Question

${}^{23}Na$ is the more stable isotope of $Na$. By which process ${}_{11}^{24}Na$ can undergo radioactive decay?

• A. ${\beta }^{-}-emission$
• B. $\alpha -emission$
• C. ${\beta }^{+}-emission$
• D. None of these

Answer 1

The correct option is D None of these

To convert ${}_{11}^{24}Na$ to ${}_{11}^{23}Na$, we need to decrease the mass number of ${}_{11}^{24}Na$ by $1$.

Since, in ${\beta }^{-}$ emission and ${\beta }^{+}$ emission, the mass number of product nucleus is same as that of parent nucleus.

Whereas, in $\alpha$ emission, the mass number of product nucleus gets decreased by $4$ compared to that of parent nucleus.

Thus ${}_{11}^{24}Na$ cannot undergo any of the radioactive decay to convert into ${}_{11}^{23}Na$.