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Q.
Of the following nuclides, the one most likely to be radioactive is
Structure of Atom
Solution:
${ }_6^{14} C n=14-6=8 $
$p=6$
$\frac{n}{p}=\frac{8}{6}=1.33 >1$
Hence, ${ }_6^{14} $ Chas $\frac{n}{p}$ ratio greater than 1. It emits particles to have $\frac{n}{p}=1$.
Thus, neutron changes to proton by emission of $\alpha$-particles
${ }_6^{14} C \rightarrow{ }_7^{14} N +{ }_{-1}^0 e$