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An excited He + ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength λ , energy E = ) nm in ( eV 1240 λ .

1. n = 4
2. n = 6
3. n = 5
4. n = 7

Correct answer: n = 5

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