Why couldn’t Rutherford explain the stability of an atom?

In Rutherford's model, electrostatic force provides the centripetal force:

$\frac{1}{4\mathrm{\pi E}{ }_{}}$_∈  

Kinetic Energy (K):

$K=\frac{1}{2}m{v}^2=\frac{e^2}{8\pi E_{0}r}$

Potential Energy (U):

$U=-\frac{1}{4\pi E_0}\cdot \frac{e^2}{r}$

Total Energy (E):

$E=K+U=-\frac{e^2}{8\pi E_{0}r}$

Note that, E is the Epsilon symbol. 

In the gold foil scattering experiment by Rutherford and his team, almost all alpha particles passed straight through the foil. There were minor deflections mostly, and only a small amount of them showed deflections at really large angles. Only a few of them rebounded. This was the main observation that led Rutherford to conclude that the positive charge and mass of the atom concentrate at a very tiny section of the atom. That’s the nucleus. The other observation was that the entire atom must be space where the alpha particles could go undeflected. 

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 = $\frac{4\times 3}{2}=6$