Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 28 - Quantum Mechanics of Atoms - Search and Learn - Page 828: 7

Answer

$2.19\times 10^{6}m/s$.

Work Step by Step

The uncertainty in position is given. Use Eq. 28–1 to find the uncertainty in the speed. Note that the uncertainty in momentum is the electron mass multiplied by the uncertainty in the velocity. $$\Delta p= m \Delta v \geq \frac{\hbar}{\Delta x}$$ $$\Delta v \geq \frac{\hbar}{m \Delta x}$$ $$=\frac{1.055\times 10^{-34}J \cdot s}{(9.11\times10^{-31}kg)(0.529\times10^{-10}m)}$$ $$= 2.19\times 10^{6}m/s $$
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