[tex]dane:\\\lambda = 0,19 \ \mu m = 0,19\cdot10^{-6} \ m = 1,9\cdot10^{-7} \ m\\W = 1,9 \ eV = 1,9\cdot1,602\cdot10^{-19} \ J = 3,044\cdot10^{-19} \ J\\1 \ eV = 1,602\cdot10^{-19} \ J\\c = 3\cdot10^{8}\frac{m}{s}h = 6,63\cdot10^{-34} \ Js\\m_{e} = 9,1\cdot10^{-31} \ kg\\szukane:\\v = ?[/tex]
Rozwiązanie
Korzystamy ze wzoru
[tex]\frac{mv^{2}}{2}=h\cdot\frac{c}{\lambda}-W \ \ /\cdot\frac{2}{m}\\\\v^{2} = \frac{2}{m}(h\cdot\frac{c}{\lambda}-W)\\\\v = \sqrt{\frac{2}{m}(h\cdot\frac{c}{\lambda}-W)}\\\\v = \sqrt{\frac{2}{9,1\cdot10^{-31}}(6,63\cdot10^{-34}\cdot\frac{3\cdot10^{8}}{1,9\cdot10^{-7}}-3,044\cdot10^{-19}}\\\\v = \sqrt{\frac{2}{9,1\cdot10^{-31}}(10,468\cdot10^{-19}-3,044\cdot10^{-19})}\\\\v = \sqrt{1,6316\cdot10^{12}}\\\\v = 1,28\cdot10^{6}\frac{m}{s}[/tex]
Analiza na jednostkach
[tex][v] = \sqrt{\frac{1}{kg}\cdot(J\cdot s\cdot\frac{\frac{m}{s}}{m}-J)} = \sqrt{\frac{1}{kg}\cdot J} = \sqrt{\frac{1}{kg}\cdot N\cdot m} = \sqrt{\frac{1}{kg}\cdot kg\cdot\frac{m}{s^{2}}\cdot m}} =\\\\=\sqrt{\frac{m^{2}}{s^{2}}} =\frac{m}{s}[/tex]
Odp. Maksymalna szybkość elektronów wynosi ok. 1,28 · 10⁶ m/s.
