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: 6.29a


The energy of a photon of this wavelength is $6.112\times10^{-19}J$.

Work Step by Step

*Strategy: The following formula would be used in dealing with this question. $$E=h\times\nu=\frac{h\times c}{\lambda}$$ $E$ : energy of a photon $h$ : Planck's constant ($h\approx6.626\times10^{-34}J.s$) $\nu$ : frequency of the UV light $c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$) $\lambda$ : wavelength of the UV light (here $\lambda=325nm=3.25\times10^{-7}m$) Therefore, the energy of a photon of this wavelength is: $E=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{3.25\times10^{-7}}\approx6.112\times10^{-19}J$
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