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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!

Qmechanic
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AsXApproches
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1 Answers1

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

https://www.cyberphysics.co.uk/Q&A/KS5/Nuclear/radius/nuclearRadiusQ4.html

different values are given in books etc. and seem to range from about 1.1 to 1.4 fm.

To get an intuitive feel for the formula, $r_0$ can be imagined as the radius of a proton or neutron.

Then the volume of a nucleus is

$\frac{4}{3} \pi R^3 = A \times \frac{4}{3}\pi r_0^3 $

and that simplifies to $R=r_0A^{1/3}$

John Hunter
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