Matter waves — de Broglie hypothesis, Davisson-Germer experiment

Question

An electron is accelerated through a potential difference of 100 V. Find the de Broglie wavelength of the electron. Also, explain how the Davisson-Germer experiment confirmed wave nature of electrons.

Given: $h = 6.626 \times 10^{-34}$ J·s, $m_e = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C

Solution — Step by Step

When an electron is accelerated through potential difference $V$ , it gains kinetic energy:

KE = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17} \text{ J}

Since $KE = \frac{p^2}{2m}$ , we get:

p = \sqrt{2m_e \cdot KE} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}

p = \sqrt{29.12 \times 10^{-48}} = 5.395 \times 10^{-24} \text{ kg·m/s}

\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.395 \times 10^{-24}} = \mathbf{1.228 \times 10^{-10} \text{ m} \approx 1.23 \text{ Å}}

This is comparable to interatomic spacing in crystals — which is exactly why electron diffraction works.

Why This Works

De Broglie proposed that every moving particle has an associated wavelength given by $\lambda = h/p$ . This was a bold hypothesis — if light (a wave) shows particle nature (photons), why not the reverse? Why can’t particles show wave nature?

graph TD
    A["de Broglie Hypothesis: λ = h/p"] --> B["Prediction: Electrons should show diffraction"]
    B --> C["Davisson-Germer Experiment 1927"]
    C --> D["Electron beam hits nickel crystal"]
    D --> E["Scattered electrons show intensity peaks"]
    E --> F["Peak angles match Bragg's law: nλ = 2d sin θ"]
    F --> G["Measured λ matches de Broglie prediction"]
    G --> H["Wave nature of matter CONFIRMED"]

The Davisson-Germer experiment fired electrons at a nickel crystal and measured the scattered electron intensity at various angles. They found sharp peaks at specific angles — exactly what you would expect from diffraction of waves off crystal planes. The wavelength calculated from the diffraction pattern matched de Broglie’s formula perfectly.

This was one of the most beautiful confirmations in physics — a theoretical prediction verified experimentally within just three years.

Alternative Method

For electrons accelerated through potential $V$ , there is a direct shortcut formula:

\lambda = \frac{12.27}{\sqrt{V}} \text{ Å}

For $V = 100$ V: $\lambda = 12.27/\sqrt{100} = 12.27/10 = 1.227$ Å

This saves a lot of calculation time in MCQs. For JEE Main, memorise this formula — it has appeared in at least 4 papers in the last 5 years.

Common Mistake

Using relativistic momentum for low-energy electrons. At 100 V, the electron’s KE is $100$ eV — far less than its rest mass energy of $0.511$ MeV. So non-relativistic treatment ( $p = \sqrt{2mKE}$ ) is perfectly valid. Students sometimes overcomplicate this. Use relativistic formulas only when $KE$ is comparable to $m_0c^2$ .

Also, do not confuse the shortcut $\lambda = 12.27/\sqrt{V}$ (for electrons) with $\lambda = 0.286/\sqrt{V}$ (for protons). The numerical constant changes because mass changes. Always check which particle the problem is about.

\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mKE}} = \frac{h}{\sqrt{2meV}}

Electron shortcut: $\lambda = \frac{12.27}{\sqrt{V}}$ Å

Davisson-Germer: $n\lambda = 2d\sin\theta$ (Bragg condition)

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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