I Uncertainty of electron energy based on confinement radius?

[excalibur313]

I am trying to calculate what we'd expect the uncertainty in energy would be for an electron in a hydrogen atom where it was confined to its usual radius (120 pm) versus if we confined it to the width of a proton (.88 fm) to try and make an argument about why the electron does not fall into the nucleus (I know there are several explanations for this, but I thought this could be an interesting approach.)

My approach is to start with the Heisenberg Uncertainty principle to calculate the corresponding uncertainty if we confine dx to be the two radii that I mentioned before.
dxdp >= hbar/2
Then, based on the momentum-energy relationship we can calculate a corresponding minimum energy based on that momentum.
E²=(pc)²
Then, I wanted to convert to eV and compare both values to the binding energy of an electron in hydrogen, which is 72 eV. The trouble is that the uncertainty in energy I get for hydrogen at 120 pm is 820 eV and at 0.88 fm it is 11 meV. Why is the 120 pm energy value about 10x higher than I'd expect? Is there something that I am missing here?

Thanks so much!

 

[PeterDonis]

I am trying to calculate what we'd expect the uncertainty in energy would be for an electron in a hydrogen atom where it was confined to its usual radius (120 pm) versus if we confined it to the width of a proton (.88 fm) to try and make an argument about why the electron does not fall into the nucleus

Have you looked at the existing literature on this subject? There is plenty of it. A good overview of the QM arguments for the stability of atoms (and of matter in general) is here:

http://ergodic.ugr.es/statphys/bibliografia/lieb3.pdf

Note in particular the second paragraph from the top in the right column of the first page.

 

[PeterDonis]

to the binding energy of an electron in hydrogen, which is 72 eV

Where are you getting this number from? The usual number is 13.6 eV.