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A hydrogen atom and a Li++ ion are each in their second excited state. If their electronic angular momenta are denoted by L_H and L_Li, and their energies by E_H and E_Li respectively, which statement is correct?

1. L_H > L_Li and |E_H| > |E_Li|
2. L_H = L_Li and |E_H| < |E_Li|
3. L_H = L_Li and |E_H| > |E_Li|
4. L_H < L_Li and |E_H| < |E_Li|

Correct answer: L_H = L_Li and |E_H| < |E_Li|

Solution

Both are in the second excited state (n=3), so angular momentum L = nh/2pi is the same: L_H = L_Li. Energy |E| = 13.6 Z^2/n^2, and Li++ has Z=3 vs Z=1 for H, so |E_H| < |E_Li|.

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