If a proton had a radius R and the charge was uniformaly fistributed , Calculate using bohr theory, the ground state energy of a H-atom When `R=10Å`

Explanation

The electrostatic fore of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.
`(mv^(2))/(r_(B)) = -(e^(2))/(r_(B)^(2)).(1)/(4piepsi_(0))`
By Bhor's postulates in ground state, we have
`mvr = h`
On solving.
`:. m (h^(2))/(m^(2)r_(B)^(2)).(1)/(r_(B)) = + ((e^(2))/(4piepsi_(0)))(1)/(r_(B)^(2))`
`:. (h^(2))/(m) . (4piepsi_(0))/(e^(2)) = r_(B) = 0.51Å` , [This is Bohr's radius]
The potential energy is given by
`- ((e^(2))/(4pir_(0))).(1)/(r_(B)) = - 27.2 eV, KE = (mv^(2))/(2)`
`= 1/2 m.(h^(2))/(m^(2)r_(B)^(2)) =(h)/(2mr_(B)^(2)) = (h)/(2mr_(B)^(2)) = + 13.6 eV`
Now, for an spherical nucleus of raidus R,
If `R lt r_(B)`, same result.
If `R gt gt r_(B)` the electron moves inside the sphere with radius `r_(B)^(')` (`r'_(B)` = new Bohr radius).
Charge inside `r_(B)^(4) = e((r_(B)^(3))/(R^(3)))`
`:. r'_(B) = (h^(2))/(m) ((4piepsi_(0))/(e^(2)))(R^(3))/(r_(B)^(3))`
`r_(B)^(4) = (0.51Å).R^(3) [R = 10Å]`
`= 510(Å)^(4)`
`:. r'_(B) ~= (510)^(1//4) Å lt R`
`KE = (1)/(2) mv^(2) = m/2 . (h)/(m^(2)r_(B)^(2)) = (h)/(2m). (1)/(r_(B)^(2))`
`= ((h^(2))/(2mr_(B)^(2)).((r_(B)^(2))/(r_(B)^(2))) = (13.6 eV) ((0.51)^(2))/((510)^(1//2)) = (3.54)/(22.6) = 0.16 eV`
`PE = + ((e^(2))/(4piepsi_(0))). ((r'_(B)^(2) - 3R^(2))/(2R^(3)))`
`= + ((e^(2))/(4piepsi_(0)).(1)/(r_(B))). ((r_(B)(r'_(B)^(2)-3R^(2)))/(R^(3)))=+(27.2eV)[(0.51(sqrt(510)-300))/(1000)]`
`= + (27.2eV).(-141)/(1000) = -3.83 eV`