 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1/section-a/a2/solution

E_{n} = n^{2} h^{2} / (8 m_{e} L^{2})

, with

n = 1, 2, \dots

. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely

n = 1 \to 2

. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap gives

\frac{h c}{λ} = E_{2} - E_{1} = \frac{3 h ^{2}}{8 m _{e} L ^{2}}, L = \frac{3 hλ}{8 m _{e} c} = 8.53 \times 1 0^{- 10} m = 0.853 nm .

Here the photon gap is approximately

1.55 eV

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