Heisenberg uncertainty principle — calculate minimum uncertainty in position of electron

Question

An electron has a velocity of $3.0 \times 10^6$ m/s with an uncertainty of 0.01%. Calculate the minimum uncertainty in its position. Given: $h = 6.626 \times 10^{-34}$ J·s, $m_e = 9.1 \times 10^{-31}$ kg.

(JEE Main 2022, similar pattern)

Solution — Step by Step

Uncertainty of 0.01% means:

\Delta v = \frac{0.01}{100} \times 3.0 \times 10^6 = 3.0 \times 10^2 = 300 \text{ m/s}

\Delta p = m_e \times \Delta v = 9.1 \times 10^{-31} \times 300 = 2.73 \times 10^{-28} \text{ kg·m/s}

The principle states:

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

For the minimum uncertainty in position:

\Delta x_{\min} = \frac{h}{4\pi \Delta p}

\Delta x_{\min} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 2.73 \times 10^{-28}}

= \frac{6.626 \times 10^{-34}}{3.43 \times 10^{-27}} = \mathbf{1.93 \times 10^{-7} \text{ m}} \approx 193 \text{ nm}

That’s roughly 193 nm — about 1000 times larger than a typical atom. Even a tiny uncertainty in velocity leads to a huge positional uncertainty for an electron.

Why This Works

The Heisenberg uncertainty principle is not about measurement limitations of our instruments. It’s a fundamental property of nature: a particle does not simultaneously possess a definite position and a definite momentum. The more precisely we know one, the less precisely we can know the other.

For macroscopic objects (like a cricket ball), $\Delta x$ calculated this way comes out to be absurdly small — like $10^{-33}$ m — which is undetectable. That’s why we never notice quantum uncertainty in daily life. But for electrons ( $m \sim 10^{-31}$ kg), the uncertainty is significant and has real physical consequences.

This is also why we talk about “electron clouds” and “probability distributions” rather than definite orbits. Bohr’s model with fixed circular orbits violates the uncertainty principle — which is one reason it was superseded by the quantum mechanical model.

Alternative Method

Some textbooks use $\frac{\hbar}{2}$ instead of $\frac{h}{4\pi}$ , where $\hbar = h/(2\pi)$ . Both give the same result:

\Delta x \cdot \Delta p \geq \frac{\hbar}{2} = \frac{h}{4\pi}

In JEE numericals, keep the value $h/(4\pi) = 0.528 \times 10^{-34}$ J·s handy. You can quickly divide this by $\Delta p$ to get $\Delta x$ without doing the full $4\pi$ calculation during the exam.

Common Mistake

The most common error: using $h/(2\pi)$ instead of $h/(4\pi)$ in the formula. The uncertainty principle has a factor of $4\pi$ in the denominator (not $2\pi$ ). Using $2\pi$ gives an answer that’s exactly double the correct minimum uncertainty. Always write out the formula before substituting — don’t rely on memory for the constant.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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