Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 40 - One-Dimensional Quantum Mechanics - Conceptual Questions - Page 1212: 1

Answer

The minimum length of the box will be 17.3 nm

Work Step by Step

The energy of the light is given by $$\Delta = \frac{hc}{\lambda}$$ The energy of a particle in the box is given by $$E_n = {n^2}\frac{h^2}{8mL^2}$$ Now the minimum energy required is $$E_n = E_2 - E_1 = \frac{4h^2}{8mL^2} - \frac{h^2}{8mL^2} = \frac{3h^2}{8mL^2}$$ The, equating the energy difference yields $$L^2=\frac{3h\lambda}{8mc}=\frac{3 \times1.38\times10^{-23}\times600\times10^{-9}}{8\times9.1\times10^{31}\times3\times10^8}=1.73\times10^{-8} \text m$$
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.