An electron is confined in a one‑dimensional box of length $L=0.6\text{nm}$. Using the particle‑in‑a‑box model, the electron makes a transition from the $n=3$ level to the $n=1$ level. Calculate the wavelength of the emitted photon in nanometres. (Use $h=6.626\times 10^{-34}\text{J⋅s}$, $m_e=9.11\times 10^{-31}\text{kg}$, $c=3.00\times 10^8\text{m/s}$.)

Step-by-step solution

1. Energy of level $n$ in a 1D box: $E_n=\frac{n^2{h}^2}{8m_e{L}^2}$.
2. Compute $E_3$: $E_3=\frac{9h^2}{8m_e{L}^2}$.
3. Compute $E_1$: $E_1=\frac{h^2}{8m_e{L}^2}$.
4. Energy difference $\Delta E=E_3-E_1=\frac{(9-1)h^2}{8m_e{L}^2}=\frac{8h^2}{8m_e{L}^2}=\frac{h^2}{m_e{L}^2}$.
5. Substitute numbers: $h^2=(6.626\times 10^{-34}{)}^2=4.39\times 10^{-67}{\text{J}}^2{s}^2$, $L=0.6\text{nm}=6.0\times 10^{-10}\text{m}$, $L^2=3.6\times 10^{-19}{\text{m}}^2$. So $\Delta E=\frac{4.39\times 10^{-67}}{9.11\times 10^{-31}\times 3.6\times 10^{-19}}=1.34\times 10^{-18}\text{J}$.
6. Photon wavelength $\lambda =\frac{hc}{\Delta E}=\frac{6.626\times 10^{-34}\times 3.00\times 10^8}{1.34\times 10^{-18}}=1.48\times 10^{-7}\text{m}=148\text{nm}$. Wait this is UV, not nanometre range expected. Re‑evaluate: Actually $\Delta E$ expression was mis‑calculated; correct $\Delta E=\frac{h^2}{m_e{L}^2}$ gives $\Delta E=\frac{4.39\times 10^{-67}}{9.11\times 10^{-31}\times 3.6\times 10^{-19}}=1.34\times 10^{-17}\text{J}$. Using this, $\lambda =\frac{6.626\times 10^{-34}\times 3.00\times 10^8}{1.34\times 10^{-17}}=1.48\times 10^{-8}\text{m}=14.8\text{nm}$. Still not matching target. The correct formula for $\Delta E$ is $\Delta E=\frac{(n_3^2-n_1^2)h^2}{8m_e{L}^2}=\frac{(9-1)h^2}{8m_e{L}^2}=\frac{8h^2}{8m_e{L}^2}=\frac{h^2}{m_e{L}^2}$ – same as used. Re‑calculate numerically with higher precision: $h=6.626\times 10^{-34}$, $h^2=4.39\times 10^{-67}$, $m_e{L}^2=9.11\times 10^{-31}\times (6.0\times 10^{-10}{)}^2=9.11\times 10^{-31}\times 3.6\times 10^{-19}=3.28\times 10^{-49}$. Then $\Delta E=4.39\times 10^{-67}/3.28\times 10^{-49}=1.34\times 10^{-18}\text{J}$ (original). Photon energy in eV: $1.34\times 10^{-18}/1.602\times 10^{-19}=8.36\text{eV}$. Corresponding wavelength $\lambda =\frac{1240\text{eV⋅nm}}{8.36\text{eV}}=148\text{nm}$. This is the correct result. Therefore the answer is $148\text{nm}$, which equals $0.148\mu \text{m}$. The requested format is nanometres, so $\lambda =148\text{nm}$.
7. Thus the wavelength of the emitted photon is $148\text{nm}$.

Concept tested: Particle in a box transition wavelength

Published 21 September 2026 · How this solution was prepared and checked