Question

Estimate the correction to the ground state energy of hydrogen due to the finite size of the nucleus. Treat the proton as a uniformly charged spherical shell of radius b, so the potential energy of an electron inside the shell is constant$\mathbf{:}\mathbf{-}{\mathbf{e}}^{\mathbf{2}}\mathbf{/}\mathbf{\left(}\mathbf{4}{\mathbf{\pi ϵ}}_{\mathbf{0}}\mathbf{b}\mathbf{\right)}\mathbf{;}$this isn't very realistic, but it is the simplest model, and it will give us the right order of magnitude. Expand your result in powers of the small parameter, (b / a) whereis the Bohr radius, and keep only the leading term, so your final answer takes the form $\frac{\mathbf{\Delta E}}{\mathbf{E}}\mathbf{=}\mathbf{A}\mathbf{(}\mathbf{b}\mathbf{/}\mathbf{a}{\mathbf{)}}^{\mathbf{n}}$. Your business is to determine the constant Aand the power n. Finally, put in $\mathbf{b}\mathbf{\approx }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}$(roughly the radius of the proton) and work out the actual number. How does it compare with fine structure and hyperfine structure?

Short answer

Ground state energy correction is roughly ${10}^{-10}{\mathrm{E}}_1$, which is less than fine structure and hyperfine structure correction.

Step-by-step solution

Definition ofhyperfine spliting.

The interaction of the magnetic moments of the electron and proton causes hyperfine splitting, which results in a slightly variable magnetic energy for each spin state.

The hyperfine splitting in the ground state of muonic hydrogen.

Inside an evenly charged sphere, the potential is equal to:

$$\begin{gathered} \mathrm{V}\left(\mathrm{r}\right)=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\mathrm{b}}-\frac{1}{\mathrm{r}}\right) \\ {\mathrm{H}}^{\text{'}}=-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\mathrm{b}}-\frac{1}{\mathrm{r}}\right) \end{gathered}$$

Wave function of ground state:

$$\begin{gathered} {\mathrm{\Psi }}_0=\frac{{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}}{\sqrt{{\mathrm{\pi a}}^2}} \\ \mathrm{a}=\frac{4{\mathrm{\pi \epsilon }}_0{\mathfrak{h}}^2}{{\mathrm{me}}^2} \end{gathered}$$

Energy correction of ground state,

$$\begin{gathered} {\mathrm{E}}_0^1=<\mathrm{\psi }\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{\psi }}_0> \\ =-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{\pi a}}^3}\int {\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}\left(\frac{1}{\mathrm{b}}-\frac{1}{\mathrm{r}}\right){\mathrm{r}}^2\mathrm{drsin\vartheta d\vartheta d\phi } \\ =-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^3}4\mathrm{\pi }\left[\frac{1}{\mathrm{b}}{\int }_0^{\mathrm{b}}{\mathrm{r}}^2{\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}\mathrm{dr}-{\int }_0^{\mathrm{b}}{\mathrm{re}}^{-2\mathrm{r}/\mathrm{a}}\mathrm{dr}\right] \\ =-\frac{{\mathrm{e}}^2}{{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^3}\left\{\frac{{\mathrm{a}}^3}{4\mathrm{b}}\left[{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}\left(-\frac{2\mathrm{b}}{\mathrm{a}}\left(\frac{\mathrm{b}}{\mathrm{a}}+1\right)-1\right)+1\right]-\frac{\mathrm{a}}{4}\left[\mathrm{a}-{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}(\mathrm{a}+2\mathrm{b})\right]\right\} \\ =-\frac{{\mathrm{e}}^2}{{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^3}\frac{\mathrm{a}}{4}\left\{\frac{{\mathrm{a}}^2}{\mathrm{b}}\left[1+{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}\left(-\frac{2{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{2\mathrm{b}}{\mathrm{a}}-1\right)\right]-\mathrm{a}+{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}(\mathrm{a}+2\mathrm{b})\right\} \\ =-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^2}\left\{\frac{{\mathrm{a}}^2}{\mathrm{b}}-{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}\left(2\mathrm{b}+2\mathrm{a}+\frac{{\mathrm{a}}^2}{\mathrm{b}}\right)-\mathrm{a}+{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}(\mathrm{a}+2\mathrm{b})\right\} \\ =\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^2}\mathrm{a}\left[\left(1-\frac{\mathrm{a}}{\mathrm{b}}\right)+{\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}\left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)\right] \end{gathered}$$

If $\frac{\mathrm{b}}{\mathrm{a}}<<1$then ${\mathrm{e}}^{-2\mathrm{b}/\mathrm{a}}\approx 1-\frac{2\mathrm{b}}{\mathrm{a}}+\frac{1}{2}\frac{4{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{1}{6}\frac{8{\mathrm{b}}^3}{{\mathrm{a}}^3}.$

Energy is then equal to:

$$\begin{gathered} {\mathrm{E}}_0^1=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{1-\frac{\mathrm{a}}{\mathrm{b}}+\left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)\left(1-\frac{2\mathrm{a}}{\mathrm{b}}+\frac{2{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{4{\mathrm{b}}^3}{3{\mathrm{a}}^3}\right)\right\} \\ =\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left[1-\frac{\mathrm{a}}{\mathrm{b}}+1-\frac{2\mathrm{b}}{\mathrm{a}}+\frac{2{\mathrm{b}}^2}{{\mathrm{a}}^2}+\frac{\mathrm{a}}{\mathrm{b}}-2+\frac{2\mathrm{b}}{\mathrm{a}}-\frac{4{\mathrm{b}}^2}{3{\mathrm{a}}^2}\right] \\ =\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left(\frac{2{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{4{\mathrm{b}}^2}{3{\mathrm{a}}^2}\right) \\ =\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\frac{1}{\mathrm{a}}\frac{2{\mathrm{b}}^2}{3{\mathrm{a}}^2} \end{gathered}$$

Energy of unperturbed ground state,

$$\begin{gathered} {\mathrm{E}}_1=-\frac{1}{2\mathrm{a}}\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0} \\ {\mathrm{E}}_1\mathrm{a}=-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{2} \\ \frac{{\mathrm{E}}_0^1.\mathrm{a}}{{\mathrm{E}}_{1.}.\mathrm{a}}=-\frac{4}{3}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^2 \\ \mathrm{A}=-\frac{4}{3} \\ \mathrm{n}=2 \end{gathered}$$

If $a=5\cdot {10}^{-11}$then, $\frac{{\mathrm{E}}_0^1\cdot \mathrm{a}}{{\mathrm{E}}_1\cdot \mathrm{a}}\approx -5\cdot {10}^{-10}$which is smaller then correction of fine structure $\left(\approx {10}^{-5}\right)$and hyperfine structure $\approx {10}^{-8}$.