 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/2/a/iii/solution

An emitted photon has positive energy

E_{γ} = \frac{h c}{λ} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 2.998 \times 1 0 ^{8} )}{18.2 \times 1 0 ^{- 9}} = 1.09147 \times 1 0^{- 17} J .

n_{f} = 2

E_{γ} = E_{n_{i}} - E_{2} = 36 R_{H} (\frac{1}{4} - \frac{1}{n _{i}^{2}}) .

Consequently

\frac{1}{n _{i}^{2}} = \frac{1}{4} - \frac{1.09147 \times 1 0 ^{- 17}}{7.848 \times 1 0 ^{- 17}} = 0.110924, n_{i} = 3.0025.

The principal quantum number must be an integer, so the transition is $n_{i} = 3 ⟶ n_{f} = 2$ . Its calculated wavelength is approximately

18.22 nm

, agreeing with the rounded observation.

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