 Solution

ID: 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} π^{2} ℏ^{2} / (2 m L^{2})

. For the stated electron well,

E_{2} = 37.6 eV = 6.02 \times 1 0^{- 18} J .

The transition emits the energy difference,

Δ E = E_{2} - E_{1} = 3 E_{1} = 28.2 eV

, so

λ = \frac{h c}{Δ E} = 44.0 nm .

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