College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1043: 49


$r_3 = 4.77\times 10^{-10}~m$

Work Step by Step

We can find the orbital radius of the electron: $r_n = \frac{n^2~h^2}{4~\pi^2~m_e~k~Z~e^2}$ $r_3 = \frac{(3)^2(6.626\times 10^{-34}~J~s)^2}{(4~\pi^2)(9.1\times 10^{-31}~kg)(9.0\times 10^9~N~m^2/C^2)(1)(1.6\times 10^{-19}~C)^2}$ $r_3 = 4.77\times 10^{-10}~m$
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