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 - Problems - Page 826: 11

Answer

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Work Step by Step

The uncertainty in the electron’s position is given. Use equation 28–1 to find the uncertainty in the electron’s momentum. $$\Delta p=m\Delta v \geq \frac{\hbar}{\Delta x}$$ From that, find the uncertainty in its speed. $$\Delta v \geq \frac{\hbar}{m\Delta x}$$ $$=\frac{(1.055\times10^{-34}J\cdot s)}{(9.11\times10^{-31}kg)(15\times10^{-9}m)}= 7.7\times10^3m/s$$ The speed is nonrelativistic. Calculate the electron's minimum kinetic energy. $$KE=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31}kg)(7720m/s)^2$$ $$=2.7\times10^{-23}J$$ This is about 0.00017 eV.
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