Practice question

**Hydrogen transitions** example n=3→n=1 ΔE=13.6(1-1/9)=12.09 eV, photon 12.09 eV, λ=1240/12.09≈102.6 nm Lyman series, n=3→n=2 ΔE=1.89 eV Balmer visible Hα 656 nm. Absorption photon energy must match difference, if atom in n=2 absorbs 1.89 eV jumps to n=3, if absorbs 12.75 eV from ground 1→4 because -13.6+12.75=-0.85 eV = -13.6/16. E₅ = -0.544 eV , E₄ = -0.85 eV . Δ E = -0.544 - (-0.85) = 0.306 eV ≈ 0.31 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 0.31 eV, consistent with