Concerning the photoelectric effect: When the intensity and applied voltage are both constant, then the current is inversely proportional to frequency $f$ (above threshold frequency). If we increase $f$ then the current decreases and vice-versa, because the number of photons from a source is inversely proportional to the incident photon's frequency at constant Intensity. So the current vs. frequency graph will be hyperbolic. Am I right?
[Physics] Is the current vs. frequency graph hyperbolic for the photoelectric effect
photoelectric-effectquantum mechanics
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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 and the work function of the metal). It is also current limited (with no load it still cannot draw more current than there are incoming photons). I think you may be slightly confusing yourself because current is charge/time so it depends on how fast the electrons are moving (i.e. the voltage). Always recall that it is a constant voltage source.
To proceed with understanding your lab you need to understand your circuit. There is a schematic diagram of the photodiode, op-amp, and associated circuitry in your pdf. I am sure you are familiar with the basic idea: the circuit acts as a capacitor which builds up voltage until the current is 0 (or approaches 0). The 'current-limited' feature of the diode power supply effects the capacitor charging only at the start of a measurement. The trick to this equipment is understanding the op-amp. There is plenty of literature out there on op-amps. Wiki is always a good place to start.
If you have done your lab by now you will have noticed that there IS some intensity dependence. How can you explain this? (as an aside... ideally you have recorded voltage vs time for each measurement and fit it to an exponential--this assumes it is capacitor charging at constant voltage) I usually tell my students to chalk it up to the finite input impedance of the amp causing the current at the input to the amp to be non-zero and changing the feedback mechanism. I got this explanation from a professor a long time ago, and I am not sure I entirely buy it (Pasco claims there is an enormous input impedance). I never came up with a better answer while I was teaching the lab though.
I would hazard that the source you have linked is incorrect. Most often when I read about the photoelectric effect the main result is that:
1) The kinetic energy of the emitted electrons depends only on the frequency of the incident light. In particular, there are no emitted electrons until the light is at least a threshold frequency, $\nu_0$ and then after that the kinetic energy of the emitted electrons increases linearly with frequency of the incident light.
$$\text{KE} = hf - \phi$$
($\text{KE}$ is the kinetic energy of the emitted electrons, $h$ and $f$ are Planck's constant and the frequency of the incident light and $\phi$ is the work function of the material under consideration, the minimum amount of energy needed to eject an electron.)
2) (For fixed incident frequency) the number of emitted electrons increases linearly with the intensity of the incident light.
$$N_{\text{electron}} = N_{\text{photon}} = \frac{IA}{h f} = \frac{P}{hf} $$
($P$ is the power in the beam, $A$ is the area of the beam so the intensity, $I$ times $A$ gives the power. It is in fact power $P$ which equals $N_{\text{photon}} hf$ where $N_{\text{photon}}$ is the number of photons impinging upon the surface per second)
These two facts are consistent with the particle or photon description of light. That is, light comes in discrete packets where each packet carries a quanta of energy. The size of this quanta of energy is linearly proportional to the frequency of light and the intensity of the light is proportional to the number flux.
I don't know if I have seen many sources that claim the photocurrent is independent of the frequency of the incident field. As you have correctly noted (and shown in my 2nd equation above) the flux of photoelectrons depends directly on the flux of photons but, because of the relationship between intensity and flux, this in turn depends on intensity and frequency.
In short, I would expect that, at constant intensity, increasing the frequency of incident light would decrease the photocurrent. Someone should please correct me if I'm mistaken.
I think it would be extremely misleading to claim the intensity, in the context of the photoelectric effect, refers only to the number of photons in the beam and not the actual energy flux per unit area of the incident beam. Particularly because the experimenters at the time of the discovery would have been quantifying the experimentally accessible quantity of real intensity.
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The number of photons from a light source is not necessarily inversely proportional to the frequency of the light emitted. It appears that this consideration followed from an artificial requirement of fixed power emitted by the source (perhaps this is what you meant by "constant intensity") across all frequencies, which is not necessarily true for a real light source. If this is the case, then the photoelectric current will indeed be inversely proportional to the frequency of light, when the latter is above the threshold frequency.
In the general (steady) case, the photoelectric current is as such directly proportional to the rate of photon incidence and independent of the frequency of incident light, with the assumption that one can independently control these two parameters. This is the case that one normally considers, and here, the current vs frequency graph (at constant rate of photon incidence) steps up to a constant value from zero when the frequency increases beyond the threshold frequency.