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! $\endgroup$ Commented Mar 15, 2016 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.