Electrons in the nucleus of an atom orbit

[garytse86]

hello people. I am kind of stuck in this problem:

electrons in the nucleus of an atom orbit around the nucleus and accelerate. They will also continously emit electromagnetic radiation and therefore they must lose energy to do so because of the conservation of energy. As a result, wouldn't electrons eventually crash into the nucleus because of their lack of energy and the electrostatic attraction between and electron and the nucleus.

What I am thinking of is that the electrons have a ground state where they cannot lose energy anymore, and thus won't crash into the nucleus? Please correct me if I am wrong, but can you give me a very detailed explanation of this?

Thank you very much.

Gary

Don't drink and derive. Alcohol and Calculus don't mix.

 

[chroot]

You essentially have the answer. Electrons in atoms cannot have ANY energy -- only specific energies. A physicist would say the energy levels are discretized. There is some minimum energy, and the electrons simply have no lower energy states to go to.

- Warren

 

[NEOclassic]

A small correction

Originally posted by chroot
You essentially have the answer. Electrons in atoms cannot have ANY energy -- only specific energies. A physicist would say the energy levels are discretized. There is some minimum energy, and the electrons simply have no lower energy states to go to.

- Warren

Hi Warren,
Implicitly,you have corrected the original post that spoke of "electrons in the nucleus". There are electrons in the nucleus as evidenced by beta +/- emissions and k-level electron captures but I believe, as I suppose you do, that the youngster's posting was concerned with the quantum orbital electrons outside the nucleus.

You failed to mention that when an xray kicks out the least energetic k-level electron,an l-level electron makes a "quantum leap" to the k-level. Of course an m-level electron jumps to the l-level, etc in a spectral cascade. The emitted photon occurs as a consequence of the abrupt stopping of each jumping electron. Cheers, Jim

 

[jimmy p]

I think the problem with this is that whatever you do to the atoms they are going to be provided with energy. If they were at absolute zero then they would sit there as there would be no energy for the electrons to collide with the nucleus, but much the same with Heisenbergs Uncertainty Principle, as soon as you look for the particle, to see it would require light which would provide the atom with energy so it would move.

I think that most radiation sources will always provide enough energy ANYWAY so electrons would never 'run out' of energy and collide with the nucleus. And if they didnt provide enough energy then you would never be able to observe it anyway

 

[deaddevilsdisguise]

there r 2 forces acting in an atom...
i) electrostatic force(repulsive force)
ii) neuclear force(attractive force) [im not sureabout this, correct me if I am wrong]

it is the electrostatic force which keeps the Elctron from falling into the neucleas and the neuclear force which keeps the Electron from falling out...

due to the Phenomenon of electron transfer it is clear that the neuclear force is not very strong in keeping the electron from falling out but it is the reason for which the neucleas stays intact!

 

[chroot]

Sorry, deaddevilsdisguise, that's all pretty much completely wrong.

The electromagnetic force attracts the negatively-charged electrons to the positively-charge nucleus. The reason it doesn't get sucked all the way in is because the nucleus+electron system only permits discrete energy levels. This is a quantum-mechanical phenomenon that exists in all kinds of systems -- if a particle moves in some potential (like an electromagnetic field), the particle can have only discrete values of energy. There is a lowest possible energy, called the ground state, and the electron cannot go lower than it.

The strong nuclear force works only inside the nucleus. Electrons do not feel the strong force at all. The strong force also operates over an incredibly small distance, much smaller than the distance between electrons and nuclei.

- Warren

 

[lethe]

Originally posted by chroot
The reason it doesn't get sucked all the way in is because the nucleus+electron system only permits discrete energy levels.

just to be pedantic, the electron nucleus system does indeed permit continuous energy levels, as well as discrete. but the bound states all have discrete energy levels (as any bound state must)

This is a quantum-mechanical phenomenon that exists in all kinds of systems -- if a particle moves in some potential (like an electromagnetic field),

not just any potential (electromagnetic or otherwise), only potentials of bound states exhibit this phenomenon.

 

[chroot]

Right, bound states.

- Warren

 

[lethe]

Originally posted by chroot
Right, bound states.

sorry, just being pedantic.

 

[garytse86]

yes electrons can only occupy certain energy levels but why?

 

[NEOclassic]

Why not?

Originally posted by garytse86
yes electrons can only occupy certain energy levels but why?

Hi Gary,
AFAIK,the energy separation between neighboring ground state 2-electron orbitals is a natural phenomenon which was evaluated experimentally, originally in the study of black-body radiation by Einstein and Planck. Offhand, I would surmise that the photon frequency resulting when an L-shell electron quantum jumps to a vacated position in the K-shell would validate Planck's constant. Cheers.

 

[NEOclassic]

Dipolar forces of a spinning electron explained

Originally posted by garytse86
hello
electrons in the atom orbit around the nucleus and accelerate. They will also continously emit electromagnetic radiation and therefore they must lose energy to do so because of the conservation of energy. As a result, wouldn't electrons eventually crash into the nucleus because of their lack of energy and the electrostatic attraction between and electron and the nucleus.

What I am thinking of is that the electrons have a ground state where they cannot lose energy anymore, and thus won't crash into the nucleus? Please correct me if I am wrong, but can you give me a very detailed explanation of this?

Hi again Gary,
An absolute and entirely natural phenomenon associated with an intrinsically spinning electron is that the dipolar magnetic force is exactly countered by the torque force attributable to the intrinsic unit mass of the electron; i.e., RPF's "up" force is opposed by the natural "down" magnetic polarity. When two electrons try to form a quantum orbital they are electrostatically repulsive and magnetically speaking, Pauli told us that they must somehow differ; RPF who has taught us that unitary mass is an insignificant force argues that inertial torque is a proper force that accrues simply by flipping one to explain why the pair is Pauli compatible. On the other hand, flipping also accrues a magnetic attractiveness that is overwhelmingly greater than the inertial attraction. Regardless of perception the quantum orbit happens and the radial acceleration of the electrons does not radiate (lest the energy lost would destroy the orbital). In this case the circular path of the electrons creates opposing "loop" forces one of which must be loop-magnetic which, again, is exactly opposed by the loop-torque. Classical physics reveals this phenomenon more clearly than does QED's referral to it as merely a "standing wave" - whatever that might mean? Cheers.

 

[SCfool_BtK]

REsponse to gary

Gary SCFOOL here to let you know the only reason that's keeping electrons from colliding into neucleus is be cause of the magnetic force between it and the protons. if anyone thinks this is a bit wiered on what i said gim me a reply

 

[garytse86]

surely the magnetic force hasn't got discrete values?

 

[chroot]

The "magnetic force" has nothing to do with it.

- Warren

 

[Orion1]

Wavefunction...



According to classical physics, an accelerated charge q, radiates at a rate:
$$\frac{dE}{dt} = - \left( \frac{1}{6 \pi \epsilon_0} \right) \left( \frac{(qa)^2}{c^3} \right)$$

An electron in a classical hydrogen atom will spiral into the nucleus at a rate:

$$\frac{dr}{dt} = - \frac{}{12c^3} \left( \frac{q^2}{\pi \epsilon_0 m_e r} \right)^2$$

An electron in an atom exists as a standing waveform existing within a probability cloud:

Radial probability density function:
$$P(r)dr = | \psi |^2 dV$$

State: (1s)
$$\psi (r) = \sqrt{ \frac{}{\pi} \left( \frac{2 \pi \alpha}{\lambda_c} \right)^3} e^- \frac{r/r_0}{}$$

$$v_e = \alpha c$$

An electron as a standing waveform existing within a probability cloud does not 'accelerate', its orbital velocity is relatively constant.