Problem — De Broglie wavelength of an electron at 100 V

Problem

An electron is accelerated through a potential difference of $V = 100$ V. Calculate its de Broglie wavelength and comment on why a crystal lattice is able to diffract it.

Solution

Step 1 — Kinetic energy gained. The electron accelerated by $V = 100$ V gains

$K = eV = 1{,}6\cdot 10^{-19}\cdot 100 = 1{,}6\cdot 10^{-17}\;\text{J}$

(since $eV = 100$ eV $\ll m_e c^2 = 511$ keV, the non-relativistic calculation is valid).

Step 2 — Momentum. From $K = p^2/(2m_e)$:

$p = \sqrt{2 m_e K} = \sqrt{2\cdot 9{,}11\cdot 10^{-31}\cdot 1{,}6\cdot 10^{-17}} = \sqrt{2{,}92\cdot 10^{-47}} \approx 5{,}4\cdot 10^{-24}\;\text{kg·m/s}$

Step 3 — De Broglie wavelength.

$\lambda = \frac{h}{p} = \frac{6{,}63\cdot 10^{-34}}{5{,}4\cdot 10^{-24}} \approx 1{,}2\cdot 10^{-10}\;\text{m}$

$\ev{\lambda \approx 1{,}2\cdot 10^{-10}\;\text{m} = 1{,}2\ \text{Å}}$

Comment. A lattice diffracts a wave only if the lattice spacing is comparable to the wavelength. Interatomic distances in crystals are of order $10^{-10}$ m, exactly like $\lambda_\text{dB}$: hence the crystal acts as a diffraction grating for electrons (Davisson–Germer experiment), direct proof of the wave nature of matter.

Collegamenti

Argomenti: Quantum physics Concetti: De Broglie wavelength · Wave-particle duality Competenze: Unit conversion · Symbolic setup