15DQCh. 38 - Prob. 38.16DQCh. 38 - Prob. 38.17DQCh. 38 - Prob. 38.1ECh. 38 - Prob. 38.2ECh. 38 - Prob. 38.3ECh. 38 - Prob. 38.4ECh. 38 - Prob. 38.5ECh. 38 - Prob. 38.6ECh. 38 - Prob. 38.7ECh. 38 - Prob. 38.8ECh. 38 - Prob. 38.9ECh. 38 - Prob. 38.10ECh. 38 - Prob. 38.11ECh. 38 - Prob. 38.12ECh. 38 - Prob. 38.13ECh. 38 - Prob. 38.14ECh. 38 - Prob. 38.15ECh. 38 - Prob. 38.16ECh. 38 - Prob. 38.17ECh. 38 - Prob. 38.18ECh. 38 - Prob. 38.19ECh. 38 - Prob. 38.20ECh. 38 - Prob. 38.21ECh. 38 - An electron and a positron are moving toward each...Ch. 38 - Prob. 38.23ECh. 38 - Prob. 38.24ECh. 38 - Prob. 38.25ECh. 38 - Prob. 38.26PCh. 38 - Prob. 38.27PCh. 38 - Prob. 38.28PCh. 38 - Prob. 38.29PCh. 38 - Prob. 38.30PCh. 38 - Prob. 38.31PCh. 38 - Prob. 38.32PCh. 38 - Prob. 38.33PCh. 38 - Prob. 38.34PCh. 38 - Prob. 38.35PCh. 38 - Prob. 38.36PCh. 38 - Prob. 38.37PCh. 38 - Prob. 38.38PCh. 38 - Prob. 38.39PCh. 38 - Prob. 38.40CPCh. 38 - Prob. 38.41PPCh. 38 - Prob. 38.42PPCh. 38 - Prob. 38.43PPCh. 38 - Prob. 38.44PPCh. 38 - Prob. 38.45PP
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- An electron accelerated in an x-ray tube hits an anode (positively charged plate) target. V = + High v m/s anode = 4ee6 kV 48 ee 464646 No No w a) Assuming all of its energy is transferred in generating x-rays, how fast is the electron moving when it reaches the anode target if it produces an x-ray with an energy of 60 keV. The mass of the electron is 9.1×10-31 kg. Hint: How is the energy of x-rays related to the KE of the electrons? How is the kinetic energy related to the speed? X-rays cathode b) What is the accelerating potential of the x-ray tube (potental difference between the anode and cathode) in part a)? Va Ponder: How large is this volatge? Remember the a regular AA battery has a voltage of 1.5 V.arrow_forwardAn electron accelerated in an x-ray tube hits an anode (positively charged plate) target. V= HOTE m/s +High V- = anode kV cer cer Ger 4848 46 N Now 484846 a) Assuming all of its energy is transferred in generating x-rays, how fast is the electron moving when it reaches the anode target if it produces an x-ray with an energy of 46 keV . The mass of the electron is 9.1 x10-31 kg. IM M Hint: How is the energy of x-rays related to the KE of the electrons? How is the kinetic energy related to the speed? X-rays cathode b) What is the accelerating potential of the x-ray tube (potental difference between the anode and cathode) in part a)? Va Ponder: How large is this volatge? Remember the a regular AA battery has a voltage of 1.5 V.arrow_forwardA particle and its anti-particle have the same mass and charge, but the charges will be of opposite polarity. A particle and its anti-particle annihilate and produce two photons. The particle has kinetic energy 2.0 MeV collides head on with its anti-particle that is at rest. Find the energy and momentum of each photon. (Given the mass of each particle is 0.632MeV/c²) Select the correct answer choice: (a) E = 1.088 MeV E = 3.264 MeV (b) (c) (d) E = 1.632 MeV E = 1.088 MeV p=1.67 MeV/c p=1.183 MeV/c p=2.366 MeV/c p=0.835 MeV/carrow_forward
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