oblicz długość i częśtotliwość przy przejściu elektronu z 5 na 3 orbitę.

[latex]E_5 =frac{-13,6eV}{25} =-0,544eV\ E_3=frac{-13,6eV}{9} =-1,511eV\ E_5-E_3=-0,544eV+1,511eV=0,967eV=0,967*1,6*10^{-19}J\ approx 1,55*10^{-19}J\ f= frac{E_f}{h} =frac{1,55*10^{-19}J}{6,63*10^{-34}J*s} =0,23*10^{15}Hz\ 2,3*10^{14}Hz\ l= frac{c}{f} = frac{3*10^8m/s}{2,3*10^{14}Hz} =1,3*10^{-6}m\[/latex]

[latex]dane:\m = 5\n = 3\E_1 = -13,6 eV = -13,6cdot1,6cdot10^{-19}J = -21,76cdot10^{-19} J\1 eV = 1,6cdot10^{-19} J\h = 6,63cdot10^{-34} Jcdot s\c = 3cdot10^{8}frac{m}{s}\szukane:\lambda = ?\ u = ?\\Rozwiazanie\E_{f} = E_{m}-E_{n} = E_1cdot(frac{1}{m^{2}}-frac{1}{n^{2}})\\E_{f} = -21,76cdot10^{-19}Jcdot(frac{1}{5^{2}}-frac{1}{3^{2}}) = -21,76cdot10^{-19}Jcdot(frac{1}{25}-frac{1}{9})=\\=-21,76cdot10^{-19}Jcdot(-frac{16}{225})=1,55cdot10^{-19} J[/latex] [latex]lambda = hcdotfrac{c}{E_{f}}\\lambda = 6,63cdot10^{-34}Jcdot scdotfrac{3cdot10^{8}frac{m}{s}}{1,55cdot10^{-19} J}=12,83cdot10^{-7} m = 1283 nm[/latex] [latex]E_{f} = hcdot u /:h\\ u = frac{E_{f}}{h}\\ u = frac{1,55cdot10^{-19}J}{6,63cdot10^{-34}Jcdot s} = 0,23cdot10^{15} Hz = 0,23 PHz (petaherca)[/latex]