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: 6.86

Answer

The number of photons that strike the CD surface in 69 minutes is $1.625\times10^{18}$.

Work Step by Step

1) First, we would calculate the energy of a photon of the output laser $(E_p)$. - The wavelength of the output laser $\lambda=780nm=7.8\times10^{-7}m$ - Planck's constant $h\approx6.626\times10^{-34}J.s$ - Speed of light in vacuum $c\approx2.998\times10^8m/s$ Then $E_p$ is calculated according to the formula $$E_p=h\frac{c}{\lambda}$$$$E_p=\frac{(6.626\times10^{-34}J.s)\times(2.998\times10^8m/s)}{7.8\times10^{-7}m}$$$$E_p\approx2.547\times10^{-19}J$$ 2) Then we calculate the energy that strikes the CD surface in 69 minutes The power level of the laser is $0.10mW=0.10mJ/s=10^{-4}J/s$ That means the laser releases $0.10\times10^{-4}J$ of energy that strikes the CD surface in $1s$. So, in 69 minutes, the amount of energy that strikes the CD surface is $$E=(10^{-4}J/s)\times(69min)\times(60s)$$$$E=(10^{-4}J/s)\times(4140s)$$$$E=0.414J$$ 3) Therefore, the number of photons that strike the CD surface in 69 minutes is $$N_p=\frac{E}{E_p}$$$$N_p=\frac{0.414J}{2.547\times10^{-19}J/photon}$$$$N_p\approx1.625\times10^{18}photons$$
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