Question

${}_{13}^{27}Al$ is a stable isotope. ${}_{13}^{29}Al$ is expected to decay by :

• A. $\alpha -$ emission
• B. $\beta -$ emission
• C. Positron emission
• D. Proton emission

Answer 1

The correct option is B $\beta -$ emission

${}_{13}A{l}^{29}$ is neutron rich isotope, will decay by $\beta -$ emission converting some of its neutron itno proton as
${}_0{n}^1{\to }_{-1}{\beta }^0{+}_1{H}^1$