A radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3\alpha$ -particles and $2$ positrons. The ratio of number of neutrons to that of protons in the final nucleus will be:

Initial: Protons $=Z$, Neutrons $=A-Z$.Emission of $3\alpha$ particles:An $\alpha$ particle is ${}_2^4\text{He}$.Reduction in $Z=3\times 2=6$.Reduction in $A=3\times 4=12$.Protons $=Z-6$, Neutrons $=(A-Z)-6$.Emission of $2$ positrons (${\beta }^{+}$):In ${\beta }^{+}$ decay, a proton converts to a neutron: $p\to n+e^{+}$.Reduction in $Z=2$.Increase in Neutrons $=2$.Mass number $A$ remains constant.Final state:Protons $=(Z-6)-2=Z-8$ Neutrons $=(A-Z-6)+2=A-Z-4$ Ratio $=\frac{A-Z-4}{Z-8}$