Question

A nucleus ${}_{11}^{24}Na$ is $\beta$-radioactive. Choose the equation representing $\beta$-decay.

• A. ${}_{11}^{24}Na{\to }_{11}^{24}Na{+}_0^{0}e$
• B. ${}_{11}^{24}Na{\to }_{12}^{24}Mg{+}_{-1}^{0}e$
• C. ${}_{11}^{24}Na{\to }_{10}^{24}Mg{+}_1^{0}e$
• D. ${}_{11}^{24}Na{\to }_{12}^{24}Na{+}_{-1}^{0}e$

Answer 1

The correct option is B ${}_{11}^{24}Na{\to }_{12}^{24}Mg{+}_{-1}^{0}e$

$\beta$ decay converts one neutron to one proton. So, the atomic number is increased by $1$, while mass number remains the same.

Thus the reaction is:

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