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In order to answer your question you must think through the entire measurement. In this case you are measuring the photoelectric effect by placing a photodiode into a circuit. What kind of circuit element is the photodiode? It is a source. And as you have pointed out above, it is a constant voltage source (with voltage determined by the energy of the photons ...

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You can increase the intensity (Power/Area) in any one of three ways: increasing the rate of incident photons, decreasing the area, or increasing the energy of the photons. (In each case we take the other two as remaining constant.) Increasing the energy per photon will increase the energy of the emitted electrons, but not the rate of emission. ...

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The intensity can only be increased by greater energy of the photon. When you increase the energy of the photon, the energy is absorbed into the electron, which consequently flies off with the photon's energy. The more energy the photon has, the more energy the electron has.

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There is an interesting view on how to relate the statistical physics distributions and the problem to arrange balls on boxes Here is the link: http://tominology.blogspot.com.br/2014/11/counting-problems-and-statistical.html Basically you can think that: Maxwell-Boltzmann: arrange distinguishable particles inside cells where the cells can contain ...

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Because the light is absorbed mostly as single photons. Two-photon absorption is also possible, but it is a much rarer event. A single photon carries energy $\hbar\omega$. If it is not sufficient to "lift" the electron to higher allowed energy, such a photon is scattered elastically, without absorption. The problem is that the electron is not alone, it is ...

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The thing is that the plates are already charged, see the picture below, hence no current flows in absence of light (charged capacitor). The sum of the voltages around the loop is already zero. Of course initially there must be some current when you connect the plates with the battery. That will charge the plates and the final charge will depend on the ...

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