Question

Calculate the binding energy in eV of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm.

$4.08 \textrm{ eV}$

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This is College Physics Answers with Shaun Dychko. The longest wavelength that can eject electrons from aluminum is 304 nanometers and we are asked to figure out, what is the binding energy of aluminum? So we know that kinetic energy of the electron that's ejected is Planck's constant times the frequency of the photon minus the binding energy of the material; that's equation [29.5]. And we'll set kinetic energy to be 0 in order to minimize the frequency, or in other words, maximize the wavelength because this is the maximum wavelength we are told. So wave equation says the speed of the wave is frequency times wavelength and so we'll divide both sides by

*λ*in order to solve for*f*then we can substitute in to this equation with*c*over*λ*in place of*f*. And we have also written 0 in place of kinetic energy. So we'll add binding energy to both sides and then we'll solve for binding energy. It is Planck's constant times speed of light divided by lambda. So we have 4.14 times 10 to the minus 15 electon volt seconds; choosing this unit for Planck's constant so that our binding energy also has electron volt units, and then multiplied by speed of light divided by 304 times 10 to the minus 9 meters and this gives 4.08 electron volts. The textbook also gives us a number for this product*hc*as a single number 1240 electron volt nanometers. So I could have gone 1240 divided by 304 and that would have given the same answer.