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: 4


$2.5\times 10^{4}m/s$

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 $q\Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ ...(1) Now, same potential difference applied for $He^{+}$. Thus, Loss in energy = Gain in energy $q\Delta V=\frac{1}{2}m_{He^{+}}v_{He^{+}}^{2}$ ...(2) From equations (1) and (2); $v_{He^{+}}=\sqrt {\frac{m_{p}}{m_{He^{+}}}}v_{p}$ $v_{He^{+}}=\sqrt {\frac{m_{p}}{4m_{p}}}v_{p}$ $v_{He^{+}}=\frac{v_{p}}{2} =\frac{ 5 \times 10^{4}}{2} = 2.5\times 10^{4}m/s$
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