It appears like black holes posses an escape velocity greater than the velocity of light. It takes an atom infinity energy to travel at speed of light.Do black holes posses that energy to retain light itself?What is the potential energy of a black hole?
Your question What is the potential energy of a black hole? doesn't make sense because energy is a somewhat tricky concept to deal with in GR.
If we treat the black hole as fixed we can study the motion of a test particle falling into it, and we find that there is a quantity analogous to total energy that is constant as the particle falls in. So in principle as the kinetic energy of the infalling particle increases its potential energy becomes more negative. But we can't easily separate this total energy into kinetic and potential. For one thing the coordinate velocity of an infalling particle tends asymptotically to zero as it approaches the horizon so it's hard to define what we mean by kinetic energy there.
Black holes are in the realm of General Relativity. In GR even the law of conservation of energy is under question when approaching singularities of the GR solution. Potential energy is a concept that comes with conservation of energy. Where the singularity in the black hole solutions is dominating, one cannot talk in terms of energy conservation and potential energy.
The behavior of particles, including photons, in space time has to be calculated using general relativity equations. These tell us that there is no escape path below the horizon.
The total energy of a testparticle in the vicinity of a Schwarzschild black hole is
$$E_{\rm total} = \frac{m \ c^2 \ \sqrt{1-r_{\rm s}/r}}{\sqrt{1-v^2/c^2}}$$
which is the sum of the rest energy, the kinetic energy and the potential energy
$$E_{\rm total} = E_{\rm rest} + E_{\rm kin} + E_{\rm pot}$$
Since the rest- and the kinetic energy are
$$E_{\rm rest} = m c^2 \ , \ \ E_{\rm kin} = \left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)m \ c^2$$
the potential energy remains
$$E_{\rm pot} = E_{\rm total} - E_{\rm rest} - E_{\rm kin}$$
which is
$$E_{\rm pot} = \frac{\sqrt{1 - r_{\rm s}/r}-1}{\sqrt{1-v^2/c^2}} m \ c^2$$
$v$ is the local (not the shapiro-delayed) velocity, $r_{\rm s}$ the Schwarzschild radius $2 \ G M/c^2$, $M$ the mass of the black hole and $m$ the rest mass of the test particle.
The total energy is, like the angular momentum, a conserved quantity throughout the whole trajectory.
For more information see Wikipedia: Schwarzschild Geodesics and Schwarzschild Trajectories
Well there is two answers: 1. It is not infinite. At a point when the black hole has taken in lots of matter, it will throw some out of the black hole to again be able to take in matter.It is like the biggest foodie in the world who can eat a lot but at a point when his tummy is full, he needs to throw up or wait until he can take in more. 2. It is infinite. According to S.Hawking , particles escape the fabric of the universe and enters another universe by black hole which is supported by string theory. However, it was also told that tearing the fabric is impossible or needs huge amount of force but even if we are able to tear it up, the strings would again stitch it. So, we must wait until we get a grand unified theory to merge gravity and quantum mechanics!
