Na pierwszej (czyli podstawowej) energia elektronu wynosi: E1 = -13.6 eV Natomiast na trzeciej: E3 = E1/3² = -13.6/9 = -1.51 eV Pochłonięta energia: ∆E = E3 - E1 = -1.51 - (-13.6) = 12.09 eV = 12.09·1.6·10⁻¹⁹ J = 19.34·10⁻¹⁹ J
[latex]dane:\E_1 = -13,6 eV\n = 1\m = 3\1 eV = 1,6cdot10^{-19} J\szukane:\Delta E = ?[/latex] [latex]Delta E = E_{m}-E_{n} = frac{E_1}{m^{2}}-frac{E_1}{n^{2}}=E_1(frac{1}{m^{2}}-frac{1}{n^{2}})\\Delta E = -13,6eVcdot(frac{1}{3^{2}}-frac{1}{1^{2}})=-13,6eVcdot(-frac{8}{9}) = 12,09 eV\\1 eV = 1,6cdot10^{-19} J\\Delta E = 12,09eVcdot1,6cdot10^{-19}frac{J}{eV} = 19,34cdot10^{-19} J[/latex]
