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: 6.52a

Answer

The uncertainty in position of the electron is $\Delta x\ge57.89nm$.

Work Step by Step

*The Heisenberg's uncertainty principle formula: $$\Delta x\times\Delta(mv)\ge\frac{h}{4\pi}$$ $\Delta x$: the uncertainty in position $\Delta(mv)$: the uncertainty in momentum $h$: Planck's constant ($h=6.626\times10^{-34}J.s$) 1) Find the known variables - Mass of electron: $m\approx9.109\times10^{-31}kg$ - The uncertainty in velocity: $\Delta v=0.01\times10^{5}m/s$ Since we only have the uncertainty in velocity, the uncertainty in momentum $\Delta(mv)=m\Delta v$ 2) Calculate the uncertainty in position of the electron $\Delta x\ge\frac{h}{4\pi m\Delta v}=\frac{6.626\times10^{-34}}{4\pi\times(9.109\times10^{-31})\times(0.01\times10^5)}\approx5.789\times10^{-8}m\approx57.89nm$
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