University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 40 - Quantum Mechanics I: Wave Functions - Problems - Exercises - Page 1357: 40.57


See explanation.

Work Step by Step

For a period of 0.500 s, the frequency is 2.00 Hz. The energy levels are given by $E_n=(n+\frac{1}{2})hf$. The ground level energy, n = 0, is $\frac{1}{2}hf$. $$ E_o = \frac{1}{2} (6.626\times10^{-34}J \cdot s)(2.00 Hz) =6.63\times10^{-34}J=4.14\times10^{-15}eV $$ The spacing between energy levels is $hf$. $$ \Delta E = (6.626\times10^{-34}J \cdot s)(2.00 Hz) =1.33\times10^{-33}J=8.28\times10^{-15}eV $$ The spacing is undetectable; it is so tiny that quantum effects are unimportant for this everyday pendulum.
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