Electrons not inside the nucleus?

[logearav]

an unstable neutron decays to electron, proton and anti neutrino.
if that being the case, why electrons are not inside the nucleus?

 

[jtbell]

The electron and antineutrino are created at the moment of decay.

Did you really mean to ask, "how do we know this is the case, instead of the electron and antineutrino previously existing inside of the nucleus?"

Once upon a time (about 1910-1930), physicists seriously considered the idea that an atomic nucleus contains Z+N protons and N electrons. (They didn't know about neutrons yet.) When an electron flies out, that's beta decay.

1. One problem comes from the Heisenberg uncertainty principle. If you use it to estimate the energies that the "nuclear electrons" might have, you get a few hundred MeV. But in beta decay, the energies of the ejected electrons are only a few (<10) MeV.

2. Another problem is that you can't get the "spin" of the nucleus to come out right, in many cases, if you assume that a nucleus has Z+N protons and N electrons, instead of Z protons and N neutrons.

3. Another problem is that for a particular kind of beta-decay, you would expect the electron to have a specific fixed energy, corresponding to the difference in mass between the nuclei before and after the decay. But it has instead a range of energy, with the maximum energy being the originally-expected "fixed" energy.

So Pauli proposed a new particle, which we now know as the antineutrino, which carries off some of the beta-decay energy, and has spin 1/2 like the electron so we can make the nuclear spin come out right. (Pauli actually called his new particle a "neutron". What we now know as the neutron hadn't been discovered yet!)

(Here's the http://www.library.ethz.ch/exhibit/pauli/neutrino_e.html.)

But then you still have problem #1 with the uncertainty principle.

In the meantime Chadwick discovered the "real" neutron, and a bit later Fermi came up with a theory that the electron and neutrino are created when a neutron decays into a proton. This was the beginning of modern weak-interaction theory.

 

[bcrowell]

I think logearav needs to clarify the question. I don't understand the scenario s/he is imagining in the second sentence.

 

[ZapperZ]

an unstable neutron decays to electron, proton and anti neutrino.
if that being the case, why electrons are not inside the nucleus?

To illustrate what jtbell has described, look at the Feynman diagram for neutron decay:

[PLAIN]http://www.revisionworld.co.uk/files/betadecay.jpg

You'll notice that the d-quark decays into a W and a u-quark. The W then this later decays into an electron and an antineutrino. So no, the original content has no electron.

Zz.

 

[physeven]

i think what he's trying to say is: since an unstable neutron decays into an electron and a proton, an electron and an antineutrino is, how come the decay does not occur in the nucleus, there being no electrons inside the nucleus, only protons and neutrons. This is because the electron is emitted as a form of energy in order to stabilize the high energy neutron, after a W- is emitted when the down quark is in the process of decaying into an up quark (neutrons are dud and protons duu) so therefore the electron is 'shot out'. but this doesn't happen in the nucleus because unstable neutrons are (correct me if I'm wrong because i did this in high school) mostly existent in nuclei with a large relative atomic mass. this decay is unlikely to occur WITHIN a nucleus. i hope that gave you a picture of the whole situation.

 

[logearav]

Thanks a lot mr. jtbell, zapperz and physeven. actually mr. jtbell have correctly interpreted my question, though my question is ambiguous. heisienberg uncertainty principle is del p, del x >= h/2 pi. am i correct? some textbooks say it is >= h/4 pi

 

[jtbell]

In the HUP, $\Delta x$ and $\Delta p$ are normally defined in terms of the mathematical "standard deviation" from statistics. This leads to the constant on the right side being $\hbar / 2 = h / 4 \pi$. This is the normal "correct" form.

 

[physalpha]

That's good and important question.

Similar question.
Protons attract electron.
Why electrons are not attached to proton?
Even electrons have no kinetic energy, they always have a distance?
Any force ?

 

[JoaoPais]

Why electrons are not attached to proton?
Even electrons have no kinetic energy, they always have a distance?
Any force ?

The electrons are not attached because they have a high kinetic energy, and when they have no kinetic energy, this being at 0 Kelvin, yes the electrons fall to the nucleus and matter as we know it cease to exist =)
Also you can’t forget that electrons in an atom don’t behave just like a particle but also as a wave.

 

[physalpha]

this being at 0 Kelvin, yes the electrons fall to the nucleus and matter as we know it cease to exist =)
.

Next, when we raise the temperature, the attached eletron-proton pairs are only able to
get energy as forms of the rotation and the translation, but not sepatated out.
Why is it impossible?

The reason is-----.
Eletron-proton pairs don't accept that kind of the energy storage.
Its come from the wave property, not separated out, only the probability of appearing outside is increased, isn't it?

 

[JoaoPais]

Well, first I don’t think that we can get anything to 0 K, so, no one knows what would happen if we heat up the “matter” again. Either way , yes, it’s because the wave property that gives a probability of where the electrons are. The point is, you can’t even really describe a movement of the electron inside an atom, because it doesn’t go from point A to B, it just can be on point A, B, C (…) at the same time, so it’s actually wrong to say they move (as far as a I know) =P

 

[physalpha]

Simply

0 K-----------------------------------------------> temperature raise
.
electron-proton pair------------------------------ >no distance (only rotation, translation energy)
.

Why is this difficult?
They are very attracted.

 

[JoaoPais]

Look, it simply isn’t that simple:
Matter at 0k (impossible because it’s wave function becomes infinite)
Close to zero – Bose-Einstein condensate formation (“superatom”)
Also, if electron-proton pair has no distance, occurs an electro capture reaction

SPECULATION
At zero ----> electrons fall to nucleus from the lack of movement -----> nuclear reaction of electron capture – protons became neutrons
Rise temperature -----> got a bunch of neutrons vibrating!

So, has you see, it isn’t difficult, but you’re so much into impossibilities that it gets ridiculous to move any further.

If you want to see that they are very attracted, it would be because the electrons are “moving” really fast and yet, the electromagnetic force is able to retain them “so” close to the nucleus.

 

[physalpha]

At zero ----> electrons fall to nucleus from the lack of movement -----> nuclear reaction of electron capture – protons became neutrons

Thanks.
Is activation energy not high at that time?

Matter 0 K.
Does the temperature influence basic particle wave - quark wave or electron wave?

 

[JoaoPais]

Thanks.
Is activation energy not high at that time?

Well, here you simply don’t have an activation energy, because the particles don’t have ANY kind of energy. Hypothetically, the electron simply fall to the attraction of electromagnetic force.

Matter 0 K.
Does the temperature influence basic particle wave - quark wave or electron wave?

Yes, that’s the problem with absolute zero. Isn’t possible, once the total lack of energy makes the quantum wave of every particle infinite. At very low temperatures, the quantum wave of particles start’s to grow, and the quantum effects become macroscopic. The atoms start to have wave functions bigger than themselves and stop existing as individual atoms, but form a new state of matter which is called the Bose-Einstein condensate. Beyond that, speculation.

 

[physalpha]

I've heard we can make 0.4K He.
How to measure and control it? It must be no easy way.

Then absolute 0 k means not no energy state, but no transferable energy existence state.
It has no energy which is transferable to any low temperature surround system without reaction.

 

[Gokul43201]

The electrons are not attached because they have a high kinetic energy, and when they have no kinetic energy, this being at 0 Kelvin, yes the electrons fall to the nucleus and matter as we know it cease to exist =)

This is false (and the entire discussion following it is meaningless). Bound atomic electrons do not lose orbital kinetic energy as the temperature drops.

 

[JoaoPais]

Bound atomic electrons do not lose orbital kinetic energy as the temperature drops.

My quoted statement don't say this. It says that, in 0 Kelvin, electrons fall to the nucleus, and yes, I’m aware that sentence is simply wrong, being a “classical” view of a non-classical event. The present discussion was mostly hypothetical, as I say so many times.
But, i wonder what is the role of electrons in a Bose-Einstein condensate... what happens to their wave function? all electrons and nucleons have overlapping wave functions or they still follow some organization?

 

[physalpha]

in 0 Kelvin, electrons fall to the nucleus

The problem between electron and proton is the assumption.
In real case we could think about this case.

neutron(separated) -----> electron + proton
At this time we can think about eletron-proton pair.
According to before discussed content, neutron, electon and proton must be independently existed.
Revese reaction is very impossible.
And then, in order to lower electron orbit, visual or ultraviolet rays should be emitted.
Lower the temperature never lower the eletron orbit.
That's the recent summary.