I Bound electrons vs. free electrons

[drl]

As I understand the photoelectric effect as given by Einstein, an electron while attached to an atom is considered to be in a bound state and remains as such until a photon carrying a quantum of energy sufficient to overcome the work function of the electron frees the electron from its bound state. It would seem logical to me to refer to the electron as being in an excited state or being in its wave aspect. In the 2 slit, the electron approaching the 2 slits is considered to be a particle but, in the process of becoming a free electron it would make more sense to consider it as an excited electron in its wave aspect and thus explain the resulting interference pattern.Go a step further and when the detector is turned on, the electron passing thru the electromagnetic field loses enough kinetic energy to revert back to its bound energy status or particle form.Yes I know that particle form and wave forms are no nos. Where did I go wrong?.

 

[Simon Bridge]

I think you have been getting confused by terminology.
The electric potential at infinity is defined to be zero. A free electron has energy bigger than zero.
A bound electron, strictly, does not have groundn and excited states: the atom (or the system it is bound to) is what has the excited etc states.
The wave description applies equally to particles that are bound and to particles that are free - same with the article description: both are incomplete.

It is possible to have an electron that is sort-of free in the sense that it is not bound to an individual atom but may be bound to a bulk material like a metal. The photoelectric effect, and the concept of a work function, applies to metals. An electron bound to a metal can be promoted to free space by absorbing energy equal or greater than the work-function.
In an atom or a molecule, the equivalent concept is "ionization energy".

Can you see now why it makes no sense to consider a free electron to be "in" an excited state of an atom?

 

[vanhees71]

https://www.physicsforums.com/insights/sins-physics-didactics/

Committing this sin leads to ideas as proposed in #1 of this thread ;-)). I hope the insights article cures the sins!

 

[drl]

Don't want to be a pain in the butt, but will try to eliminate confusing terminology. The free electron has more kinetic energy than a bound electron and translates into the interference pattern in the 2 slit experiment. "Go a step further" explains the detector results. Any chance any of this makes sense?

 

[vanhees71]

I've no clue what you are talking about. A free electron has a higher total energy than a bound electron. The interference pattern in the two-slit experiment is found by solving for the Schrödinger equation for this case. It's a similar calculation as you evaluate the interference pattern for light waves going through a double slit.

 

[drl]

I've no clue what you are talking about. A free electron has a higher total energy than a bound electron. The interference pattern in the two-slit experiment is found by solving for the Schrödinger equation for this case. It's a similar calculation as you evaluate the interference pattern for light waves going through a double slit.

Lets try this again.A free electron has more kinetic energy than a bound electron and consider that in this state shows its wave form. When the two slit exp. is done,the electron is considered to be in its particle form since a wave can only come from a particle. O.K. but as I postulated, the wave started from a free electron( which has more energy than a bound electron) and passes thru the 2 slits producing an interference pattern as a result. When the detector is turned on, an electric field is produced which drains off some of the kinetic energy of the electron changing it back from its wave aspect to its bound energy status resulting in the loss of the interference pattern.What it comes down to is that a free electron forms a wave because it has more kinetic energy than a bound electron.There seems to be a world of difference in how wave particle duality is interpreted.

 

[ZapperZ]

Lets try this again.A free electron has more kinetic energy than a bound electron and consider that in this state shows its wave form. When the two slit exp. is done,the electron is considered to be in its particle form since a wave can only come from a particle. O.K. but as I postulated, the wave started from a free electron( which has more energy than a bound electron) and passes thru the 2 slits producing an interference pattern as a result. When the detector is turned on, an electric field is produced which drains off some of the kinetic energy of the electron changing it back from its wave aspect to its bound energy status resulting in the loss of the interference pattern.What it comes down to is that a free electron forms a wave because it has more kinetic energy than a bound electron.There seems to be a world of difference in how wave particle duality is interpreted.

Then you will have a heck of a time trying to explain why I can also get a similar interference pattern in a superconducting quantum interference device. Here, these electrons are STILL inside a conductor, and they never came out. Where are you "free electron" there, since you already considered them as "bound" inside the solid?

Zz.

 

[drl]

Am not knowledgible of this exp. but let's go back to the 2 slit . Cover one slit and fire electrons individually. We would get one bar on the screen. Now introduce a detector to the exp. and we would still get one bar forming. Is there any way to determine if the electrons in the second case had less kinetic energy than the electrons of the first case where no detector was used? Can it be that an electron in its free form acts as a wave and as a particle when its kinetic energy is reduced to the level it had in its bound state?Thanks for your time and efforts.

 

[ZapperZ]

Am not knowledgible of this exp. but let's go back to the 2 slit . Cover one slit and fire electrons individually. We would get one bar on the screen. Now introduce a detector to the exp. and we would still get one bar forming. Is there any way to determine if the electrons in the second case had less kinetic energy than the electrons of the first case where no detector was used? Can it be that an electron in its free form acts as a wave and as a particle when its kinetic energy is reduced to the level it had in its bound state?Thanks for your time and efforts.

This makes no sense. I've given you an example that smashed your scenario. Claiming ignorance doesn't cut it. You should have gone and dig it up and figure out what it is rather than stubbornly continuing along this line. It gives the impression that you do not wish to learn!

Zz.

 

[Dale]

Thread closed due to personal speculation.