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

Answer: (A)

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