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?