Would a particle the size of a neutron, if it had enough mass, collapse into a blackhole? For example, a neutron is a particle that occupies a certain volume. If you pack enough mass into that volume, it would collapse into a black hole (I assume there is not enough mass now). At least if you don't consider quantum effects. Now, if QM is considered would that prevent a singularity from occurring? 
Another example is the electron. It's a point particle so you'd predict a collapse. But it's also spread out due to quantum mechanics. So, no collapse. 
The reason for this question is to create an analog to the prediction by classical EM that an electron should lose energy and spiral into the nucleus. Quantum mechanics comes to the rescue.
 A: If you believe classical parameters for black holes down to a neutron's scale, then it turns out that a neutron has too much angular momentum to form a black hole, and would be best interpreted as a naked singularity.
The reason why is that the radius of a black hole's event horizon is predicted to be:
$$M \pm \sqrt{M^{2} - a^{2}}$$
where $M$ is the mass of the black hole, and $a$ is the angular momentum per unit mass of the black hole.  If you put in the parameters for a neutron, $a > M$, and there are no real values for the horizon.  
A: The general relativity model for an electron is the Kerr-Newman solution, that is a charged rotating black holes. Unfortunately the radius of the exterior horizon in geometric units is given by: 
$$r_{+}=M + \sqrt{M^2-Q^2-P^2-\left(\frac{J}{M}\right)^2}$$
Here P is the magnetic charge, zero in this case. If you convert this relation to standard units ( for instance insert the mass in Kg and multiply for the conversion factor $G/c^2$) you will find that this radius is imaginary. That is, elementary particle are naked singularities from the point of view of general relativity.
As an aside, inspired from string theory and the fuzzball proposal it seems possible to find a completely regular (no singularity!) solution describing microstates of what you would call a naked singularity in general relativity. Basically, a solution of the equation of motion with the same asymptotics charges of a naked singularity, but without pathologies! This is still working in progress anyway. 
This is the classical picture. Roughly, the quantum picture tells you that when the Compton wavelenght is equal to the Schwarzschild radius (that is, at the Planck mass) you have reached the mininum mass for a black hole. So from this point of view there is no possibility for a particle to collapse .
The full answer of course cannot be given without the full quantum gravity theory in hand.
A: A black hole in the Schwarzschild case has a radius 
$$
r=\frac{2GM}{c^2}
$$
Let us solve for the mass with a radius of about Fermi or $10^{-15}$m, or the radius of a baryon such as a neutron, 
$$
M = \frac{rc^2}{2G}=\frac{10^{-15}m\times 9\times 10^{16}m^2/s^2}{2\times 6.67\times 10^{-11}Nm^2/kg^2}
$$
$$
=6.7\times 10^{11}kg.
$$
This is close to a billion tons! This would be a quantum black hole, and a pretty hot one with Hawking radiation and is racing towards it final evaporation.
