A) \[{{\beta }^{-}}\]emission
B) \[\alpha \]emission
C) \[{{\beta }^{+}}\]emission
D) K electron capture
Correct Answer: A
Solution :
\[_{11}^{23}Na\to \frac{n}{p}\text{ratio}=12/11\] \[_{11}^{24}Na\to \frac{n}{p}\text{ratio}=23/11\] so decrease in \[\frac{n}{p}\]ratio gives out \[\beta \]-particle \[n\to p+e\,({{\beta }^{-}})\].You need to login to perform this action.
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