1. ## emission question

if 13 Al 27 isotope is stable which emission can 13 Al 29 undergo

2. For future reference you don't need to give us the atomic number since it's implied in the element name. In this case every Al atom has 13 protons - if it didn't it wouldn't be Al!

Back on topic, $^{29}\text{Al}$ needs to lose two atomic mass units (amu) without changing atomic mass number. It's a weird one since to lose 2 amu without changing atomic number implies neutron emission twice which would be $^{29}\text{Al} \rightarrow ^{28}\text{Al} + n \rightarrow ^{27}\text{Al} + n$

Yet google and wolfram say beta minus decay: $^{29}\text{Al} \rightarrow ^{29}\text{Si} + e^- + \bar{\nu_e}$ and $^{29}\text{Si}$ is stable.

I would be inclined to believe google if this is a real world problem since neutron emission usually occurs when there are more than two excess neutrons compared to the stable isotope