The velocity of an electron is known to an accuracy of $1\%$ at $300\ \text{m/s}$. Calculate the minimum uncertainty in specifying its position.

The velocity of an electron is known to an accuracy of 1\% at 300m/s. Calculate the minimum uncertainty in spe

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Concept understanding — Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle: Why Nature Won't Let You Be Too Precise

Imagine you're trying to photograph a speeding car at night. To see where it is, you need light — but light carries energy. If you use a dim flashlight, the car barely gets lit, and your photo is blurry. If you use a bright flash, you see the car sharply — but the flash itself pushes the car slightly, changing its speed. You can never know both exactly where it was and exactly how fast it was going after the flash.

That's the intuition. But for an electron, the problem isn't a clumsy flashlight — it's built into the fabric of reality.

The core idea

An electron is not a tiny billiard ball with a definite position and speed. It's a quantum object that behaves like a wave. A wave, by its nature, is spread out — you cannot point to a single point and say "the wave is exactly here." The more you try to pin down where the wave is, the less you know about its wavelength (which tells you its momentum). And the more precisely you measure its momentum, the more the wave spreads out in space.

This trade-off is not a limitation of your instruments. It is a fundamental law of nature: you cannot simultaneously know both the exact position and the exact momentum of a particle.

The precise statement

For any particle, the uncertainties in its position ( $\Delta x$ ) and its momentum ( $\Delta p$ ) satisfy:

\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

where $h$ is Planck's constant ( $6.626 \times 10^{-34} \ \text{J·s}$ ).

Important

$\Delta x$ and $\Delta p$ are standard deviations of many measurements on identical systems — not errors in a single measurement. The principle says: if you prepare many electrons in the same state, the spread in their positions times the spread in their momenta can never be smaller than $h/4\pi$ .

What it means for an electron

If you try to measure an electron's position very precisely ( $\Delta x$ very small), the uncertainty in its momentum becomes huge ( $\Delta p$ very large). The electron's speed becomes wildly unpredictable. Conversely, if you measure its momentum precisely, you lose track of where it is.

Note

For everyday objects (a cricket ball, a car), $h$ is so tiny that $\Delta x \cdot \Delta p$ is always much larger than $h/4\pi$ — so the uncertainty is negligible. The principle only matters for particles as small as electrons, protons, and photons.

Why it's not about measurement

A common misunderstanding: "We can't measure both because the measurement disturbs the particle." That's part of the story, but not the whole truth. Even in principle, with perfect instruments, the uncertainty remains. The electron does not have a simultaneous definite position and momentum — the question itself is meaningless in the quantum world.

Watch out

Do not think: "The electron has a definite position and momentum, but we just can't know them." That is wrong. The electron's state is described by a wavefunction, which gives probabilities — not hidden definite values. The uncertainty is ontological, not epistemological.

The energy-time version

There is also an uncertainty relation between energy and time:

$\Delta E \cdot \Delta t \geq \frac{h}{4\pi}$ …

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The uncertainty in velocity is $1\%$ of $300\ \text{m/s}$ :

\Delta v = \frac{1}{100} \times 300 = 3\ \text{m/s}

So the uncertainty in momentum is

\Delta p = m\,\Delta v = 9.11 \times 10^{-31} \times 3 = 2.733 \times 10^{-30}\ \text{kg m/s}

By Heisenberg's principle, $\Delta x \cdot \Delta p \geq h/4\pi$ , so

$\Delta x \geq \frac{h/4\pi}{\Delta p} = \frac{5.273 \times 10^{-35}}{2.733 \times 10^{-30}} = 1.93 \times 10^{-5}\ \text{m}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 mark
$Q. Write the mathematical expression of Heisenberg's Uncertainty Principle.$
›Reveal solutionSolution

Heisenberg's Uncertainty Principle: it is impossible to simultaneously determine, with perfect precision, both the position and the momentum of a microscopic particle (like an electron); mathematically, $\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}$ .

…

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CBSE 2023Set ANNUAL1 mark
$Q. Match the column: Column I item 'Principle of uncertainty' is to be matched with the correct term from Column II:$
(a) Sharing of electron,
(b) increasing,
(c) To find out molecular mass of organic compound,
(d) SF6,
(e) Transition element,
(f) Molarity,
(g) Mulliken,
(h) Heisenberg,
(i) Molality. Which Column II term matches this Column I item?
›Reveal solutionSolution

The Principle of Uncertainty is Heisenberg's Uncertainty Principle (option h), a cornerstone of quantum mechanics for microscopic particles.

Werner Heisenberg proposed, in 1927, that for microscopic (sub-atomic) particles, it is fundamentally impossible to simultaneously know both the exact position (Dx) and the exact momentum (Dp) with unlimited precision. Mathematically, Dx . Dp >= h/4(pi), where h is Planck's constant.

…

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CBSE 2019Set ANN1 mark
$Q. The minimum value for the product of uncertainties in position and momentum of a moving microscopic particle is equal to ..........$
›Reveal solutionSolution

Heisenberg's Uncertainty Principle sets h/4π as the minimum possible value for the product of the uncertainties in a microscopic particle's position and momentum.

Werner Heisenberg's Uncertainty Principle states that it is impossible to simultaneously determine, with unlimited precision, both the exact position and the exact momentum of a microscopic (quantum-scale) particle such as an electron. If Δx is the uncertainty in position and Δp is the uncertainty in momentum, the principle is expressed mathematically as:

Δx · Δp ≥ h / 4π

…

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Q.The velocity of an electron is known to an accuracy of 1%1\%1% at 300 m/s300\ \text{m/s}300 m/s. Calculate the minimum uncertainty in specifying its position.

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Q.The velocity of an electron is known to an accuracy of $1\%$ at $300\ \text{m/s}$ . Calculate the minimum uncertainty in specifying its position.

Unlock everything free for 14 days