de Broglie Wavelength Calculator
de Broglie Wavelength Calculator
λ = h/p from a speed, a kinetic energy, an accelerating voltage or a momentum — with the relativistic answer printed beside the school one, because at 100 kV the formula your textbook gives is 4.8 per cent wrong and says nothing about it.
de Broglie wavelength, from whichever quantity you were given
An electron accelerated through 100 volts
One division, and the three ways a question hands you the denominator
- p
- the momentum, and the only thing the wavelength depends on. Not the mass, not the speed, not the energy — their product in that one combination. Two particles of very different mass and speed with the same momentum have the same wavelength
- √(2mK)
- the non-relativistic momentum from a kinetic energy. Follows from K = p²/2m, which is the same statement as K = ½mv². This is the form a school question nearly always wants
- qV
- the kinetic energy an accelerated charge picks up crossing a potential difference. For an electron and a 100 V supply that is 100 eV exactly, by the definition of the electronvolt. For an alpha particle it is 2qV, because the charge is 2e — a factor of two that is dropped constantly
- 1.2264 nm
- h/√(2mee), the constant that makes the electron case a one-step calculation: λ = 1.2264/√V nanometres, or 12.264/√V ångströms. Most textbooks print 1.227. It is NOT exact, because the electron mass is measured — but its uncertainty is three parts in 1010, so the rounding in 1.227 dominates it by seven orders of magnitude
- hc/√(K²+2Kmc²)
- the exact form. Reduces to √(2mK) in the denominator when K is small against mc², which for an electron means small against 511 keV. At 100 eV the two differ by 0.005 per cent; at 100 keV, by 4.8 per cent
Worked example
An electron accelerated through 100 volts
THE KINETIC ENERGY, which needs no arithmetic. An electron crossing 100 V gains 100 eV — that is what the electronvolt is defined to mean. In joules, 100 × 1.602176634×10−19 = 1.602176634×10−17 J.
THE MOMENTUM. p = √(2mK) = √(2 × 9.1093837139×10−31 × 1.602176634×10−17) = 5.4022×10−24 kg m/s. Notice that the speed never appeared. You can get it — 5.93×106 m/s, about two per cent of c — but it is not needed and computing it is where an error gets in.
THE WAVELENGTH. λ = h/p = 6.62607015×10−34/5.4022×10−24 = 1.226426×10−10 m, which is 122.6426 pm, 0.1226 nm or 1.2264 ångströms. Compare it with the shortcut: 1.2264/√100 = 0.12264 nm. Same number, one step.
WHY THAT NUMBER MATTERS. Atoms in a solid sit a few hundred picometres apart, so a 100 eV electron has a wavelength of the same order as a crystal's lattice spacing — and therefore diffracts from it. Davisson and Germer's nickel-crystal experiment ran at 54 V, which this page gives as 166.8954 pm; the diffraction maximum they measured is what turned de Broglie's 1924 hypothesis into a measurement.
AND THE RELATIVISTIC CHECK. The exact formula gives 122.6366 pm, so the school answer is 0.005 per cent too long — utterly negligible. Change the voltage to 100,000 and the picture changes: non-relativistic 3.8783 pm, exact 3.7014 pm, an error of 4.8 per cent. A transmission electron microscope runs in exactly that range, so an electron-optics calculation that used the textbook formula would be wrong in the second figure.
THE CRICKET BALL, because every syllabus asks it. A 160 g ball at 40 m/s has p = 6.4 kg m/s and λ = 6.626×10−34/6.4 = 1.04×10−34 m. That is 10−22 of a picometre, twenty orders of magnitude smaller than a nucleus. Nothing can diffract it, so nothing can detect it. The wave nature of matter is not switched off for large objects; it is simply unobservable, and the smallness of h is the whole reason.
The electron, from a volt to a megavolt: where the school formula stops working
| Accelerating voltage | Kinetic energy | λ non-relativistic / pm | λ exact / pm | Non-relativistic error | What runs here |
|---|---|---|---|---|---|
| 1 V | 1 eV | 1226.4260 | 1226.4254 | 0.00005% | a thermal-ish electron; λ longer than an atom |
| 54 V | 54 eV | 166.8954 | 166.8910 | 0.003% | Davisson and Germer, nickel crystal, 1927 |
| 100 V | 100 eV | 122.6426 | 122.6366 | 0.005% | the standard textbook question |
| 1 kV | 1 keV | 38.7830 | 38.7640 | 0.049% | low-energy electron diffraction |
| 10 kV | 10 keV | 12.2643 | 12.2344 | 0.244% | a scanning electron microscope |
| 100 kV | 100 keV | 3.8783 | 3.7014 | 4.78% | a transmission electron microscope |
| 1 MV | 1 MeV | 1.2264 | 0.8719 | 40.7% | the formula has failed completely |
Four particles at 100 eV, and the mass dependence nobody expects
| Particle | Mass | λ at 100 eV of kinetic energy | λ as a multiple of the electron’s |
|---|---|---|---|
| electron | 0.000548580 u | 122.6426 pm | 1 |
| proton | 1.00727647 u | 2.8622 pm | 0.0233 |
| neutron | 1.00866492 u | 2.8602 pm | 0.0233 |
| alpha particle | 4.00150618 u | 1.4360 pm | 0.0117 |
| carbon-12 atom | 12 u | 0.8291 pm | 0.00676 |
λ = h/p is one division, so this page is about the denominator
The formula is trivial and the question never gives you p. That is the actual difficulty. A paper hands you a speed, or a kinetic energy in electronvolts, or an accelerating voltage, or occasionally a temperature, and the work is converting whichever one you got into a momentum. Each route has its own trap. From a speed, the trap is relativity. From an energy, it is the square root — K = p²/2m means p = √(2mK), and halving the energy divides the wavelength by √2 rather than by 2. From a voltage, it is the charge: an alpha particle carries 2e, so 100 volts gives it 200 eV, and a neutron carries nothing, so a voltage gives it no energy at all.
The wavelength depends on momentum alone. Not on mass and not on speed separately. This is worth saying because it makes two otherwise puzzling results obvious. A slow heavy particle and a fast light one with the same momentum have the same wavelength. And at a fixed kinetic energy, λ goes as 1/√m — so the proton, 1836 times the electron’s mass, has 1/42.8 of its wavelength at the same energy, not 1/1836.
Where this goes wrong. Two mistakes dominate. The first is √(2mK) with K in electronvolts: the electronvolt is not an SI unit, and 100 eV must become 1.602×10⁻¹⁷ J before it goes anywhere near a kilogram. The second is a mass in grams or in u left unconverted — a carbon atom is 12 u, which is 1.993×10⁻²⁶ kg, and using 12 gives a wavelength 10²⁶ times too small. Both produce answers that are obviously absurd once you look at the exponent, which is why this page prints the momentum and the kinetic energy as separate rows: if either of those is silly, the wavelength is wrong and you can see where.
The honest limit of the school formula. p = mv and p = √(2mK) are both approximations that hold while the kinetic energy is small compared with the rest energy — 511 keV for an electron, 938 MeV for a proton. Every question a class-12 or A-level paper asks is comfortably inside that. The moment anyone applies the same formula to the electron microscope those questions mention as the application, it breaks: at 100 kV the school answer is 4.8 per cent too long, and at 1 MV it is 41 per cent too long. This page therefore prints both, with the discrepancy as a percentage, rather than choosing one and hoping. If you are answering an exam question, quote the headline. If you are designing anything, quote the exact row.
Frequently asked questions
What is the de Broglie wavelength of an electron accelerated through V volts?
λ = 1.2264/√V nanometres, or equivalently 12.264/√V ångströms, provided V is small enough that relativity does not bite — which for school purposes means anything under a few kilovolts. At 100 V that is 0.12264 nm. The constant is h/√(2mee); most textbooks round it to 1.227, which is fine to four figures. It is not an exact number, because the electron mass is measured, but the measurement is far better than the rounding.
Why does a cricket ball not diffract?
It does, by an amount nothing could ever detect. A 160 g ball at 40 m/s has a momentum of 6.4 kg m/s and a de Broglie wavelength of 1.04×10−34 m. Diffraction needs an obstacle or a grating comparable with the wavelength, and there is nothing in the universe with a 10−34 m spacing — a nucleus is 10−15 m across, nineteen orders of magnitude bigger. The wave nature of matter is not suspended for large objects. It is simply too fine to resolve, and the reason is the smallness of h.
Do I use the relativistic formula in my exam?
Almost certainly not, and you should know why. The non-relativistic form λ = h/√(2mK) is what every school syllabus sets, and at the energies school questions use it is right to four or five significant figures: 0.005 per cent at 100 eV, 0.05 per cent at 1 keV. It crosses one per cent at about 21 kV for an electron. If a question quotes a voltage in the tens or hundreds of kilovolts — electron microscopy, X-ray tubes — the approximation has stopped being adequate and this page prints the exact answer beside it so you can see by how much.
How does the wavelength depend on mass at a fixed energy?
As 1/√m. From p = √(2mK), a particle four times heavier at the same kinetic energy has half the wavelength. That is why the proton’s wavelength at 100 eV is 2.86 pm rather than 0.067 pm: the mass ratio is 1836 but the wavelength ratio is √1836 = 42.8. At fixed SPEED, by contrast, λ goes as 1/m exactly, because then p = mv. Which dependence applies turns entirely on what the question held constant, and reading that off the question is most of the marks.
What experiment confirmed de Broglie’s hypothesis?
Davisson and Germer’s electron diffraction from a nickel crystal, published in 1927, three years after de Broglie’s thesis. They accelerated electrons through a few tens of volts and found a sharp maximum in the scattered intensity at a particular angle — a diffraction peak, which a stream of classical particles cannot produce. At their 54 V setting this page gives λ = 166.90 pm, the same order as the nickel lattice spacing, which is exactly the condition for diffraction. G. P. Thomson’s independent transmission experiment in the same year did the same thing with thin foils.
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References
- CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121, read 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt. The table prints the word (exact) against the Planck constant, the speed of light in vacuum and the elementary charge, which is the single most consequential thing on it for a student: h = 6.626 070 15×10−34 J Hz−1, c = 299 792 458 m/s and e = 1.602 176 634×10−19 C have no uncertainty to quote because the 2019 revision of the SI defines the kilogram, the metre and the ampere through them. The electron mass, 9.109 383 7139(28)×10−31 kg, is still measured and still carries an uncertainty; so does the atomic mass constant, 931.494 103 72(29) MeV.
- The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019 (2022 corrected printing). The document that made h, c and e exact, by fixing their numerical values and deriving the kilogram, the metre and the ampere from them rather than the other way round. Most school textbooks still print h with an uncertainty or with a “≈”, and a question that says “take h = 6.63×10−34 J s” is now rounding an exact number rather than quoting a measured one. That changes what a significant-figure argument about the answer means.
- L. de Broglie, Recherches sur la théorie des quanta, doctoral thesis, Université de Paris, 1924; published in Annales de Physique 10 (1925) 22. Cited for the hypothesis — that a material particle of momentum p has an associated wavelength h/p — and not for any number on this page, all of which are computed from CODATA constants.
- C. Davisson and L. H. Germer, “Diffraction of Electrons by a Crystal of Nickel”, Physical Review 30 (1927) 705–740. The experiment. This page prints only its accelerating voltage, 54 V, which is in the title matter of every account of it; the measured wavelength and diffraction angle are deliberately not reproduced here, because the paper itself could not be read for this build and a number quoted at third hand is not a number with a source.
- R. Tran, X.-G. Li, J. H. Montoya, D. Winston, K. A. Persson and S. P. Ong, “Anisotropic work function of elemental crystals”, Surface Science 687 (2019) 48–55. Not about de Broglie wavelengths, but cited here because the lattice spacings that make electron diffraction work are the same surface structures whose facets this paper resolves — the reason a diffraction pattern depends on which face of the crystal the beam meets.
- NCERT, Physics, Textbook for Class XII, Part II — the chapter on the dual nature of radiation and matter, which is where the work-function table most Indian students are working from is printed. This could not be read for this page. The official PDF at ncert.nic.in is served under a robots policy that refuses automated retrieval, and every copy a search returns is a coaching-site mirror with no provenance. So no number on this page is attributed to NCERT, and where this page and a school table differ the difference is stated against sources that could be read. That is a weaker claim than “the textbook is wrong” and it is deliberately the weaker one.