Question

${}^{23}Na$ is the more stable isotope of Na. ${}_{11}^{24}Na$ can undergo radioactive decay by :

• A. ${\beta }^{-}-emission$
• B. $\alpha -emission$
• C. ${\beta }^{+}-emission$
• D. K-electron capture

Answer 1

The correct option is A ${\beta }^{-}-emission$

Reaction:
${}_{11}^{24}Na{\stackrel{-\beta }{\to }}_{12}^{24}Mg$