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: 49



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: $\lambda=\frac{6.626\times 10^{-34}\times 2.9979\times 10^8}{4.645\times 10^{-19}}$ $\lambda=4.277\times 10^{-7}m=427.7nm$
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