Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 28 - The Electric Potential - Exercises and Problems - Page 833: 13


$8.35 \times 10^{4}$ volt

Work Step by Step

Initial speed of an $He^{+}$ ion = 0 Thus; Loss in energy = Gain in energy $e\Delta V=KE_{f}-KE_{i}$ $e\Delta |V|=\frac{1}{2}m_{He^{+}}v_{He^{+}}^{2}-0$ $\Delta |V|=\frac{\frac{1}{2}m_{He^{+}}v_{He^{+}}^{2}}{e}$ $=\frac{{\frac{1}{2}4m_{p}v_{He^{+}}^{2}}}{1.6 \times 10^{-19}}$ $=\frac{1}{2}\times \frac{{ 4\times 1.67 \times 10^{-27} \times(2\times 10^{6})^{2}}}{1.6 \times 10^{-19}}$ $=8.35 \times 10^{4} volt$
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