Question

${}^{23}Na$ is the more stable isotope of 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

In stable isotope of Na, there are 11 protons and 12 neutrons. In the given radioactive isotope of sodium $\left(N{a}^{24}\right)$, there are 13 neutrons, one neutrons more than that required for stablility. A neutrons rich isotope always decay by $\beta -$ emission as
${}_0{n}^1{\to }_{-1}{\beta }^0{+}_1{H}^1$