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In Photoelectric Effect of Theory of Spectral Lines , an electron takes the entire or none of the energy of the Photon ( it absorbs the entire quanta not its fractionS resulting in the disappearance of the Photon).

But in Compton Effect the Electron takes only a fraction of the Photon Energy and the Photon stays alive.

What I specifically want to know is that Quantum theory says that an electron will either take the whole or none of the Photon energy .

Why then in Compton Effect the electron takes a fraction of the Photon energy Is that allowed by Quantum theory ?

Or is not so ? Or the electron takes up 1 photon only when in an energy level . When an electron is free , can it absorb half the energy of a photon ?

Edit:

I still don't get whether the electrons are allowed to take the energy of the whole photon when they are in energy levels and can they absorb a fraction of the photon energy when they are free

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  • $\begingroup$ Related: physics.stackexchange.com/questions/231173/… and physics.stackexchange.com/questions/31581/… $\endgroup$ – user146020 Mar 6 '17 at 21:10
  • $\begingroup$ @Countto10 But the question is why for a bound electron it is Photoelectric Effect and for free electron its Compton Effect $\endgroup$ – Shashaank Mar 6 '17 at 21:13
  • $\begingroup$ Hi, my references are not meant as duplicates,sorry, just as background. What is annoying me is I wrote an answer to almost this question some time ago, but the search engine here does not list it, if I can find it on Google I will list it. $\endgroup$ – user146020 Mar 6 '17 at 21:20
  • $\begingroup$ @Countto10 Yes please . If you can find the answer please list it. What I specifically want to know is that Quantum theory says that an electron will either take the whole or none of the energy . Why then here it takes a fraction of the Photon energy $\endgroup$ – Shashaank Mar 6 '17 at 21:22
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    $\begingroup$ I don't know if you can justify it that way, but you could go on and say that in a compton scattering the photon absorbs the whole photon, then re-emits a photon but in a different direction and with different energy. This is not actually how it happens but I guess it kind of justify it. $\endgroup$ – Frotaur Mar 7 '17 at 7:29
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I still don't get whether the electrons are allowed to take the energy of the whole photon when they are in energy levels and can they absorb a fraction of the photon energy when they are free

Electrons may be free, or bound within an atom due to the relative potential between the nucleus and the electron, for a simple case, i.e. the hydrogen atom.

When bound in an atom a photon with an exact energy difference ( within the width) to the ionization level can be completely absorbed by the atomic system and an electron will go free and the atom will recoil. This kinematically is a two body situation. Before it is "photon + atom" after "electron + atom". There exists a physical center of mass system.

In two body interactions energy and momentum are conserved by the end particles, in this case an ionized atom and an electron.

When an electron is free , as in compton scattering, the system is "Photon1 + electron" as initial state, and the final state is "photon2+electron" There exists a consistent physical center of mass system in both cases.

Energy and momentum are conserved from initial to final state.

If photon2 is a zero vector, the initial center of mass will have the invariant mass of "photon1+electron" which will be larger than just the mass of the electron, and the final one will have just the invariant mass of the electron. This is inconsistent with energy conservation and cannot be physical.

And how energy and Momentum will not be conserved when the bound electron is hit by a photon of energy less than the energy difference between any two consecutive cells and then the electron gets out of its shell

In quantum mechanics, the atom is a bound system to which energy can be given in quanta, corresponding to the energy levels.

If the transition is "Photon + atom" going to "excited atom", i.e. the electron changed energy level, energy is conserved because now the atom has more energy, the one the photon gave up. In the center of mass before, momentum is zero, and after momentum is zero and the energy before is equal to the energy after because the atom is at an excited level carrying the photon energy.

The case of photon +electron ---> electron cannot conserve energy, because at the center of mass there is the energy of the photon +the energy of the mass of the electron, incoming, but only the energy of the mass of the electron outgoing, the electron does not have quantum mechanical excited states to increase its rest mass energy, as the excited atom does.

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  • $\begingroup$ You mean to say that an electron will absorb a part of the photon's energy only when it is free. When it is bound in an atomic shell it will absorb the all photon's energy ( the the difference being equal to the difference in two. successive energy State , again energy conservation). This is all because of the conservation laws. So the electron which is in a shell can absorb a full photon only because then the conservation laws will be true . So it's necessary for an electron in an orbit to absorb a full photon so that energy conservation can be true is that right ? $\endgroup$ – Shashaank Apr 11 '17 at 11:37
  • $\begingroup$ If yes....Then when x rays are allowed to fall on a metal like in a compton scattering experiment then there should be both Compton scattering and Photoelectric effect taking place because in a metal there will be both bound and free electrons.... $\endgroup$ – Shashaank Apr 11 '17 at 11:39
  • $\begingroup$ Actually it is the system that absorbs the energyy in the bound case, not just the electron .The atom is a whole system, the absorbed photon energy kicks out the electron and the (diminshed)atom recoils ( though the difference in the masses makes this recoil very small). $\endgroup$ – anna v Apr 11 '17 at 13:36
  • $\begingroup$ electrons are bound in the metal, they are bound to the lattice as a whole, in energy levels see hyperphysics.phy-astr.gsu.edu/hbase/Solids/band.html . hyperphysics.phy-astr.gsu.edu/hbase/mod2.html $\endgroup$ – anna v Apr 11 '17 at 13:43
  • $\begingroup$ If you look at this table colby.edu/chemistry/PChem/notes/AOIE.pdf you will see that the atomic binding energies are many electron volts, but the work function in the second link above is less than 1 ev. in the beginning of the line. It is the binding in the conduction band of the electrons. $\endgroup$ – anna v Apr 11 '17 at 13:47
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There is a simple reason why a free election can't absorb a photon completely: you can't conserve both energy and momentum for the system of you start off with an electron and a photon and and up with only an electron. You need a final photon as well for conservation of energy and momentum to be satisfied.

For the photoelectric effect things are different. We don't have a free electron ; it is bound to a nucleus. Consequently it can only have certain precise energy values (and won't interact with photons that would move it to a non existent energy level, which is the basic argument for quantum behaviour, but not really comparable to Compton scattering).

In the situation analagous to Compton scattering, the photon has more energy than the binding energy of the electron, and so we wind up with a free electron. But in this case energy and momentum can be conserved without a final photon since the nucleus is also involved in the interaction. The initial state is {photon, bound electron, nucleus } and the final state {free electron, nucleus}.

It is possible to construct Feynman diagrams with a final photon present too, but since they have an extra vertex they happen less often by a factor of roughly the fine structure constant $\alpha\sim\frac{1}{137}.$

Or to put it another way, sometimes in the photoelectric effect where you end up with a free electron you do get a final photon, but it is in less than 1% of cases (unless I've overlooked some reason why it can't happen).

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  • $\begingroup$ 2 questions - Firstly the absorption of a full photon ( the basis of Bohr theory) is only valid for the Electron bound in a shell because it can be only in certain energy levels. And a free electron need not take the full energy of the photon because because it is not constrained to Move in certain energy levels. Is my understanding is right ? Secondly Compton effect is for free electron and photo electric for bound one. But in the photo electric , the Electron ultimately becomes free whereas in Compton the Electron is bound . So the 1 reasoning still ok $\endgroup$ – Shashaank Mar 7 '17 at 14:18
  • $\begingroup$ @Phillis Isn't the graph of absorbance of photon vs photon frequency going to have some troughs, for the material's quantum behaviour? $\endgroup$ – Mockingbird Mar 8 '17 at 10:05

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