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 343: 47


$1.50\times 10^{23}atoms$

Work Step by Step

We know that $E=\frac{hc}{\lambda}$ We plug in the known values to obtain: $E=\frac{6.626\times 10^{-34}\times 2.9979\times 10^8}{150\times 10^{-9}}$ $E=1.32\times 10^{-18}\frac{J}{photon}$ Now $\frac{1.98\times 10^5J}{1.32\times 10^{-18}\frac{J}{photon}}=1.50\times 10^{23}photons$ Similarly $1.50\times 10^{23}photons\times \frac{atom}{photon}=1.50\times 10^{23}atoms$
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