[tex]Dane:\\E_1 = -13,6 \ eV\\1 \ eV = 1,6\cdot10^{-19} \ J\\k = 4\\n = 2\\h = 6,63\cdot10^{-34} \ J\cdot s\\c = 3\cdot10^{8}\frac{m}{s}\\Szukane:\\E = ?\\\lambda = ?[/tex]
Rozwiązanie
Obliczam energię, jaką emituje elektron:
[tex]E = E_{k}-E_{n} = E_1\cdot(\frac{1}{k^{2}}-\frac{1}{n^{2}})\\\\E = -13,6 \ eV\cdot(\frac{1}{4^{2}}-\frac{1}{2^{2}})\\\\E = -13,6 \ eV\cdot(\frac{1}{16}-\frac{1}{4})\\\\E = -13,6 \ eV\cdot(\frac{1}{16}-\frac{4}{16})\\\\E = -13,6 \ eV\cdot(-\frac{3}{16})\\\\E = 2,55 \ eV\\\\2,55 \ eV =2,55\cdot1,6\cdot10^{-19} \ J = 4,08\cdot10^{-19} \ J\\\\\boxed{E = 2,55 \ eV = 4,08\cdot10^{-19} \ J}[/tex]
Obliczam długość fali:
[tex]E = h\cdot\frac{c}{\lambda} \ \ \rightarrow \ \ \lambda = h\cdot\frac{c}{E}\\\\\lambda = 6,63\cdot10^{-34} \ Js\cdot\frac{3\cdot10^{8}\frac{m}{s}}{4,08\cdot10^{-19} \ J}\\\\\boxed{\lambda = 4,88\cdot10^{-7} \ m = 488 \ nm}[/tex]
