Shift in hydrogen ground state energy due to proton size
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In the hydrogen atom, the proton is not really a point charge but has a finite size. Assume that the proton behaves as a uniformly-charged sphere of radius R=10^(-15) m. Calculate the shift this produces in the ground-state energy of hydrogen.
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Solution Summary
We explain how the finite size of the proton leads to a shift in the ground state energy of hydrogen. We work out this shift to first order in perturbation theory.
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From perturbation theory, we know that an energy eigenvalue shifts to first order by:
<psi|V|psi>
were |psi> is the unperturbed state and V is the perturbation term in the Hamiltonian. In this problem, the unperturbed Hamilonian H_0 is:
H_0 = p^2/(2m) - e^2/(4 pi epsilon_0 r)
If we treat the proton as a uniformly charged sphere of radius R, the potential energy term in H_0 is not correct inside this sphere. So, let's evaluate the correct potential energy function.
Consider a sphere of radius R with a volume charge of rho. Then by Gauss' law, the radial component of the electric field at distance r will be:
E(r) 4 pi r^2 = Q(r)/epsilon_0 (1)
where Q(r) is the charge enclosed within a distance r from the center of the sphere. By symmetry, the electric fleld only has a radial component, so E(r) is also the magnitude of the electric field. And it then also follows that the potential only depends on r. If r < R, we have
Q(r) = 4/3 pi r^3 rho,
inserting this in (1) gives:
E(r) = rho/(3 epsilon_0) r (2)
If r > R, then Q(r) is the total charge ...
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