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Mass of mercury, formula of time taken to evaporate by a black hole

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In order to find out how long does it take for an black hole to evaporate of mass "m", we first need to find the total radiated power in a timeframe which happens via hawking radiation which is basically the black hole analogue of thermal blackbody radiation as if the black hole is a perfect blackbody (due to antimatter-matter pairs escaping, negative energy destroying the singularity).

Step 1:

So first we need to find the radiated power using the hawking radiation equation:

$P = c^5ħ/15360πG^2M^2$

Wherein c is the speed of light constant

ħ is the reduced plank constant

π is the Pi constant

G is the Newtonian Gravitational constant

M is the Mass variable

In the case of mercury where M = 3.285 × 10^23 kg or 0.055 M⊕ the total radiated power by hawking radiation is 3.3×10^-15 watts

Note: It is inversely proportional to the mass

Step 2:

Now we have to convert the total radiated power to the mass of the energy, via the mass-energy equivalence equation which is

$E = mc^2$

and since we want to find out the mass not energy we have to do:

$M = e/c^2$

and by plugging in our value of e we get the mass losing per second = 3.672×10^-32 kilograms per second

Step 3:

Now all we have to do is divide the total mass of the body by mass losing per second in order to find out the total evaporation time, which in this case is $3.285 × 10^23 kg/3.672×10^-32 kg/s$.

Voila! Via this method we find out the total evaporation time of a black hole of the mass of mercury is 8.946×10^54 seconds or 2.1*10^37 gigayears or the lifetime of the universe (13.8 billion years)

Furthermore we can make a equation by combining all these steps which will be Evaporation time or Et = $M/((c^5ħ/15360πG^2M^2)/c^2)$.

Now by cancelling out common terms we can simplify this equation in a simple and elegant way (Will modify it)

Thank you!

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    $\begingroup$ This isn't exactly right because as the black hole loses mass, the rate at which it loses mass increases. $\endgroup$
    – Sten
    Commented Jul 26, 2023 at 13:03
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    $\begingroup$ Thank you, @Sten. I will incorporate the iteration because of the increasing rate of total radiant mass $\endgroup$
    – Arjun
    Commented Jul 26, 2023 at 16:15

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