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: 6.82b

Answer

The energy of $0.10mol$ of the emitted photons is $$E\approx2.031\times10^4J$$

Work Step by Step

We would use the following formula $$E_p=h\nu$$ to calculate the energy of an emitted photon. $E_p$: energy of an emitted photon $h$: Planck's constant $(h\approx6.626\times10^{-34}J.s)$ $\nu$: frequency of the emitted light 1) The frequency of the emitted light is $\nu\approx5.09\times10^{14}s^{-1}$ So, the energy of an emitted photon is $$E_p=h\nu$$$$E_p=(6.626\times10^{-34}J.s)\times(5.09\times10^{14}s^{-1})$$$$E_p\approx3.373\times10^{-19}J$$ 2) $1mol$ equals to $6.02\times10^{23}$ photons. So, $0.10mol$ equals to $0.10\times(6.02\times10^{23}photons)=6.02\times10^{22}$ photons. Therefore, the energy of $0.10mol$ of the emitted photons is $$E=(6.02\times10^{22}photons)\times E_p$$$$E=(6.02\times10^{22}photons)\times(3.373\times10^{-19}J/photon)$$$$E\approx2.031\times10^4J$$
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