This might be a bit of a speculative question.

They say that if an object falls into a black hole, it would appear
to an observer at a great distance relative to the black hole that
the object slows down but never actually reaches the event horizon.

I'll refrain from pondering the idea that for an object near the
event horizon, quantum mechanically due to the uncertainty principle,
the probability of being inside the event horizon could already be
greater than 0. I'm sure experts have thought about that as well.

So, where was I? Ah yes, for an outside observer, the object reaches
the event horizon only after an infinite amount of time has passed.

On the other hand, from an outside perspective, the black hole
itself could have evaporated after a finite amount of time AFAIK.

What happens to the object, which from the outside appears
to be waiting just short of the event horizon, after the black
hole has evaporated from the point of view far from the black hole?
Would we find the object outside the event horizon then? - TIA!

[[Mod. note --
A few comments (I suspect the author already knows these things):

for an outside observer, the object reaches the event horizon only
after an infinite amount of time has passed

It's useful to think of the infalling object as sending out continuous
light (or radio) signals.

As the object approaches the event horizon, these signals take longer
and longer to reach a (fixed) outside observer, and the signals are more
and more redshifted. This results in the effect the author described,
and means that the outside observer never sees any signals emitted on
or inside the event horizon.

It may be useful to rephrase the above-quoted statement as:

Light/radio signals from the moment the infalling object crosses
the event horizon will take infinitely long to reach, and will
be infinitely redshifted when they arrive at, a (fixed) outside
observer.

This phrasing makes it clear(er) that the effects we're describing are
caused by light/radio signal propagation near the event horizon, rather
than the infalling object's motion being in any way arrested.

Indeed, an observer riding the infalling object itself would measure it
passing through the event horizon very quickly; it's "just" that after
passing through the event horizon her light/radio signals would never
reach the outside observer.

The time scales are interesting: For a 1-solar-mass black hole
(event horizon is a few kilometers in diameter):
* outside observer sees the object's radio signals infinitely redshift
and cut off on a time scale of microseconds; the last photon received
by the outside observer will be very low-enegy (upon reception) and
have originated just before the infalling object crossed the event
horizon
* infalling observer reaches r=0 singularity in microseconds
(but any messages from her about what it's like, will never
get to the outside observer)
* outside observer sees the black hole evaporate in something
on the order of 10^67 years

As for your actual question, what happens to things inside a black
hole when it (eventually) evaporates, it's a good question. Alas, I
don't think we (yet) understand quantum gravity well enough to know
the answer. :( There's a brief introduction to this topic at:
https://en.wikipedia.org/wiki/Hawking_radiation#Black_hole_evaporation
-- jt]]

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