Question

${}^{23}\text{Na}$ is the more stable isotope of Na. Find out the process by which ${}_{11}^{24}\text{Na}$ 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

$n/p$ ratio of ${}^{24}\text{Na}$ is 13/11, greater than one so it is unstable and to convert it into stable nuclei n/p should be one and to decreases this ratio, one neutron has to be converted into proton and it will happen through $\beta$ - emission as:

${}_0^{1}n{⟶}_1^{1}p{+}_{-1}^{0}e$
neutron proton beta particles

${}_{11}^{24}\text{Na}{⟶}_{12}^{24}\text{Mg}{+}_{-1}^{0}e$