Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 7 - Atomic Structure and Periodicity - Exercises - Page 342: 42

Answer

$4.8\times 10^6\frac{J}{mol}$

Work Step by Step

We know that $E=\frac{hc}{\lambda}$ We plug in the known values to obtain: $E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{25nm\times \frac{1}{1\times 10^9nm}}=8.0\times 10^{-18}\frac{J}{photon}$ We can convert the energy into $\frac{J}{mol}$ as follows: $E=8.0\times 10^{-18}\frac{J}{photon}\times \frac{6.02\times 10^{23}photons}{mol}=4.8\times 10^6\frac{J}{mol}$
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