WBJEE · Chemistry · Chemical Kinetics
\({ }_{11} Na \)^{24}$ is radioactive and it decays to
- A \({ }_{9} \mathrm{F}^{20}\) and \(\alpha\) -particles
- B \({ }_{13} \mathrm{Al}^{24}\) and positron
- C \({ }_{11} \mathrm{Na}^{23}\) and neutron
- D \({ }_{12} \mathrm{Mg}^{24}\) and \(\beta\) - particles
Answer & Solution
Correct Answer
(D) \({ }_{12} \mathrm{Mg}^{24}\) and \(\beta\) - particles
Step-by-step Solution
Detailed explanation
\(_{11} \mathrm{Na}^{24} \longrightarrow _{12} \mathrm{Mg}^{24}+_{-1}\beta^{0}\)
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