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Theoretically we can convert a body into black hole by compressing its mass below some radius (known as Schwarzschild Radius).

Suppose such object after becoming black hole has a radius which is comparable to atom and nucleus radius. Now my question is:

what happens if we run this object into double slit experiment setup (single black hole at a time)?

Shall we get interference pattern on the detector wall because its radius is in quantum domain (say we somehow know when on detector screen this black hole will hit) or simple-particle pattern (i.e. classical way) because of the mass associated with that black hole?

If you have a really tiny black hole, will it behave as a classical particle, or will it exhibit quantum behavior?

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    $\begingroup$ Great question! Physicists would be foaming at the mouth for chance to actually perform this experiment. $\endgroup$ – David H Mar 21 '14 at 13:30
  • $\begingroup$ As an aside, using the Schwarzschild radius of 2 fm (approximately the hydrogen nucleus radius), the mass would be something like $10^{15}$ grams (approximately the global biomass of fish). It would also last about $10^{22}$ years before finally evaporating. $\endgroup$ – Kyle Kanos Mar 21 '14 at 15:55
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    $\begingroup$ Why the downvote? $\endgroup$ – Hunter Mar 21 '14 at 17:14
  • $\begingroup$ I get something like 40000kg and 2ms for the 2fm blackhole. If it was stable I guess it would behave classical because of its mass. $\endgroup$ – Mirc Breitschuh Mar 21 '14 at 23:59
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    $\begingroup$ @MircBreitschuh Not sure how you got those, but I used the Schwarzschild radius and solved for $M$: $r_s=2GM/c^2\to M=rc^2/2G=(2\cdot10^{-13}{\rm cm}\cdot(2.9979\cdot10^{10}{\rm cm/s})^2/(2\cdot6.674\cdot10^{-8}{\rm cm^3/g/s^2}=1346629369193887{\rm g}\sim10^{15}{\rm g}$. Using that in the evaporation timescale gives you the $10^{22}$ years I got. $\endgroup$ – Kyle Kanos Mar 22 '14 at 0:16
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I'd like to start by addressing some common misconceptions that might be influencing your question.

Firstly quantum mechanics holds at every scale, it is not that quantum mechanics fails at some level, it is just that sometimes its stranger properties are producing very small effects. For instance the de Broglie wavelength of a baseball is quite small, so you'd need a very very very small slit to get diffraction of a baseball and the spacing between the bright fringes and dark fridges would be so small you'd just see a gray smudge even if you did manage it so practically you just don't worry about it.

Secondly, the mass of a black hole is not the sum of the masses of all the things in it. The mass of a black hole is a parameter describing what mass in newtonian gravity it looks like from far away. If you have to press an object hard to compress it, then its mass is going to go up (all that energy from pushing hard went somewhere and it goes somewhere in a way that increases that parameter). But once you get it small enough that it will compress the rest of the way on its own then you can try to steal some of that energy back if you wanted to, maybe even more than you put in. So it's not clear how the mass of the final black hole is related to the mass of the object you started with.

So there isn't an obvious size to aim for, and even if I knew the mass of your object to begin with, I don't know how you made the black hole, so I don't know the mass of the black hole you end up with.

Now to address your question. One way to estimate the quantum properties of a black hole, is to compare the compton wavelength $h/(mc)$ to other lengths. If the compton wavelength $h/(mc)$ is much smaller than the other lengths of interest then quantum affects can often be ignored. For a large everyday mass that length is quite small indeed.

So your black hole probably won't have interesting quantum effects that are easy to detect.

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    $\begingroup$ "Firstly quantum mechanics holds at every scale": while I agree with the sentiment here, this has not been verified experimentally. $\endgroup$ – DanielSank Jan 4 '15 at 5:29

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