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: 6.25b

Answer

The energy of a photon of radiation is $4.81\times10^{-19}J$.

Work Step by Step

*Strategy: - First, we would use the formula $\lambda\times\nu = c$ to find out the frequency of this radiation. - Then, we would use the Planck's theory formula $E=h\times\nu$ to find the energy of a photon of 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 the radiation has wavelength of $413nm$; so $\lambda=413nm=4.13\times10^{-7}m$. Step 2: Find the frequency of this radiation $\nu=\frac{c}{\lambda}=\frac{2.998\times10^8}{4.13\times10^{-7}}\approx7.259\times10^{14}s^{-1}$ Step 3: Find the energy of a photon of radiation $E=h\times\nu=(6.626\times10^{-34})\times(7.259\times10^{14})\approx4.81\times10^{-19}J$
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