An electron is confined in a cubic box of side $L=1.0\text{nm}$. Using the uncertainty principle $\Delta x,\Delta pge\frac{\hslash }{2}$, 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. Step 1: Take the position uncertainty $\Delta x$ as the box side: $\Delta x=L=1.0\times 10^{-9}\text{m}$.
2. Step 2: From $\Delta pge\frac{\hslash }{2\Delta x}$, compute the minimum momentum uncertainty: $\Delta p_{min}=\frac{1.055\times 10^{-34}}{2\times 1.0\times 10^{-9}}=5.275\times 10^{-26}\text{kg⋅m/s}$.
3. Step 3: Estimate the minimum kinetic energy using $K_{min}=\frac{(\Delta p_{min}{)}^2}{2m_e}$.
4. Step 4: Compute $(\Delta p_{min}{)}^2=(5.275\times 10^{-26}{)}^2=2.783\times 10^{-51}{\text{kg}}^2{\text{m}}^2{\text{/s}}^2$.
5. Step 5: Divide by $2m_e=1.822\times 10^{-30}\text{kg}$: $K_{min}=2.783\times 10^{-51}/1.822\times 10^{-30}=1.527\times 10^{-21}\text{J}$.
6. Step 6: Convert to eV: $K_{min}=1.527\times 10^{-21}/1.602\times 10^{-19}\approx 9.5\times 10^{-3}\text{eV}$, round to $0.01\text{eV}$.

Concept tested: Uncertainty principle minimum kinetic energy

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked