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Lithium has an atomic number of 3, so the neutral atom has three
protons (Z=3) and 3 electrions. The Li2+ ion, therefore, has only one
electron, and would be called a "hydrogen-like" ion.
If we ignore effects such as spin-orbit coupling, the Lamb shift, and hyperfine splitting, the energy of an electronic orbital in a hydrogen-like ion depends only on the value of the principal quantum number, n. The excited electron in this case has n = 3, and the only way it can emit two photons as it decays to the ground state (n=1) is if it does this in two steps:
n=3 -> n=2
n=2 -> n=1
The energy (or equivalently, the wavelength) of the photon emitted by the transition between two electronic energy levels in a hydrgen-like ion or atom is given by the Rydberg formula:
1/wavelength = R*(Z^2)*(1/(n_1)^2 - 1/(n_2)^2)
where Z is the atomic number of the ion, R is the Rydberg constant for that element, and the wavelength is of the photon emitted by the transition from the state n_2 -> n_1 (n_2 > n_1).
The Rydberg constant for a given element is equal to:
R = R_i/(1 + (me)/M)
where R_i is the "infinite" Rydberg constant = 1.0974*10^7 meters^-1,
(me) is the mass of an electron = 9.10938*10^-31 kg
M is the mass of all the protons in the nucleus of the ion. Here, Z = 3, and the mass of a proton = 1.67262*10^-27 kg, so M = 5.01786*10^-27 kg (neglecting the mass defect due to the binding energy of the nucleons in the Li nucleus)
Putting these together gives
R_Li = (1.0974*10^7 meters^-1)/(1 + (9.10938*10^-31)/(5.019*10^-27)) = 1.0972 * 10^7 meters^-1
We then have for Li2+
1/wavelength = (3^2)* (1.0972 * 10^7 meters^-1)*(1/(n_1)^2 - 1/(n_2)^2)
1/wavelength = (9.87481*10^7 meters^-1)*)*(1/(n_1)^2 - 1/(n_2)^2)
Plug in n_1 = 2 and n_2 = 3, for the first transition to get:
wavelength = 72.91 nm
Plug in n_1 = 1 and n_2=2 for the second transition to get:
wavelength = 13.50 nm
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