Solution
= Solution
Substitution into the <Schrödinger equation> gives $E_n=V_0+n^2\pi^2\hbar^2/(2mL^2)$. For the stated <electron> well,
$$
\boxed{E_2=37.6\,\mathrm{eV}=6.02\times10^{-18}\,\mathrm J.}
$$
The transition emits the energy difference, $\Delta E=E_2-E_1=3E_1=28.2\,\mathrm{eV}$, so
$$
\boxed{\lambda=\frac{hc}{\Delta E}=44.0\,\mathrm{nm}.}
$$
Using the full excited-state energy as the <photon> energy would neglect the nonzero ground-state energy.