[latex]E=frac{hc}{lambda}[/latex] [latex]A(1-frac{1}{n^{2}})=frac{6,63cdot10^{-34}Jcdot scdot3cdot10^{8}frac{m}{s}}{1,026cdot10^{-7}m}=19,39cdot10^{-19}J[/latex] [latex]1-frac{1}{n^{2}}=frac{19,39cdot10^{-19}J}{21,76cdot10^{-19}J}=0,891[/latex] [latex]frac{1}{n^{2}}=0,109[/latex] [latex]n^{2}=9,17[/latex] [latex]n=sqrt{9,17}approx 3[/latex] Odp. Elektron przeszedł z orbity n = 3.
[latex]Dane:\ lambda=1,026*10^{-7}m\ h=6,63*10^{-34}Js\ c=3*10^8frac{m}{s}\ E_1=-21,76*10^{-19}J\ Oblicz:\ n=?\ Rozwiazanie:\ E=frac{h*c}{lambda}\\ E=frac{6,63*10^{-34}*3*10^8}{1,026*10^{-7}}\\ E=19,386*10^{-19}J\\ E_n=E_1+E\ E_n=-21,76*10^{-19}+19,386*10^{-19}\ E_n=-2,374*10^{-19}J\\ E_n=frac{E_1}{n^2}Rightarrow n=sqrt{frac{E_1}{E_n}}\\ n=sqrt{frac{-21,76*10^{-19}}{-2,374*10^{-19}}}\\ n=sqrt{9,16}\\ n=3,02, elektron znajdowal sie na 3 orbicie[/latex]
