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?

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.

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 $\approx {10}^{-8}$.