\simeq e^2\lambda_P^4M^2 R^2\omega_0^6 \left(\frac{R_0^3}{2GM_0}\right)^{1/2}F(\eta_0).$$

Thus we see that for $2GM_0/R_0 =  1, F(\eta_0)$ vanishes, as it should, since the collapse takes place at the beginning. If instead $2GM_0/R_0$ is smaller than $1$ but of that order, then $F(\eta_0)$ is also of the same order: e.g., if $2GM_0/R_0 = 1/2$ then $F(\eta_0) = 7/15$. Moreover, if we take as before $M_0 \simeq M \simeq 10^{56} GeV = 10^{37} M_P \simeq \cE _0$ and we recall that for $\hbar  = c = 1$ we have $G=M_P^{-2}$, we deduce $R_0  = 4GM_0 \simeq 10^{37} M_P^{-1} \simeq 10^{-1}\,km$, so that, assuming again $R \simeq 10^3 km \simeq 10^{41} M_P^{-1}$ and $T  \simeq 10^{-2} s  \simeq  10^{42} \tau_P$, we get $$\cE    \simeq e^2  \lambda_P^4 M^2 R^2R_0 \omega_0^6    \simeq  e^2  \lambda_P^4 M R^2R_0 \omega_0^6 \cE _0
 \simeq 10^{-96}  \cE _0 \simeq 10^{-40}\,GeV.$$

Note that using the same figures and multiplying the fraction of the total energy emitted as electromagnetic interaction per second, as given by the previous more brutal computation, Eq. , by the collapse time $t_{collapse} = t(\eta_0) = \frac1{\sqrt{2}}(\frac\pi 2-1)R_0 \simeq 10^{37} \tau_P \simeq 10^{-7} s$, we get an estimate of exactly the same order of magnitude.

Thus, as noticed earlier on in this 