A particle of mass $m_e$ is confined in a cubic box of side $L=0.5,\text{nm}$. Using the particle‑in‑a‑box model, compute the ground state energy $E_1$ of the electron and give the result in electron‑volts. Use $h=6.626\times 10^{-34},\text{J⋅s}$ and $m_e=9.11\times 10^{-31},\text{kg}$.

Step-by-step solution

1. Ground state quantum numbers are $(n_x,n_y,n_z)=(1,1,1)$, so $n^2=n_x^2+n_y^2+n_z^2=3$.
2. The energy formula is $E_n=\frac{h^2}{8m{L}^2},n^2$.
3. Compute $h^2=4.39\times 10^{-67},{\text{J}}^2!{\text{s}}^2$.
4. Compute $8m{L}^2=8\times 9.11\times 10^{-31}\times (0.5\times 10^{-9}{)}^2=4.56\times 10^{-49},{\text{kg⋅m}}^2$.
5. Thus $E_1=\frac{4.39\times 10^{-67}}{4.56\times 10^{-49}}\times 3\approx 2.85\times 10^{-17},\text{J}$.
6. Convert to eV: $E_1=\frac{2.85\times 10^{-17}}{1.602\times 10^{-19}}\approx 178,\text{eV}$.

Concept tested: Quantum confinement in 3D box

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