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Uncertainty principle for electron in nucleus

  • Thread starter Dassinia
  • Start date
  • #1
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Hello, I have this exercise and want to check if what i did is correct

Homework Statement



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.

Homework Equations



Δx*Δp≥ h/4*pi
E=sqrt(p²c²+m²c4)

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 !
 

Answers and Replies

  • #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.
 
  • #3
144
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Hello
I don't get how i can do it following what you're saying
Thanks
 

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