de Broglie wavelength of an electron at 300 V

Problem

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of $300\;\text{V}$. ($m_e = 9{,}11\cdot 10^{-31}\;\text{kg}$)

Solution

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

Step 2 — Momentum (non-relativistic regime):

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

Step 3 — de Broglie wavelength.

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

About $0{,}07\;\text{nm}$: of the same order as interatomic distances, so an electron accelerated this way is suitable for crystal-diffraction experiments.

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Topics: Quantum physics Concepts: de Broglie wavelength