# Why can't a nucleus become a black hole?

Let's take the carbon nucleus as a convenient example. Its mass is $1.99 \times 10^{-26}$ kg and its radius is about $2.7 \times 10^{-15}$ m, so the density is about $2.4 \times 10^{17}$ kg/m$^3$. Your density is ten orders of magnitude too high.

The Schwarzschild radius of a black hole is given by:

$$ r_s = \frac{2GM}{c^2} $$

and for a mass of $1.99 \times 10^{-26}$ kg this gives us:

$$ r_s = 2.95 \times 10^{-53} \,\text{m} $$

This is far below the Planck length, so it is unlikely matter could be squeezed into a region that small i.e. a single carbon nucleus cannot form a black hole.

If we take the Planck length as $r_s$ and calculate the associated black hole mass the result is half the Planck mass $\tfrac{1}{2}\sqrt{hc/2\pi G}$, which is about $11\,\mu\text{g}$ or about $5.5 \times 10^{17}$ times larger than the mass of the carbon nucleus. This is the smallest mass that we expect could form a black hole.

## The minimum radius of a spherical object to be a black hole is given by :

## r = 2Gm/ (c^2)

From this, I think we may be able to calculate the minimum density for the object to be a black hole, which is:

```
d = (21/704)((c^6)/((G^3)(m^2)) (Assuming pi = 22/7)
```

## it is, d = (7.37 x 10^79 )/(m^2)

So, you can now guess how high it is.
For URANIUM, it is **4.61 x 10^128 kg/metervolume**

as you can see, it is in the order of 10 raised to power of 128 !!!

For hydrogen, it is **2.8 x 10^133 kg/metervolume**

In the order of 10 raised to the power of 133 !!!

But, the average density of nucleus is much lower than the above values.

Thus, we can never expect nucleus to become a black hole.

Nature is always remarkable!!! It always helps us.....