Electron at 300 V and crystal diffraction

Problem

Electron and de Broglie. An electron is accelerated through a potential difference $V = 300\;\text{V}$. Calculate the associated de Broglie wavelength. Sketch the experiment, showing the acceleration, the crystal target and the detector.

Solution

Step 1 — Kinetic energy. The electron gains $K = eV = 1{,}6\cdot 10^{-19}\cdot 300 = 4{,}8\cdot 10^{-17}\;\text{J}$.

Step 2 — Momentum (non-relativistic, $K \ll m_e c^2$):

$p = \sqrt{2 m_e K} = \sqrt{2\cdot 9{,}11\cdot 10^{-31}\cdot 4{,}8\cdot 10^{-17}} = \sqrt{8{,}75\cdot 10^{-47}} \approx 9{,}35\cdot 10^{-24}\;\text{kg·m/s}$

Step 3 — de Broglie wavelength.

$\ev{\lambda = \frac{h}{p} = \frac{6{,}63\cdot 10^{-34}}{9{,}35\cdot 10^{-24}} \approx 7{,}1\cdot 10^{-11}\;\text{m} \approx 0{,}071\;\text{nm}}$

Comment (experiment). The electron, accelerated by the potential difference $V$, strikes a crystal target; since $\lambda \approx 0{,}07\;\text{nm}$ is comparable to the lattice spacing ($\sim 0{,}1\;\text{nm}$), the crystal diffracts it and the detector picks up maxima at certain angles (Bragg condition). This is the Davisson–Germer experiment, direct proof of the wave nature of the electron.

Numerical note $\lambda_\text{dB} = 1{,}226\,\text{nm}/\sqrt{V}$ we get $\lambda = 1{,}226/\sqrt{300} \approx 0{,}0708\;\text{nm} \approx 7{,}1\cdot 10^{-11}\;\text{m}$.

With

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Topics: Quantum physics Concepts: de Broglie wavelength · Diffraction