photoelectric effects(2)

(a) Electrode A

(b) From the graph, a potential of -2 v would stop the electronsfrom reaching the electrode A. Hence, the emitted eelctrons have energy of 2 eV. In other words, the energy requred to remove an electron from the metal = (5 - 2 ) eV = 3 eV

(c) All kinetic energy of the emitted electrons turns into electrostatic potential energy before reaching electrode A. Thus no electrons can be caught by A, which leads to no current flow.