I have the following exercise:
Use Heisenberg's uncertainty principle and the relation Δu=√⟨u2⟩−⟨u⟩2 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≥ℏ/2. Therefore Δp≥ℏ/2Δx.
Without considering relativistic corrections (don't know if this is OK), Ec=p2/2m
From the definition of standard deviation Δp=√⟨p2⟩−⟨p⟩2. Then Δp2=⟨p2⟩−⟨p⟩2. Therefore ⟨p2⟩=Δp2+⟨p⟩2
The energy of the electron will be the kinetic energy minus the potential energy:
E=p2/2m−e2/r
And so the average energy will be
⟨E⟩=⟨p2⟩/2m−e2/⟨r⟩=(Δp2+⟨p⟩2)/2m−e2/⟨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 1∗10−10m (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⟩/2m−e2/4πϵ0⟨|r|⟩=⟨Δp⟩2/2m−e2/4πϵ0⟨|r|⟩≈ℏ2/8ma0−e2/4πϵ0a0≈−24eV
which is pretty close to the actual binding energy −13.6eV.
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