How can beta decay have energy of 3-4 MeV

In summary: The uncertainty principle states that there is always a limit to how well we can measure both the position and energy of a particle at the same time. This means that the electron is not always in the nucleus, and even if it is, we cannot accurately measure its energy at that moment. Therefore, the energy of the electron in beta decay is not affected by its temporary presence in the nucleus. In summary, the Heisenberg's uncertainty principle states that there is a limit to how well we can measure both the position and energy of a particle at the same time. Therefore, the energy of the electron in beta decay is not affected by its temporary presence in the nucleus. The link provided explains this concept in more detail. The electron does not gain or
  • #1
Moazin Khatri
29
1
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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  • #2
Why do you think the electron is localized to the nucleus in beta decay? Isn't it outside the nucleus when you detect it?
 
  • #3
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.
 
  • #4
No, it will not. Your view of how the decay occurs is too classical.
 
  • #5
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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  • #6
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.
 
  • #7
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.
 
  • #8
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.
 

Related to How can beta decay have energy of 3-4 MeV

1. What is beta decay?

Beta decay is a type of radioactive decay in which a nucleus emits a beta particle, which can either be an electron or a positron, resulting in the transformation of a neutron into a proton or vice versa.

2. How does beta decay have energy?

In beta decay, the energy is released from the nucleus in the form of a beta particle. This energy is a result of the difference in mass between the original nucleus and the decay products. The energy of beta decay can range from a few kiloelectron volts (keV) to several megaelectron volts (MeV).

3. How can the energy of beta decay be 3-4 MeV?

The energy of beta decay is determined by the mass difference between the parent nucleus and the products of decay. The larger the mass difference, the higher the energy released. In the case of beta decay with an energy of 3-4 MeV, the parent nucleus must have a relatively large mass compared to the decay products.

4. What factors affect the energy of beta decay?

The energy of beta decay is influenced by several factors, including the type of beta decay (beta minus or beta plus), the mass of the parent nucleus, and the stability of the decay products. Additionally, the nuclear forces and the conservation of energy and momentum also play a role in determining the energy of beta decay.

5. Can beta decay energy be harnessed for practical use?

Yes, beta decay energy can be utilized in various practical applications. For example, in nuclear power plants, the energy from beta decay is used to generate electricity. It is also used in medical imaging techniques, such as positron emission tomography (PET), and in the production of radiopharmaceuticals for cancer treatment.

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