De Broglie Wavelength Calculator

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How to Use the De Broglie Wavelength Calculator

1. 1. Choose the particle. Electron, muon, proton, neutron and alpha particle masses are the 2022 CODATA values. Choose Another particle or object to type any mass in kg, g or atomic mass units.
2. 2. Say what you know. Speed (in m/s, km/h or as a fraction of c), kinetic energy (µeV to GeV or joules), momentum, or the accelerating voltage for a charged particle.
3. 3. Choose the method. Exact uses relativistic kinematics and is right at any speed. Classical uses p = mv and K = ½mv², which is accurate only well below the speed of light.
4. 4. Read the wavelength. The result comes with momentum, energy, speed, β and γ, the classical formula's error, and a scale strip showing where the wavelength falls between a proton and visible light.

The de Broglie Relation

In his 1924 doctoral thesis Louis de Broglie proposed that matter has wave properties, with a wavelength λ = h ÷ p, where h is the Planck constant and p the particle's momentum. It is the same relation as for a photon, p = h ÷ λ, applied to particles with mass. The Davisson–Germer experiment of 1927, in which electrons scattered from a nickel crystal formed a diffraction pattern, confirmed it; de Broglie received the 1929 Nobel Prize in Physics, and Davisson and G. P. Thomson shared the 1937 prize.

For an electron accelerated from rest through V volts, the classical result is λ = h ÷ √(2meV) = 1.2264 nm ÷ √V, a handy rule that gives 0.1669 nm at 54 V. It is accurate to under 1% up to about 10 kV, but at 100 kV the classical formula is 4.8% too long, and at higher energies it fails: a relativistic electron's wavelength shrinks as 1/γ, and at very high energy tends to the photon value hc ÷ E.

Worked Examples

• The Davisson–Germer electron, 54 V (the starting example): K = 54 eV, p = 3.970 × 10⁻²⁴ kg·m/s and λ = 166.9 pm = 0.1669 nm, about the spacing of atoms in a crystal. It moves at 4.358 × 10⁶ m/s, 1.45% of c, so the classical formula is off by only 0.003%.
• An electron microscope at 100 kV: the exact wavelength is 3.701 pm (γ = 1.196, β = 0.548). The classical formula gives 3.878 pm, 4.8% too long. That is far smaller than an atom, which is why electron microscopes can resolve atoms where light, at about 550 nm, cannot.
• A thermal neutron, 0.025 eV: λ = 0.1808 nm at 2.19 km/s, matching atomic spacings, which is why neutron diffraction reveals crystal structure.
• A 5 MeV alpha particle: λ = 6.42 fm, similar to the size of an atomic nucleus.
• A C₆₀ molecule at 220 m/s (mass 720 u): λ = 2.52 pm. Interference of such molecules was observed by Arndt and colleagues in 1999, showing matter waves for objects far heavier than an electron.
• A 0.145 kg baseball at 40 m/s: λ = 1.14 × 10⁻³⁴ m, about 10⁻¹⁹ of a proton's size. Everyday objects have wavelengths far too small to show any wave behaviour.

Frequently Asked Questions

What is the de Broglie wavelength formula?

λ = h ÷ p, where h = 6.626 070 15 × 10⁻³⁴ J·s and p is the particle's momentum. For a slow particle p = mv, so λ = h ÷ (mv); for relativistic speeds use p = γmv.

How do you find the wavelength of an electron accelerated through a voltage?

The electron gains K = eV, and p = √(2mK) classically, so λ = h ÷ √(2meV), or about 1.226 nm ÷ √V. At 150 V that is 0.100 nm. Above roughly 10 kV use the relativistic form, which this calculator applies in its exact mode.

Why can't we see the wavelength of a baseball?

Its momentum is huge compared with h, so the wavelength is around 10⁻³⁴ m, trillions of trillions of times smaller than anything it could diffract around. Wave behaviour needs a wavelength comparable to the size of the obstacle, which never happens for everyday objects.

When do I need the relativistic formula?

Once the speed passes about 10% of the speed of light, which for an electron is a kinetic energy of about 2.6 keV. There the classical wavelength is already off by about 0.5% if you give the speed, and the error grows quickly: for an electron given its energy it is 4.8% at 100 keV, 41% at 1 MeV, and a factor of two near 3 MeV.

Do all particles have a de Broglie wavelength?

Yes: electrons, neutrons, atoms and molecules all show interference. Heavier objects have shorter wavelengths for the same speed, which makes the effects harder to detect, but interference has been seen with molecules of hundreds of atoms.

What is the difference between a photon's wavelength and a particle's?

Both obey λ = h ÷ p. A photon has no mass and always moves at c, so its energy is pc and λ = hc ÷ E. A particle with mass can have any speed below c, so the same energy gives a longer wavelength than a photon of that energy would have.