de Broglie Wavelength and Matter Waves

Every moving particle has a wavelength

If light (normally a wave) can behave like a stream of particles (photons), de Broglie proposed the reverse should hold too: every moving particle has an associated wavelength, set by its momentum $p=mv$ :

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

For everyday objects this wavelength is absurdly small and unobservable. It only becomes significant for particles with very small mass, like electrons, whose wavelength at laboratory energies is comparable to the spacing of atoms. Electron microscopes rely on this: an electron beam's wavelength is far shorter than visible light's, so it resolves much finer detail.

The same formula in other forms

Questions rarely hand you the momentum directly. Rewrite $p$ in terms of what you're given, using $K = \dfrac{p^2}{2m}$ for kinetic energy and $K = qV$ for a charge accelerated from rest through a potential difference $V$ :

de Broglie wavelength from different starting quantities

Given	Wavelength
Momentum $p$	$\lambda = \dfrac{h}{p}$
Mass $m$ and speed $v$	$\lambda = \dfrac{h}{mv}$
Kinetic energy $K$	$\lambda = \dfrac{h}{\sqrt{2mK}}$
Charge $q$ accelerated through $V$	$\lambda = \dfrac{h}{\sqrt{2mqV}}$
Electron accelerated through $V$	$\lambda \approx \dfrac{1.227}{\sqrt{V}}$ nm

Worked example

Find the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V.

Solution: The accelerating field does work $eV$ on the electron, which becomes its kinetic energy:

eV = \dfrac{1}{2}mv^2 \;\Rightarrow\; mv = \sqrt{2meV}

Substituting $p=mv=\sqrt{2meV}$ into the de Broglie relation:

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

Plugging in $h=6.626\times10^{-34}$ J·s, $m=9.11\times10^{-31}$ kg, and $e=1.6\times10^{-19}$ C gives a clean, commonly-used shortcut for electrons specifically:

\lambda \approx \dfrac{1.227}{\sqrt{V}}\text{ nm} \qquad (V \text{ in volts})

For $V=100$ V:

\lambda \approx \dfrac{1.227}{\sqrt{100}} = \dfrac{1.227}{10} \approx 0.123\text{ nm}

That's comparable to the spacing between atoms in a crystal, which is exactly why fast electrons diffract off crystal lattices the same way X-rays do.

Worked example: why a cricket ball doesn't diffract

Find the de Broglie wavelength of a 0.15 kg cricket ball moving at 30 m/s.

Solution:

\lambda = \dfrac{h}{mv} = \dfrac{6.626\times10^{-34}}{0.15 \times 30} = \dfrac{6.626\times10^{-34}}{4.5} \approx 1.5\times10^{-34}\text{ m}

That's around $10^{19}$ times smaller than a proton. No gap or obstacle is anywhere near that size, so the ball shows no wave behaviour at all.

Worked example: comparing two particles

A proton and an alpha particle are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths.

Solution: From $\lambda = \dfrac{h}{\sqrt{2mqV}}$ with the same $V$ , $\lambda \propto \dfrac{1}{\sqrt{mq}}$ . An alpha particle has about 4 times the proton's mass and twice its charge:

\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4 \times 2} = 2\sqrt{2}

If instead the two particles had the same kinetic energy, the charge would drop out and the ratio would be $\sqrt{\dfrac{m_\alpha}{m_p}} = 2$ .

When the simple formula stops working

$K = \dfrac{p^2}{2m}$ is the non-relativistic kinetic energy, so every formula above assumes the particle is much slower than light. For an electron that holds well at ordinary voltages: the error in $\lambda$ is about 0.005% at 100 V and about 0.5% at 10 kV, but it grows to about 5% at 100 kV.

At high voltages, use the relativistic momentum instead, where $mc^2$ is the particle's rest energy (511 keV for an electron):

\lambda = \dfrac{h}{\sqrt{2mK\left(1 + \dfrac{K}{2mc^2}\right)}}

Common mistakes

Using an energy in eV without converting. In $\lambda = \dfrac{h}{\sqrt{2mK}}$ with SI values, $K$ must be in joules: multiply eV by $1.6\times10^{-19}$ .
Forgetting the charge. An alpha particle carries $2e$ , so it gains $2eV$ of energy through a potential difference $V$ , not $eV$ .
Writing $\lambda \propto \dfrac{1}{V}$ . The wavelength goes as $\dfrac{1}{\sqrt{V}}$ . Four times the voltage halves the wavelength.
Using the electron shortcut for other particles. $\dfrac{1.227}{\sqrt{V}}$ nm has the electron's mass built in. For anything else, use the full formula.

Where this comes up in exams

Matter waves and the de Broglie relation are listed in the Dual Nature of Matter and Radiation units of the JEE Main 2026 and NEET (UG) 2026 syllabi, and the JEE Advanced 2026 syllabus lists the de Broglie wavelength of matter waves.

Checked against JEE (Main) 2026 syllabus (NTA), NEET (UG) 2026 syllabus (NMC, published by NTA) and JEE (Advanced) 2026 syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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