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I Beta radiation

  1. Apr 9, 2016 #1
    my understanding of beta radiation is that an up quark in a proton changes to a down quark, forming a neutron and emitting an electron as the result of the change in charge.
    My questions are,
    1. Why does the quark change?
    2. How does it change and how does it change charge?
  2. jcsd
  3. Apr 10, 2016 #2


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    Staff: Mentor

    We don't know.
    Quantum mechanics predicts the probability of interactions like this one, but it doesn't provide a detailed picture of what's happening during the interaction.
  4. Apr 10, 2016 #3


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    Hello Stan,

    Just to make sure: you now have a correct answer from Nugatory, but was that what you wanted to know or could you be helped with a more down-to earth answer in the category: beta decay happens because it's energetically favorable ?
  5. Apr 10, 2016 #4
    Why would it be more energetically favourable?
  6. Apr 10, 2016 #5
    Ok thank you for helping
  7. Apr 10, 2016 #6


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    The decay you are describing (proton to neutron) is actually an anti-beta decay - it emits a positron instead of an electron. You can see this just by considering charge conservation; the particles going in have a net charge of +1 so the particles coming out must also have that charge.

    This process happens in a nucleus that is proton-rich, meaning that it takes more energy to hold the nucleus together than it would if it had one more neutron and one fewer proton. Thus, if a proton were to be converted into a neutron, energy would be released. If the amount of energy released is greater than the amount of energy required to create a positron plus the amount of energy required to turn a proton into a neutron we say that the reaction is "energetically favorable" - it can happen without violating conservation of energy or requiring us to add energy to the system.

    Generally if a reaction is energetically favorable it will happen eventually.
  8. Apr 10, 2016 #7
    Ok thank you that helps alot
  9. Apr 10, 2016 #8


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    Staff: Mentor

    and violates no other conservation laws (charge, baryon number, lepton number...)
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