Analysis about the force of electrons revolve around the nucleus

1, Let's compare the difference of two algorithms: the electrostatic force between protons and electrons, F1 $=$ ke$^{\mathrm{2}}$ / r$^{\mathrm{2}}$, r is the radius of the electron around the nucleus movement -- within 10$^{\mathrm{-10}}$ meters; Electronic movement speed is close to the light- about 10$^{\mathrm{7\thinspace }}$meters per second, the size of the centripetal force F2 $=$ v$^{\mathrm{2}}$m/r. F1 should be approximately equal to F2,calculate the ratio of F1 and F2, F2 / F1 $=$ (v$^{\mathrm{2}}$m/r) (ke$^{\mathrm{2}}$ / r$^{\mathrm{2}})$ / $=$ (10$^{\mathrm{7}}$ * 10$^{\mathrm{7}}$ * 0.91 * 10$^{\mathrm{-30}}$ / r)/(9 * 10$^{\mathrm{9}}$ * 1.6* 10$^{\mathrm{-19}}$*1.6*10$^{\mathrm{-19}}$ / r$^{\mathrm{2}}) \quad =$ 4 x 10$^{\mathrm{3}}$.The calculation shows that not only the electrostatic force and other force. 2, The radius of the electron orbiting around the nucleus named r, F $=$ Ke$^{\mathrm{2}}$ / r$^{\mathrm{2}} \quad =$ 9 x 10$^{\mathrm{9\thinspace }}$x £¨1.6 x 10 $^{\mathrm{-19}}) \quad^{\mathrm{2}}$ / r$^{\mathrm{2\thinspace }}=$ v$^{\mathrm{2}}$m/r, r $=$ 2.5 x 10$^{\mathrm{-14}}$ meters, namely that the radius of hydrogen atom is about 2.5 x 10$^{\mathrm{-14\thinspace }}$meters, that is different with the observed result (10$^{\mathrm{-10}}$ meters).Electrons revolve around the nucleus may faster than 10$^{\mathrm{7\thinspace }}$m/s, can almost reach 10$^{\mathrm{8}}$ meters per second, if the electronic moves by 10$^{\mathrm{8\thinspace }}$meters per second, hydrogen atom radius is approximately 2. 5 x 10 $^{\mathrm{-16}}$ meters, has converged in the interior of the nucleus, it is not possible. Use density to instead of electricity, can solve this problem. Author: hanyongquan TEL: 15611860790