I googled it a bit and found that photoelectric current is independent of frequency(of incident light). Some further look revealed that actually "saturation current" is independent of frequency.I could not find about the instantaneous current(current other than saturation current).
Speculation 1: If saturation current is not reached, then radiation with higher frequency will give greater photoelectric current.
Reason 1: Greater frequency means greater velocity of electrons, which will help them to counteract "space charge" and more electrons can reach the anode.
Problem 1:Let's say the intensity (W/m²) of the radiation remains the same, and only the frequency increases. The intensity multiplied by the Area of the plate results in the total energy that arrives the plate in each second. So IA = E/Δt, where E = nhf (n photons of frequency f) Let's say each photon is able to pull out one electron from the plate, so the current i = ne/Δt, where e is the charge of the electron. That gives n/Δt = i/e, and IAe=hfi -> i = IAe/hf, so when we increase the frequency, if the intensity of the radiation remains the same, the current decreases. Is this right?
Are my speculation and reason correct? And please help me to resolve my problem ?
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