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What is the kinetic energy of an electron in a TEM having a 0.0100-nm wavelength?
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Final Answer
$15.0 \textrm{ keV}$
Solution Video

OpenStax College Physics Solution, Chapter 29, Problem 57 (Problems & Exercises) (1:34)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We want to know the kinetic energy of an electron in a tunneling electron microscope given that the electron has a de Broglie wavelength of 0.0100 nanometers. So kinetic energy is one-half mass times velocity squared and so we are going to need to substitute somehow for this velocity <i>v</i>. Well, we know the de Broglie wavelength of the electron is Planck's constant divided by momentum and momentum is mass times velocity. So we substitute <i>mv</i> in place of <i>p</i> here for the de Broglie wavelength and this can be rearranged to solve for <i>v</i> by multiplying both sides by <i>v</i> over <i>λ</i>. And so <i>v</i> is <i>h</i> over <i>mλ</i> and this can be substituted in for <i>v</i> in our kinetic energy formula. So we have kinetic energy of the electron is one-half times the mass times Planck's constant over mass times lambda squared and this works out to <i>h</i> squared over 2<i>m λ</i> squared. So here's Planck's constant squared divided by 2 times the mass of an electron times the wavelength of 0.01 nanometers, which is times 10 to the minus 9 meters, and we square that. And then this is gonna give us units of joules but we are told to find the energy in kiloelectron volts. So we multiply by 1 electron volt for every 1.602 times 10 to the minus 19 joules and then convert further by multiplying by 1 kiloelectron volt for every 10 to the 3 electron volts giving us 15.0 kiloelectron volts of kinetic energy for these electrons.