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: 6.3.4


(e) 122 nm

Work Step by Step

$n_{i}=2$ and $n_{f}=1$ 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, $\frac{1}{\lambda}=\frac{2.18\times10^{-18}\,J}{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}(\frac{1}{1^{2}}-\frac{1}{2^{2}})$ $=8.225\times10^{6}\,m^{-1}$ $\lambda=\frac{1}{8.225\times10^{6}\,m^{-1}}=122\times10^{-9}\,m$ $=122\,nm$
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