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: 40.36


0.0976 eV.

Work Step by Step

The energy levels are given by $E_n=(n+\frac{1}{2})\hbar\omega$. The ground state energy is when n=0. The energy of the photon equals the energy difference between the two states. Find the energy. $$\Delta E=\hbar \omega=\frac{hc}{\lambda}$$ $$\Delta E =\frac{1.24\times10^{-6}eV\cdot m}{6.35\times10^{-6}m }=0.195eV$$ The ground state energy is $ \frac{1}{2}\hbar\omega$, which is half the spacing between energy levels. $$ \frac{1}{2}\hbar\omega=0.0976 eV $$
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.