Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 7 - Atomic Structure and Periodicity - Exercises - Page 343: 50

Answer

$574.0nm$

Work Step by Step

We know that $E=\frac{hc}{\lambda}$ This can be rearranged as: $\lambda=\frac{hc}{E}$ We plug in the known values to obtain: $\lambda=\frac{6.626\times 10^{-34}\times 2.9979\times 10^8}{3.461\times 10^{-19}}$ $\lambda=5.740\times 10^{-7}m=574.0nm$
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.