Another way of looking at the Hydrogen atom, other than the Rydberg Multiplier method, is as follows: Because the proton is that much smaller than the Planck mass the theoretical surface g is likely to be higher than c(equivalence) necessitating a rise in the event horizon bringing its diameter nearer to the size of its Compton wavelength. This results in a Lorentz transformation of the proton's Gm product to 29.6906036. This multiplied by the proton mass, 1.672623x10^-27kg, is 4.966118653x10^-26, which is equal to hc/4 and also equal to the Gm^2 product of G times the Planck mass squared. The electron speed is 2.187691412x10^6 ms so the electron Compton wavelength will be 3.324918617x10^-10m; divide this into hc/4 and you get 1.49360607x10^-16. Divide this by the Rydberg energy and you get 137.035989/2, the reciprocal of the fine structure constant/2. We can bypass this with hv, using the electron speed, above, we get 1.4495808x10^-27.
THE DEUTERIUM ATOM
We can apply the same reasoning to the deuterium atom as follows: One nucleon is ascribed an energy of 1.78204x10-13 J, a speed equivalent to 1.03219017x10^7ms and a wavelength of 1.91896991x10^-14m. (hc/4)/1.91896991x10^-14m equals 2.587908558x10^-12. Divide this by the Deuterium nucleon energy of 1.78204x10^-13 J and you get 14.52216869. Using the hv principle above, hx1.03219017x10^7ms=6.839369997x10-27. Divide this by 1.44958084x10^-27, above, and you get 4.718170781. Multiply this by 14.52216869, above, and you get 137.035989. This is because c/v=29.04427537 and 29.6906036 divided by 29.04427537=1.022253206. Multiply this by two and you get 2.044506412; multiply by the proton mass and divide by the wavelength 1.9189699x10-14m and you get 1.7820438x10^-13 J, the nucleon energy.