Correct option: A.$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1$
a
$\mathrm{Z}_1 Z^{A_4} \rightarrow Z_2 Y^{A_1}+N_{12} \mathrm{He}^4+N_{21} e^0+N_{3-1} e^0$
Conservation of charge
$\mathrm{Z}_1=\mathrm{Z}_2+2 \mathrm{~N}_1+\mathrm{N}_2-\mathrm{N}_3$ $. . . . . .(1)$
Conservation of nucleons.
$A_1=A_2+4 N_1$ $. . . . . (2)$
$N_1=\frac{A_1-A_2}{4}$
From ($i$) and ($ii$)
$\mathrm{N}_2-\mathrm{N}_3=\mathrm{Z}_1-\mathrm{Z}_2-\left(\frac{\mathrm{A}_1-\mathrm{A}_2}{2}\right)$
$(P)$
${ }_{92} \mathrm{U}^{238} \rightarrow{ }_{91} \mathrm{~Pa}^{234}$
$\mathrm{~N}_1=\frac{238-234}{4}=1 \rightarrow 1 \alpha$
$\mathrm{N}_2-\mathrm{N}_3=(92-91)-\left(\frac{4}{2}\right)=-1 \rightarrow 1 \beta^{-}(Q)$
${ }_{\mathrm{k} 2} \mathrm{~Pb}^{214} \rightarrow_{32} \mathrm{~Pb}^{210}$
$\mathrm{~N}_1=\frac{214-210}{4}=1 \rightarrow 1 \alpha$
$\mathrm{N}_2-\mathrm{N}_3=(82-82)-\left(\frac{4}{2}\right)=-2 \rightarrow 2\beta^{(\mathrm{R})}$
${ }_{81} T \ell^{210} \rightarrow{ }_{\mathrm{k} 2} \mathrm{~Pb}^{206}$
$\mathrm{~N}_1=\frac{210-206}{4}=1 \rightarrow 1 \alpha $
$\mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 \rightarrow 3 \beta^{-}$
$(\mathrm{S})$
${ }_{91} \mathrm{~Pa}^{22 k} \rightarrow_{k 8} \mathrm{Ra}^{224}$
$\mathrm{~N}_1=\frac{228-224}{4}=1 \alpha$
$\mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}$