Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - Search and Learn - Page 770: 6

Answer

The electron is not relativistic.

Work Step by Step

The electrostatic attraction provides the centripetal force keeping the electron in a circular orbit. $$\frac{1}{4 \pi \epsilon_o}\frac{e^2}{r^2}=\frac{m_{electron}v^2}{r}$$ Solve for the speed of the electron. $$v=\sqrt{\frac{1}{4 \pi \epsilon_o}\frac{e^2}{ m_{electron} r}}$$ $$v=\sqrt{\frac{(8.99\times10^9)(1.60\times10^{-19})^2}{(9.11\times10^{-31}kg)(0.53\times10^{-10}m)}}=2.17\times10^6 m/s$$ This is equal to 0.0072c, which is far less than 0.1c. The electron is not relativistic.
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