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Question: Answered & Verified by Expert
${ }_{11} Na $^{24}$ is radioactive and it decays to
ChemistryChemical KineticsWBJEEWBJEE 2012
Options:
  • 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
Solution:
1494 Upvotes Verified Answer
The correct answer is: ${ }_{12} \mathrm{Mg}^{24}$ and $\beta$ - particles
$_{11} \mathrm{Na}^{24} \longrightarrow _{12} \mathrm{Mg}^{24}+_{-1}\beta^{0}$

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