Electrons and protons

They do not. The revolution was a simple analogy to describe the probability distribution of an electron near a nucleon.

 

Nor that :)
Electrons, just like any other atomic particle, don't just exist in one specific point in space. They exist as a "cloud", and it's that cloud that envelops the nucleus.
Also note that the shapes of the clouds aren't spherical in most cases. Only the first "orbital" (i.e. the orbital of the electron in a hydrogen atom) is spherical. The more electrons around a nucleus, the more elaborate those shapes get. (e.g. http://chemwiki.ucdavis.edu/@api/deki/files/8855/Single_electron_orbitals.jpg )

 

A classical analogy (although a poor one) is the solar system. The planets go around the sun becaue the sun is much heavier. Electrons are much lighter than nuclei.

 

Quantum mechanics is only really prepared to tell us what we are likely to find if we measure something. What is really going on underneath it all is quantum metaphysics, which, though interesting, is far from being resolved.

The best we can do to describe what an electron is doing is to know what its wavefunction is. That will give us the probability of any outcome of any measurement we could to to look at the electron.

Oddly, enough, we can have wavefunctions of electrons which orbit the nucleus like planets in a solar system. These are highly excited states, where the electrons can be described classically. Hydrogen atoms in such highly excited states are known as Rydberg atoms.

 

Electrons are stationary wave
They are not noving

 

Not true in general. You can see this by calculating the expectation value of the momentum of the electron from its wavefunction, and finding that the momentum is in general nonzero.

 

The expectation value of the momentum [itex]\langle p \rangle[/itex] of an electron in a single orbital [itex](n, \ell, m_{\ell})[/itex] is indeed zero. This can be seen from Ehrenfest's theorem, and that the expectation value of the position is a constant for one of these eigenstates.

However, the variance of the momentum [itex]\sigma_{p}^{2}[/itex] is nonzero even in these states, which means that you are very likely to find the electron having a nonzero momentum if you actually measure it (it's only zero "on average").

Also, most electrons (I would think) exist in a superposition of different orbitals (i.e., their wavefunctions are not stationary, even if they are still concentrated around the nucleus). In this case, the expectation value of the momentum may change in time all sorts of ways.

 

Thank you for the clarification.

 

Okk

i will check
But your question is really awesom

 

This may be nitpicking, but even in eigenstates [itex]\langle p \rangle[/itex] is only zero in one inertial frame---the one in which the nuclear core to which the electron belongs is at rest. If the entire atom is moving (core and all), the electrons have non-zero expectation values of momentum, too.

This is of course not very enlightening: In classical mechanics, the expectation value of momentum of an electron revolving around an stationary infinite mass is zero, too (both in the time-average and statistical average sense momentum averages to zero). This does not mean that there is no instantaneous movement. But, unlike classical mechanics, quantum mechanics does not actually tell us anything beyond expectation values. Thus, the best answer to the question of "does the quantum mechanical electron move?" was given in post #6 (jtbell) ND #7 (jfizzix): This question cannot be answered within the model of quantum mechanics.