Question 11.4: Use first-order perturbation theory to estimate the energy s......
Use first-order perturbation theory to estimate the energy shifts of the hydrogen 2s and 2p states due to the fact that the proton is not a point charge, treating it (for simplicity) as a uniformly charged hollow spherical shell of radius b=5\times 10^{-14} cm. Comment on these results. Compare the results with the expected order of magnitude of the correction due to the finite proton mass. Explain why measurements of such energy shifts is not a good way of studying the proton charge distribution and suggest, with reasons, a more effective way.
The wave functions for the hydrogen atom are of the form
\psi _{nlm} =R_{nl} (r)\Theta^{m}_{l}(\theta )\Phi _{m} (\phi ).
The 1s state for a hydrogenic state of charge eZ has wave function
\psi _{100} =\left(\frac{1}{4\pi } \right)^{1/2} \left(\frac{Z}{a_{B} } \right)^{3/2} 2\exp \left(-\frac{Zr}{a_{B} } \right)The 2s state for a hydrogenic state of charge eZ has a wave function given by
\psi _{200} =\left(\frac{1}{4\pi } \right)^{1/2} \left(\frac{Z}{a_{B} } \right)^{3/2} 2\left(1-\frac{Zr}{2a_{B} } \right) \exp \left(-\frac{Zr}{2a_{B} } \right)and the 2p wave function is
\psi _{210} =\frac{1}{2} \left(\frac{3}{\pi } \right)^{1/2} \cos (\theta )\left(\frac{Z}{2a_{B} } \right)^{3/2} \frac{1}{\sqrt{3} } \frac{Zr}{a_{B} } \exp \left(-\frac{Zr}{2a_{B} } \right)The Bohr radius is a_{B} =0.0529 nm.
If we consider that the proton is not a point charge but a finite size of radius b, then the wave function has nonzero probability of being within b. There is a change in potential seen by the electron.
\Phi =\frac{e^{2} }{4\pi \varepsilon _{0}r }for r > b and
\Phi =\frac{e^{2} }{4\pi \varepsilon _{0}b}for r ≤ b.
Assuming \hat{H} =\hat{H}_{0} +\Delta \hat{H}, then to first-order, the change in energy is given by
\Delta E=\left\langle\psi \left|\Delta \hat{H} \right|\psi \right\ranglewhere
\hat{H}=-\frac{\hbar ^{2} }{2m} \nabla^2 +\frac{e^{2} }{4\pi \varepsilon _{0}r }for all r, and
\Delta \hat{H}=-\frac{e^{2} }{4\pi \varepsilon _{0}r } +\frac{e^{2} }{4\pi \varepsilon _{0}b }with
\Delta E=\frac{e^{2} }{4\pi \varepsilon _{0} }\left\langle\psi\mid \frac{1}{b}-\frac{1}{r} \mid \psi \right\ranglefor r < b and
\Delta \hat{H} =0for r > b.
Hence, for the 2s state
Since b\ll a_{B}, we may approximate \exp (-Zr/2a_{B} )\cong 1 and
\Delta E_{2s} =\frac{e^{2} }{\pi \varepsilon _{0} }\left(\frac{Z}{2a_{B} } \right)^{3} \int\limits_{0}^{b}{r^{2} \left(\frac{1}{b}-\frac{1}{r} \right)dr } =\frac{e^{2} }{\pi \varepsilon _{0} }\left(\frac{Z}{2a_{B} } \right)^{3} \int\limits_{0}^{b}{ \left(\frac{r^{2} }{b}-r \right)dr }\Delta E_{2s} =\frac{e^{2} }{\pi \varepsilon _{0} }\left(\frac{Z}{2a_{B} } \right)^{3}\left[\frac{1}{3}\frac{r^{3} }{b} -\frac{1}{2} r^{2} \right] ^{b}_{0} =-\frac{e^{2} }{8\pi \varepsilon _{0} }\left(\frac{Z}{a_{B} } \right)^{3} \frac{1}{6} b^{2}
\Delta E_{2s} =-\frac{e^{2} }{48\pi \varepsilon _{0} }\left(\frac{Z}{a_{B} } \right)^{3} b^{2}
For hydrogen Z = 1 and, putting in the numbers,
\Delta E_{2s} =-\frac{e^{2} }{48\pi \varepsilon _{0} }\left(\frac{1}{a_{B} } \right)^{3} b^{2} =\frac{-(1.602\times 10^{-19} )^{2} }{48\pi \times 8.854\times 10^{-12} } \left(\frac{1}{0.5292\times 10^{-10} } \right)^{3} (5\times 10^{-16} )^{2}giving
\Delta E_{2s} =2.0\times 10^{-10} eVFor the 2p state one obtains
\Delta E_{2p} =1.5\times 10^{-21} eVWe expect \Delta E_{2p} to be smaller than \Delta E_{2s} as the electron spends less time within the region r < b.
The eigenenergies of the hydrogenic atom are
E_{n} =-\frac{m_{r}e^{4} }{2(4\pi \varepsilon _{0} )^{2} \hbar ^{2} } \left(\frac{Z}{n} \right)^{2}where the reduced mass is
m_{r} =\frac{m_{0}m_{p} }{m_{0}+m_{p} }The change in energy eigenvalue introduced by assuming, as Bohr did, an infinite proton mass is given by the normalized energy
\frac{\Delta E}{E_{n} } =1-\frac{m_{0} }{m_{r} } =1-\frac{m_{0} (m_{0}+m_{p} )}{m_{0}m_{p} } =1-\frac{m_{0} }{m_{p} }-1=\frac{-m_{0} }{m_{p} }For the 2s state with eigenenergy E_{2s} \cong \frac{-Ry}{2^{2} } =\frac{-13.6}{4}=-3.4 eV the energy shift due to the finite mass of the proton is
\Delta E=\frac{m_{0} }{m_{p} } E_{2s}\sim \frac{9\times 10^{-31} }{1.6\times 10^{-27} } E_{2s}\sim 5\times 10^{-4}\times 3.4 eV=1.5\times 10^{-3} eVTherefore, correction for finite mass of the proton is a much more important energy correction (\sim 10^{-3} eV) than the first-order energy shift due to the finite size of the nucleus (\sim 10^{-10} eV).