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Yes, because the energy per photon, E=(h)(frequency), so light with lower wavelength/higher frequency has higher energy per photon. Each photon interacts with one electron and must have enough energy to eject one electron, which is why light with higher frequency ejects more electrons.
The one to one relationship between photons and electrons accounts for this. By looking at the equation we can determine that a shorter wavelength means higher frequency and thus more energy per photon, allowing it to eject one electron. It doesn't matter if the overall energy is more; if the energy per photon is not high enough, electrons will not be ejected.
Photons with longer wavelength (lower frequency) do not have enough energy to meet the threshold energy, which is the energy required to remove an electron from a metal surface. Unless the E (photon) is greater than or equal to E (energy to remove an electron), then an electron is not emitted, even for high intensity light. The only way to eject an electron from a metal surface is by increasing the frequency and energy of the photon, NOT by increasing the intensity of the light. This is because increasing the intensity of the light would mean to increase the number of photons shining on the metal. However, having more photons with wavelengths that still do not have the right amount of energy to reach threshold will never result in an electron ejecting from the metal surface.
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