College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 27 - Problems - Page 1043: 47


$v = 2.2\times 10^6~m/s$

Work Step by Step

We can find the speed of the electron: $\frac{mv^2}{r} = \frac{k~q^2}{r^2}$ $v^2 = \frac{k~q^2}{m~r}$ $v = \sqrt{\frac{k~q^2}{m~r}}$ $v = \sqrt{\frac{k~q^2}{m~(\frac{\hbar^2}{m~k~q^2})}}$ $v = \sqrt{\frac{k^2~q^4}{\hbar^2}}$ $v = \frac{k~q^2}{\hbar}$ $v = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)^2}{1.0546\times 10^{-34}~J~s}$ $v = 2.2\times 10^6~m/s$
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