de Broglie Waves and Heisenberg Uncertainty
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Direct answer
Every moving particle carries an associated matter wave of wavelength lambda = h/p = h/(mv), about 12.27/sqrt(V) angstrom for an electron accelerated through V volts; a cricket ball's wavelength near 10^-34 m is why buses refuse to diffract. Davisson and Germer confirmed the wave in 1927 by diffracting 54 V electrons from a nickel crystal, observing a peak at 50 degrees matching the predicted 1.67 angstrom. Heisenberg's uncertainty principle caps simultaneous knowledge: delta-x × delta-p at least h/(4 pi), and delta-E × delta-t at least h/(4 pi) — confining an electron to a nucleus-sized box would demand kinetic energies far beyond beta-decay energies. Bohr's quantisation falls out as a standing-wave condition: 2 pi r = n lambda.
What you must remember
- de Broglie relation: lambda = h/p; for an electron through potential difference V, lambda = h/sqrt(2 m e V) = 12.27/sqrt(V) angstrom (V in volts); at 100 V the electron's wavelength is 1.227 angstrom.
- Scale argument: Planck's constant (6.63 × 10^-34 J s) is so tiny that macroscopic wavelengths are unmeasurably small — quantum effects are a small-mass, small-length phenomenon.
- Davisson-Germer numbers: 54 V electrons, nickel crystal, strong diffraction peak at 50 degrees scattering angle; measured wavelength 1.65-1.67 angstrom against de Broglie's prediction — the first matter-wave confirmation.
- Uncertainty relations: delta-x delta-p at least h/(4 pi); delta-E delta-t at least h/(4 pi); the bound is on the product, so sharpening position necessarily blurs momentum.
- Bohr quantisation as waves: 2 pi r = n lambda makes m v r = n h/(2 pi) — the angular momentum postulate becomes a standing-wave requirement on the orbit's circumference.
- Estimation power: electron confined to a nucleus (delta-x ~ 10^-14 m) implies kinetic energy of order tens of MeV, incompatible with the few-MeV electrons nuclear physics observes — the standard "why electrons are not nuclear constituents" argument.
- Mass dependence: heavier particle, shorter wavelength at the same speed; an alpha particle's wavelength is far below an electron's at comparable energies.
- Pattern note: Main computes wavelengths and applies h/4 pi directly; Advanced builds estimation arguments and connects uncertainty to ground-state energies.
One formula, two doors
Compute the electron's wavelength at 54 V first as an instrument: lambda = 12.27/sqrt(54) = 1.67 angstrom. Nickel's crystal planes (d = 0.91 angstrom for the principal reflection) then satisfy Bragg's law 2 d sin(theta) = lambda with theta the glancing angle — sin(theta) = 1.67/1.82 ≈ 0.92, consistent with the observed 50-degree scattering peak. The deeper door is estimation. Squeeze an electron into delta-x = 10^-14 m and delta-p of order h/(4 pi delta-x) implies kinetic energies in the tens-of-MeV range; nuclear beta decay ejects electrons of at most a few MeV, so free-residing nuclear electrons are ruled out — the uncertainty principle used as an exclusion argument, which is exactly how Advanced phrases it.
Between those doors sits the everyday scale check: a 150 g cricket ball at 40 m/s carries lambda near 10^-34 m, a million million times below a nucleus — quantum mechanics politely exits the stadium.
Where students slip
The 12.27 formula is for electrons only — applying it to a proton or alpha particle (different mass and charge) gives wavelengths off by factors of 40-plus; recompute h/sqrt(2 m q V) from scratch for each particle. Second, the uncertainty bound is h/(4 pi), not h; questions set the trap of both options, and h alone is 2 pi times too generous. Third, uncertainty is not a measurement deficiency — it is nature's floor on simultaneous definiteness of conjugate variables, and "better instruments will remove it" is a permanently wrong statement. Fourth, in the Bohr standing-wave derivation, the circumference fits whole wavelengths (2 pi r = n lambda), not the diameter — using pi r = n lambda halves the angular momentum and breaks the correspondence with n h/2pi. And wavelength falls as momentum rises: fast electrons diffract less.
Frequently asked questions
What is the de Broglie wavelength of an electron accelerated through 100 V?
lambda = 12.27/sqrt(100) = 1.227 angstrom, from lambda = h/sqrt(2 m e V) — the working formula for all electron-wavelength numericals.
What did the Davisson-Germer experiment establish?
Electrons of 54 V diffracted from a nickel crystal with a 50-degree peak matching lambda = 1.67 angstrom, confirming de Broglie's matter waves experimentally.
What does Heisenberg's uncertainty principle state?
delta-x delta-p is at least h/(4 pi): position and momentum cannot both be known exactly at once; similarly delta-E delta-t is at least h/(4 pi).
How does de Broglie's idea explain Bohr's angular momentum postulate?
Requiring a whole number of matter waves around the orbit, 2 pi r = n lambda, immediately yields m v r = n h/(2 pi) — quantisation becomes geometry.
Can the uncertainty principle be beaten with perfect instruments?
No — it is not experimental clumsiness but an intrinsic property of conjugate variables; nature itself does not assign exact simultaneous values to position and momentum.