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

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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 $\left(1/2\right){\text{mv}}^{2}\text{.}$

Solution for (a)

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

$\lambda =\frac{h}{p}=\frac{h}{\text{mv}}.$

Solving for $v$ gives

$v=\frac{h}{\mathrm{m\lambda }}.$

Substituting known values yields

$v=\frac{6\text{.}\text{63}×{\text{10}}^{\text{–34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot \text{s}}{\left(\text{9.11}×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(0\text{.}\text{167}×{\text{10}}^{\text{–9}}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}=\text{4}\text{.}\text{36}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s}.$

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.

$\begin{array}{lll}\text{KE}& =& \frac{1}{2}{\text{mv}}^{2}\\ & =& \frac{1}{2}\left(9.11×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{4.36}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}\\ & =& \text{(86.4}×{\text{10}}^{\text{–18}}\phantom{\rule{0.25em}{0ex}}\text{J)}\left(\frac{\text{1 eV}}{1.602×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J}}\right)\\ & =& \text{54.0 eV}\end{array}$

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.

Electron microscopes

One consequence or use of the wave nature of matter is found in the electron microscope. As we have discussed, there is a limit to the detail observed with any probe having a wavelength. Resolution, or observable detail, is limited to about one wavelength. Since a potential of only 54 V can produce electrons with sub-nanometer wavelengths, it is easy to get electrons with much smaller wavelengths than those of visible light (hundreds of nanometers). Electron microscopes can, thus, be constructed to detect much smaller details than optical microscopes. (See [link] .)

There are basically two types of electron microscopes. The transmission electron microscope (TEM) accelerates electrons that are emitted from a hot filament (the cathode). The beam is broadened and then passes through the sample. A magnetic lens focuses the beam image onto a fluorescent screen, a photographic plate, or (most probably) a CCD (light sensitive camera), from which it is transferred to a computer. The TEM is similar to the optical microscope, but it requires a thin sample examined in a vacuum. However it can resolve details as small as 0.1 nm ( ${\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}m$ ), providing magnifications of 100 million times the size of the original object. The TEM has allowed us to see individual atoms and structure of cell nuclei.

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