Question

$^{23}_{}Na$ is the more stable isotope of Na. Find out the process by
which $^{24}_{11}Na$ 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 $(Na^{24})$, 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 \longrightarrow \,{}^{}_{-1}\beta^0 + \,{}^{}_{1}H^1$