From what locations do radiated electrons start their movement in Beta decay?

In summary: If everything was that uncertain any natural(for example Photosynthesis: electron transport chain) or man-made mechanisms(any electrical devices from laptops to photodetectors) that require the movement of electrons in a certain way would not exist, so, it is clear that we can assume that electrons are moving along the highest probability path.
  • #1
cemtu
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TL;DR Summary
I mean is there a possibility of a beta particle from a nuclear decay to come into existence at the far end of the galaxy for example? What is the limiting condition for its position?
I know that the Heisenberg Uncertainty principle states that the position of an electron is uncertain, however, if an electron is created due to beta decay, then at what location is it more likely to begin its movement?

Is it right inside the proton? Is it the outer edge of the proton? Is it adjacent to the proton? Is it near the proton?

I mean is there a possibility of a beta particle from a nuclear decay to come into existence at the far end of the galaxy for example?
 
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  • #2
cemtu said:
I know that the Heisenberg Uncertainty principle states that the position of an electron is uncertain,
So isn't that your answer?
 
  • #3
Vanadium 50 said:
So isn't that your answer?
But I asked for a more likely situation, the highest possibility...
 
  • #4
I don't see how this can be answered. As you say yourself, the electron doesn't have a well-defined position. So how can you ask where exactly it is?
 
  • #5
Vanadium 50 said:
I don't see how this can be answered. As you say yourself, the electron doesn't have a well-defined position. So how can you ask where exactly it is?
But I asked for a more likely situation, the highest possibility...
 
  • #6
Does the electron, or for the matter antineutrino, have a well defined lever arm relative to the nucleus?
 
  • #7
snorkack said:
Does the electron, or for the matter antineutrino, have a well defined lever arm relative to the nucleus?
We know orbital-wise the approximate locations of electrons with respect to the nucleus or the orbital locations where electrons are most likely to be found around the nucleus, I think that's why there is something called an outer shell - inner shell - valance shell, etc.
 
  • #8
You can ask and ask and ask but it won't change the answer: the electron can't be localized better than what QM predicts, which is large compared to a nucleus.
 
  • #9
cemtu said:
We know orbital-wise the approximate locations of electrons with respect to the nucleus or the orbital locations where electrons are most likely to be found around the nucleus, I think that's why there is something called an outer shell - inner shell - valance shell, etc.
Sea shells by the sea shore...

1691094127835.png

https://energywavetheory.com/atoms/orbital-shapes/
 
  • #10
Is the angular momentum of electron relative to the nucleus defined? That is, an electron can be on a s orbital (angular momentum 0), p orbital (angular momentum 1), d orbital (angular momentum 2)... does quantization rule out electron having an angular momentum of, say, 0,8?
 
  • #11
Perhaps what the OP is asking about is this:

A proton and the resultant neutron it decays into are pretty confidently localized (in the nucleus). How does that state transition to the next state - with the electron, which is not so confidently localized?

Which diagram in Berkeman's post, above, defines that electron's probability for where it is found when measured?
 
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  • #12
DaveC426913 said:
Which diagram in Berkeman's post
None, as the electron is unbound.
 
  • #13
Vanadium 50 said:
None, as the electron is unbound.
That seems to reinforce the validity of the OPs question.

If we observed a bazillion beta decay events, would we not build up a probability map of where to expect the unbound electron to be found?
 
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  • #14
Only if you can figure out where the electrons are relative to a bezillion nuclei.

Look, you're not going to beat QM and the Heisenberg Uncertainty principle this way. (Or any other way)
 
  • #15
Vanadium 50 said:
Look, you're not going to beat QM and the Heisenberg Uncertainty principle this way. (Or any other way)
That wasn't my thought, no. I just thought you'd still get a probability cloud.
 
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  • #16
DaveC426913 said:
Perhaps what the OP is asking about is this:

A proton and the resultant neutron it decays into are pretty confidently localized (in the nucleus). How does that state transition to the next state - with the electron, which is not so confidently localized?

Which diagram in Berkeman's post, above, defines that electron's probability for where it is found when measured?
Exactly!
 
  • #17
DaveC426913 said:
That wasn't my thought, no. I just thought you'd still get a probability cloud.
Yes you can get the probability cloud and there are higher chances that an electron can be found and the lower chances also exist so electrons act according to it, just I say apply this to Negatron(beta) decay.
Vanadium 50 said:
Only if you can figure out where the electrons are relative to a bezillion nuclei.

Look, you're not going to beat QM and the Heisenberg Uncertainty principle this way. (Or any other way)
If everything was that uncertain any natural(for example Photosynthesis: electron transport chain) or man-made mechanisms(any electrical devices from laptops to photodetectors) that require the movement of electrons in a certain way would not exist, so, it is clear that we can assume that electrons are moving along the highest probability path.
 
  • #18
Vanadium50 is one of the smartest guys I know so, personally, I'm gonna take him at his word.
 
  • #19
DaveC426913 said:
Vanadium50 is one of the smartest guys I know so, personally, I'm gonna take him at his word.
He is also unstable, and has a serious spin angular momentum problem.

His two conceivable daughters are Cr-50, energy of beta decay to ground state 1821 keV, and Ti-50, energy of electron capture to ground state 3761 keV, thus positron emission to ground state 2739 keV.

The problem is that both are forbidden. V-50 has spin 6; the daughters´ ground states do not have spin (duh, they are even-even).

1 unit of angular momentum can be handled by the spin of electron or positron and antineutrino or neutrino; since they must have opposite helicities, their spins add to 1 when they are emitted in the opposite directions but that can be handled. This still leaves 5 spin units.
Both Cr-50 and Ti-50 have excited states with spin. Cr-50 at 783 keV, leaving 1038 keV for beta decay; Ti-50 at 1554 keV, leaving 2208 keV for electron capture, thus 1186 keV for positron emission.

But both of these states have spin just 2. Still forbidden. 3 angular momentum units unaccounted for.

Precisely how does V-50 dispose of his spin angular momentum when he decays? Does it go to orbital angular momentum of electron, orbital angular momentum of antineutrino, or some into both?
 
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  • #20
DaveC426913 said:
Vanadium50 is one of the smartest guys I know so, personally, I'm gonna take him at his word.
Do whatever you want sir but it is not about taking anybody's word. For me just there are things about electron position and decay right now that I do not get.
 
  • #21
Numbers:

Beta decay tops out at maybe 5 MeV. Assume ΔE is all 5 MeV.
Planck's constant is 200 MeV-fm.
Therefore, a beta electron can be localized to ~40 fm.
A proton is a composite object a little smaller than 1 frm,
Since 40 >> 1, "what part of the proton" is an unsnaswweable question.
 
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  • #22
And in case of beta emission, measuring the electron´s lever arm has some practical difficulties.
How about electron capture, though?
V-50 has electrons, in ground state: 1s22s23s24s22p63p63d3
When V-50 nucleus captures one electron, can you track x-ray and UV/visible photons emitted by the deexcitation of the state of Ti-50 atom formed by capture to find out which of the V-50 electrons it was that was captured?
 
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