In the Bohr model of the hydrogen atom, the ground-state electron energy is -13.6 eV. What is the energy of the photon emitted when an electron transitions from the n = 3 state to the n = 2 state (the H-alpha line)?
Answer and explanation
Correct answer: B. 1.89 eV
Energy levels in hydrogen are given by E_n = -13.6 eV / n^2. For n = 3: E3 = -13.6 / 9 ≈ -1.51 eV. For n = 2: E2 = -13.6 / 4 = -3.40 eV. The energy of the emitted photon is E_photon = E3 - E2 = -1.51 eV - (-3.40 eV) = +1.89 eV.
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