Can black holes evaporate into Neutron stars? If adding mass to a neutron star eventually makes it a black hole, why do black holes after losing mass through hawking radiation not evaporate into Neutron stars?
 A: I will add some details on what joseph h means by "once a black hole forms, it stays a black hole"
Our current understanding of black hole formation is that once a black hole is formed, the light cones flip at 90° inside and point out toward the singularity. In fact in that Schwarzschild metric
\begin{equation}
ds^2=\left(1-\frac{r_s}{r}\right)dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2 -r^2 d\Omega^2
\end{equation}
you can easily see that for $r<r_s$, the signs of the radial part and the time part are switched, and this is what we usually mean by "inside a black hole time becomes space and space becomes time". It means that if you add some matter to a neutron star, the star will collapse at some point. As the black hole grows, matter inside cannot do anything but falling toward the singularity because this is the only possible motion inside a black hole, just as moving forward in time is the only possible motion in time.
Once the matter of the neutron star has completely fallen to the insularity, it is impossible to reverse the situation go back to some state where one has a stable neutron star inside an event horizon (which is an impossible state). Finally, the black hole horizon will evaporate as time flows and we have the two senarii exposed by joseph h.
A: Because the evaporation just shrinks the black hole.
In short, while there might be a minimum size for a star to become a black hole, there's no minimum size for a black hole itself; as a result, the evaporation of a black hole through Hawking radiation will simply result in the black hole growing steadily smaller. Since the intensity of Hawking radiation increases as the black hole decreases in size, this is a process that would accelerate until the black hole basically explodes into a burst of Hawking radiation.
A: Entropy argument:
A black hole has the maximum possible entropy for a given volume (or, by extension, mass - see here https://en.wikipedia.org/wiki/Black_hole_thermodynamics )
This holds true for any size of a black hole, including the mass interval where the neutron stars are stable.
In order to decay into a neutron star, a black hole has to dispose off some entropy and it has no imaginable means to do so.
A: Once a black hole forms, it stays a black hole. There are two possible final states of a black hole which emits Hawking radiation:

*

*At some point it stops emitting Hawking radiation and becomes a permanent object called a "remnant"$^1$, with a very small mass that's roughly the Planck mass $m_{pl}$  where $$m_{pl} =\sqrt{\frac{\hbar c}{G}}\approx 2.2\times 10^{-8}kg$$
or


*The black hole completely evaporates leaving particles whose combined mass is even much smaller i.e., $m \ll m_{pl}$
This means there will not be a further state of a black hole that is consistent with a neutron star since the black hole with its Hawking evaporation leads to one of the above two possible final states.
$^1$ A black hole remnant is the stable or meta-stable end state of Hawking evaporation. That is, the Hawking radiation may stop when the mass of the black hole gets to the Planck scale.
