What energy photon is emitted when an electron in hydrogen falls from n = 3 to n = 2 (Eₙ = −13.6/n² eV)?
Correct answer B. 1.89 eV
Explanation
ΔE = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV. This is the red H-alpha line of the Balmer series.
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