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: 6.47c


The wavelength of the lithium atom is $\lambda=2.279\times10^{-13}m$.

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 the lithium atom: The mass of a lithium atom is $7amu$. $1amu=\frac{1}{6.022\times10^{23}}g\approx1.661\times10^{-24}g\approx1.661\times10^{-27}kg$ Therefore, $7amu=7\times(1.661\times10^{-27}kg)\approx1.163\times10^{-26}kg$ In other words, $m=1.163\times10^{-26}kg$ - Velocity of the lithium atom: $v=2.5\times10^5m/s$ 2) Calculate the wavelength of the lithium atom $\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}}{(1.163\times10^{-26})\times(2.5\times10^5)}\approx2.279\times10^{-13}m$
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