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In photo-electric effect Einstein said that photons incidents on material and gives their energy which will gives kinetic energy to electrons. But i also want to know that why Compton's effect not works in this situation. In my view when photon incident on material it should eject a electron as well as a photon of less energy than incident photon.

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In some cases it does eject a photon with a lower wavelength, if it did not do this then the laws of conservation of momentum would not be supported thus disproving many aspects of modern physics. The problem with this is without the experimental evidence or data, it is hard for someone to calculate or predict the new photons wavelength, let alone detect it. The photon would be emitted with such a low energy that the problem would be "How would you detect it?"

You could test the discrepancy of how much energy the electron should have compared to what it does have but even that would be difficult.

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  • $\begingroup$ "if it did not do this then the laws of conservation of momentum would not be supported thus disproving many aspects of modern physics." How so? A photon scattering having the same wavelength before and after would just be inelastic scattering, which definitely does not violate conservation of momentum. $\endgroup$ Feb 7 '16 at 9:40
  • $\begingroup$ If no photon was scattered then conservation of momentum would not be conserved. The problem with the photon keeping the same wavelength is then it did not lose any energy meaning that conservation of energy was not conserved. $\endgroup$ Feb 7 '16 at 17:31
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Compton scattering occurs on free electrons, i.e. not in a bound state. The equivalent to a Compton scatter would be a scattering of a photon off the field of a solid, momentum and energy balance happening collectively with the total mass of the solid. In the best case it would be a whole atom that the photon would scatter inelastically off, the atom taking the momentum balance and a reduced frequency photon leaving. This would give a continuous spectrum, no cut offs. This could happen only if the Thomson model of the atom were correct , the electrons freely distributed in a continuum . It would not give rise to the photoelectric effect, which gives a discrete electron momentum and implies quantized states for the electron.

In fact the photoelectric effect is one of the lynch pins of proof of the quantized condition of nature in the microcosm. The energy of the photon is constrained by the quantized binding of the electrons, as with a lower energy of the photon no electrons appear.

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    $\begingroup$ I don't think "Compton Scattering" necessarily has to be off of free particles or electrons specificially. It's just an effect seen in the scattering off of charged particles. I think that it's just usually treated in the special case of a free particle as an easy simplification. $\endgroup$ Feb 7 '16 at 9:38
  • $\begingroup$ @aquirdturtle If the particle it scatters off is bound, one needs a photon with the ionization energy, which means quantized steps and the photon is absorbed in the process. It is a specific feynman diagram. see the link $\endgroup$
    – anna v
    Feb 7 '16 at 10:46
  • $\begingroup$ that's not true. The photon needs not have the ionization energy and it needs not be absorbed in the process. Your comment seems to imply that photons that aren't resonant with specific transitions can't scatter off of or interact with atoms. Your link definitely refers to the bound case, but that doesn't mean there aren't other cases involving bound particles which involve the same phenomena. $\endgroup$ Feb 7 '16 at 10:49
  • $\begingroup$ @aquirdturtle my comment addresses the photoeletric effect, which is not compton scattering. Of course photons can scatter off the field of atoms, that is not what the question is about, imo $\endgroup$
    – anna v
    Feb 7 '16 at 10:53
  • $\begingroup$ I'm referring to your statement "Compton scattering occurs on free electrons, i.e. not in a bound state." And the link you gave when saying "If the particle it scatters off is bound, one needs a photon with the ionization energy, which means quantized steps and the photon is absorbed in the process" clearly refers to the Compton effect. $\endgroup$ Feb 7 '16 at 10:54

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