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If the potential energy of a hydrogen electron is -3.02 eV, in which excited state is the electron?

1. 1st excited state
2. 2nd excited state
3. 3rd excited state
4. 4th excited state

Correct answer: 2nd excited state

Solution

For hydrogen atom, kinetic energy = -Total Energy and PE = 2 * Total Energy. So Total Energy E = PE/2 = -3.02/2 = -1.51 eV. Using Eₙ = -13.6/n²: -13.6/n² = -1.51 => n² = 13.6/1.51 ≈ 9 => n = 3. n=1 is ground state, n=2 is 1st excited state, n=3 is 2nd excited state.

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