Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2: Worked examples

This page is the drill room for the parent topic. We take the single inequality $Δ x Δ p \geq \frac{ℏ}{2}$ and push it through every kind of situation it can face — tiny masses, huge masses, energy–time form, a "know one exactly" limiting case, a real-world word problem, and an exam trap. Nothing here uses a symbol the parent did not already build, but let us re-anchor the three we lean on constantly.

The scenario matrix

Below is every case class this topic can throw at you. Each of the worked examples that follows is tagged with the cell it fills.

Cell	Case class	What makes it distinct	Example
A	Tiny mass, tiny box (electron in atom)	Uncertainty is enormous → destroys classical orbits	Ex 1
B	Huge mass, tiny box (macroscopic object)	Uncertainty is negligible → classical world survives	Ex 2
C	Energy–time form	Different conjugate pair, $Δ E Δ t \geq ℏ/2$	Ex 3
D	Degenerate / limiting: "know $x$ exactly" ( $Δ x \to 0$ )	Forces $Δ p \to \infty$ — the boundary of the rule	Ex 4
E	Wrong-axis trap: $Δ x Δ p_{y}$	No constraint — you must recognise it	Ex 5
F	Real-world word problem (nucleus confinement)	Translate physics prose into $Δ x$	Ex 6
G	Exam twist: given $h /4 π$ form, find minimum KE	Combine two ideas, unit gymnastics	Ex 7
H	Sanity/scale comparison (thermal electron)	Is uncertainty comparable to actual speed?	Ex 8

The figures visualise the two extremes (cells A and B) and the boundary (cell D), because those are the ones where geometry/intuition, not just arithmetic, carries the meaning.

Worked Examples

Worked example Ex 1 — Cell A: electron trapped in an atom (

$Δ x = 1 \overset{˚}{A} = 1.0 \times 1 0^{- 10}$ m) Find the minimum uncertainty in the electron's velocity. Electron mass $m_{e} = 9.11 \times 1 0^{- 31}$ kg.

Forecast: guess first — will $Δ v$ be around walking speed, jet speed, or a fraction of light speed? Write down your gut number.

Step 1. Use the minimum-momentum bound: $Δ p = \frac{ℏ}{2 Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 10} )} = 5.28 \times 1 0^{- 25} kg\cdotm/s$ . Why this step? At the minimum the inequality becomes equality, so we set $Δ x Δ p = ℏ/2$ and solve for the smallest allowed $Δ p$ .

Step 2. Convert momentum fuzz to velocity fuzz: $Δ v = \frac{Δ p}{m _{e}} = \frac{5.28 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}} \approx 5.8 \times 1 0^{5} m/s$ . Why this step? $p = m v$ with fixed $m$ , so dividing the momentum spread by mass gives the velocity spread.

Verify: $Δ v \approx 580, 000$ m/s — about $0.2%$ of light speed. Units check: $\frac{kg\cdotm/s}{kg} = m/s$ . ✓ This is a ferocious spread for something meant to sit in a fixed orbit — exactly why the Bohr fixed-orbit picture collapses and we need probability clouds.

Figure — Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2

Worked example Ex 2 — Cell B: a cricket ball (

$m = 0.16$ kg) pinned to $Δ x = 1 μ m = 1.0 \times 1 0^{- 6}$ m) Find the minimum $Δ v$ and compare with Ex 1.

Forecast: same $ℏ$ , but the mass is $\sim 1 0^{30}$ times bigger. Will $Δ v$ be measurable at all?

Step 1. Use the one-line move directly: $Δ v = \frac{ℏ}{2 m Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 0.16 ) ( 1.0 \times 1 0 ^{- 6} )}$ . Why this step? We already have $Δ x$ and $m$ ; the mnemonic formula skips the intermediate momentum step.

Step 2. Compute: denominator $= 2 \times 0.16 \times 1 0^{- 6} = 3.2 \times 1 0^{- 7}$ , so $Δ v = \frac{1.055 \times 1 0 ^{- 34}}{3.2 \times 1 0 ^{- 7}} \approx 3.3 \times 1 0^{- 28} m/s$ . Why this step? Pure arithmetic; the giant mass in the denominator crushes the answer.

Verify: At $3.3 \times 1 0^{- 28}$ m/s the ball would take $\sim 1 0^{20}$ years to drift one metre — utterly unmeasurable. Compared to Ex 1, the answer shrank by roughly $\frac{5.8 \times 1 0 ^{5}}{3.3 \times 1 0 ^{- 28}} \approx 1.8 \times 1 0^{33}$ . ✓ Uncertainty only bites for tiny masses.

Worked example Ex 3 — Cell C: natural line width from lifetime (energy–time form)

An excited atom lives $Δ t = 1.0 \times 1 0^{- 8}$ s before emitting. Find the minimum energy spread $Δ E$ .

Forecast: energy and time are the other conjugate pair. Will $Δ E$ be a big or tiny fraction of a typical photon energy ( $\sim 1 0^{- 19}$ J)?

Step 1. Apply the energy–time bound at minimum: $Δ E = \frac{ℏ}{2 Δ t} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 8} )}$ . Why this step? Same structure as position–momentum, but the conjugate pair is $(E, t)$ : a short-lived state cannot have a razor-sharp energy.

Step 2. Compute: $Δ E = \frac{1.055 \times 1 0 ^{- 34}}{2.0 \times 1 0 ^{- 8}} \approx 5.3 \times 1 0^{- 27} J$ . Why this step? Direct division.

Verify: Units: $\frac{J\cdots}{s} = J$ . ✓ This $Δ E$ is the natural line width — the reason spectral lines are never infinitely thin. It is $\sim 1 0^{- 8}$ of a typical optical photon energy, i.e. small but real.

Worked example Ex 4 — Cell D (degenerate limit): "we measure position perfectly,

$Δ x \to 0$ " A student claims to know an electron's position exactly. What does the principle force for $Δ p$ ?

Forecast: the rule is $Δ x Δ p \geq ℏ/2$ . If $Δ x$ heads to zero while the product stays $\geq$ a fixed number, what must $Δ p$ do?

Step 1. Solve the bound for $Δ p$ : $Δ p \geq \frac{ℏ}{2 Δ x}$ . Why this step? Isolate the quantity that must respond to the shrinking $Δ x$ .

Step 2. Take the limit $Δ x \to 0$ : the denominator $\to 0$ , so $Δ p \to + \infty$ . Why this step? Dividing a fixed positive number by something approaching zero blows up without bound.

Verify — numerically: at $Δ x = 1 0^{- 10}$ m, $Δ p \approx 5.3 \times 1 0^{- 25}$ ; at $Δ x = 1 0^{- 15}$ m (nuclear scale), $Δ p \approx 5.3 \times 1 0^{- 20}$ — a hundred-thousand-fold jump for a hundred-thousand-fold tighter box. ✓ A perfectly localized particle would need infinite momentum spread (infinitely many wavelengths in its wave packet) — physically impossible. This is the boundary of the rule, not a contradiction of it.

Worked example Ex 5 — Cell E (wrong-axis trap): can you know

$x$ and $p_{y}$ together? You know an electron's $x$ -position to $Δ x = 1 0^{- 12}$ m. A friend asks: "So its $y$ -momentum $p_{y}$ must be super fuzzy, right?"

Forecast: the parent note warned about conjugate pairs on the same axis. Is $(x, p_{y})$ such a pair?

Step 1. Write which pairs are constrained: only $Δ x Δ p_{x}$ , $Δ y Δ p_{y}$ , $Δ z Δ p_{z}$ . Why this step? The uncertainty relation links a coordinate to the momentum along the same axis, because those are the mathematically "conjugate" quantities.

Step 2. Since $x$ and $p_{y}$ are on different axes, there is no lower bound on $Δ x Δ p_{y}$ : $Δ p_{y}$ can be zero while $Δ x$ is tiny. Why this step? Non-conjugate quantities commute — the theory places no trade-off between them.

Verify: The answer is "your friend is wrong — you can know $x$ and $p_{y}$ simultaneously to any precision." No number to compute; the check is logical: applying the same-axis rule off-axis is the classic error. ✓

Worked example Ex 6 — Cell F (real-world word problem): is the electron

inside the nucleus? A nucleus has radius $\approx 5.0 \times 1 0^{- 15}$ m. Suppose an electron were confined inside it, so $Δ x \approx 5.0 \times 1 0^{- 15}$ m. Find the resulting $Δ v$ and argue whether an electron can live there.

Forecast: the box is $\sim 1 0^{5}$ times smaller than an atom (Ex 1). Predict roughly how much bigger $Δ v$ becomes.

Step 1. Translate the prose: "confined inside the nucleus" ⇒ position spread ≈ nuclear radius, $Δ x = 5.0 \times 1 0^{- 15}$ m. Why this step? Word problems hide $Δ x$ inside physical language; naming it is the whole skill.

Step 2. Apply the one-line move: $Δ v = \frac{ℏ}{2 m _{e} Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 9.11 \times 1 0 ^{- 31} ) ( 5.0 \times 1 0 ^{- 15} )}$ . Why this step? Same electron, far smaller box, so we expect an even wilder speed.

Step 3. Compute: denominator $= 2 (9.11 \times 1 0^{- 31}) (5.0 \times 1 0^{- 15}) = 9.11 \times 1 0^{- 45}$ , so $Δ v = \frac{1.055 \times 1 0 ^{- 34}}{9.11 \times 1 0 ^{- 45}} \approx 1.16 \times 1 0^{10} m/s$ . Why this step? Pure arithmetic.

Verify: $1.16 \times 1 0^{10}$ m/s is ~39 times the speed of light — physically forbidden. ✓ Conclusion: an electron cannot be confined inside a nucleus; the uncertainty principle forbids it. (This is a genuine argument physicists used against "nuclear electrons.") The de Broglie link $p = h / λ$ underlies why such tight confinement demands such huge momentum.

Worked example Ex 7 — Cell G (exam twist): given the

$h /4 π$ form, find minimum kinetic energy An electron is confined to $Δ x = 2.0 \times 1 0^{- 10}$ m. Using the textbook form $Δ x Δ p \geq \frac{h}{4 π}$ with $h = 6.626 \times 1 0^{- 34}$ J·s, estimate the electron's minimum kinetic energy, taking $p \approx Δ p$ .

Forecast: two twists here — (i) it uses $h /4 π$ not $ℏ/2$ (same number!), and (ii) it asks for energy, not velocity. Guess: femto-joules? atto-joules?

Step 1. Minimum momentum: $Δ p = \frac{h}{4 π Δ x} = \frac{6.626 \times 1 0 ^{- 34}}{4 π ( 2.0 \times 1 0 ^{- 10} )}$ . Why this step? $\frac{h}{4 π} = \frac{ℏ}{2}$ , so this is the identical bound the parent note derived — the exam just wrote it with $h$ .

Step 2. Compute $Δ p$ : $4 π (2.0 \times 1 0^{- 10}) = 2.513 \times 1 0^{- 9}$ , so $Δ p = \frac{6.626 \times 1 0 ^{- 34}}{2.513 \times 1 0 ^{- 9}} \approx 2.64 \times 1 0^{- 25} kg\cdotm/s$ . Why this step? Isolate the momentum before touching energy.

Step 3. Minimum kinetic energy $KE = \frac{p ^{2}}{2 m _{e}}$ with $p \approx Δ p$ : $KE = \frac{( 2.64 \times 1 0 ^{- 25} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )}$ . Why this step? Kinetic energy relates to momentum by $KE = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ (since $p = m v \Rightarrow v = p / m$ ); this avoids computing $v$ separately.

Step 4. Compute: numerator $= (2.64 \times 1 0^{- 25})^{2} = 6.95 \times 1 0^{- 50}$ ; denominator $= 1.822 \times 1 0^{- 30}$ ; $KE \approx 3.8 \times 1 0^{- 20} J$ . Why this step? Arithmetic on the squared momentum.

Verify: $3.8 \times 1 0^{- 20}$ J $\approx 0.24$ eV (dividing by $1.602 \times 1 0^{- 19}$ ). Units: $\frac{( kg\cdotm/s ) ^{2}}{kg} = \frac{kg ^{2} m ^{2} / s ^{2}}{kg} = kg\cdotm^{2} / s^{2} = J$ . ✓ A confined electron carries irreducible kinetic energy — it can never sit perfectly still. This is zero-point energy.

Worked example Ex 8 — Cell H (sanity/scale check): is the fuzz bigger than the real speed?

A conduction electron drifts at $v \approx 1.0 \times 1 0^{5}$ m/s and is localized to $Δ x = 1.0 \times 1 0^{- 9}$ m. Is its velocity uncertainty larger or smaller than its actual velocity?

Forecast: if $Δ v > v$ , the very idea of "the electron's velocity" is meaningless. Guess before computing.

Step 1. Compute $Δ v = \frac{ℏ}{2 m _{e} Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 9.11 \times 1 0 ^{- 31} ) ( 1.0 \times 1 0 ^{- 9} )}$ . Why this step? We need the fuzz to compare against the stated $v$ .

Step 2. Denominator $= 1.822 \times 1 0^{- 39}$ , so $Δ v = \frac{1.055 \times 1 0 ^{- 34}}{1.822 \times 1 0 ^{- 39}} \approx 5.8 \times 1 0^{4} m/s$ . Why this step? Arithmetic.

Verify: $Δ v \approx 5.8 \times 1 0^{4}$ m/s vs $v = 1.0 \times 1 0^{5}$ m/s — the fuzz is ~58% of the drift speed! ✓ Comparable size ⇒ you genuinely cannot pretend the electron has a well-defined velocity; the wave nature dominates. Contrast with Ex 2, where the fuzz was $1 0^{- 33}$ of any real speed.

Recall Self-test before you leave

Which cell has no lower bound at all? ::: Cell E — different-axis pair $(x, p_{y})$ . As $Δ x \to 0$ , what happens to $Δ p$ ? ::: It diverges to $+ \infty$ (Cell D). Why is the cricket ball's uncertainty invisible? ::: Its huge mass makes $Δ v = ℏ/ (2 m Δ x)$ absurdly tiny (Cell B). In Ex 6, what impossible speed rules out electrons in the nucleus? ::: ~ $1.2 \times 1 0^{10}$ m/s, about 39× the speed of light. Which form did Ex 7 use, and is it different from $ℏ/2$ ? ::: $h /4 π$ — identical, since $h /4 π = ℏ/2$ .