Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 6 - Section 6.3 - Bohr’s Theory of the Hydrogen Atom - Checkpoint - Page 250: 6.3.3


(c) $1.28\times10^{-6}\,m$

Work Step by Step

$n_{i}=5$ and $n_{f}=3$ Recall that $\frac{1}{\lambda}=\frac{2.18\times10^{-18}\,J}{hc}(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$ where $h=6.626\times10^{-34}\,J\cdot s$ and $c=3.00\times10^{8}\,m/s$ Then, $\frac{1}{\lambda}=\frac{2.18\times10^{-18}\,J}{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}(\frac{1}{3^{2}}-\frac{1}{5^{2}})$ $=7.79868\times10^{5}\,m^{-1}$ $\lambda=\frac{1}{7.79868\times10^{5}\,m^{-1}}=1.28\times10^{-6}\,m$
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