De Broglie Wavelength Calculator
Compute the de Broglie wavelength λ = h/(m·v) of a massive particle from its mass and velocity. Demonstrates wave-particle duality: every massive particle has an associated quantum wavelength.
Last updated: May 2026
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About this calculator
The de Broglie hypothesis (1924) assigns a wavelength λ = h/p = h/(mv) to any moving particle, where h = 6.62607015 × 10⁻³⁴ J·s is the Planck constant (CODATA exact value), m is mass in kg, and v is velocity in m/s. This wavelength is observable through diffraction and interference of particle beams — the Davisson-Germer experiment (1927) confirmed electron wave behaviour by demonstrating Bragg diffraction from nickel crystals, and electron microscopes and neutron diffraction exploit it daily. The calculator uses the non-relativistic form p = mv, which is accurate for v ≪ c; for relativistic particles, replace with relativistic momentum p = γmv. Edge cases: at low velocities, λ becomes very large — a 1 g object moving at 1 m/s has λ ≈ 7 × 10⁻³¹ m, far smaller than any feature it could interact with, which is why macroscopic objects do not exhibit wave behaviour. Electrons at v = 2 × 10⁶ m/s have λ ≈ 0.36 nm, comparable to atomic spacing — diffractable from crystals. Photons are not described by this formula (they are massless); use λ = hc/E or λ = h/p with photon momentum p = E/c. Including thermal motion: at temperature T, the thermal de Broglie wavelength λ_th = h/√(2π·m·kB·T) gives a characteristic quantum length scale; at room temperature it's about 0.07 nm for electrons, 4 pm for hydrogen atoms.
How to use
Example 1 — electron in a typical electron microscope. m = 9.109 × 10⁻³¹ kg (electron), v = 2 × 10⁶ m/s (about 6.7 × 10⁻³ c, non-relativistic regime). Step 1: p = m·v = 9.109e−31 × 2e6 = 1.822e−24 kg·m/s. Step 2: λ = h/p = 6.62607e−34 / 1.822e−24 ≈ 3.64e−10 m = 0.364 nm. Verify: published electron-microscope wavelengths at this velocity are around 0.3–0.5 nm, comparable to atomic spacing in metals (~0.25 nm) — enough resolution to image atoms in modern TEMs ✓. Example 2 — neutron used in neutron scattering. m = 1.675 × 10⁻²⁷ kg, v = 2,200 m/s (thermal neutron from a reactor at 293 K). Step 1: p = 1.675e−27 × 2200 ≈ 3.685e−24 kg·m/s. Step 2: λ = 6.62607e−34 / 3.685e−24 ≈ 1.80e−10 m = 0.18 nm = 1.8 Å. Verify: thermal neutron wavelengths are conventionally ~1.5–2 Å, matching this result ✓. This is the wavelength that makes neutron diffraction useful for crystallography — comparable to atomic spacing, but with much weaker electronic scattering than X-rays, making neutrons sensitive to light atoms (H, Li, B) that X-rays miss.
Frequently asked questions
Why don't I see macroscopic objects diffract?
Their de Broglie wavelengths are astronomically small relative to any feature they could interact with. A 100 g baseball moving at 30 m/s has λ ≈ 2 × 10⁻³⁴ m — about 10²¹ times smaller than a proton's radius and 10²⁵ times smaller than the atomic spacing of any potential 'diffraction grating'. For diffraction to be observable, the wavelength must be comparable to the slit or grating spacing. Macroscopic objects can only diffract from features many orders of magnitude smaller than any physical structure could practically be (single atoms, nuclei). For a wavelength comparable to a typical 1-μm-sized slit, you would need v on the order of 10⁻²² m/s, slower than the slowest reasonable measurement. This is the practical reason quantum effects become invisible at macroscopic scales — not because they don't exist, but because they're invisibly small relative to the structures involved. The wave nature is genuinely there; it just doesn't manifest as observable interference.
How does the de Broglie wavelength scale with mass and velocity?
Inversely with momentum: λ = h/p = h/(mv). Doubling mass at fixed velocity halves λ; doubling velocity at fixed mass also halves λ. Higher-energy particles (faster or heavier) have shorter wavelengths. This is why high-energy particle accelerators can probe very small structures: the LHC's protons at γ ≈ 7,000 have momentum ~7 TeV/c, giving λ ≈ 2 × 10⁻¹⁹ m, smaller than a proton — that's how the Higgs boson was probed. Conversely, ultracold-atom experiments slow atoms to near-zero velocity, producing thermal de Broglie wavelengths of micrometres or more — when this exceeds the inter-atomic spacing, the atoms become quantum-degenerate and form a Bose-Einstein condensate (or fermionic equivalent), with collective wave behaviour at macroscopic scale. The interplay of mass and velocity (or temperature) controls when quantum effects become dominant; the thermal de Broglie wavelength λ_th = h/√(2π·m·k_B·T) is the natural scale.
When does relativity matter for the de Broglie wavelength?
When v becomes comparable to c. The classical de Broglie formula uses p = mv, but the correct momentum is p = γmv for any speed. At v = 0.1c, γ ≈ 1.005 — relativistic correction of 0.5%, often negligible. At v = 0.5c, γ ≈ 1.155 — 15% correction; the non-relativistic λ overestimates by ~15%. At v = 0.99c, γ ≈ 7.09 — the actual wavelength is 7× smaller than the non-relativistic estimate. For ultra-relativistic particles like LHC protons (γ ≈ 7000), the non-relativistic formula is off by 4 orders of magnitude. Always check: if v > 0.1c, switch to p = γmv. The relativistic formula can also be written as λ = h/p = hc/√(E² − m²c⁴), useful when energy is known. For photons (m = 0), the relativistic relation reduces to E = pc, giving λ = hc/E — consistent with the Planck-Einstein relation E = hν and ν = c/λ.
What are the common mistakes when computing de Broglie wavelength?
The biggest mistake is using non-SI units without conversion: m in kg, v in m/s, h in J·s. Inputting m in g or v in km/h gives wrong magnitudes. The second is forgetting that the formula is non-relativistic — at v > 0.1c, use γmv instead of mv. The third is applying the formula to massless particles like photons; they have m = 0 and the formula gives λ = ∞, which is wrong. For photons use λ = hc/E (Planck-Einstein) or λ = h/p with p = E/c. People also conflate the de Broglie wavelength with thermal de Broglie wavelength (the latter is a characteristic length scale at temperature T, not a single-particle wavelength). Mixing m as relativistic mass (γm) with v gives p = γm·v but then dividing h by that gives the right answer accidentally — better practice is to keep m as rest mass throughout. Finally, the de Broglie wavelength describes the central wavelength of a free-particle wavepacket; localised particles have a range of wavelengths (uncertainty principle) — the formula is an idealisation.
When should I not use this calculator?
Do not use it for photons or other massless particles — they have m = 0, λ = h/p with p = E/c, not p = mv. For visible photons use λ = hc/E directly. It is not appropriate for relativistic particles (v > 0.1c) without using γmv instead of mv; for ultra-relativistic particles the non-relativistic formula is off by orders of magnitude. Do not use it for collections of particles or thermodynamic systems — for those use the thermal de Broglie wavelength λ_th = h/√(2πmk_BT). It is unsuitable for bound or confined particles where momentum is not sharply defined (electrons in atoms have a momentum distribution, not a single λ). Avoid it for non-massive 'particles' (quasiparticles in solid-state physics) without careful definition of effective mass and velocity. For very slow particles (v near 0), λ becomes very large and the particle's wavefunction extends over macroscopic distances — quantum many-body effects may dominate and a single-particle de Broglie wavelength is no longer the relevant scale.