In photoelectric effect the scattered photon have less energy than in Compton effect , in photoelectric effect the photon hit the electron in inner shell but in Compton the photon hit the electron in outer shell. How that?
closed as unclear what you're asking by Red Act, Pieter, Jon Custer, sammy gerbil, Ben Crowell Apr 3 '18 at 19:45
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Seems to me a confusion of two experimental set ups.
a) the phototelectric effect, one of the experimental observations that made quantum mechanical theory inevitable
The minimum energy required to eject an electron from the surface is called the photoelectric work function.
The fact of a minimum energy needed for ejecting electrons from a surface agreed with the necessity of a particle nature, quantization , of light and energy levels in materials.
b)The generic term "photoelectric" , i.e. interactions of photons with bound electrons, which is seen in the answer by Floris in the proposed duplicate.
The K etc states excited are found with x-rays with the appropriate energy for the K etc bound states.
The Compton effect is not quantized, it is on free electrons, or assumed to be as free; note it is xrays used in that first Compton experiment. The photoelectric effect is with visible or ultravioelet light, much lower energies.