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New Theories / New solution to the Double slit experiment and to the photoelectric effect
« on: 04/01/2017 16:00:20 »
In the double slit experiments, in terms of Quantum Mechanics (QM) , it is often said that the measurement collapses the wavefunction or more exactly, the measurement destroys the interference. However, we can also think, that is not the measurement that destroys the interference ( except for some experiments). The delayed choice quantum eraser can be used as a proof. Scientists developed the DCQE (Delayed Choice Quantum Eraser by Kim et. al) experiment, where there is no measurement involved at the slits, but they went forward with QM believing that even simply watching the results destroys the interference. If QM is all correct, this is a rational conclusion. However, I think it is not the case. The theory is probably wrong. Apparently, the detectors cannot detect the energy of a photon, if the energy that reaches the detector is by a certain degree less than the electron bind energy . QM says, that there is a significant probability of detection when the wave passes through both slits to have a photoelectron release on either detector A or B. That must be wrong. The probability is almost zero. Unless the other part of the wave hits the orbiting electron that still holds the energy from the previous electromagnetic wave in a very short time, the atom will not release the electron. Let's suppose we have a photon with energy Eϒ. After it passes the slits we get two waves of EϒL and EϒR. Say, the energy required for an atom in ground state to release an electron is Ee- ( this is the workfunction). If a wave with energy EL hits the atom, it will energize it to an intermediate unstable electron orbital, but the atom will not release the electron yet. Since the intermediate state does not correspond to a stable electron orbital, it will quickly loose energy. However, if the wave ER comes quickly enough, there will still be enough energy left in the intermediate unstable orbital and ER+EL>Ee-. Hence the electron will be released.
Notice, that the energy of a photon when it was just released will always exceeds the workfunction (the energy Ee-),otherwise the frequency of the wave is not enough to trigger an photo-electron release. We can, then say that this mechanism should work at any frequency of the photon. More exactly, a wave of any frequency could trigger an electron release, because photons can add up energies to the atom and excite it until releasing the electrons. It is clear that it doesn't since the experiments show that the frequency matters. However, we can then understand that for a photon to interfere with an electron, it needs a minimum frequency. It is possible to explain how this happens. A low frequency wave interferes with an electron attached to an atom as well, just like a high frequency does. However the energised electron , looses energy quicker than the low frequency wave is able to provide the energy for releasing an electron.
Supposing we have a Hydrogen atom.
We can write: EA(t)=EA0+Eϒ- PA*t where, EA0 is the energy of the atom when the electron is a stable configuration, like say first orbital
t is measured in absolute seconds, according to this concept. All symbols for time denote absolute time E(t) is the energy stored in the orbiting electron, which is a function of time. EA is the energy received from an electromagnetic wave. PA is the radiation power of the atom in the unstable state.
Eϒ = Pϒ tA
tA is the emission time of an electromagnetic wave. Eϒ is the energy produced by an electromagnetic wave.
We can also write: Eϒ(t) = Pϒ t
Pϒ- power produced by an electromagnetic wave This power depends on the frequency:
Pϒ=kυ k is a constant and υ is the frequency of the wave. We can now write:
Eϒ(t) = kυ t To produce a photo-electron, the energy EA needs to reach Ee
EA(t)=EA0+ kυ t -PA t
In conclusion, to produce photo-electron, the electromagnetic wave must have a frequency that satisfies the following inequality: υ >= PA /k
These equations need to be expressed in absolute time. Applying relativistic equations to them will generate wrong results in certain situations, because the relativistic effect are generated automatically by the wave behaviour of all particles.
It is important to know that photon emissions are in fixed packets of energy, coresponding
to electron orbital transitions.
The transition times are very small.
A more thorough description implies power densities and wave interference. However these effects seem negligible since it doesn't seem to be possible to create a high amplitude low frequency wave that can extract a photoelectron.
In the DCQE experiment, to get a detection at D3 or D4 the wave must recombine quick enough so that there is no time for the energised electron to revert to initial state. It all has to do with the coherence length.
Notice, that the energy of a photon when it was just released will always exceeds the workfunction (the energy Ee-),otherwise the frequency of the wave is not enough to trigger an photo-electron release. We can, then say that this mechanism should work at any frequency of the photon. More exactly, a wave of any frequency could trigger an electron release, because photons can add up energies to the atom and excite it until releasing the electrons. It is clear that it doesn't since the experiments show that the frequency matters. However, we can then understand that for a photon to interfere with an electron, it needs a minimum frequency. It is possible to explain how this happens. A low frequency wave interferes with an electron attached to an atom as well, just like a high frequency does. However the energised electron , looses energy quicker than the low frequency wave is able to provide the energy for releasing an electron.
Supposing we have a Hydrogen atom.
We can write: EA(t)=EA0+Eϒ- PA*t where, EA0 is the energy of the atom when the electron is a stable configuration, like say first orbital
t is measured in absolute seconds, according to this concept. All symbols for time denote absolute time E(t) is the energy stored in the orbiting electron, which is a function of time. EA is the energy received from an electromagnetic wave. PA is the radiation power of the atom in the unstable state.
Eϒ = Pϒ tA
tA is the emission time of an electromagnetic wave. Eϒ is the energy produced by an electromagnetic wave.
We can also write: Eϒ(t) = Pϒ t
Pϒ- power produced by an electromagnetic wave This power depends on the frequency:
Pϒ=kυ k is a constant and υ is the frequency of the wave. We can now write:
Eϒ(t) = kυ t To produce a photo-electron, the energy EA needs to reach Ee
EA(t)=EA0+ kυ t -PA t
In conclusion, to produce photo-electron, the electromagnetic wave must have a frequency that satisfies the following inequality: υ >= PA /k
These equations need to be expressed in absolute time. Applying relativistic equations to them will generate wrong results in certain situations, because the relativistic effect are generated automatically by the wave behaviour of all particles.
It is important to know that photon emissions are in fixed packets of energy, coresponding
to electron orbital transitions.
The transition times are very small.
A more thorough description implies power densities and wave interference. However these effects seem negligible since it doesn't seem to be possible to create a high amplitude low frequency wave that can extract a photoelectron.
In the DCQE experiment, to get a detection at D3 or D4 the wave must recombine quick enough so that there is no time for the energised electron to revert to initial state. It all has to do with the coherence length.