Recent content by Eric71

Thanks blue_leaf77. I think I understand now. I simply calculated the kinetic energy since v<<c. I forgot to add the potential energy, which is twice the kinetic energy but negative, hence the sign error in my result.

Thanks jfizzix But that's what surprises me so much. The energy difference due to this relativistic effect of "only" 3 in 100,000 (to be more precise 2.66 in 100,000) is the same as the energy level of the electron from the Bohr model: 2.66/100000 * mc2 = 13.5 eV

From Bohr's model of the Hydrogen atom in the ground state v = e2/(2εh) = 2.18x106 m/s Using m=m0 / √(1-v2/c2) and E=(m-m0)*c2 leads to E = 13.5 eV Using Bohr's equations E = m*e4/(8ε2h2) also gives 13.5 eV I tried to see what can be derived by substituting the first equation in the second...

From the semi-classical Bohr model of the hydrogen atom the velocity of the electron in a certain orbit can be determined. With these velocities the electron's relativistic masses can be determined. With E=mc2 the energy levels are in agreement with those from the Bohr model. I know Bohr's model...