E1=λ11240=108.51240≈11.43eV
E2=λ21240=30.41240≈40.79eV
ETotal=E1+E1=52.22eV
52.22=13.6(2)2[1−n21]
52.22=54.4(1−n21)
0.96=1−n21
n21=0.04
n2=0.041=4100=25
n=5
An excited He+ ion emits two photons in succession, with wavelengths 108.5nm and 30.4nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is
(for a photon of wavelength λ, energy E=λ(innm)1240eV)
Held on 12 Apr 2019 · Verified 6 Jul 2026.
n=6
n=5
n=7
n=4
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