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: 10

Answer

$4.37 \times 10 ^{4} m/s$

Work Step by Step

By the conservation of energy, Loss in energy = Gain in energy Thus: $q \Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ $v_{p}=[\frac{2q\Delta V}{m}]^{1/2}$ $v_{p}=[\frac{2\times 1.6 \times 10^{-19} \times 10^{3}}{1.67\times 10^{-27}}]^{1/2}= 4.37 \times 10 ^{4} m/s$
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