Heisenberg Uncertainty Principle Calculator

Pin a particle down to a width you choose and this gives the smallest spread of momentum, speed and energy it can then have — in whichever of the three conventions your board writes, ħ/2, ħ or h. It also says plainly what the principle does not mean: it is not a statement about clumsy apparatus.

Last updated 11 Nov 2026

Confine a particle to Δx — what is the least it can be uncertain by?

Δx → minimum Δp, Δv and confinement energy, plus the ΔE·Δt form

ParticleMasses are CODATA 2022. The mass does not change Δp at all — that depends only on Δx and the constant you pick below. It changes the speed uncertainty Δv = Δp/m and the confinement energy, both of which fall as the mass rises. That is the whole reason a cricket ball has no quantum blur worth drawing and an electron has nothing else.

Mass, if you are typing oneIn unified atomic mass units, 1 u = 1.66053906892×10⁻²⁷ kg. The default here is 10²⁶ u, which is 166 grams, about a cricket ball. Type it and watch Δv collapse to 3×10⁻²⁴ m/s. Nothing in the maths switched off; the answer simply became unmeasurable.

Position spread ΔxIn nanometres. 0.1 nm is one ångström, roughly the size of an atom, and is the standard textbook case. 1×10⁻⁶ nm is one femtometre, about the width of a nucleus — try it and the answer comes back faster than light, which is the formula telling you honestly that it has left its own range of validity.

Which form of the inequalityNot three different physics — three different definitions of the word “uncertainty”. ħ/2 is the exact bound when both spreads are standard deviations; the looser forms come from rougher definitions and are right for an estimate. They differ by a factor of 12.6, so quote the one your board sets and say which you used.

Lifetime Δt of a state, for the energy–time formIn nanoseconds, and separate from Δx above — this field drives the ΔE·Δt rows only. Excited atomic states typically live for nanoseconds, so 10 ns is a fair default; a metastable one can live for seconds. The shorter the life, the fuzzier the energy and the broader the line.

Your position spread and the momentum spread it forces. Widen one and the other narrows in front of you — that is the whole content of the inequality. Both widths are drawn on a logarithmic scale so that a factor of a thousand still fits on a phone, so read the shapes and the printed numbers, not the pixels.

578,838.18m/sExample

An electron confined to Δx = 0.1 nm — about the size of an atom — with the exact ħ/2 bound, and a state lifetime of 10 ns

Two inequalities, one constant, and the step that is not an inequality at all

Δx · Δp ≥ ħ/2 · ΔE · Δt ≥ ħ/2 · ħ = h/2π = 1.054 571 817×10⁻³⁴ J s · Δv = Δp/m · E_conf = Δp²/2m

Δx, Δp: standard deviations of position and of momentum, taken over many identically prepared particles. Not error bars on one measurement, and not the resolution of an instrument. Change the definition of “spread” and the constant on the right changes with it, which is why three versions of this inequality are in circulation
ħ/2: half the reduced Planck constant, 5.272 859 09×10⁻³⁵ J s. Exact, because ħ is h/2π and h is exact in the SI. This is the bound Kennard proved in 1927 and the one that is a theorem rather than an estimate
≥: greater than or equal to. The equality is reached by one shape of state only — a Gaussian — so a real particle is usually fuzzier than this page says. Everything here is a floor
Δv = Δp/m: the only step on this page that is not quantum mechanics. It is ordinary Newtonian momentum, and it is the step that breaks first: when the answer approaches c it is this line, not the inequality, that has failed
Δp²/2m: the kinetic energy that a momentum spread of Δp implies, called the confinement or zero-point energy. Compare it with a binding energy to see whether a particle can be held in a box that size at all. For an electron in 0.1 nm it is 0.95 eV, the right order for atomic physics
ΔE·Δt: the energy–time form. Δt is the lifetime of a state, not a reading on a clock, because time is not an observable the way position is. A state living 10 ns cannot have an energy sharper than 33 neV, which is why spectral lines have a natural width

Worked example

An electron confined to Δx = 0.1 nm — about the size of an atom — with the exact ħ/2 bound, and a state lifetime of 10 ns
THE CONSTANT. ħ = h/2π = 6.626 070 15×10⁻³⁴/6.283 185 31 = 1.054 571 817×10⁻³⁴ J s, so ħ/2 = 5.272 859 09×10⁻³⁵ J s. Both are exact, because h is exact in the SI.
THE MOMENTUM SPREAD. Δp = ħ/(2Δx) = 5.272 859 09×10⁻³⁵ ÷ 1.0×10⁻¹⁰ = 5.2729×10⁻²⁵ kg m/s. Notice what did not appear: the mass. Δp depends on the size of the box and nothing else.
THE SPEED SPREAD. Δv = Δp/m = 5.2729×10⁻²⁵ ÷ 9.1093837139×10⁻³¹ = 578,838 m/s, which is 0.193 per cent of the speed of light. Here the mass does enter, and it is why the same calculation for a proton in the same box gives only 315 m/s.
THE CONFINEMENT ENERGY. Δp²/2m = (5.2729×10⁻²⁵)² ÷ (2 × 9.1094×10⁻³¹) = 1.526×10⁻¹⁹ J = 0.9525 eV. Compare that with hydrogen’s 13.6 eV binding energy: the same order of magnitude, from one inequality and no atomic physics at all. That agreement is the reason anybody believes this.
THE ENERGY–TIME FORM, which uses the other input. ΔE ≥ ħ/(2Δt) = 5.272 859×10⁻³⁵ ÷ 1.0×10⁻⁸ = 5.273×10⁻²⁷ J = 32.91 neV. Divided by h that is a natural linewidth of 7.958 MHz. A state that does not last forever cannot have a sharp energy, and that is measurable: it is the floor on how narrow a spectral line can be.
WHAT THIS DOES NOT SAY. It does not say the electron is somewhere definite and we are too clumsy to find out. It does not say a measurement of position kicks the momentum by 5.3×10⁻²⁵ kg m/s. It says that if you prepare many electrons the same way and confine them to 0.1 nm, the spread of momenta you will find cannot be narrower than that — however good the apparatus.
AND THE LIMIT OF THE FORMULA. Change Δx to 1×10⁻⁶ nm, one femtometre, and the page returns 5.8×10¹⁰ m/s — 193 times the speed of light. The inequality has not broken; Δv = Δp/m has, and the page bands that case in red rather than printing it quietly.

The three forms in circulation, all for the same electron in the same 0.1 nm box

Computed on this page’s own expressions. The factor of 12.57 between the top and bottom rows is not a disagreement about physics: it is three different definitions of the word “uncertainty”. Only the first row uses standard deviations and only the first row is a theorem. If a question says “estimate”, the third is what it wants; if it asks for the minimum, the first is. Boards differ, so state which form you used.
Written as	Constant on the right	Δp	Δv	Confinement energy	Where you meet it
Δx·Δp ≥ ħ/2	5.2729×10⁻³⁵ J s	986.63 eV/c	578,838 m/s	0.9525 eV	the exact bound, proved by Kennard in 1927; standard in university quantum mechanics
Δx·Δp ≥ ħ	1.0546×10⁻³⁴ J s	1,973.27 eV/c	1,157,676 m/s	3.8100 eV	many school and first-year texts, usually written h/2π; a looser but always-true bound
Δx·Δp ≈ h	6.6261×10⁻³⁴ J s	12,398.42 eV/c	7,273,895 m/s	150.41 eV	Heisenberg’s own 1927 estimate from the gamma-ray microscope; an order of magnitude, not a theorem

An electron, from a nucleus to a millimetre — and where Δv = Δp/m gives up

The first two rows are printed to show you a failure, not a result: Δv larger than c is impossible, because Δv = Δp/m assumes p = mv. The bound on Δp itself is fine in every row. Read the last column instead. Confining an electron to an atom costs about an electronvolt, which an atom can pay; confining it to a nucleus costs 9.5 GeV, which nothing in an atom can — and that is the principle’s own answer to why electrons are not found inside nuclei.
Δx	What is that size	Δp	Δv	Δv/c	Confinement energy
1 fm	across a nucleus	9.87×10⁷ eV/c	5.79×10¹⁰ m/s	193	9.52×10⁹ eV
0.1 pm	a hard X-ray wavelength	9.87×10⁵ eV/c	5.79×10⁸ m/s	1.93	9.52×10⁵ eV
10 pm	a tenth of an atom	9,866 eV/c	5.79×10⁶ m/s	0.0193	95.25 eV
0.1 nm	an atom	986.6 eV/c	578,838 m/s	0.00193	0.9525 eV
1 nm	a small quantum dot	98.66 eV/c	57,884 m/s	0.000193	0.009525 eV
1 µm	a bacterium	0.0987 eV/c	57.88 m/s	1.93×10⁻⁷	9.52×10⁻⁹ eV
1 mm	a pencil line	9.87×10⁻⁵ eV/c	0.0579 m/s	1.93×10⁻¹⁰	9.52×10⁻¹⁵ eV

What the principle says, and the four things it is constantly taken to mean instead

The inequality is about preparation, not about clumsiness. Take a very large number of particles, prepare every one of them in the same way, and measure the position of half of them and the momentum of the other half. You get two distributions. Their standard deviations — call them Δx and Δp — satisfy Δx·Δp ≥ ħ/2, and no improvement in apparatus changes that. The spread was there in the state before anything was measured. That is the version Kennard proved as a theorem in 1927, and it is what this page computes.

Heisenberg’s own argument was a different one. He imagined locating an electron with a microscope: the shorter the wavelength the better the position and the harder the photon kicks the electron — a Compton-scattering argument. From that he got δp·δq ∼ h. Fine physical reasoning, and not the same statement: it is about disturbance by measurement, it uses no definition of “uncertainty” at all, and Bohr objected to it at the time. Whether a disturbance relation of that kind holds in general is still argued about. The preparation relation is not.

So four readings to avoid. It is not “the observer affects the system”. It is not “we cannot measure both accurately” — you can measure either as accurately as you like, just not on the same particle. It is not a limit set by current technology, so a better microscope does not help. And it is not a claim that the particle has a definite position and momentum merely hidden from us; quantum mechanics does not supply those values for the inequality to be about.

Where this goes wrong in exams. Four places. The first is units: Δx in nanometres or ångströms must become metres before it meets ħ, and a stray factor of 10⁹ is the commonest single error on this topic. The second is h for ħ, a factor of 6.28, or ħ for ħ/2, a factor of 2 — which is why this page makes you choose. The third is dividing by the mass in grams or in u instead of kilograms. The fourth is quoting a speed uncertainty larger than c and not noticing; the page bands that case in red rather than letting it pass.

And one thing it genuinely explains. Why the electron does not fall into the nucleus. Squeeze it closer and the Coulomb energy drops as 1/r, but the confinement energy rises as 1/r² and wins. Put r = 0.1 nm into this page and you get 0.95 eV; the balance between that rise and the Coulomb fall is what fixes the size of an atom, and it lands within a factor of a few of the real Bohr radius. One inequality, no orbits, and the right answer to the order of magnitude.

Frequently asked questions

Does the uncertainty principle mean the measurement disturbs the particle?

No — or at least, that is a different statement and a much more contested one. The inequality this page computes is about the spreads of the probability distributions belonging to a prepared state. It holds before you measure anything, and it would hold for a perfect, non-disturbing instrument. Heisenberg’s 1927 paper did argue from disturbance, using a thought experiment about scattering light off an electron, and that argument is where the “clumsy apparatus” picture comes from. The Stanford Encyclopedia of Philosophy entry cited below sets out why the two are not interchangeable and notes that the disturbance version is still disputed in the literature. For an exam, the safe sentence is: it limits how sharply both quantities can be defined for a state, not how carefully they can be measured.

Should I use h/4π or h/2π in my answer?

Use whichever your syllabus prints, and say so. h/4π is ħ/2, the exact bound, and is what you want if a question asks for the minimum. h/2π is ħ, a looser bound that is also always true and is what many school texts write. Heisenberg’s own h is an order-of-magnitude estimate and is fine when a question says “estimate”. The three differ by a factor of 12.57 overall, so a numerical answer that does not name its convention cannot be marked. This page lets you switch between them so you can see exactly what changes.

Why can a confined particle never have zero energy?

Because zero kinetic energy means zero momentum with no spread at all, Δp = 0, and the inequality then demands Δx be infinite — the particle could be anywhere in the universe. If it is held inside a box of width Δx it must carry a momentum spread of at least ħ/2Δx, and a spread of momenta means a non-zero average of p², which means energy. That floor is the zero-point energy, and this page prints it as the confinement energy. It is not a convention or a correction; it is forced. The particle in a box page works the same idea out exactly for the simplest case and gets h²/8mL², which is the same order as Δp²/2m here.

What is Δv for a cricket ball?

Choose “something else” and leave the mass at its default of 10²⁶ u, which is 166 grams, and confine it to 0.1 nm: Δv comes out at 3.2×10⁻²⁴ m/s. At that rate the ball “drifts” by one micrometre, the width of a bacterium, in about ten billion years — comparable with the age of the universe. Nothing in the formula has changed: the same Δp of 5.3×10⁻²⁵ kg m/s is divided by a mass nearly 2×10²⁹ times larger. The quantum limit on a cricket ball is real and utterly unmeasurable, and the smallness of ħ is the whole reason.

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References

CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121. Read on 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt. Every constant on this page comes from that file. It marks 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 as (exact), because the 2019 SI defines the kilogram, the metre and the ampere through them; the masses are measured, the electron being 9.109 383 7139(28)×10⁻³¹ kg.
Jan Hilgevoord and Jos Uffink, “The Uncertainty Principle”, Stanford Encyclopedia of Philosophy, first published 8 October 2001, substantive revision 12 July 2016; read 11 November 2026 at plato.stanford.edu/entries/qt-uncertainty/. The source for this page’s central distinction. It states that “the first mathematically exact formulation of the uncertainty relations is due to Kennard”, with the uncertainties as standard deviations, and that Kennard’s result “is a theorem of the formalism, not an empirical principle like Heisenberg’s”. It also notes that Heisenberg “never relied on standard deviations” and that the measurement-disturbance question remains unresolved. Heisenberg’s and Kennard’s 1927 papers are cited here at second hand through this entry; neither original was read, and no number here comes from either.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019 (the copy read on 11 November 2026 is marked V4.01, June 2026). §2.2 fixes the numerical values of h, c and e, which is why those three carry no uncertainty here while every mass does. A school question that says “take h = 6.63×10⁻³⁴ J s” is now rounding an exact number rather than quoting a measured one.
NCERT, Physics, Textbook for Class XII, Part II — the syllabus most Indian readers of this page are working to. This could not be read for this build: the chapter PDFs at ncert.nic.in covering these topics refused automated retrieval, and every other copy a search returns is an unattributed mirror. Nothing here is attributed to NCERT, and where this page and a school table differ the difference is stated against sources that could be read — a weaker claim than “the textbook is wrong”, and deliberately the weaker one.