Oblicz energię i długość fali światła wysłanego przez atom, gdy elektron przechodzi z 4 na 3 orbite

[latex]E_4= frac{-13,6eV}{16}= -0,85eV\ E_3= frac{-13,6eV}{9}= -1,51eV\ E_4-E_3=-0,85eV+1,51eV=0,66eV\ 0,66*1,6*10^{-19}J=1,056*10^{-19}J\ l= frac{hc}{E} =frac{6,63*10^{-34}J*s*3*10^{8}m/s}{1,056*10^{-19}J} approx 18,8*10^{-7}m [/latex]

[latex]dane:\m = 4\n = 3\E_1 = -13,6 eV\1 eV = 1,6cdot10^{-19} J\h = 6,63cdot10^{-34} Jcdot s\c = 3cdot10^{8}frac{m}{s}\szukane:\E = ?\lambda = ?[/latex] [latex]E = E_{m} - E_{n} = E_1(frac{1}{m^{2}}-frac{1}{n^{2}})\\E = -13,6eVcdot(frac{1}{4^{2}}-frac{1}{3^{2}})\\E = -13,6eVcdot(frac{1}{16}-frac{1}{9})\\E = -13,6eVcdot(-frac{7}{144})\\E = 0,66 eV=0,66cdot1,6cdot10^{-19} J = 1,056cdot10^{-19} J[/latex] [latex]lambda = hcdotfrac{c}{h}\\lambda = 6,63cdot10^{-34}Jcdot scdotfrac{3cdot10^{8}frac{m}{s}}{1,056cdot10^{-19} J}\\lambda = 18,83cdot10^{-7} m = 1 883 nm[/latex]