Sep 28 2021

## Diameter of the Electron

It has been surmised that with conjugate wave gravitons, the number of gravitons in a free electron in the earth’s locale may be the reciprocal of the fine structure constant. We may then use this number to determine a possible diameter of the electron.

First, start with the energy:

(5.011 x 10^{-11} J) x 137 = 6.8651 x 10^{-9 }J

Then use the electron’s spin angular momentum:

ħ/2 = (½) mvr = (½) mr^{2}ω

Rearranging:

ω = ħ / (mr^{2})

Considering rotational kinetic energy, we have:

E_{k} = (½) [Iω^{2}] = (½) [ (½) mr^{2} ] ω^{2} = 6.8651 x 10^{-9 }J

Substituting ω from above:

E_{k} = (½) [ (½) mr^{2} ] [ħ / (mr^{2}]^{2} = (½) (½) [ħ^{2} / (mr^{2})]

1/r^{2} = [(4) (6.8651 x 10^{-9 }J) (9.1095 x 10^{-31} kg)] / ħ^{2}

r = 6.6676 x 10^{-16} m

D = 2r = 1.3335 x 10^{-15} m

ω = ħ / (mr^{2}) = 2.6040 x 10^{26} rad/sec

This angular velocity is fictitious, though it can be used in calculations due to Green’s Theorem (see April 2007 paper).

Of course the number of gravitons in an electron in an atomic orbital varies, and thus the electron diameter would vary also.