[latex]1.\dane:\r_1 = 0,53*10^{-10} m\n = 3\szukane:\r_3 = ?[/latex] [latex]r_{n} = r_1*n^{2}\\r_3 = r_1*3^{2}\\r_3 = 0,53*10^{-10}*9\\r_3 = 4,77*10^{-10} m[/latex] Odp. Promień 3 orbity wynosi 4,77 * 10⁻¹⁰ m. 2. dane: λ = 550 nm = 550 * 10⁻⁹ m = 5,5 * 10⁻⁷ m c = 3 * 10⁸ m/s szukane: f = ? [latex]f = frac{c}{lambda}[/latex] [latex]f = frac{3*10^{8}frac{m}{s}}{5,5*10^{-7}m} = 0,5*10^{15} Hz = 0,5 PHz (petaherca)[/latex] Odp. Częstotliwość promieniowania wynosi 0,5 * 10¹⁵ Hz (0,5 PHz) [latex]3.[/latex] [latex]E_{n} = -3,4 eV[/latex] [latex]E_1 = -13,6 eV[/latex] [latex]szukane:[/latex] [latex]n = ?[/latex] [latex]E_{n} = E_1*frac{1}{n^{2}}[/latex] [latex]n^{2} = frac{E_1}{E_{n}} = frac{-13,6eV}{-3,4eV}=4[/latex] [latex]n = 2[/latex] Odp. To orbita druga. [latex]4.[/latex] [latex]dane:[/latex] [latex]k = 4[/latex] [latex]n = 1[/latex] [latex]szukane:[/latex] [latex]E_{f} = ?[/latex] Korzystamy z wzoru: [latex]E_{f} = E_{k}-E_{n}=E_1(frac{1}{k^{2}}-frac{1}{n^{2}})[/latex] [latex]E_1 = -13,6 eV[/latex] [latex]E_{f} = -13,6eV*(frac{1}{4^{2}}-1)= -13,6*(frac{1}{16}-1) eV =-13,6*(-frac{15}{16}) eV[/latex] [latex]E_{f} = frac{204}{16} eV[/latex] [latex]E_{f} = 12,75 eV[/latex] Odp. Energia fotonu wynosi 12,75 eV. [latex]5.[/latex] [latex]dane:[/latex] [latex]E_{f} = 3 eV[/latex] [latex]E_{k} = 0,86 eV[/latex] [latex]szukane:[/latex] [latex]W = ?[/latex] Ze wzoru Millikana - Einsteina [latex]E_{k} = E_{f} - W[/latex] [latex]W = E_{f}-E_{k}[/latex] [latex]W = 3eV - 0,86eV[/latex] [latex]W = 2,14 eV[/latex] Odp. Praca wyjścia wynosi 2,14 eV.
