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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 ? Graph

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  • $\begingroup$ There are a number of questions on this site about photoelectric saturation. Could you provide a bit more detail as to why those don't answer your question? Perhaps an explanation of what saturation you are talking about - saturation with respect to light intensity, applied voltage, ...? $\endgroup$ – Floris Apr 28 '17 at 21:16
  • $\begingroup$ The wavelength of the electromagnetic radiation is kept constant. However the variable on x axis (which I mention as stopping voltage) is the potential difference across the cell. When the pd is varied, a graph for photoelectric current is obtained. Should I send the graphs too? $\endgroup$ – mathnoob123 Apr 28 '17 at 21:18
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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.

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  • $\begingroup$ Can you explain when the emitting electrode is connected to positive terminal of the cell, why does the current not reduces when pd across cell is positive? $\endgroup$ – mathnoob123 Apr 28 '17 at 21:44
  • $\begingroup$ It should reduce but it does depend on relative energy of incident photons, work function of metal, and potential difference applied. $\endgroup$ – Floris Apr 28 '17 at 21:45
  • $\begingroup$ The intensity of the EM radiation is kept constant. Work function is also kept constant. Potential difference applied is varied. Will it reduce in this case too? $\endgroup$ – mathnoob123 Apr 28 '17 at 21:52
  • $\begingroup$ Yes - when the potential is large enough to pull the electrons back to the emitting electrode the current will be cut off (zero). When it is biased in the other direction the largest current (saturation) is achieved when all electrons make it to the other electrode. $\endgroup$ – Floris Apr 28 '17 at 23:24
  • $\begingroup$ Can you please check the edit? $\endgroup$ – mathnoob123 Apr 29 '17 at 9:38

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