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UV radiation having a 300-nm wavelength falls on uranium metal, ejecting 0.500-eV electrons. What is the binding energy of electrons to uranium metal?
Question by OpenStax is licensed under CC BY 4.0.
$3.64 \textrm{ eV}$
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OpenStax College Physics Solution, Chapter 29, Problem 11 (Problems & Exercises) (1:34)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Photons with a wavelength of 300 nanometers incident on Uranium eject electrons that have kinetic energy of 0.500 electron volts. And we assume this is the maximum kinetic energy they can have because that's what this formula is all about—the maximum kinetic energy. This is [29.5]. So maximum kinetic energy of the electrons ejected are Planck's constant times the frequency of the photon minus the binding energy. Now we need to replace frequency, with something in terms of lambda because we are given the wavelength. So the wave equation says that the speed of the wave is frequency times lambda and then we divide both sides by lambda and we solve for frequency is c over λ. So f gets replaced with c over λ here. Then we are gonna add binding energy to both sides. And then also subtract kinetic energy from both sides. Then we have binding energy then is hc over λ minus kinetic energy of the electrons. So the binding energy of Uranium is Planck's constant times the speed of light divided by this wavelength—300 times 10 to the minus 9 meters— minus 0.5 electron volts giving 3.64 electron volts. And if you prefer less button pushing in your calculations, you could replace this numerator with 1240 electron volt nanometers which is equation [29.14] and then divide by 300 nanometers and you would get the same answer.