Answer
$1.21\times10^{-43}\,m$
Work Step by Step
We find:
$\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}\,kg\,m^{2}\,s^{-1}}{(91\,kg)(\frac{1}{5}\times3.00\times10^{8}\,m/s)}$
$=1.21\times10^{-43}\,m$
General Chemistry: Principles and Modern Applications (10th Edition)
by Petrucci, Ralph H.; Herring, F. Geoffrey; Madura, Jeffry D.; Bissonnette, Carey
Published by
Pearson Prentice Hal
ISBN 10:
0132064529
ISBN 13:
978-0-13206-452-1
Chapter 8 - Electrons in Atoms - Example 8-6 - Calculating the Wavelength Associated with a Beam of Particles - Page 315: Practice Example A
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