Can an electron stand in the same place as a proton?

[Godparicle]

The atomic orbital refers to the physical region where the electron can be calculated to be present, as defined by the particular mathematical form of the orbital (the statement is extracted from atomic orbitals-wiki).

The picture of 1s orbital seems to signify that electron can exist in the same place as the protons. Is this the case?

When the distance between proton and electron becomes 0, the gravitational force becomes ∞(considering limits). So, for the electron to escape from the proton, electron needs to have high velocity nearly ∞, which means temperature to be ∞. I don't think we have ever observed such a situation where we have a chunk of matter with ∞ temperature. This condition doesn't seem to allow electron to stand in place of proton, but the QM (if I am not wrong) says it is possible.

What do you say, is it possible for the ghostly electron to stand in the same place as the proton?

 

[Dale]

The picture of 1s orbital seems to signify that electron can exist in the same place as the protons. Is this the case?

Yes. The probability is non-zero in the region where the nucleus is located.

 

[Godparicle]

Yes. The probability is non-zero in the region where the nucleus is located.

Ok. Proceed.

 

[jtbell]

The probability is non-zero in the region where the nucleus is located.

And this is what enables the nuclear decay process called electron capture (a close relative of beta- and beta+ decay) to occur.

When the distance between proton and electron becomes 0, the gravitational force becomes ∞(considering limits).

You're trying to combine a classical description of gravity with the quantum-mechanical picture of the electron, which is problematical.

Also note that even in a classical picture, the gravitational force between electron and nucleus is always much smaller than the electrical force between them, no matter how close together they are. This suggests that gravitational effects ought to be negligible in the quantum picture also, except maybe at extremely high energies. So far as I know, we haven't yet observed anything that requires "quantum gravity" effects in situations like this.

 

[Dale]

Ok. Proceed.

There isn't really much to proceed with. The probability is non-zero. The classical concept of force is not terribly relevant in QM, and the electron cannot go below the ground state.

 

[Godparicle]

And this is what enables the nuclear decay process called electron capture (a close relative of beta- and beta+ decay) to occur.

I read the wiki page on electron capture. Here is a sentence from it - "Electron capture is a process in which a proton-rich nuclide absorbs an inner atomic electron, thereby changing a nuclear proton to a neutron and simultaneously causing the emission of anelectron neutrino."

The wiki quote assumes the process to occur in the proton-rich nuclide. In #9 of this thread: https://www.physicsforums.com/threads/why-dont-hydrogen-atoms-neutron-neutrino.23830/, Janitor claims that electron capture doesn't work for hydrogen isotope. Even though it is not clear me, whether it is the rarest for hydrogen isotope or it is not possible for ever, it has left me in confusion to accept your above quote.

You're trying to combine a classical description of gravity with the quantum-mechanical picture of the electron, which is problematical.

Also note that even in a classical picture, the gravitational force between electron and nucleus is always much smaller than the electrical force between them, no matter how close together they are. This suggests that gravitational effects ought to be negligible in the quantum picture also, except maybe at extremely high energies. So far as I know, we haven't yet observed anything that requires "quantum gravity" effects in situations like this.

If it is true that electron has some probability to be inside nucleus or inside proton or inside quarks, then the gravitational force will become infinity according to Newton's equation. The gravitational force will not be negligible.

 

[mfb]

The wiki quote assumes the process to occur in the proton-rich nuclide. In #9 of this thread: https://www.physicsforums.com/threads/why-dont-hydrogen-atoms-neutron-neutrino.23830/, Janitor claims that electron capture doesn't work for hydrogen isotope. Even though it is not clear me, whether it is the rarest for hydrogen isotope or it is not possible for ever, it has left me in confusion to accept your above quote.

It is possible in some proton-rich nuclei. Hydrogen is not one of them, it cannot occur there. A neutron is too heavy, an isolated proton plus a bound electron (=hydrogen) do not have enough energy to make one.

If it is true that electron has some probability to be inside nucleus or inside proton or inside quarks, then the gravitational force will become infinity according to Newton's equation. The gravitational force will not be negligible.

Again, you cannot use classical equations here. The probability for a particle to be at a single specific point is always zero anyway.

 

[jtbell]

I didn't intend to imply that electron capture is possible for hydrogen. The masses of the nuclides before and after the capture process have to be such that energy can be conserved.

[added: mfb beat me to it]

 

[Dale]

If it is true that electron has some probability to be inside nucleus or inside proton or inside quarks, then the gravitational force will become infinity according to Newton's equation. The gravitational force will not be negligible.

First, quantum mechanics doesn't work that way. As I said before, the classical concept of force is essentially irrelevant for quantum mechanics. What is important in QM is the potential and the state. The potential does have a sharp bend at r=0, which is what corresponds to an infinite force for a classical potential, but the quantum state cannot go below the ground state despite that fact.

Second, forget gravity, it is many orders of magnitude smaller than the Coulomb potential which also becomes infinite at r=0.

 

[Godparicle]

It is possible in some proton-rich nuclei. Hydrogen is not one of them, it cannot occur there. A neutron is too heavy, an isolated proton plus a bound electron (=hydrogen) do not have enough energy to make one.

So, electron doesn't always gets converted into neutron when it comes in place of nucleus. Is this the case?
Although, this doesn't relate to my original question, I like to know this.

Again, you cannot use classical equations here.

Ok. I read the page Strong gravitational constant-wikiversity. (link: http://en.wikiversity.org/wiki/Strong_gravitational_constant) Here is the statement from that page:
"The strong gravitational constant, denoted

or [PLAIN]http://upload.wikimedia.org/math/7/3/b/73bba714fbb01bb916d9690c91c73a06.png, is a grand unified physical constant of strong gravitation, involved in the calculation of the gravitational attraction at the level of elementary particles and atoms.

According to http://en.wikiversity.org/wiki/Isaac_Newton's law of universal gravitation, the force of gravitational attraction between two massive points with masses

and [PLAIN]http://upload.wikimedia.org/math/4/8/a/48af40241cac88e3498c04f6dc267cd0.png, located at a distance http://upload.wikimedia.org/math/5/a/4/5a4d81b4f7d3b2207410b3ba0c12c029.png between them, is:

The coefficient of proportionality

in this expression is called gravitational constant. It is assumed, that in contrast to the usual force of gravity, at the level of elementary particles acts strong gravitation. In order to describe it

in the formula for gravitational force must be replaced on [PLAIN]http://upload.wikimedia.org/math/6/b/9/6b9169c1fcbff554de21c771cb6d851d.png:

"


I hope that the above equation can be used here. Even here gravitational force becomes infinity at r=0.

 

[mfb]

So, electron doesn't always gets converted into neutron when it comes in place of nucleus. Is this the case?

An electron never gets converted to a neutron. "Electron+proton" don't always form a neutron, yes. Actually, this is quite an unusual process and requires very special conditions.

Ok. I read the page Strong gravitational constant-wikiversity. (link: http://en.wikiversity.org/wiki/Strong_gravitational_constant)

Keep in mind this is a purely hypothetical concept. There is no single measurement indicating that this concept is true.

I hope that the above equation can be used here. Even here gravitational force becomes infinity at r=0.

Again not in the classical way. You have to consider quantum mechanics, and then nothing is infinite.

 

[Dale]

Even here gravitational force becomes infinity at r=0.

The classical concept of force is irrelevant in QM.

Since you persist in ignoring this, the thread is closed.

 

[Nugatory]

Ok. I read the page Strong gravitational constant-wikiversity. (link: http://en.wikiversity.org/wiki/Strong_gravitational_constant) Here is the statement from that page:

Please remember the PhysicsForums policy about acceptable sources. A pretty good rule of thumb is that if something in a disallowed source leaves you confused, perplexed, or disagreeing with what a PF science advisor is trying to tell you... Chances are that the problem is with the disallowed source.

Here the problem is that your source has failed to mention that this "strong gravity" concept was proposed more than fifty years ago, and although it was a good idea it has proved to be a dead end. Furthermore, to properly apply it you still have to use the methods of quantum mechanics; the analogy to classical gravity doesn't work the way that article might lead you to think.