# C-15 spin and parity at ground state

Assuming the extreme independent particle model:

When we draw the energy levels for neutrons (9) and protons (6) for Carbon 15 all protons energy levels are filled and there is one uncoupled neutron at 1d(5/2). So the nucleus is descbribed by the properties of its uncoupled nucleon but I checked online and the nuclear spin is 1/2 and parity +.

My question is this, why isn't the nuclear spin 5/2, and how do I determine the parity since (-1)^(5/2)=i doesn't make sense.

The $d$ states have orbital angular momentum $L=2$. The parity is related to orbital angular momentum, not total angular momentum $j=5/2$.
The previous closed shell contains $8$ neutrons. Given it’s closed, you can deduce the value of total spin $S$ for the first $8$ neutrons. Since neutrons have $s=1/2$, it’s not hard to figure out what are the possible values of $S_{tot}$ for the system of $8+1$ neutrons. As a final check you can verify that $j=5/2$ is indeed a possible in the coupling of this $S_{tot}$ to orbital angular momentum $L$.
• @user3636673 I don't understand your comment:there must be a typo. One way or the other, 1. What is the total spin of a set of (here $8$) nucleons in a closed shell? 2. The spin of one neutron is $1/2$. If you can answer 1. and 2. you're done. Oct 26 '17 at 12:48