---
title: "Uncertainty in momentum in a 0.1 nm well"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:well-uncertainty-momentum"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-uncertainty-momentum
site: BCT Engineering
---

# Uncertainty in momentum in a 0.1 nm well

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2070 Asar, 4 marks; 2069 Asar, 4 marks; 2068 Shrawan, 8 marks (asked 3 times)
- Video: [particle in a box (quantum mechanics)](https://www.youtube.com/watch?v=7TVdth6U4j4) — Eduphile
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-uncertainty-momentum
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

For an electron confined to an infinite potential well of width 0.1 nm, determine the uncertainty in momentum and kinetic energy.

## Short answer

Take Δx ≈ well width 0.1 nm. Then Δp ≈ ℏ/Δx = 1.055×10⁻³⁴ / 1×10⁻¹⁰ = 1.06×10⁻²⁴ kg m s⁻¹. Kinetic energy KE ≈ (Δp)²/(2m_e) = 6.11×10⁻¹⁹ J ≈ 3.8 eV.

## Answer

The **Heisenberg uncertainty principle** says that position and momentum cannot both be known exactly. In the form used in Kasap's textbook,

$$
\Delta x\,\Delta p_x \ge \hbar
$$

For an electron confined in a well of width $L$, its position is uncertain by at most the width of the well, so take $\Delta x = L$.

**Given:** $L = 0.1\ \text{nm} = 1\times10^{-10}$ m, $\hbar = 1.055\times10^{-34}$ J s, $m = 9.1\times10^{-31}$ kg.

#### Uncertainty in momentum

$$
\begin{aligned}
\Delta p_x &\approx \frac{\hbar}{\Delta x} = \frac{1.055\times10^{-34}}{1\times10^{-10}} \\
&= 1.055\times10^{-24}\ \text{kg m s}^{-1}
\end{aligned}
$$

#### Kinetic energy
The momentum of the electron must be at least of the order of its uncertainty, so the corresponding kinetic energy is

$$
\begin{aligned}
KE &\approx \frac{(\Delta p_x)^2}{2m} = \frac{(1.055\times10^{-24})^2}{2\times9.1\times10^{-31}} \\
&= 6.11\times10^{-19}\ \text{J} = \frac{6.11\times10^{-19}}{1.6\times10^{-19}} = 3.82\ \text{eV}
\end{aligned}
$$

#### Comment
- A confined electron can never be at rest: confining it to 0.1 nm (about the size of an atom) forces it to have a kinetic energy of a few eV. This is why electrons in atoms have energies in the eV range.
- This is an order-of-magnitude estimate. With the stricter form $\Delta x\,\Delta p \ge \hbar/2$, the values are half and one quarter: $\Delta p = 5.27\times10^{-25}$ kg m/s and $KE \approx 0.95$ eV. The exact ground-state energy of a 0.1 nm infinite well is $h^2/8mL^2 = 37.7$ eV.

**Answer:** $\Delta p \approx 1.06\times10^{-24}$ kg m s⁻¹, $KE \approx 6.11\times10^{-19}$ J $\approx 3.8$ eV.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is taken as Delta x for an electron in the well?
   A. Zero
   B. The width of the well, L
   C. The Bohr radius always
   D. Half the electron wavelength
   Answer: B. The electron is somewhere in the well, so its position uncertainty is at most L = 0.1 nm. (Key point: Delta x choice)

2. What is Delta p for L = 0.1 nm?
   A. 1.06e-34 kg m/s
   B. 1.06e-24 kg m/s
   C. 1.06e-10 kg m/s
   D. 6.6e-24 kg m/s
   Answer: B. Delta p = hbar / L = 1.055e-34 / 1e-10 = 1.055e-24 kg m/s. (Key point: Momentum value)

3. How is the kinetic energy estimated and what is its value?
   A. KE = (Delta p)^2 / 2m, about 3.8 eV
   B. KE = Delta p x m, about 3.8 eV
   C. KE = (Delta p)^2 / 2m, about 38 keV
   D. KE = Delta p / 2m, about 0.38 eV
   Answer: A. KE = (1.055e-24)^2 / (2 x 9.1e-31) = 6.11e-19 J = 3.8 eV. (Key point: Kinetic energy)

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