# What $r_0$ in the nucleus radius equation?

I'm doing some self-study and I'm attempting to calculate the energies needed to overcome the Coulomb barrier. I stumbled upon $$R = r_0A^{1/3}$$ for the radius of nuclei. I've been looking around online and I've seen that r0 is equal to 1.3E-15.

But I can't find out why it's equal to that, and where it comes from, any help would be fantastic!

• – rob
Mar 27 at 1:39

$$r_0$$ would be determined by experiment as in this link

To get an intuitive feel for the formula, $$r_0$$ can be imagined as the radius of a proton or neutron.
$$\frac{4}{3} \pi R^3 = A \times \frac{4}{3}\pi r_0^3$$
and that simplifies to $$R=r_0A^{1/3}$$