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Uncertainty principle for electron in nucleus

  1. Sep 20, 2013 #1
    Hello, I have this exercise and want to check if what i did is correct
    1. The problem statement, all variables and given/known data

    Nuclei, typically of size $10^{-14}$ m, frequently emit electrons with energies of 1-10 MeV. Use the uncertainty principle to show that electrons of energy 1 MeV could not be contained in the nucleus before the decay.

    2. Relevant equations

    Δx*Δp≥ h/4*pi
    E=sqrt(p²c²+m²c4)

    3. The attempt at a solution
    Searching for the minimum value of the energy
    Δx ≈ x
    Δp ≈ p
    x*p≈h/4*pi
    so x=10-14
    p=h/(4*pi*x)
    sqrt(p²c²+m²c4)=sqrt(h²/(4²*pi²*x²)+m²c4)
    E=sqrt(h²/(4²*pi²*x²)+m²c4)
    I found ≈ 5 Mev for the minimum energy for an electron in a nucleus, is my method correct ?
    Thanks !
     
  2. jcsd
  3. Sep 20, 2013 #2

    vela

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    I think you did it backwards. You want to assume the electron has an energy of 1 MeV and show that this implies ##\Delta x > 10^{-14}\text{ m}##.

    Off the top of my head, I'd expect your calculation to give E ~ 10 MeV.
     
  4. Sep 21, 2013 #3
    Hello
    I don't get how i can do it following what you're saying
    Thanks
     
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