Najpierw musimy policzyć ile energii wydobywa się podczas przejścia między orbitami. Do tego stosujemy ogólny wzór: [latex]E_n=E_1( frac{1}{n^2} - frac{1}{m^2} ) \\ E_{4 o 1}=E_1( frac{1}{4^2} - frac{1}{1^2} )=-13,6eVcdot (- frac{15}{16} )=12,75eV\\ E_{4 o 3}=E_1( frac{1}{4^2} - frac{1}{3^2} )=-13,6eVcdot (- frac{7}{16cdot 9} )approx0,66eV\\ E_{3 o 2}=E_1( frac{1}{3^2} - frac{1}{2^2} )=-13,6eVcdot (- frac{5}{36} )approx1,9eV\\ E_{2 o 1}=E_1( frac{1}{2^2} - frac{1}{1^2} )=-13,6eVcdot (- frac{3}{4} )approx10,2eV\[/latex] Teraz możemy wprost ze wzoru obliczyć długość fali, pamiętając o zamianie jednostek (elektronowoltów na dżule) [latex]E=hcdot frac{c}{lambda} \ lambda-dlugosc fali\ c-predkosc swiatla\ h-stala Plancka[/latex] [latex]E=hcdot frac{c}{lambda} \\ lambda= frac{hc}{E}\ \ lambda_{4 o 1} = frac{6,63cdot 10^{-34}cdot 3cdot 10^8}{12,75cdot 1,6cdot 10^{-19}} =0,975cdot 10^{-7}=97,5nm\ \ lambda_{4 o 3} = frac{6,63cdot 10^{-34}cdot 3cdot 10^8}{0,66cdot 1,6cdot 10^{-19}} =18,83cdot 10^{-7}=1883nm\ \ lambda_{3 o 2} = frac{6,63cdot 10^{-34}cdot 3cdot 10^8}{1,9cdot 1,6cdot 10^{-19}} =6,54cdot 10^{-7}=654nm\ [/latex] [latex]lambda_{2 o 1} = frac{6,63cdot 10^{-34}cdot 3cdot 10^8}{10,2cdot 1,6cdot 10^{-19}} =1,22cdot 10^{-7}=122nm\[/latex]
[latex]dane:\1.\n = 4\m = 1\2.\n = 4\m = 3\3.\n = 4\m = 2\4.\n = 2\m = 1\szukane:\E = ?\\lambda = ?[/latex] [latex]E_1 = -13,6 eV = -13,6*1,6*10^{-19}frac{J}{eV} = -21,76*10^{-19} J\1 eV = 1,6*10^{-19} J\h = 6,63*10^{-34} J*s\c = 3*10^{8}frac{m}{s}[/latex] [latex]1.\\E = E_1*(frac{1}{n^{2}}-frac{1}{m^{2}}) = -21,76*10^{-19}J*(frac{1}{4^{2}}-1) =- 20,4*10^{-19} J[/latex] [latex]lambda = h*frac{c}{E}=6,63*10^{-34}Js*frac{3*10^{8}m/s}{20,4*10^{-19}J} =0,975*10^{-7}m = 97,5 nm[/latex] [latex]2.\\E = E_1*(frac{1}{4^{2}}-frac{1}{3^{2}}) = -21,76*10^{-19}J*(frac{1}{16}-frac{1}{9}) =1,06*10^{-19} J[/latex] [latex]lambda = h*frac{c}{E} = 6,63*10^{-34}*frac{3*10^{8}}{1,06*10^{-19}} m = 18,8*10^{-7}m = 1880 nm[/latex] [latex]3.\\E = E_1*(frac{1}{4^{2}}-frac{1}{3^{2}}) = -21,76*10^{-19}J*(frac{1}{16}-frac{1}{9}) = 4,8*10^{-19} J[/latex] [latex]lambda = h*frac{c}{E} = 6,63*10^{-34}*frac{3*10^{8}}{4,8*10^{-19}} m = 6,52*10^{-7}m = 652 nm[/latex] [latex]4.\\E = E_1*(frac{1}{2^{2}}-1) = -21,76*10^{-19}J * (-0,75) = 16,32*10^{-19} J[/latex] [latex]lambda = h*frac{c}{E} = 6,63*10^{-34}*frac{3*10^{8}}{16,32*10^{-19}} m = 1,22*10^{-7} m = 122 nm[/latex]
