Oblicz częstotliwość światła emitowanego podczas przeskoku z orbity 5 na 2

k = 2 n = 5 A = 2,2 *10⁻¹⁸ J h = 6,63 *10⁻³⁴ J*s [latex]E = A * ( frac{1}{k^2} - frac{1}{n^2} ) [/latex] [latex]E = 2,2 *10 ^{-18 } J * ( frac{1}{2^2} - frac{1}{5^2} ) = 0,462 *10 ^{-18} J [/latex] [latex]E = h * f [/latex] [latex]f = frac{E}{h} [/latex] [latex]f = frac{0,462 *10 ^{-18} J }{6,63 *10 ^{-34}J * s } = 6,97 *10 ^{14} Hz[/latex]

[latex] frac{1}{lambda} =R_H( frac{1}{k^2}- frac{1}{n^2})[/latex] [latex]R_H=1,097*10^7 frac{1}{m} [/latex] - stała Rydberga k=2, n=5 [latex]frac{1}{lambda} =1,097*10^7 frac{1}{m}( frac{1}{2^2}- frac{1}{5^2})=2303700 frac{1}{m} \ lambda= frac{1}{2303700} mapprox4,34*10^{-7}m \ lambda= frac{c}{ u} \ u=frac{c}{lambda}= frac{3*10^8 frac{m}{s}}{4,34*10^{-7}m} approx6,9111*10^{14} frac{1}{s} =6,9111*10^{14}Hz[/latex]