_{11}^{23}Na is the more stable isotope of Na | vedclass

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Question

(A) For $_{11}^{23}Na$,the neutron-to-proton ratio is $\frac{n}{p} = \frac{12}{11} \approx 1.09$.For $_{11}^{24}Na$,the neutron-to-proton ratio is $\f

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$\beta ^ -$ emission

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(A) For $_{11}^{23}Na$,the neutron-to-proton ratio is $\frac{n}{p} = \frac{12}{11} \approx 1.09$.For $_{11}^{24}Na$,the neutron-to-proton ratio is $\frac{n}{p} = \frac{13}{11} \approx 1.18$.Since $_{11}^{24}Na$ has a higher $\frac{n}{p}$ ratio than the stable isotope,it undergoes radioactive decay to decrease this ratio.This is achieved by the emission of a $\beta^-$-particle,where a neutron converts into a proton: $n 	o p + e^- + \bar{
u}$ ($\beta^-$ emission).

$_{11}^{23}Na$ is the more stable isotope of $Na$. Find out the process by which $_{11}^{24}Na$ can undergo radioactive decay.

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