Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 25 - The Electric Potential - Exercises and Problems - Page 709: 3


$ 2.1\times 10^{6}m/s$

Work Step by Step

Proton velocity, $v_{p}=5 \times 10^{4} m/s$ For proton: By the conservation of energy; Loss in energy = Gain in energy Thus; $q\Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ ...(1) Same potential difference applied for an electron. Thus, Loss in energy = Gain in energy $q\Delta V=\frac{1}{2}m_{e}v_{e}^{2}$ ...(2) From equations (1) and (2): $v_{e}=\sqrt {\frac{m_{p}}{m_{e}}}v_{p}$ $v_{e}=\sqrt {\frac{1.67\times 10^{-27}}{9.11 \times 10^{-31}}} \times 5 \times 10^{4}$ $v_{e}= 2.1\times 10^{6}m/s$
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