Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises - Page 250: 6.20b

Answer

$4.684\times10^{-4}m$ or $468400nm$

Work Step by Step

*Strategy: We would use the formula $$\lambda\times\nu=c$$ in which $\lambda$ : wavelength of radiation $\nu$: frequency of radiation $c$: speed of light (we would take the speed of light in vacuum, which means $c\approx2.998\times10^8m/s$) - Step 1: Find the known variables The radiation has a frequency of $6.4\times10^{11}s^{-1}$, therefore $\nu=6.4\times10^{11}s^{-1}$ - Step 2: Do the calculation The wavelength of radiation is $\lambda=\frac{c}{\nu}=\frac{2.998\times10^8}{6.4\times10^{11}}\approx4.684\times10^{-4}m\approx468400nm$
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