How can beta decay have energy of 3-4 MeV

AI Thread Summary
The discussion centers on the energy of electrons emitted during beta decay, which is observed to be around 3-4 MeV, despite theoretical predictions suggesting much higher energies based on the Heisenberg uncertainty principle. Participants clarify that while electrons can be in the vicinity of the nucleus, their energy does not significantly increase just because they are localized there. The conversation emphasizes that the energy of an electron is not solely determined by its position but also by its quantum state. Misunderstandings about the nature of electron behavior in quantum mechanics are addressed, highlighting that the wave function's presence in the nucleus does not equate to high energy. Overall, the discussion seeks to reconcile theoretical concepts with experimental observations in particle physics.
Moazin Khatri
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First of all let me tell that I am a high school student and I have no background in particle physics.
If we apply the Heisenberg's uncertainty principle to an electron confined in the nucleus it tells us that the energy of the electron must be something like 3.8X10^8eV .. but in a beta decay the energy of the electron is 3-4 MeV.
The moment a neutron decays into a proton and an electron and a neutrino particle the electron will be in the region of nucleus for some small time at least. While in that region, according to Heisenberg's principle it must have such a large energy. But experimentally we find that beta particles have an energy of 3-4 MeV. Why?
Please clarify me as much as possible and correct me wherever i am wrong.
I read about this from this link: http://elearning.vtu.ac.in/e-con/Phys/html/0018.htm
 
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Why do you think the electron is localized to the nucleus in beta decay? Isn't it outside the nucleus when you detect it?
 
If it comes from the decay of a down quark it has to come from the nucleus. And for some small time it will be in that region. And yes that's true that we detect it when it is outside the nucleus but still, the question remains, does it loose energy as it comes out of the nucleus because for some time it would be in the region of nucleus and must have high energy.
 
No, it will not. Your view of how the decay occurs is too classical.
 
Yes sir. I understand that. But if any simpler basic level explanation can be provided I will be very grateful.
I do understand some of the basic ideas of quantum mechanics. So if a simple explanation based on few basic ideas of quantum mechanics can be given I will welcome it.
 
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Moazin Khatri said:
some time it would be in the region of nucleus and must have high energy.

That's simply not true. Just being "in the region" doesn't do anything. Forget radioactive decay for a second and just consider atoms. Electrons are sometimes in the region of the nucleus, but there energy doesn't skyrocket when this happens.
 
Vanadium 50 said:
Forget radioactive decay for a second and just consider atoms. Electrons are sometimes in the region of the nucleus, but there energy doesn't skyrocket when this happens.
This depends on what you mean by "sometimes in the region of the nucleus". If you would actually locate the electron to the nucleus this would add a large amount of energy to the electron. It is not the same thing as the wave function having a non-zero value within the nucleus, which is a large part of the reason we have electron capture in some elements.
 
Orodruin said:
If you would actually locate the electron to the nucleus this would add a large amount of energy to the electron.

That's true, but it doesn't happen in either case.
 
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