Solutions/Answers to Exercises
Exercise 1. An X-ray has a wavelength of 2.64 nm. What is the energy of one photon?
2.64 nm = 2.64 x 10-9 m
\(\displaystyle E\;=\;\frac{hc}{\lambda}\;=\;=\frac{(6.626\times 10^{-34}J⋅s)\times (2.998\times 10^8\;m/s)}{2.64\times 10^{-9}\;m}\;=\;\mathbf{7.52\times 10^{-17}\;J}\)
Back to Particle Properties of Electromagnetic Radiation
Exercise 2. What is the energy of the following photons in kJ/mol?
First, find the energy for one photon.
\(\displaystyle E\;=\;hν\;=\;(6.626\times 10^{-34}J⋅s)\times (4.98\times 10^{19}\;s^{-1})\;=\;3.30\times 10^{-19}J/photon\)
\(\displaystyle 3.30\times 10^{-19}\frac{J}{photon}\times\frac{6.02\times 10^{23}\;photons}{mol}\times\frac{1\;kJ}{1000\;J}\;=\;\mathbf{198.7\;kJ/mol}\)
b) ν = 3.24 x 107 s-1
\(\displaystyle E\;=\;hν\;=\;(6.626\times 10^{-34}J⋅s)\times (3.24\times 10^{7}\;s^{-1})\;=\;2.15\times 10^{-26}J/photon\)
\(\displaystyle 2.15\times 10^{-26}\frac{J}{photon}\times\frac{6.02\times 10^{23}\;photons}{mol}\times\frac{1\;kJ}{1000\;J}\;=\;\mathbf{1.29\times 10^{-5}\;kJ/mol}\)
c) λ = 4.67 x 10-6 m
\(\displaystyle E\;=\;\frac{hc}{\lambda}\;=\;\frac{(6.626\times 10^{-34}\;J⋅s)\times (2.998\times 10^8\;m/s)}{4.67\times 10^{-6}\;m}\;=\;4.32\times 10^{-20}\;J/photon\)
\(\displaystyle 4.32\times 10^{-20}\frac{J}{photon}\times\frac{6.02\times 10^{23}\;photons}{mol}\times\frac{1\;kJ}{1000\;J}\;=\;\mathbf{26.0\;kJ/mol}\)
Back to Particle Properties of Electromagnetic Radiation
Exercise 3. What frequency of light is needed to eject electrons from a silver solid? The work function for silver is 436 kJ/mol.
First find the energy, in J, for one photon
\(\displaystyle 436\;\frac{kJ}{mol}\times\frac{1000\;J}{1\;kJ}\times\frac{1\;mol}{6.02\times 10^{23}\;photons}\;=\;7.24\times 10^{-19}\;\frac{J}{photon}\)
\(\displaystyle \nu\;=\;\frac{E}{h}\;=\;\frac{7.24\times 10^{-19}\;J}{6.626\times 10^{-34}\;J⋅s}\;=\;\mathbf{1.09\times10^{15}\;s^{-1}}\)
Back to Particle Properties of Electromagnetic Radiation
Exercise 4. Sunburn can be caused by exposure to UV light that is about 326 nm.
First, find the energy of one photon of this light.
\(\displaystyle E\;=\;\frac{hc}{\lambda}\;=\;\frac{6.626\times 10^{-34}\;J⋅s\times (2.998\times 10^8\;m/s)}{3.26\times 10^{-7}\;m}\;=\;\mathbf{6.09\times 10^{-19}\;J}\)
b) What is the energy of 3.5 moles of these photons?
\(\displaystyle 3.5\;mol\;photons\times\frac{6.02\times 10^{23}\;photons}{1\;mol\;photons}\times\frac{6.09\times 10^{-19}\;J}{photon}\;=\;\mathbf{1.3\times 10^6\;J}\)
c) How many photons are in a 1.25 mJ burst of this radiation?
\(\displaystyle 1.25\;mJ\times\frac{1\;J}{1000\;mJ}\times\frac{1\;photon}{6.09\times 10^{-19}\;J}\;=\;\mathbf{2.05\times 10^{15}\;photons}\)
Back to Particle Properties of Electromagnetic Radiation
Exercise 5. A laser emits a wavelength of 944 nm. The total energy measured by the detector is 0.48 J over a 29.5 second time period. How many photons per second are being emitted by the laser?
Find the energy of one photon of this light.
\(\displaystyle E\;=\;\frac{6.626\times 10^{-34}\;J⋅s\times (2.998\times 10^8\;m/s)}{9.44\times 10^{-7}\;m}\;=\;2.10\times 10^{-19}\;J/photon\)
\(\displaystyle \frac{0.48\;J}{29.5\;s}\times\frac{1\;photon}{2.10\times 10^{-19}\;J}\;=\;\mathbf{7.83\times 10^{16}\;photons/s}\)
Back to Particle Properties of Electromagnetic Radiation