# Derivation of nuclear spin $I$ for $^{87}$Rb and $^{40}$K

So I know that $$^{87}$$Rb has $$I=3/2$$, and has 37 protons and 50 neutrons.

I try to make sense of it form the nuclear shell model:

taken from here.

• 50 neutrons: they are a closed shell, so no net angular momentum
• 37 neutrons: 28 get stacked in a closed shell, the remaining 4 fill the 2p$$_{3/2}$$ sub-shell and contribute no net angular momentum, and the remaining 5 go into the 1f$$_{5/2}$$ sub-shell.
4 of them are paired, so zero net angular momentum, and the fifth should have net 5/2?

What is wrong in the above argument?

How would I go on proving that $$^{40}$$K has $$I = 4$$?