# Uncertainty principle for electron in nucleus

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

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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.

Hello
I don't get how i can do it following what you're saying
Thanks