Heisenberg's Uncertainty Principle

If an electron behaves like a wave, can you still speak of its exact position and exact speed at one instant, the way you can for a cricket ball? Werner Heisenberg's answer, in 1927, was no — and that single answer is what finally broke Bohr's picture of the electron circling the nucleus in a fixed, precisely defined orbit. The uncertainty principle says position and momentum cannot both be pinned down with unlimited precision; the more sharply you know one, the fuzzier the other must become. This is not a flaw in our instruments — it is a fundamental property of nature at the atomic scale. For a cricket ball the effect is too small to ever notice, but for an electron it is large enough to matter, and it is the reason chemistry replaced "the electron is here, moving like this" with "the electron is probably here" — the idea of an orbital.

The core insight

Position and momentum of a microscopic particle cannot both be known exactly at the same time, because Δx Δp≥h/4π\Delta x\,\Delta p \geq h/4\pi, and this rules out Bohr's fixed circular orbit for the electron.

The core idea

The statement. It is impossible to simultaneously determine, with unlimited precision, both the exact position and the exact momentum (or velocity) of a microscopic particle such as an electron. Improving the precision of one measurement necessarily worsens the precision of the other — the trade-off is built into nature, not into the quality of the apparatus.

The inequality. If Δx\Delta x is the uncertainty in position and Δp\Delta p is the uncertainty in momentum, Heisenberg's principle states:

Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

where: Δx\Delta x is the position uncertainty (m), Δp\Delta p is the momentum uncertainty (kg m/s), and hh is Planck's constant (6.626×10−34 J s6.626\times10^{-34}\ \text{J s}). Since momentum p=mvp = mv, and mass mm is fixed for a given particle, the inequality is equally often written in terms of velocity:

Δx⋅Δv≥h4πm\Delta x \cdot \Delta v \geq \frac{h}{4\pi m}

where: Δv\Delta v is the velocity uncertainty (m/s) and mm is the particle's mass (kg). Both forms say the same thing: the product of the two uncertainties has a fixed minimum, so narrowing one uncertainty forces the other to widen.

Why mass decides who feels it. Because hh is so tiny (order 10−34 J s10^{-34}\ \text{J s}), the minimum uncertainty product h/4πh/4\pi is negligible next to the mass and everyday precision of macroscopic objects — a cricket ball's position and velocity can both be known to exquisite accuracy without ever bumping into this limit. An electron's mass, only about 9.1×10−31 kg9.1\times10^{-31}\ \text{kg}, sits in the denominator of Δv=h/(4πm Δx)\Delta v = h/(4\pi m\,\Delta x), so the same small hh now forces a large, physically significant Δv\Delta v for any reasonably small Δx\Delta x.

Why Bohr's orbit falls. A "trajectory" — a definite path — needs a simultaneously well-defined position and well-defined velocity at every instant, exactly what the uncertainty principle forbids for an electron confined to atomic dimensions. Bohr's model assumed the electron moves in a fixed circular orbit of known radius and known speed; the uncertainty principle shows this picture cannot be physically correct. What survives is a description of the electron's whereabouts only in terms of probability — the basis of the quantum mechanical model and the idea of an orbital, taken up next in this unit.

Figure
Bar chart comparing the minimum velocity uncertainty for an electron and a cricket ball for the same position uncertainty, electron bar far taller than cricket ball bar
Same principle, very different consequence minimum velocity uncertainty from delta x times delta v is greater than or equal to h over 4 pi m electron delta v approx 5.8 times 10 to the 5 m per s delta x = 1.0 times 10 to the minus 10 m cricket ball delta v approx 3.5 times 10 to the minus 28 m per s delta x = 1.0 times 10 to the minus 6 m, mass 0.15 kg velocity uncertainty (not to scale)
Derivation AMinimum velocity uncertainty from position uncertainty

The goal is to start from Heisenberg's inequality Δx Δp≥h/4π\Delta x\,\Delta p \geq h/4\pi, rewrite it in terms of velocity using p=mvp = mv, and solve for the smallest possible Δv\Delta v once Δx\Delta x and the particle's mass mm are fixed, using nothing beyond algebraic rearrangement of the given inequality.

Step 1. Start from the momentum form of the inequality, Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}. Since momentum is p=mvp = mv, the uncertainty in momentum for a particle of essentially fixed mass is Δp=m Δv\Delta p = m\,\Delta v.

Step 2. Substitute Δp=m Δv\Delta p = m\,\Delta v into the inequality: Δx⋅m Δv≥h4π\Delta x \cdot m\,\Delta v \geq \dfrac{h}{4\pi}.

Step 3. Divide both sides by m Δxm\,\Delta x to isolate Δv\Delta v: Δv≥h4πm Δx\Delta v \geq \dfrac{h}{4\pi m\,\Delta x}. The minimum uncertainty in velocity, the smallest value consistent with the principle, is obtained by taking the equality:

Δvmin⁡=h4πm Δx\Delta v_{\min} = \frac{h}{4\pi m\,\Delta x}

Step 4. Read the result. Δvmin⁡\Delta v_{\min} grows as Δx\Delta x shrinks (pinning down position more tightly forces velocity to be less certain) and shrinks as mm grows (heavier particles feel the effect far less). This one formula is what both worked examples below plug numbers into.

Worked example

Question. The uncertainty in the position of an electron is Δx=1.0×10−10 m\Delta x = 1.0\times10^{-10}\ \text{m} (about the size of an atom). Calculate the minimum uncertainty in its velocity, given me=9.1×10−31 kgm_e = 9.1\times10^{-31}\ \text{kg} and h=6.626×10−34 J sh = 6.626\times10^{-34}\ \text{J s}.

Step 1. Write down the formula derived above: Δvmin⁡=h4πm Δx\Delta v_{\min} = \dfrac{h}{4\pi m\,\Delta x}.

Step 2. Evaluate the denominator first: 4πm Δx=4×3.1416×(9.1×10−31)×(1.0×10−10)≈1.144×10−394\pi m\,\Delta x = 4 \times 3.1416 \times (9.1\times10^{-31}) \times (1.0\times10^{-10}) \approx 1.144\times10^{-39}.

Step 3. Divide: Δvmin⁡=6.626×10−341.144×10−39≈5.79×105 m s−1\Delta v_{\min} = \dfrac{6.626\times10^{-34}}{1.144\times10^{-39}} \approx 5.79\times10^{5}\ \text{m s}^{-1}.

The minimum uncertainty in velocity is about 5.8×105 m s−15.8\times10^{5}\ \text{m s}^{-1} — comparable to typical electron speeds inside an atom, so the uncertainty is not a rounding error; it is as large as the quantity itself, confirming the electron cannot be said to follow a fixed orbit.

Competitive stretch

For a 0.15 kg0.15\ \text{kg} cricket ball with the same style of calculation but Δx=1.0×10−6 m\Delta x = 1.0\times10^{-6}\ \text{m} (one micrometre, an extraordinarily precise position measurement): 4πm Δx=4×3.1416×0.15×1.0×10−6≈1.885×10−64\pi m\,\Delta x = 4\times3.1416\times0.15\times1.0\times10^{-6} \approx 1.885\times10^{-6}, so Δvmin⁡=6.626×10−34/1.885×10−6≈3.5×10−28 m s−1\Delta v_{\min} = 6.626\times10^{-34}/1.885\times10^{-6} \approx 3.5\times10^{-28}\ \text{m s}^{-1}. That is unmeasurably, physically irrelevantly small — exactly why the uncertainty principle never shows up in the mechanics of everyday objects, only for particles as light as the electron.

Common traps

  • Treating the uncertainty principle as a statement about clumsy instruments — it is a fundamental limit of nature, not something better technology could someday remove.
  • Expecting a noticeable effect for macroscopic objects — as the cricket-ball case shows, Δvmin⁡\Delta v_{\min} is absurdly small unless the mass is as tiny as an electron's.
  • Dropping the factor of 4π4\pi, or writing the inequality the wrong way round — it is Δx Δp≥h/4π\Delta x\,\Delta p \geq h/4\pi, a lower bound on the product, never an upper bound.
  • Concluding the electron "does not exist" at any instant — the principle only says we cannot simultaneously know both its exact position and exact momentum; this is why the fixed-orbit picture gives way to a probability-based description.

Ask the tutor: want to see how the 4π4\pi and the mass together decide whether an uncertainty is chemically significant or utterly negligible — ask the AI tutor to walk through the algebra with a mass and Δx\Delta x of your choosing.

Does the uncertainty principle mean the electron does not really have a fixed position at any given moment?

Not quite. The electron does have some definite state of affairs at each instant, but the principle says we cannot simultaneously measure or know both its exact position and exact momentum — trying to pin one down more precisely necessarily blurs the other. That is a limit on knowledge and on what "a definite trajectory" can even mean at atomic scale, not a claim that the electron is smeared out physically. This is exactly why chemists describe the electron's location using probability rather than a traced path.

Why does the uncertainty principle matter for electrons but not for a thrown ball or a moving car?

Because Δvmin⁡=h/(4πm Δx)\Delta v_{\min} = h/(4\pi m\,\Delta x) has mass in the denominator, and hh itself is extremely small (about 10−34 J s10^{-34}\ \text{J s}). For a cricket ball of mass 0.15 kg0.15\ \text{kg}, even a micrometre-level position precision gives a velocity uncertainty of only about 3.5×10−28 m s−13.5\times10^{-28}\ \text{m s}^{-1} — utterly unmeasurable. An electron's mass is about 10−31 kg10^{-31}\ \text{kg}, so the same hh divided by a far smaller mass gives a velocity uncertainty of order 105 m s−110^{5}\ \text{m s}^{-1}, comparable to the electron's actual speed in an atom. Small mass is what turns a mathematically universal principle into a chemically significant one.

If Bohr's orbits are wrong, why do we still draw electrons in circular shells in class 11 chemistry?

Bohr's model is kept early on because it correctly predicts hydrogen's energy levels and gives a simple, visualizable starting picture — but the uncertainty principle shows the "fixed circular path" part cannot be physically real for the electron. The syllabus uses Bohr's orbits as a stepping stone, then replaces the fixed-orbit idea with the quantum mechanical model, where an orbital is a region of space describing the probability of finding the electron, not a traced-out path with a definite radius and speed at every instant.

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Where this leads