There are two diagrams for photoelectric experiment. One in which emission is done from the electrode that is connected to positive terminal of cell and one in which emission is due the electrode that is connected to negative terminal of the cell.
However, the graphs of photoelectric current against stopping potential vary too much in each condition. In one a saturation current is obtained whereas in another current doesn't get saturated.
Can somebody please explain the discrepancy in the graphs?
Edit
Is this graph wrong for circuit in which emitting electrode is connected to positive terminal ?
Best Answer
Usually we look at just one graph - current as a function of voltage. When the voltage is positive, a certain amount of energy is needed in each incident photon to give the electron enough energy to escape. However, since the velocity of the electron is random, it may or may not have sufficient momentum perpendicular to the surface to escape. The lower the potential, the less energy is needed - and therefore the greater the probability of escape.
Once the potential is positive, the chances of an electron escaping tend to be independent of this initial direction and thus the current is only determined by the number of electrons knocked out of the conductor - and this number is independent of the voltage.
In summary: same number of electrons is knocked out per unit time. Voltage determines whether they will escape or return to the plate.