Figure 7
Black holes will turn into white holes
Every black hole loses mass due to Hawking radiation. In fact all the mass it has swallowed will eventually be released as photon energy. The evaporation rate is depending on its mass. This plot shows that the evaporation rate of a black hole, due to Hawking radiation, is higher for small black holes with less mass. The lower the mass the higher the evaporation rate. Consequently, the lower the mass the more intense and energetic the emitted radiation.
A shortlived micro white hole of 1.0 metric Tons
A black hole which has a luminosity which is equal to L_{Sun} has a mass of 1.0 metric Tons and is 10^{8} times smaller than a proton. However, it has a lifetime of about 100 nanoseconds. The evaporation rate is astronomically high: 10 metric Mega Tons per second. So it seems that when black holes are starting to be observable as a white hole they will be be observed as a burst of Gamma radiation with rapidly increasing intensity and energy over a time period of only 100 nanoseconds.

The luminosity of the smallest back hole / white hole
The smallest black hole, which is possible theoretically, has a mass of 2.177 x 10^{8} kg. But it is the brightest one with a luminosity of 2.36 x 10^{48} Watt. But because of its extremely short lifetime of 8.68 x 10^{40} seconds it will only release 2.05 x 10^{9} Joule. That is what a one gigawatt powerplant would produce in two seconds.
So the swan song of this micro black hole is a 8.68 x 10^{40} seconds burst of Gamma radiation, with a total energy of about 1.28 x 10^{22} MeV.
Personal reminder:
It requires some research on my side to find out whether there are models which describe the possble radiation spectra of such Gamma radiation bursts. In other words what is the expected energy distribution of the Gamma photons which are emitted during the burst? And what are the factors which might influence such a distribution?
Another fact what confuses me has to do with the calculated photon energy:
h·ν = h·c^{3}/(4·G·M)
The mass (M) of our micro black hole is 2.177 x 10^{8} kg.
So the calculated photon energy is 3.073 x 10^{9} Joule
We know that the luminosity is 2.36 x 10^{48} Watt.
That means that it takes 1.3 x 10^{39} seconds to emit that 3.073 x 10^{9} Joule, which is, however, longer than the remaining 8,68 x 10^{40} seconds of lifetime.
We should however keep in mind that in these very last moments of 10^{40} seconds both the emitted energy and therefore luminosity is increasing in an extremely fast rate. The classical mathematical relationship between energy and luminosity (E = L x Δt) is probably not that straightforward.
The luminosity is derived from the evaporation rate: dM/dt. The energy, in this consideration, is the emitted photon energy whose frequency of radiation can be calculated with expression 4 shown at the top of this page. That is another reason why a description of the relationship between emitted energy and luminosity needs a more sophisticated approach.
Although a black hole of mass equal to the Planck mass is a rather extreme case, this case seems nevertheless relevant to me. After all it is the final phase of any black hole which is no longer fed.
These formulas seem also to suggest that for such a small micro black hole of absolute minimum mass there is no energy distribution of emitted Gamma photons, only one extremely high energy Gamma photon is emitted. How about conservation of momentum, I might ask, for which at least two photons are required?
To put this huge amount of energy in perspective we have to calculate how much matter represents the same amount of energy.
The rest energy of a proton: E_{p,0} = m_{p}·c^{2} = 1.504 x 10^{10} Joule.
When this rest energy is expressed in eV instead of Joule: 938.6 MeV.
So 1.28 x 10^{22} MeV of energy is equivalent to the total rest mass of 1.36 x 10^{19} protons (which is about 22 microgram).
It is to be expected that a fraction of the emitted Gamma photons result in the production of matter and antimatter through protonantiproton and electronantielectron pair production for example. After a short while (nanoseconds?) matter and antimatter might recombine resulting in a secondary burst of Gamma radiation.
Personal speculation:
It would not surprise me if this process of pair production is repeated multiple times resulting in multiple generations of Gamma ray bursts with decreasing intensity and energy (over a time period of about several nanoseconds). When the energy of each Gamma photon has dropped below a threshold value of about 1 MeV the pair production will stop. If however all previous generations of Gamma Ray burst have resulted in pair production it will probably mean that the black hole dissipation event will eventually result in a Gamma ray burst with a photon energy of 1 MeV or less. We will probably never be able to detect the previous generations of Gamma ray bursts.
But there might be a condition which can influence the recombination process. Maybe there is a possiblity that some of the produced particles can leak outwards.
