Question
LEED spectroscopy records the intensities and locations of electrons that are diffracted from a surface. For an electron to diffract, its de Broglie wavelength must be less than twice the distance between the atomic planes in the solid (see Section 29-9). Show that the de Broglie wavelength of an electron accelerated through a potential difference of $\phi$ volts is given by$$\lambda / \mathrm{pm}=\left(\frac{1.504 \times 10^{6} \mathrm{~V}}{\phi}\right)^{1 / 2}$$
Step 1
First, recall the de Broglie wavelength formula: $$ \lambda = \frac{h}{p} $$ where $\lambda$ is the de Broglie wavelength, $h$ is the Planck constant, and $p$ is the momentum of the electron. Show more…
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