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 - Additional Exercises: 6.87b

Answer

The energy of the photon is $$E\approx4.366\times10^{-19}J$$

Work Step by Step

The energy of the photon is calculated according to the formula $$E=h\frac{c}{\lambda}$$ Here, Planck's constant $h\approx6.626\times10^{-34}J.s$ The speed of light in a vacuum $c\approx2.998\times10^8m/s$ The wavelength of the absorbed photons $\lambda=455nm=4.55\times10^{-7}m$ Therefore, the energy of the photon is $$E=\frac{(6.626\times10^{-34}J.s)\times(2.998\times10^8m/s)}{4.55\times10^{-7}m}$$$$E\approx4.366\times10^{-19}J$$
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