Heisenberg Uncertainty Principle Calculator
Apply the Heisenberg uncertainty principle, Δx·Δp ≥ ħ/2: from a position uncertainty find the minimum momentum and velocity uncertainty of an electron, proton, neutron or any particle, and the matching kinetic energy.
How it works
Δx · Δp ≥ ħ ÷ 2, with ħ = h ÷ 2π = 1.054571817 × 10⁻³⁴ J·s · Δp_min = ħ ÷ (2Δx) · Δv_min = Δp_min ÷ m · E ≈ Δp² ÷ (2m)
Heisenberg’s uncertainty principle (1927, in the exact form of Kennard) says that the standard deviations of position and momentum of any quantum particle cannot both be small: their product is at least ħ/2. It is a property of waves, not a limit of instruments. Confining a light particle such as an electron to an atom-sized region forces a velocity spread of hundreds of kilometres per second and a kinetic energy of the order of an electronvolt, which is why atoms have a minimum size and energy.
Worked example
An electron confined to Δx = 0.1 nm (1 × 10⁻¹⁰ m) has Δp ≥ 1.0546 × 10⁻³⁴ ÷ (2 × 10⁻¹⁰) = 5.273 × 10⁻²⁵ kg·m/s. Dividing by its mass, 9.109 × 10⁻³¹ kg, the velocity is uncertain by at least 578,838 m/s, and a momentum of that size means a kinetic energy of about 0.9525 eV.
Sources: NIST CODATA — Fundamental physical constants · HyperPhysics — Uncertainty principle
Frequently asked questions
What does the Heisenberg uncertainty principle say?
The more precisely a particle’s position is pinned down, the less precisely its momentum can be known, and vice versa: Δx·Δp ≥ ħ/2. The same kind of relation links energy and time, ΔE·Δt ≳ ħ/2.
Should I use h/4π or h/2?
The rigorous form for standard deviations is Δx·Δp ≥ ħ/2 = h/4π, which this calculator uses. Some textbooks use Δx·Δp ≈ h or ≥ ħ as a rough estimate, which gives answers 2 to 4π times larger.
Why does the uncertainty principle not matter for everyday objects?
ħ is tiny. A 0.1 kg ball located to within a micrometre has a velocity uncertainty of only about 5 × 10⁻²⁸ m/s — far too small to ever notice.
Why can an electron not sit in the nucleus?
Confining an electron to a nucleus-sized 1 fm region requires a momentum spread whose energy is tens of MeV, far more than the electric attraction can hold, so electrons occupy atom-sized orbitals instead.
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This calculator is for educational purposes. Double-check important results before acting on them.