Question

What is the kinetic energy of an electron in a TEM having a 0.0100-nm wavelength?

$15.0 \textrm{ keV}$

Solution Video

#### Sign up to view this solution video!

## Calculator Screenshots

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

*v*. 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*mv*in place of*p*here for the de Broglie wavelength and this can be rearranged to solve for*v*by multiplying both sides by*v*over*λ*. And so*v*is*h*over*mλ*and this can be substituted in for*v*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*h*squared over 2*m λ*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.