Does, and if so, why does the frequency of light and wavelength of light affect the photoelectric current? It makes sense that intensity of light affects the photoelectric current, but what about the frequency and wavelength, given that intensity remains constant?
The formula for intensity would be I = nhf/tA (where n is the number of photons and hf the energy of one photon, t the time, and A the area).
Now, let's say that we double the frequency, we get 2f, but since I is constant, n/t must halve, hence the current must be smaller. But apparently it increases or has no effect according to my research in the internet, what went wrong in my calculation? If we reduce the wavelength, the current decreases, right? Many thanks in advance!
 A: Your reasoning is correct. Assuming the frequency is high enough that the photon energy is greater than the work function, if the intensity is kept constant, increasing the frequency must decrease the photon flux, and thus, the photoelectric current.
A: We need to be really careful in our definitions - as @Stealth849 points out, it is very easy to confuse what is photon flux, and intensity.
Flux is the number of photons which strike a unit area within a given time interval.
Intensity is the amount of energy per unit area within a given time, and the two are related by individual photon energy $hf$. We can say (where $\Phi$ is the flux, and $I$ the intensity):
$$\Phi = \frac{n}{A\Delta t} \space\space\space \rightarrow \space\space\space I = \Phi hf=\frac{nhf}{A\Delta t}$$
The flux describing the number of incident photons dictates the resulting photocurrent so long as the photon energy is high enough to free an electron from the surface.
Now with the intensity, you're making the connection that $f\rightarrow2f$, and thusly, $\Phi\rightarrow\frac{\Phi}{2}$. You are making the assumption that when you increase the frequency of photons, you will only be sending half as many photons in order to preserve your desired constant intensity.
Given that your photocurrent is $I_P=ne$ where $e$ is the electron charge, we then notice that the photocurrent would decrease if you decrease the number of photons (flux).
But given the scenario where you do not require the intensity $I$ to be constant, and you increased the frequency of the incident photons while retaining the original flux, you would not notice a change in the photocurrent so long as the incident photons are above the cutoff frequency.
