My textbook says that there will be no impact but that does not feel right to me. The energy of each photon must increase so the no. of photons will decrease and the no. of ejected electrons will decrease as well. So, shouldn't the photoelectric current decrease?
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$\begingroup$ Just a slight correction to your headline. Photons don't have frequencies. The frequency we are talking about in the photoelectric effect is the frequency of the classical electromagnetic wave. An individual photon only has energy, momentum and angular momentum. Many photons can form classical waves and those waves can have a frequency, if they are sufficiently monochromatic (otherwise they are a mix of many waves with different frequencies). I know, it's pedantic, but it's necessary to understand the difference between one photon and a wave that consists of many photons. $\endgroup$– FlatterMannCommented Feb 1 at 14:36
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$\begingroup$ Their definition of intensity tries to imply the number of photons per second is the same ..... but the correct definition of intensity includes energy ... as you noticed. $\endgroup$– PhysicsDaveCommented Feb 1 at 18:32
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For a textbook idealization, you're right. In reality, the efficiency of the process tends to increase with photon energy. And with enough photon energy, the photoelectron can interact with bound electrons in the material, producing multiple free electrons per photon.
