Using the photoelectric experiment,it is known that if light of same intensity but different frequency is used, stopping potential is changed and current changed. For instance,same intensity but frequency increased. E=hf, energy of photon increased which leads to a higher K.E of photoelectrons,hence stopping potential increased. But why is the current affected,in this case decreased. I can deduce it from the formula Intensity= (Number of photons)(Energy of each photon)/( Time times area). For I to be constant, number of photons must decrease since energy of each photon increases. But i do not understand the concept behind it. My arguement is that K.E of photoelectrons increases,but the distance between each photoelectrons is still the same,which meant that number of photoelectrons arriving at the other end per area is still the same,meaning that intensity is constant even if frequency is increased?