An electron is confined in a one‑dimensional box of length $L=0.5,\text{nm}$. Using the uncertainty principle $\Delta x,\Delta pgehbar/2$, estimate the minimum kinetic energy $K$ of the electron. Take $m_e=9.11\times 10^{-31},\text{kg}$ and $\hslash =1.055\times 10^{-34},\text{J⋅s}$. Express $K$ in electron‑volts.

Step-by-step solution

1. Take $\Delta x\approx L=0.5\times 10^{-9},\text{m}$.
2. Use $\Delta p\approx \hslash /(2\Delta x)=1.055\times 10^{-34}/(2\times 0.5\times 10^{-9})=1.055\times 10^{-34}/1\times 10^{-9}=1.055\times 10^{-25},\text{kg⋅m/s}$.
3. Compute the minimum kinetic energy: $K_{min}=\frac{(\Delta p{)}^2}{2m_e}=\frac{(1.055\times 10^{-25}{)}^2}{2\times 9.11\times 10^{-31}}\text{J}\approx 2.41\times 10^{-19},\text{J}$.
4. Convert to electron‑volts: $K_{min}=\frac{2.41\times 10^{-19}}{1.602\times 10^{-19}}\text{eV}\approx 1.50,\text{eV}$.

Concept tested: Heisenberg uncertainty in confined systems

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked