29.6 The wave nature of matter  (Page 2/4)

Electron wavelength versus velocity and energy

For an electron having a de Broglie wavelength of 0.167 nm (appropriate for interacting with crystal lattice structures that are about this size): (a) Calculate the electron’s velocity, assuming it is nonrelativistic. (b) Calculate the electron’s kinetic energy in eV.

Strategy

For part (a), since the de Broglie wavelength is given, the electron’s velocity can be obtained from $\lambda =h/p$ by using the nonrelativistic formula for momentum, $p=\text{mv.}$ For part (b), once $v$ is obtained (and it has been verified that $v$ is nonrelativistic), the classical kinetic energy is simply $(1/2){\text{mv}}^2\text{.}$

Solution for (a)

Substituting the nonrelativistic formula for momentum ( $p=\text{mv}$ ) into the de Broglie wavelength gives

Solving for $v$ gives

Substituting known values yields

Solution for (b)

While fast compared with a car, this electron’s speed is not highly relativistic, and so we can comfortably use the classical formula to find the electron’s kinetic energy and convert it to eV as requested.

Discussion

This low energy means that these 0.167-nm electrons could be obtained by accelerating them through a 54.0-V electrostatic potential, an easy task. The results also confirm the assumption that the electrons are nonrelativistic, since their velocity is just over 1% of the speed of light and the kinetic energy is about 0.01% of the rest energy of an electron (0.511 MeV). If the electrons had turned out to be relativistic, we would have had to use more involved calculations employing relativistic formulas.