Heisenberg Uncertainty Principle — Δx·Δp ≥ ℏ/2

medium CBSE JEE-MAIN • JEE Main 2024 • 3 min read
Tags Atomic Structure

Question

An electron has a position uncertainty of Δx=1×10−10\Delta x = 1 \times 10^{-10} m (roughly the size of one atom). Find the minimum uncertainty in its momentum, Δp\Delta p.

Given: ℏ=1.055×10−34\hbar = 1.055 \times 10^{-34} J·s

Solution — Step by Step

The Heisenberg Uncertainty Principle states:

Δx⋅Δp≥ℏ2\Delta x \cdot \Delta p \geq \frac{\hbar}{2}

For minimum uncertainty, we use the equality sign. This is the best-case scenario — in practice, Δp\Delta p can only be larger.

 Δp≥ℏ2⋅Δx\Delta p \geq \frac{\hbar}{2 \cdot \Delta x}

We’re isolating Δp\Delta p by dividing both sides by Δx\Delta x. Nothing tricky here — just algebra.

 Δp=1.055×10−342×1×10−10\Delta p = \frac{1.055 \times 10^{-34}}{2 \times 1 \times 10^{-10}} Δp=1.055×10−342×10−10\Delta p = \frac{1.055 \times 10^{-34}}{2 \times 10^{-10}} Δp=5.275×10−25 kg⋅m/s\Delta p = 5.275 \times 10^{-25} \text{ kg·m/s}

Minimum momentum uncertainty = 5.275×10−255.275 \times 10^{-25} kg·m/s

Why This Works

The uncertainty principle isn’t a measurement limitation — it’s a fundamental property of quantum objects. An electron doesn’t have a precise position and momentum simultaneously. The more sharply we define where it is, the more “spread out” its momentum state becomes, and vice versa.

Think of it this way: to locate a particle precisely, you need a wave with a very short wavelength. But a short-wavelength wave is made by superposing many different wavelengths together — and each wavelength corresponds to a different momentum via de Broglie’s relation (p=h/λp = h/\lambda). So pinning down position automatically blurs momentum.

This is why electrons in atoms can’t spiral into the nucleus. Confining an electron to nuclear dimensions (∼10−15\sim 10^{-15} m) would give it an enormous Δp\Delta p, and therefore enormous kinetic energy — far too much to stay bound.

Alternative Method — Using h instead of ℏ

Some textbooks (especially NCERT and older CBSE material) write the relation as:

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

This is the same thing, since ℏ=h/2π\hbar = h/2\pi, so ℏ/2=h/4π\hbar/2 = h/4\pi.

Using h=6.626×10−34h = 6.626 \times 10^{-34} J·s:

Δp=6.626×10−344π×10−10=6.626×10−341.257×10−9\Delta p = \frac{6.626 \times 10^{-34}}{4\pi \times 10^{-10}} = \frac{6.626 \times 10^{-34}}{1.257 \times 10^{-9}} Δp≈5.27×10−25 kg⋅m/s\Delta p \approx 5.27 \times 10^{-25} \text{ kg·m/s}

Same answer. Use whichever form is given in your exam — JEE Main tends to give ℏ\hbar directly, CBSE questions often give hh.

In JEE Main 2024, the uncertainty principle appeared as a one-mark MCQ asking which form of the inequality is correct. The answer was Δx⋅Δp≥ℏ/2\Delta x \cdot \Delta p \geq \hbar/2 — not ≥ℏ\geq \hbar, not ≥h/4π\geq h/4\pi (both are equivalent, but the exact symbolic form matters for the options).

Common Mistake

The single most common error: using Δx⋅Δp≥h\Delta x \cdot \Delta p \geq h (without the 4π4\pi in the denominator or the factor of 2 on ℏ\hbar).

This gives an answer that’s off by a factor of 4π≈12.64\pi \approx 12.6. The correct relation is Δx⋅Δp≥h/4π=ℏ/2\Delta x \cdot \Delta p \geq h/4\pi = \hbar/2.

The confusion comes from de Broglie’s relation (λ=h/p\lambda = h/p), where hh appears without any 2π2\pi or 4π4\pi. Students mix the two. Keep them separate: de Broglie uses hh directly; Heisenberg’s minimum uncertainty uses ℏ/2\hbar/2.

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