Q.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

For a point-charge proton the electron moves in a $1/r$ Coulomb potential, giving the ground state $-13.6\,\text{eV}$ at the Bohr radius $a_0 = 0.53\,\text{Å}$.

(i) $R = 0.1\,\text{Å}$. Here $R \ll a_0$, so the electron's orbit lies entirely outside the charged sphere, where the potential is the ordinary Coulomb form. The finite size changes almost nothing:

(ii) $R = 10\,\text{Å}$. Here $R \gg a_0$, so the electron orbits inside the uniformly charged proton. There the field is $E(r) = \dfrac{e}{4\pi\epsilon_0}\dfrac{r}{R^3}$ — a linear (harmonic) restoring force. Bohr's condition $mvr = \hbar$ with the force balance $\dfrac{mv^2}{r} = \dfrac{e^2}{4\pi\epsilon_0}\dfrac{r}{R^3}$ gives

$r^4 = a_0 R^3 \;\Rightarrow\; r = (0.53\times10^3)^{1/4} \approx 4.8\,\text{Å}\;(<R),$ …

1. Potential of a uniformly charged sphere (total charge $+e$, radius $R$):

Outside the sphere the field is the usual Coulomb field; inside it grows linearly with $r$.

2. Case (i): $R = 0.1\,\text{Å}$.

The Bohr radius is $a_0 = 0.53\,\text{Å}$, so $R \ll a_0$: the electron's orbit lies well outside the proton, where the potential is exactly Coulombic. The result is essentially unchanged:

3. Case (ii): $R = 10\,\text{Å}$.

Now $R \gg a_0$, so the electron orbits inside the charged sphere. The inside field

produces a linear restoring force $F = \dfrac{e^2}{4\pi\epsilon_0}\dfrac{r}{R^3}$ (harmonic, like a spring).

4. Find the orbit radius from Bohr's condition.

With $mvr = \hbar$ (ground state, $n=1$) and force balance

substitute $v = \hbar/(mr)$:

$\frac{\hbar^2}{mr^3} = \frac{e^2}{4\pi\epsilon_0}\frac{r}{R^3} \;\Rightarrow\; r^4 = \frac{4\pi\epsilon_0\hbar^2}{me^2}\,R^3 = a_0 R^3.$ …

Method: Compare $R$ to $a_0$ First, Then Use the Virial Theorem Inside the Sphere

Rather than blindly computing kinetic and potential energy by integration in every case, first check which regime you're in — this tells you immediately whether any calculation is even needed, and which shortcut applies if it is.

Steps

Step 1: Compare the given radius to the Bohr radius $a_0 = 0.53\,\text{Å}$

• If $R \ll a_0$: the electron's orbit lies entirely outside the charged sphere, where the field is ordinary $1/r^2$ Coulomb. No new calculation is needed — the ground-state energy is just the standard $-13.6\,\text{eV}$.
• If $R \gg a_0$: the electron orbits inside the sphere, where the field is linear in $r$ (a Hooke's-law-type restoring force) — a genuinely different problem.

Step 2: For the inside-sphere case, recognise it as a power-law-force problem

Inside a uniformly charged sphere, $F(r) \propto r$ (like a spring). For any circular orbit under a central force $F(r)$, the centripetal condition alone gives kinetic energy directly:

This is a general shortcut — you don't need to separately solve for $v$ and square it once you know $r$ and $F(r)$.

Step 3: Find $r$ from Bohr quantization, then get $K$ in one line …