If a proton had a radius $R$ and the charge was uniformly distributed,calculate using Bohr theory,the ground state energy of a $H$-atom when $(i) R = 0.1 \mathring{A}$ and $(ii) R = 10 \mathring{A}$.

(A) The orbital radius of an electron in the ground state of a $H$-atom is the Bohr radius $a_0 = 0.53 \mathring{A} = \frac{\epsilon_0 h^2}{\pi m e^2}$.
For case $(i)$,$R = 0.1 \mathring{A}$. Since $R < a_0$,the proton can be treated as a point charge. The ground state energy is $E_1 = -13.6 \text{ eV}$.
For case $(ii)$,$R = 10 \mathring{A}$. Since $R > a_0$,the electron moves inside the proton. The effective charge $e'$ at radius $b_0$ is $e' = e \left( \frac{b_0^3}{R^3} \right)$.
The Bohr radius condition is $b_0 = \frac{\epsilon_0 h^2}{\pi m e e'} = \frac{\epsilon_0 h^2}{\pi m e^2} \left( \frac{R^3}{b_0^3} \right) = a_0 \frac{R^3}{b_0^3}$.
Thus,$b_0^4 = a_0 R^3 \Rightarrow b_0 = (a_0 R^3)^{1/4} = (0.53 \times 10^3)^{1/4} \approx 4.8 \mathring{A}$.
The potential energy inside a uniform charge sphere is $U = \frac{e}{4\pi\epsilon_0 R^3} \left( \frac{3R^2 - b_0^2}{2} \right) (-e)$.
The total energy $E = K + U = \frac{e^2}{8\pi\epsilon_0 R^3} (3R^2 - b_0^2) - \frac{e^2}{4\pi\epsilon_0 R^3} (3R^2 - b_0^2) = -\frac{e^2}{8\pi\epsilon_0 R^3} (3R^2 - b_0^2)$.