# Why is the spin-orbit interaction for a nucleus so much more important than the spin-orbit interaction in atomic physics?

In atomic physics, the spin-orbit is a small correction between 1/1000 and 10ppm, so fairly small. In contrast, in nuclear physics the inclusion of the spin-orbit interaction is necessary to reproduce the basic level scheme, see picture below (from Kenneth Krane, Introductory Nuclear physics). Why is that?

I imagine it has to do with the $$r$$-dependence: In a semi-classical calculation, the energy correction due to the spin-orbit interaction has a $$1/r^3$$ dependency. However, I have only seen this calculation for the atomic case and not for the nuclear one.

Yes, it is indeed because of the $$1/r^3$$ dependency of the spin-orbit interaction. Intuitively speaking, in the atomic case the electron orbits the nucleus on the order of $$10^{-10}$$m. In the nuclear case, one nucleon orbits another one with a distance on the order of $$10^{-15}$$m.
Another more mathematical point of view is that considering a series expansion of all the forces involved, the spin-orbit-interaction is a higher order term and thus scales with $$\nabla^n F$$. Since $$F$$ is much more close-ranged for the nucleus, $$\nabla^n F$$ is much higher and contributes more strongly.