Thursday, 19 October 2017

quantum mechanics - Uncertainty Principle and Energy range for an electron in an atom


I have the following exercise:


Use Heisenberg's uncertainty principle and the relation Δu=u2u2 to find the range of energy an electron has in an atom of diameter 1 amstrong.


This is the attempt at a solution:




  • From the uncertainty principle: ΔpΔx/2. Therefore Δp/2Δx.




  • Without considering relativistic corrections (don't know if this is OK), Ec=p2/2m





  • From the definition of standard deviation Δp=p2p2. Then Δp2=p2p2. Therefore p2=Δp2+p2




  • The energy of the electron will be the kinetic energy minus the potential energy:




E=p2/2me2/r


And so the average energy will be



E=p2/2me2/r=(Δp2+p2)/2me2/r


Here I don't know how to continue. Do I have to assume p=0? Why? And even when I assume that, replacing r with Δx, which I suppose is 11010m (why?) I don't get a value of energy similar to the ground level energy of a hydrogen atom (which has roughly the same diameter than this one).


What am I doing wrong?



Answer



p=0 because momentum is a vector and it points in all directions with equal probability assuming the wavefunction ψ(r) is spherically symmetrical.


|r|a0.


Putting these together gives the lower bound


E=p2/2me2/4πϵ0|r|=Δp2/2me2/4πϵ0|r|2/8ma0e2/4πϵ0a024eV


which is pretty close to the actual binding energy 13.6eV.


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