Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises - Page 251: 6.47a


The wavelength of the skiing person is $\lambda=5.613\times10^{-37}m$.

Work Step by Step

*The de Broglie relationship: $$\lambda=\frac{h}{mv}$$ $\lambda$: wavelength of object $h$: Planck's constant ($h=6.626\times10^{-34} J.s$) $m$: mass of object $v$: velocity of object 1) Find the known variables - Mass of person: $m=85kg$ - Velocity of the person: $v=50km/h=5\times10^4m/h\approx13.889m/s$ 2) Calculate the wavelength of the skiing person $$\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}}{85\times13.889}\approx5.613\times10^{-37}m$$
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