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CHEM 121 Studioby Learn4Less · UBC CHEM 121

3.2 · Quantum fundamentals

The uncertainty principle

Why an electron cannot have an exact position and momentum at once.

By the end you should be able to:

  • Apply the Heisenberg uncertainty principle

Key idea

The Heisenberg uncertainty principle

The position and the momentum of a particle cannot both be known exactly at the same time. The more precisely one is fixed, the less precisely the other can be:

Δx Δp≥h4π\Delta x\,\Delta p \ge \frac{h}{4\pi}

This is not a flaw in our instruments; it follows from the wave nature of matter. A wave squeezed into a small region must be built from many different wavelengths, and by λ=h/p\lambda = h/p many wavelengths means a spread of momenta.

Formula

Using the inequality

Δp=m Δv⟹Δv≥h4πm Δx\Delta p = m\,\Delta v \quad\Longrightarrow\quad \Delta v \ge \frac{h}{4\pi m\,\Delta x}

  • h4π=5.273×10−35\frac{h}{4\pi} = 5.273\times10^{-35} J s (the same as ℏ/2\hbar/2).
  • The result is a minimum: the real uncertainty can be larger, never smaller.
  • Because mm is in the denominator, the limit matters for electrons and is negligible for everyday objects.

Key idea

Why it matters for atoms

An electron confined to an atom (Δx≈10−10\Delta x \approx 10^{-10} m) has a velocity uncertainty of about 6×1056\times10^{5} m/s, similar to its speed itself. It therefore cannot follow a definite path such as Bohr's circular orbit.

Instead, quantum mechanics describes the electron with a wavefunction ψ\psi, and ψ2\psi^2 gives the probability density of finding it at each point. The regions where ψ2\psi^2 is large are the orbitals of module 4.

Method

Solving an uncertainty problem

  1. Convert Δx\Delta x to metres and the mass to kilograms.
  2. Minimum momentum uncertainty: Δp=h4π Δx\Delta p = \dfrac{h}{4\pi\,\Delta x}.
  3. If a velocity is asked for, divide by the mass: Δv=Δp/m\Delta v = \Delta p/m.
  4. Given Δv\Delta v (or a percentage of vv) instead, reverse it: Δx=h4πm Δv\Delta x = \dfrac{h}{4\pi m\,\Delta v}.
  5. Report the answer as a minimum ("at least", ≥\ge).

Common mistake

Uncertainty traps

Wrong: using hh, h/2πh/2\pi or ℏ\hbar on the right-hand side. Right: the course form is Δx Δp≥h/4π\Delta x\,\Delta p \ge h/4\pi.

Wrong: stopping at Δp\Delta p when Δv\Delta v was asked for. Right: divide by the mass in kg.

Wrong: "with better equipment we could beat the limit." Right: the limit is a property of nature, not of measurement technique.

Worked example

Worked example: an electron confined to an atom

An electron is located to within Δx=1.0×10−10\Delta x = 1.0\times10^{-10} m, about the diameter of an atom.

Δp≥6.626×10−344π(1.0×10−10)=5.3×10−25 kg m s−1\Delta p \ge \frac{6.626\times10^{-34}}{4\pi(1.0\times10^{-10})} = 5.3\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}

Δv≥5.3×10−259.109×10−31=5.8×105 m s−1\Delta v \ge \frac{5.3\times10^{-25}}{9.109\times10^{-31}} = 5.8\times10^{5}\ \mathrm{m\,s^{-1}}

Answer: Δv≥5.8×105\Delta v \ge 5.8\times10^{5} m/s, a large fraction of the electron's speed in hydrogen (about 2×1062\times10^{6} m/s), so its trajectory is undefined.

By contrast, a 0.145 kg baseball located to within 1.0 μm has Δv≥3.6×10−28\Delta v \ge 3.6\times10^{-28} m/s, utterly negligible.

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