# Low-energy excited states of $^{13}C$

While doing some studying for an exam in introductory nuclear physics, I stumbled upon a question I can't answer. I'm supposed to explain the ground state and the three lowest energy excited states of $$^{13}C$$ using the shell model. The ground state has spin and parity $$1/2^-$$. The excited state with $$E = 3.09 \text{ MeV}$$ corresponds to $$1/2^+$$, the $$3.68 \text{ MeV}$$ state to $$3/2^-$$ and $$3.85 \text{ MeV}$$ to $$5/2^+$$. I know that $$^{13}C$$ has one unpaired neutron, which will determine the spin and parity of the nucleus. $$1/2^-$$ must then correspond to a $$p_{1/2}$$-level, $$1/2^+$$ to an $$s_{1/2}$$-level, $$3/2^-$$ to $$p_{3/2}$$ and $$5/2^+$$ to $$d_{5/2}$$. (See table below.) The unpaired nucleon must reside in the level corresponding to each state.

\begin{align} 0 \text{ MeV} && 1/2^- && p_{1/2} \\ 3.09 \text{ MeV} && 1/2^+ && s_{1/2} \\ 3.68 \text{ MeV} && 3/2^- && p_{3/2} \\ 3.85 \text{ MeV} && 5/2^+ && d_{5/2} \\ \end{align}

In order to transition from $$1/2^-$$ to $$1/2^+$$, we can either let the unpaired neutron in $$1p_{1/2}$$ jump to $$2s_{1/2}$$ or let a paired neutron jump from $$1s_{1/2}$$ to $$1p_{1/2}$$. Other options seem out of the question, since the $$3s_{1/2}$$-level lies much higher and would therefore require too much energy to achieve. How would I determine which of the two would happen?