How to determine whether a nuclear transition would be electric octupole, or hexadecapole? The transition from one nuclear state to another is classified as quadrupole/octupole, etc, depending on the units on angular momentum transferred. But depending on the angular momentum of the two states involved, the net J can take different values. So what decides whether a nuclear transition would be electric octupole, or hexadecapole?
 A: A way to decide whether a gamma photon emition is electrical or magnetic is from the photon's parity.
Parity conservation requires: $π_i = π_f π_{ph}  $ where the parity at the left side is the initial parity of the nuclei, and at the right the final parity of the nuclei and of the photon. 
Thus, an electrical transitions happens when: $π_{ph}=(-1)^l $ and a magnetic when $π_{ph}=-(-1)^l $, where l is the the angular momentum of the photon.
From the conservation of momentum demands in general: $\bar j_i -\bar j_f = \bar j_{ph} $ where the subscripts $i$ and $f$ mean the initial and final state of the nuclei. Thus, it is $$\left|\bar j_i - \bar j_f\right| \leq j_{ph} \leq \bar j_i + \bar j_f  $$. 
Thus, to your question, by defining the order of the polypole by the angular momentum of the photon we have:
Radiation |Symbol|   l photon momentum  | photon parity |


*

*E dipole, E1, $\ell=1$,    $\pi_\gamma=-1$

*B dipole,  M1, $\ell=1$,    $\pi_\gamma=+1$

*E quadrupole, E2, $\ell=2$, $\pi_\gamma=+1$

*B quadrupole, M2, $\ell =2$, $\pi_\gamma=-1$

*E octapole ,E3,$\ell=3$, $\pi_\gamma=-1$

*B octapole ,B3,$\ell=3$, $\pi_\gamma=+1$

*E hexadecapole, E4 $\ell=4$, $\pi_\gamma=+1$

*B hexadecapole, B4, $\ell=4$, $\pi_\gamma=-1$


and so on...
I hope this helps with your question.
A: The short answer is that all transitions which are not forbidden by parity or angular momentum conservation happen, though not necessarily at the same rate. When several multipolarities are allowed, the one with the lowest multipolarity dominates. 
