### Ground State Potential Energy

 From the home page, the velocity-to-light speed ratio for the ground state is:$\frac{v}{c}=\alpha Z\approx \frac{Z}{137}$ Making the relativistic Lorentz factor, gamma: $\gamma =\frac{1}{\sqrt{1-\alpha ^{2}Z^{2}}}$ Another property of the single-electron ion's ground state is that its shell circumference must be equal to the electron's de Broglie wavelength. The ground state shell circumferences are, therefore: $2\pi r=\lambda _{d}=\frac{1}{\gamma }\lambda _{c}\frac{c}{v}=\frac{\lambda _{c}}{\gamma \alpha Z}$ So that the ratio of the Compton wavelength to shell circumference is: $\frac{\lambda _{c}}{2\pi r}=\gamma \alpha Z$ From the definitions of alpha and the classical electron radius, the potential energy at a shell circumference of one Compton wavelength is: $E_{p(2\pi r=\lambda _{c})}=-\alpha Zm_{e}c^{2}$ Since potential energy is inversely proportional to shell radius, (and circumference), the potential at any allowed circumference is the product of the two previous equations. $E_{p}=\frac{\lambda _{c}}{2\pi r}E_{p(2\pi r=\lambda _{c})}=-\gamma \alpha ^{2}Z^{2}m_{e}c^{2}$ Since, $\alpha ^{2}Z^{2}=1-\frac{1}{\gamma ^{2}}$, $E_{p}=-(\gamma -\frac{1}{\gamma })m_{e}c^{2}$