For hydrogen, E_n = −13.6 eV/n². Calculate photon wavelengths for transitions 3→2 and 4→2 and state their spectral region. Determine the ionization energy from n = 2. Explain why the Bohr model predicts discrete lines, why absorption requires matching photon energies, and identify two phenomena that need a more complete quantum model. Use hc = 1240 eV nm.

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3→2 gives 1.89 eV and 656 nm; 4→2 gives 2.55 eV and 486 nm, both visible. Ionization from n = 2 requires 3.40 eV, corresponding to a threshold wavelength of 365 nm. For 3→2, ΔE = 13.6(1/2²−1/3²) = 1.89 eV. Wavelength is 1240/1.89 = 656 nm. For 4→2, ΔE = 13.6(1/4−1/16) = 2.55 eV, giving λ = 1240/2.55 = 486 nm. Both lie in the visible Balmer series. At n = 2 the electron energy is −3.40 eV, so reaching the zero-energy continuum requires 3.40 eV. Threshold wavelength is 1240/3.40 = 365 nm. Stationary states have discrete energies, so emitted photon energy equals one allowed energy difference. Absorption occurs only when photon energy matches an upward transition or exceeds an ionization threshold. The simple Bohr orbit picture does not correctly describe multi-electron atoms, transition intensities, fine structure, spin or Zeeman splitting without additional rules. Wavefunctions and quantum numbers replace definite classical orbits in a complete model.

Hydrogen spectrum and model limitations

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