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 - Exercises: 6.27a

Answer

- The energy of photon of wavelength $3.3\mu m$ is $6.02\times10^{-20}J$. - The energy of photon of wavelength $0.154nm$ is $1.29\times10^{-15}J$. - The energy of photon of wavelength $0.154nm$ is higher than the energy of photon of wavelength $3.3\mu m$.

Work Step by Step

*Strategy: We would calculate the energy of a photon of each wavelength according to the following formula $$E=h\times\nu=\frac{h\times c}{\lambda}$$ then compare the energy in each case with each other. $E$ : the energy of a photon $h$ : Planck constant ($h\approx6.626\times10^{-34}J.s$) $\nu$ : frequency of photon $\lambda$ : wavelength of photon $c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$) - Step 1: Find the known variables There is one photon (let's call it $p1$) of wavelength $3.3\mu m$, which means $\lambda_{p1} = 3.3\mu m= 3.3\times10^{-6}m$. The other photon ($p2$) has wavelength $0.154nm$; so $\lambda_{p2}= 0.154nm=1.54\times10^{-10}m$. - Step 2: Calculate the energy of photon $p1$ and photon $p2$ $E_{p1}=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{3.3\times10^{-6}}\approx6.02\times10^{-20}J$ $E_{p2}=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{1.54\times10^{-10}}\approx1.29\times10^{-15}J$ We see that the energy of photon $p2$ (wavelength $0.154nm$) is higher than the energy of photon $p1$ (wavelength $3.3\mu m$).
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