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: 6.25c

Answer

The wavelength of this radiation is $327.8nm$.

Work Step by Step

*Strategy: This exercise is actually a reverse of the exercise in part b). Therefore, we would employ the same formulas, only in reversed order. - First, we would use the Planck's theory formula $E=h\times\nu$ to find the frequency of radiation. - Then, we use the formula $\lambda\times\nu = c$ to find the wavelength of this radiation. $\lambda$ : wavelength of radiation $\nu$ : frequency of radiation $c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$) $E$ : the energy of a single quantum (or in this case, photon) $h$ : Planck constant ($h\approx6.626\times10^{-34}J.s$) Step 1: Find the known variables We know each photon of this radiation has an energy of $6.06\times10^{-19}J$; so $E=6.06\times10^{-19}J$. Step 2: Find the frequency of this radiation $\nu=\frac{E}{h}=\frac{6.06\times10^{-19}}{6.626\times10^{-34}}\approx9.146\times10^{14}s^{-1}$ Step 3: Find the wavelength of radiation $\lambda=\frac{c}{\nu}=\frac{2.998\times10^8}{9.146\times10^{14}}\approx3.278\times10^{-7}m\approx327.8nm$
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