Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 7 - Quantum Theory and the Electronic Structure of Atoms - Questions & Problems - Page 317: 7.40

Answer

$1.37\times10^{-6}\,nm$

Work Step by Step

Mass $m=1.673\times10^{-27}\,kg$ Velocity $u=2.90\times10^{8}\,m/s$ Wavelength $\lambda=\frac{h}{mu}$ where $h$ is the Planck's constant equal to $6.63\times10^{-34}\,J\cdot s$ Then, $\lambda=\frac{6.63\times10^{-34}\,J\cdot s}{(1.673\times10^{-27}\,kg)(2.90\times10^{8}\,m/s)}=1.37\times10^{-15}\,m$ $=1.37\times10^{-6}\times10^{-9}\,m=1.37\times10^{-6}\,nm$
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