Chapter 29 - The Nucleus - Learning Path Questions and Exercises - Multiple Choice Questions - Page 995: 8

(c) neither type of decay would be an option for $^{26}Al$.

For $^{26}Al$ to be stable, it should transform to $^{27}Al$. There is a change in the mass number involving this process. In the case of beta decay, the mass number doesn't change. Therefore, neither type of decay would be an option for $^{26}Al$.