- #1
Quantum_man
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The typical energy needed to ionise an atom is around 5 eV. Use the Uncertainty principle to estimate the size of an atom.
E=mc^2
E = p^2/2m
Δx.Δp ≥ h/4π
So I got the mass rearranging E = mc^2
m = 5*1.6*10^-19 / (3*10^8)^2
m = 8.88*10^-36 kg
then Δp = √(2mE)
which = 3.77*10^-27 kg m s^-1
Finally
Δx ≥ h/(4*pi*3.77*10^-27)
= 14nm
My only problem is that this nucleus is greater than the hydrogen atom. So shouldn't its ionisation energy be greater than 13.6 eV?
Thanks for anyone that can help.
Homework Equations
E=mc^2
E = p^2/2m
Δx.Δp ≥ h/4π
The Attempt at a Solution
So I got the mass rearranging E = mc^2
m = 5*1.6*10^-19 / (3*10^8)^2
m = 8.88*10^-36 kg
then Δp = √(2mE)
which = 3.77*10^-27 kg m s^-1
Finally
Δx ≥ h/(4*pi*3.77*10^-27)
= 14nm
My only problem is that this nucleus is greater than the hydrogen atom. So shouldn't its ionisation energy be greater than 13.6 eV?
Thanks for anyone that can help.