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 Å and (ii) R=10 Å.

Hint: The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.
Step 1: Find the value of the Bohr radius.
mv2rB=-e2rB2·14πε0
By Bohr's postulates in the ground state, we have; 
mvr=h2π
On solving,
mh24π2m2rB2·1rB=+(e24πε0)1rB2h24π2m·4πε0e2=rB=0.51Å            [This is Bohr's radius]
Step 2: Find the potential energy and kinetic energy.
The potential energy is given by;
-(e24πε0)·1rB=-27.2eV;KE=mv22=12m·h24π2m2rB2=h8π2mrB2=+13.6 eV
Now, for a spherical nucleus of radius R.
Step 3: Find the value of the Bohr radius for R.
If R<rB, same result.
If R>>rB, the electron moves inside the sphere with radius r'B (r'B=new Bohr radius).
Charge inside r'B=e(r'B3R3) r'B=h24π2m(4πε0e2)R3r'B3r'B4=(0.51Å)·R3            [R=10Å]=510(Å)4r'B(510)1/4Å<RStep 4: Find the potential energy and kinetic energy for R.KE=12mv2=m2·h4π2m2r'B2=h8π2m·1r'B2=(h28π2mrB2)·(rB2r'B2)=(13.6 eV)(0.51)2(510)1/2=3.5422.6=0.16 eVPE=+(e24πε0)·(r'B2-3R22R3)=+(e24πε0)·(rB(r'B2-3R2)R3)=+(27.2eV)[0.51(500-300)1000]=+(27.2eV)·-1411000=-3.83 eV