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 - Additional Exercises: 6.81b

Answer

The frequency of wave A is $\nu_A\approx8.431\times10^{15}s^{-1}$ The frequency of wave B is $\nu_B\approx3.748\times10^{15}s^{-1}$

Work Step by Step

Since these 2 waves are both electromagnetic radiations, we would find their frequencies according to the following formula $$\nu=\frac{c}{\lambda}$$ $\nu$: frequency of radiation $c$: speed of light in a vacuum $(c\approx2.998\times10^8m/s)$ $\lambda$: wavelength of radiation *Wave A: We already know from part a) that $\lambda_A=3.556\times10^{-8}m$ Therefore, the frequency of wave A is $$\nu_A=\frac{c}{\lambda_A}=\frac{2.998\times10^8m/s}{3.556\times10^{-8}m}\approx8.431\times10^{15}s^{-1}$$ *Wave B: We already know from part a) that $\lambda_B=8\times10^{-8}m$ Therefore, the frequency of wave B is $$\nu_B=\frac{c}{\lambda_B}=\frac{2.998\times10^8m/s}{8\times10^{-8}m}\approx3.748\times10^{15}s^{-1}$$
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