[tex]Dane:\\\lambda = 350 \ nm = 350\cdot10^{-9} \ m = 3,5\cdot10^{-7} \ m\\W = 2,75 \ eV= 2,75\cdot1,6\cdot10^{-19} \ J = 4,4\cdot10^{-19} \ J\\1 \ eV = 1,6\cdot10^{-19} \ J\\h = 6,63\cdot10^{-34} \ Js \ - \ stala \ Plancka\\c = 3\cdot10^{8}\frac{m}{s}\\m = 9,1\cdot10^{-31} \ kg\\Szukane:\\E_{k} = ?\\v = ?[/tex]
Rozwiązanie
Zjawisko fotoelektryczne zachodzi wtedy, gdy:
[tex]E_{f} \geq W[/tex]
[tex]E_{f} = h\cdot\frac{c}{\lambda} = 6,63\cdot10^{-34} \ Js \cdot\frac{3\cdot10^{8}\frac{m}{s}}{3,5\cdot10^{-7} \ m} = 5,68\cdot10^{-19} \ J\\\\W = 4,4\cdot10^{-19} \ J\\\\E_{f} > W, \ czyli \ efekt \ fotoelektryczny \ zachodzi[/tex]
[tex]E_{f} = W+E_{k}\\\\E_{k} = E_{f} - W\\\\E_{k} = 5,68\cdot10^{-19} \ J - 4,4\cdot10^{-19} \ J\\\\\boxed{E_{k} = 1,28\cdot10^{-19} \ J}[/tex]
[tex]E_{k} = \frac{mv^{2}}{2} \ \ \ |\cdot\frac{2}{m}\\\\v^{2} = \frac{2E_{k}}{m}\\\\v = \sqrt{\frac{2E_{k}}{m}}}\\\\v = \sqrt{\frac{2\cdot1,28\cdot10^{-19} \ J}{9,1\cdot10^{-31} \ kg}}}=\sqrt{\frac{2,56\cdot10^{-19} \ kg\cdot\frac{m^{2}}{s^{2}}}{9,1\cdot10^{-31} \ kg}}=\sqrt{28,13\cdot10^{10}\frac{m^{2}}{s^{2}}}\\\\\boxed{v = 5,3\cdot10^{5}\frac{m}{s}}[/tex]