Are the below observation valid?

Shortly after developing his theory of General Relativity, Einstein developed his Quantum Theory of Radiation. If we converge these two theories we can develop Einstein’s A and B Gravitational Coefficient. For the moment we will focus on the B Gravitational Coefficient, which blends the classical and quantum relationships.

\(B=\frac{l}{m}=\frac{c^3}{h\;v^2}\)

As such, Einstein’s B_{e} coefficient for the electron is;

\(B_e=\frac{r_e}{m_e}=\frac{2.8\times 10^{-15}\;m}{9.1\times 10^{-31}\;kg}= 3.1\times 10^{15} \frac{m}{kg}\)

In Einstein's field equations, Einstein’s gravitational constant kappa is proportional to Einstein's B_{k} gravitational coefficient, (at the Planck scale).

\(B_\kappa=\frac{\kappa}{8 \pi}=\frac{G}{c^2}=B_P= \frac{l_P}{m_P}=7.4\times 10^{-28} \frac{m}{kg}\)

Comparing the ratio of B_{k} to B_{e},

\(\frac{B_\kappa}{B_e}=2.4\times 10^{-43} \)

This ratio is exactly equal to the relative strength of the gravitational to the electrostatic force of two electrons. All values of B_{x} are frequency squared scaled versions of Einstein's B_{k} gravitational coefficient;

\(B_x=B_{\kappa} \times \frac{v^2_{\kappa} }{v^2_x}\)

As the B coefficient is proportional to the inverse frequency squared the following holds:

\(\frac{G\;{m^2_e}}{k_e\;q_e^2}=\frac{B_\kappa}{B_e}=\frac{\upsilon_{Be}^2}{\upsilon_{B\kappa}^2}=2.4\times 10^{-43} \)

From the above we can also conclude the weakness of the gravitational force is related to a frequency squared and the probability of stimulated emission/absorption of energy due to curvature.