Matter waves — de Broglie hypothesis, Davisson-Germer experiment

Question

An electron is accelerated through a potential difference of $100$ V. Find its de Broglie wavelength. Why don’t we observe wave nature for everyday objects like a cricket ball?

(CBSE 12 & JEE Main pattern)

Solution — Step by Step

de Broglie proposed that every moving particle has an associated wavelength:

\lambda = \frac{h}{p} = \frac{h}{mv}

where $h = 6.63 \times 10^{-34}$ J s is Planck’s constant and $p$ is momentum.

When an electron is accelerated through potential $V$ , its KE equals $eV$ :

\frac{1}{2}mv^2 = eV \implies p = mv = \sqrt{2meV}

\lambda = \frac{h}{\sqrt{2meV}}

\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}

= \frac{6.63 \times 10^{-34}}{\sqrt{2.912 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{5.396 \times 10^{-24}} = \mathbf{1.23 \times 10^{-10} \text{ m} = 1.23 \text{ A}}

Shortcut: for electrons, $\lambda(\text{A}) = \dfrac{12.27}{\sqrt{V}}$ . For $V = 100$ : $\lambda = 12.27/10 = 1.227$ A.

A cricket ball ( $m = 0.15$ kg) moving at $v = 30$ m/s:

\lambda = \frac{6.63 \times 10^{-34}}{0.15 \times 30} = 1.47 \times 10^{-34} \text{ m}

This is about $10^{-19}$ times smaller than a proton. No instrument can detect this wavelength. Wave nature is observable only when $\lambda$ is comparable to the size of obstacles (atomic scale), which happens only for subatomic particles.

Why This Works

de Broglie extended wave-particle duality to matter: if light (a wave) can behave as particles (photons), then particles should also have wave properties. The Davisson-Germer experiment confirmed this by showing electron diffraction from nickel crystals — the diffraction pattern matched the wavelength predicted by $\lambda = h/p$ .

graph TD
    A["Wave-Particle Duality"] --> B["Light"]
    A --> C["Matter"]
    B --> B1["Wave: interference,<br/>diffraction"]
    B --> B2["Particle: photoelectric<br/>effect, Compton"]
    C --> C1["Particle: tracks in<br/>cloud chamber"]
    C --> C2["Wave: electron diffraction<br/>(Davisson-Germer)"]
    C2 --> D["λ = h/p = h/mv"]
    D --> E{"Mass large?"}
    E -->|"Yes (macro)"| F["λ ≈ 0<br/>Wave nature undetectable"]
    E -->|"No (electron, proton)"| G["λ ~ atomic size<br/>Diffraction observable"]

Alternative Method — Using Energy in eV Directly

For any charged particle accelerated through $V$ volts:

\lambda = \frac{h}{\sqrt{2mqV}}

For a proton ( $m_p \approx 1836 m_e$ ), the wavelength is $\sqrt{1836}$ times smaller than for an electron at the same $V$ . This means electron diffraction is far easier to observe than proton diffraction.

The shortcut $\lambda = 12.27/\sqrt{V}$ angstroms works only for electrons. For protons, multiply by $1/\sqrt{1836} \approx 1/42.8$ . JEE occasionally asks to compare wavelengths of different particles at the same KE: $\lambda \propto 1/\sqrt{m}$ , so lighter particles have longer wavelengths.

Common Mistake

Students use the non-relativistic formula even when the electron energy is very high (above ~ $10$ keV). At high energies, the relativistic momentum $p = \sqrt{2mKE + KE^2/c^2}$ must be used instead of $p = \sqrt{2mKE}$ . For CBSE and NEET, problems stay non-relativistic. But JEE Advanced may push into relativistic territory — watch for very high accelerating voltages.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Energy in eV Directly

Common Mistake

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