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If the ground-state energy of He+ is -54.4 eV, what is the energy of Li2+ in its first excited state?

1. -30.6 eV
2. 27.2 eV
3. -13.6 eV
4. -27.2 eV

Correct answer: -30.6 eV

Solution

E = -13.6*Z^2/n^2 eV. For Li2+ (Z = 3) in the first excited state (n = 2): E = -13.6*9/4 = -30.6 eV.

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