De Broglie Wavelength Calculator
Calculate a particle's de Broglie wavelength from its mass and velocity, using Planck's constant of 6.626 x 10^-34 J s.
De Broglie Wavelength Calculator
Calculator
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Formula: λ = h/p = h/(mv)
Worked example — 0.727 nm
Formula
λ = h/p = h/(mv)
Every particle has a wave nature. h = 6.626×10⁻³⁴ J·s.
Worked Examples
Example 1: Electron
Problem:mₑ=9.109×10⁻³¹ kg, v=10⁶ m/s
Solution:λ=6.626e-34/(9.109e-31×10⁶)=7.27×10⁻¹⁰ m
Result:0.727 nm
Frequently Asked Questions
What is the de Broglie wavelength?
The de Broglie wavelength is the quantum mechanical wave associated with any moving particle. It is given by λ = h/p = h/(mv), where h = 6.626×10⁻³⁴ J·s is Planck's constant, m is mass in kg, and v is velocity in m/s. This wave-particle duality was proposed by Louis de Broglie in 1924.
How do you calculate de Broglie wavelength?
Use λ = h/(mv). For an electron (m = 9.109×10⁻³¹ kg) moving at 10⁶ m/s: p = 9.109×10⁻³¹ × 10⁶ = 9.109×10⁻²⁵ kg·m/s, and λ = 6.626×10⁻³⁴ / 9.109×10⁻²⁵ ≈ 7.27×10⁻¹⁰ m = 0.727 nm, comparable to atomic spacings.
What units does the de Broglie calculator use?
Mass is entered in kilograms (kg) and velocity in meters per second (m/s). The output wavelength is in meters (m) and also converted to nanometers (nm). Momentum p = mv is shown in kg·m/s, and kinetic energy KE = ½mv² is shown in joules (J) and electron-volts (eV), where 1 eV = 1.602×10⁻¹⁹ J.
Can we observe de Broglie wavelength of everyday objects?
No — a 0.1 kg tennis ball at 50 m/s has λ = 6.626×10⁻³⁴ / (0.1 × 50) ≈ 1.3×10⁻³⁴ m, about 10¹⁹ times smaller than a proton. Wave effects become physically measurable only for particles lighter than about 10⁻²⁵ kg, such as electrons, protons, and neutrons.
How is the de Broglie wavelength related to kinetic energy?
For non-relativistic particles the momentum is the square root of twice the mass times the kinetic energy, so the wavelength is Planck's constant divided by that quantity. For an electron accelerated through a potential difference of V volts, the kinetic energy is eV and the wavelength comes to about 1.226 nanometres divided by the square root of V, giving roughly 0.123 nm at 100 V.
What experiments confirmed that matter behaves as a wave?
Davisson and Germer scattered low-energy electrons from a nickel crystal in 1927 and found intensity peaks at angles predicted by the Bragg condition applied to the de Broglie wavelength. G. P. Thomson independently observed ring patterns from electrons passing through thin metal foils. Later work extended matter-wave diffraction to neutrons, atoms and even large molecules, all following the same relation.
Why does an electron microscope resolve far finer detail than a light microscope?
Resolving power is limited by the wavelength of the probing wave. Visible light spans roughly 400 to 700 nanometres, whereas electrons accelerated through tens or hundreds of kilovolts have de Broglie wavelengths in the picometre range, thousands of times shorter. In practice lens aberrations, not the wavelength, set the achievable resolution, but the wavelength is what makes atomic-scale imaging possible at all.
How does the de Broglie relation connect to the uncertainty principle?
A single wavelength corresponds to a perfectly definite momentum but a wave that extends everywhere, so the position is completely unknown. Localising a particle requires superposing a spread of wavelengths, which spreads the momentum. That trade-off is quantified by the Heisenberg relation, in which the product of the position and momentum uncertainties cannot be smaller than a constant of order Planck's constant divided by two pi.
Do I need relativistic corrections at high particle speeds?
Yes, once the speed becomes an appreciable fraction of the speed of light. The correct momentum is the Lorentz factor times mass times velocity, so the wavelength is Planck's constant divided by that. At one tenth of the speed of light the Lorentz factor is about 1.005 and the non-relativistic formula overestimates the wavelength by roughly half a per cent; the error grows rapidly beyond that.
How does the de Broglie wavelength explain quantised electron orbits?
Treat the electron as a standing wave around the nucleus. A stable orbit is one whose circumference holds a whole number of wavelengths, so 2 pi r equals n lambda. Substituting lambda equal to h divided by mv rearranges to angular momentum equal to n times h divided by 2 pi, which is exactly the quantisation condition Bohr had earlier assumed without justification.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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