From what I understand, due to Hawking radiation, black holes lose mass in the form of energy (electromagnetic radiation), with these characteristics:

  • The larger the black hole, the less energy it gives off and the slower it evaporates.
  • As it loses mass and shrinks, it begins evaporating faster.
  • At the end of its life, it approaches infinite evaporation due to its mass approaching zero.
  • Thus a black hole dies in a burst of radiation as the last bit of mass inside it explodes into radiative energy.

Given that a Schwarzschild black hole's evaporation rate and lifetime is dependent only on it's mass, all black holes will die identically.

However, I have no idea how fast a black hole actually evaporates.

How much energy is released in the final second of a black hole's life? Equivalently, how much does a black hole one second from death weigh?

And how does a black hole's death compare to a nuclear weapon?


1 Answer 1


The timescale for black hole evaporation by Hawking radiation is

$$t = 5120 \frac{\pi G^2 M^3}{\hbar c^4}$$

Turning this around. If $t=1$ s, then $$ M = \left(\frac{\hbar c^4}{5120 \pi G^2}\right)^{1/3} = 2.3\times 10^{5}\ kg $$ Thus the energy released is $Mc^2 = 2\times 10^{22}$ J.

This is 5 million megatons of TNT.

NB: All this is available on the wikipedia page on Hawking radiation.

  • 1
    $\begingroup$ I hadn't seen that, it answers my exact question. :S Thanks though! $\endgroup$ Mar 15, 2016 at 21:31

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