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 - General Problems - Page 828: 59

Answer

$3.7\times 10^{-37}m$

Work Step by Step

The uncertainty in the velocity is given in the problem. Calculate the lowest uncertainty (maximum precision) in the position by using equation 28–1. $$\Delta x \geq \frac{\hbar}{\Delta p}=\frac{\hbar}{m \Delta v}$$ $$=\frac{1.055\times 10^{-34}J \cdot s}{(1300kg)(0.22m/s)}$$ $$=3.7\times 10^{-37}m $$
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