Are Buckyball-sized black holes possible?

Question

The first item is the basic question; the subsequent items build upon it if it's possible. If these need to be broken into separate questions, I can do that, but they're pretty tightly related.

1. Is a non-rotating negatively-charged singularity small enough to be contained by a C₆₀ Buckyball theoretically possible?
2. Would the charge repulsion of the carbon electrons be stronger than the gravitational attraction, keeping the singularity from consuming them?
3. Is it theoretically possible for a singularity to be too small to absorb hadrons or even elementary particles?
4. Anything this small would doubtless evaporate very quickly, but just how long could they last?

Thanks!

Answer 1

1. A buckyball is about a nanometre ($10^{-9}$ m) across. If you limit the charge on the black hole to something like that of an electron or a few electrons, then this would mean the event horizon(s) of a charged, spinless, Reissner-Nordstrom black hole would be almost indistinguishable from that of a Schwarzschild black hole. The mass of this black hole would therefore be around $r_s c^2/2G \simeq 10^{18}$ kg. Yes, this is theoretically possible and maybe such black holes were produced during the big bang and are still around today.

2. For Coulomb forces to outweigh gravitational forces then you need $$ \frac{Q^2}{4\pi \epsilon_0} > G M_{\rm BH}{m_C}\ .$$ In this case $Q \sim 10^{-18}$ Coulombs, $M_{\rm BH}\sim 10^{18}$ kg, $m_C = 12\times 1.67\times 10^{-27}$ kg. The LHS is $\sim 10^{-26}$ Nm$^2$ and the RHS is $\sim 10^{-18}$ Nm$^2$. So perhaps surprisingly, gravity will win and the buckyball will almost instantly be incorporated into the black hole.

3. We don't have a theory governing the quantum behaviour of black holes. A singularity doesn't really have a size in a non-spinning black hole.

4. A $10^{18}$ kg black hole would evaporate via Hawking radiation in about $10^{30}$ years.