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

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