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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? What is the limiting condition for its position?

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I know that the Heisenberg Uncertainty principle states that the position of an electron is uncertain,

The exact phrase is "there is a probability of measuring the beta decay electron, that can be calculated quantum mechanically giving probability values for points in space (x,y,z,t).

however, if an electron is created due to beta decay, then at what location is it more likely to begin its movement?

In mainstream physics, at the (x,y,z,t) of the decay.

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which, as far as nuclear dimensions go can be considered a point with dimensions of the nucleus.

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?

the proton obeying momentum conservation rules is an independent track and has nothing to do with the track of the beta electron.

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?

If one has computed the wavefunction of the system , one could say there is a probability, but it will be ridiculously small, meaningless.

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  • $\begingroup$ I don't quite get what you are talking about here in relation to what my question asks essentially. $\endgroup$ Commented Aug 4, 2023 at 7:39
  • $\begingroup$ do you mean the question in the title? "as far as nuclear dimensions go can be considered a point with dimensions of the nucleus." refers to that also. The diagram shown, for nuclear reactions can be considered to represent a point from which two straight tracks leave, obeying energy and momentum conservation rules. $\endgroup$
    – anna v
    Commented Aug 4, 2023 at 10:03
  • $\begingroup$ the probability of getting a hit from such a particle depends on the wavefunction of the system ,the point will be along the line of the track $\endgroup$
    – anna v
    Commented Aug 4, 2023 at 10:14
  • $\begingroup$ but I did not ask anyting about "getting a hit by particle" $\endgroup$ Commented Aug 4, 2023 at 11:26
  • $\begingroup$ I dont know what you are talking about. $\endgroup$ Commented Aug 4, 2023 at 11:28

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