$$_{13}^{27}Al$$  is a stable isotope, $$_{13}^{29}Al$$  is expected to disintegrate by

Solution :
The species $$_{13}A{l^{29}}$$  ( No. of neutrons = 16 ) contains more neutrons than the stable isotope $$_{13}A{l^{27}}$$  ( No. of neutrons= 14 ) .
Neutron on decomposition shows $$\beta $$ -emission.
$$\eqalign{ & _0{n^1}{ \to _{ + 1}}{p^1}{ + _{\, - 1}}{e^0} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\beta - {\text{particle}} \cr} $$