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 25 - Electric Charges and Forces - Exercises and Problems - Page 747: 51

Answer

$N = 6.56\times 10^{15}~rev/s$

Work Step by Step

The force between the proton and the electron provides the centripetal force to keep the electron moving in a circle. We can find the electron's speed: $\frac{kq^2}{r^2} = \frac{mv^2}{r}$ $v^2 = \frac{kq^2}{m~r}$ $v = \sqrt{\frac{kq^2}{m~r}}$ $v = \sqrt{\frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)^2}{(9.109\times 10^{-31}~kg)~(0.053\times 10^{-9}~m)}}$ $v = 2.185\times 10^6~m/s$ We can find the number of revolutions per second the electron makes: $N = \frac{v}{2\pi~r}$ $N = \frac{2.185\times 10^6~m/s}{(2\pi)~(0.053\times 10^{-9}~m)}$ $N = 6.56\times 10^{15}~rev/s$
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