[tex]Dane:\\k = 4\\n = 1\\E_1 = -13,6 \ eV\\1 \ eV = 1,602\cdot10^{-19} \ J\\Szukane:\\E_{f} = ?[/tex]
Rozwiązanie
Korzystamy ze wzoru:
[tex]E_{f} = E_{k}-E_{n} = E_1(\frac{1}{k^{2}}-\frac{1}{n^{2}})\\\\E_{f} = -13,6 \ eV\cdot(\frac{1}{4^{2}}-\frac{1}{1^{2}})\\\\E_{f} = -13,6 \ eV\cdot(\frac{1}{16}-1)\\\\E_{f} = -13,6 \ eV\cdot(\frac{1}{16}-\frac{16}{16})\\\\E_{f} = -13,6 \ eV\cdot(-\frac{15}{16})\\\\E_{f} = 12,75 \ eV\\\\E_{f}=12,75 \ eV=12,75\cdot1,602\cdot10^{-19} \ J = 20,43\cdot10^{-19} \ J\\\\\boxed{E_{f} = 12,75 \ eV = 20,43\cdot10^{-19} \ J}[/tex]
