Hydrogen emission: n = 3 to n = 2
Using Eₙ = −13.6 eV/n², what is the energy of the photon emitted when a hydrogen electron drops from n = 3 to n = 2?
Answer B: explanation for every choice
A
1.51 eV is the magnitude of the n = 3 energy level.
B
Correct. ΔE = 13.6(1/4 − 1/9) = 13.6 × 5/36 ≈ 1.89 eV, which is red light (Balmer series).
C
3.40 eV is the magnitude of the n = 2 energy level.
D
12.1 eV is the energy of the n = 3 → n = 1 transition.
TakeawayPhoton energy = difference between the two energy levels.
Equation usedHydrogen energy levels
Next question →Reviewed 2026-11-05 by a medical student. Spot an error? Report it with question ID cp95.