Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 9 / c / Solution

Substitution into the Schrödinger equation gives $E_n=V_0+n^2{\pi }^2{\hslash }^2/(2m{L}^2)$. For the stated electron well,The transition emits the energy difference, $\Delta E=E_2-E_1=3E_1=28.2\mathrm{eV}$, soUsing the full excited-state energy as the photon energy would neglect the nonzero ground-state energy.