De Broglie wavelength of an electron at 70 V

Problem

An electron is accelerated through a potential difference of $V = 70\;\text{V}$. Calculate its de Broglie wavelength (neglect relativistic effects). Comment: can this wave be diffracted by the lattice planes of a crystal?

Solution

Step 1 — Kinetic energy gained. The electron gains $K = eV$:

$K = 1{,}6\cdot 10^{-19}\cdot 70 = 1{,}12\cdot 10^{-17}\;\text{J}$

Step 2 — Momentum. Non-relativistic, $p = \sqrt{2 m_e K}$:

$p = \sqrt{2\cdot 9{,}11\cdot 10^{-31}\cdot 1{,}12\cdot 10^{-17}} = \sqrt{2{,}04\cdot 10^{-47}} \approx 4{,}52\cdot 10^{-24}\;\text{kg·m/s}$

Step 3 — De Broglie wavelength.

$\ev{\lambda = \frac{h}{p} = \frac{6{,}63\cdot 10^{-34}}{4{,}52\cdot 10^{-24}} \approx 1{,}45\cdot 10^{-10}\;\text{m}}$

Comment. This gives $\lambda \approx 0{,}145\;\text{nm}$, the same order of magnitude as the spacing between atomic planes in a crystal ($\sim 10^{-10}\;\text{m}$). The lattice therefore acts as a diffraction grating: yes, the electron wave is diffracted. This is the basis of the Davisson–Germer experiment and of electron diffraction (LEED).

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Topics: Fisica quantistica Concepts: Lunghezza d’onda di de Broglie · Diffrazione