The Bohr model for the H-atom relies on the Coulomb 's law of electrostatics. Coulomb's law has not directly been verified for very
short distances of the order of angstroms. Supposing Coulomb's law between two opposite charge +q₁, – q₂modified to

F= $\frac{{\mathit{k}\mathit{q}}_1{\mathit{q}}_2}{{\mathit{r}}^2}$ r>R₀

  = $\frac{{\mathit{k}\mathit{q}}_1{\mathit{q}}_2}{{\mathit{R}}_0^2}$ ( ${\frac{{\mathit{R}}_{\mathit{o}}}{\mathit{r}})}^{\mathit{\epsilon }}$  r0

Calculate in such vase the ground state energy of a H atom , if E= 0.1 R₀= 1A⁰

Kindly go through the solution 

Change in surface energy = work done

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}\times 10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63\times 10^{-34}\times 10.2}{12400\times 10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8\times 10^{-27}$           

$v=\frac{6.63\times 10.2\times 10^{-34}}{12400\times 10^{-10}}$

$v=\frac{6.63\times 10.2}{12400\times 1.8}\times 10^3$            

$=\frac{6.63\times 102}{124\times 1.8}=3.02$ ]

= 3 m/s

$-0.85=\frac{-13.6}{n^2}$  

n = 4

Number of transitions = $\frac{4\times 3}{2}=6$  

Kinetic energy: Potential energy = 1 : –2