Question

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

• A. $\alpha$-emission
• B. $\beta$-emission
• C. positron emission
• D. proton emission

Answer 1

Answer: (B)

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