Well, according to Einstein's explanation of photoelectric effect, I know that above cutoff frequency, the high frequency the incident photon is, the more electrons will be strike out. Now we add the batter on the photoelectric apparatus. At some negative voltage (stop voltage) there is no photoelectric current. At that voltage we increase the voltage gradually, we will also see the current increase (nonlinearly), and if we keep increase the voltage and over some big value, the current almost doesn't change. I wonder why in the beginning the current change rapidly and nonlinearly and later it is also flat?(adsbygoogle = window.adsbygoogle || []).push({});

In addition, if we compare the I-V curve of two different incident light with different frequency but same intensity. According to experimental data, we see that the curves start from different stop voltage and the currents increases gradually. Simple analysis tells us the higher the frequency of the photon, the larger (absolute value) the stop voltage is. Since the intensity of these two lights are the same, and intensity is defined as

[tex]

I = \frac{\textnormal{number of photon}}{\textnormal{sec}\cdot\textnormal{m}^2}h\nu

[/tex]

we see that if I unchanged, higher frequency will lead to less number of photon in unit time. So higher-frequency light will strike out less photoelectrons in unit time. If we consider the I-V curve at V=0, the current corresponding to high frequency should be lower that that corresponding to low frequency, but why the actual case is just the opposite?

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# IV curve on photoelectric effect

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