Question

${}_{11}^{23}Na$ is the more stable isotope of Na than ${}_{11}^{24}Na$. Find out the process by which ${}_{11}^{24}Na$ can undergo radioactive decay.

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

Answer 1

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

${}_{11}^{24}Na$ isotope can undergo radioactive ${\beta }^{-}$ emission decay.

The given equation is:

${}_{11}^{24}Na{⟶}_{12}^{24}Mg+{\beta }^{-}$