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 1355: 40.33


$2.22\times10^{-33}J, 1.39\times10^{-14}eV$.

Work Step by Step

Calculate the angular frequency. $$\omega=\sqrt{\frac{k’}{m}}=\sqrt{\frac{110N/m}{0.250kg}}=21.0\frac{rad}{s}$$ The energy levels are given by $E_n=(n+\frac{1}{2})\hbar\omega$. The ground state energy is when n=0. $E_0=\frac{1}{2}\hbar\omega=1.11\times10^{-33}J=6.93\times10^{-15}eV $ The spacing between adjacent levels is now calculated. $\Delta E=\hbar\omega=2.22\times10^{-33}J= 1.39\times10^{-14}eV $ The spacing is so tiny that quantum effects are unimportant.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.