An electron is confined in a cubic region of length $L=1.0\times 10^{-10}\text{m}$. Using Heisenberg’s uncertainty principle $\Delta x\Delta p\ge \hslash /2$, estimate the minimum kinetic energy of the electron. Take $m_e=9.11\times 10^{-31}\text{kg}$ and $\hslash =1.055\times 10^{-34}\text{J⋅s}$. Express your answer in electron‑volts.

Step-by-step solution

1. Apply the uncertainty principle: $\Delta p\ge \hslash /(2\Delta x)$ with $\Delta x=L$.
2. Compute $\Delta p=1.055\times 10^{-34}\text{J⋅s}/(2\times 1.0\times 10^{-10}\text{m})=5.275\times 10^{-25}\text{kg⋅m/s}$.
3. Assume $\Delta p$ is approximately the electron momentum $p$ in the lowest energy state.
4. Use kinetic energy formula $E=p^2/(2m_e)$.
5. Calculate $E=(5.275\times 10^{-25}{)}^2/(2\times 9.11\times 10^{-31})$ J.
6. Numerator $(5.275\times 10^{-25}{)}^2=2.78\times 10^{-49}$.
7. Denominator $2\times 9.11\times 10^{-31}=1.822\times 10^{-30}$.
8. Divide to get $E\approx 1.52\times 10^{-19}$ J.
9. Convert joules to electron‑volts: $1\text{eV}=1.602\times 10^{-19}\text{J}$.
10. Thus $E\approx 1.52\times 10^{-19}/1.602\times 10^{-19}=0.95$ eV.

Concept tested: Heisenberg uncertainty principle applied to electrons in atoms

Published 5 September 2026 · How this solution was prepared and checked