How does the photoelectric current vary with increase in amplitude of the incident photon? In Photoelectrc effect, increasing the amplitude of the incident light increases the amplitude of a single energy packet. So, increasing the amplitude will not increase the number of photons. But intensity is proportional to amplitude squared . And increasing the intensity increaes the number of incoming photons . In a nutshell, does increasing the amplitude incerase the number of photons?
What am i missing? 
Thank you for your answers
 A: The way you are thinking has got a tiny bit of problem. What you are doing is mixing waves with particles. You are defining the Amplitude in terms of waves, but the intensity in terms of particles. 
Since photoelectric effect is basically explained using particle nature of light, let's do it that way.
If we use a very crude formulation of quantum mechanics, then we can say that the photons have a wave function associated with them, and the amplitude is just the maxima or minima of the function, and is not physical. But what is physical is the square of the amplitude which corresponds to the photons being found at a place (according to the Born interpretation). So, when you increase the amplitude, you are basically increasing the probability of finding a photon, or crudely, increasing the "number of photons" (although the last words makes almost no sense in general).
Anyways, once we increase the number of photons, the number of ejected photoelectrons increase (considering a frequency greater than the cutoff frequency), and this in turn increases the photocurrent.
