Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 7 - Quantum Mechanics and The Atom - Exercises - Page 330: 40

Answer

(a) 2.99 m (b) $2.80\times10^{2}\,m$ (c) 0.359 m

Work Step by Step

Using the relation $\lambda=\frac{c}{\nu}$ where $c$ is the speed of light, we have (a) $\lambda= \frac{3.00\times10^{8}\,m/s}{100.2\,MHz}$ $=\frac{3.00\times10^{8}\,m/s}{100.2\times10^{6}\,s^{-1}}$ $=2.99\,m$ (b) $\lambda= \frac{3.00\times10^{8}\,m/s}{1070\,kHz}$ $=\frac{3.00\times10^{8}\,m/s}{1070\times10^{3}\,s^{-1}}$ $=2.80\times10^{2}\,m$ (c) $\lambda= \frac{3.00\times10^{8}\,m/s}{835.6\,MHz}$ $=\frac{3.00\times10^{8}\,m/s}{835.6\times10^{6}\,s^{-1}}$ $=0.359\,m$
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