Question
List-$I$ shows different radioactive decay processes and List-$II$ provides possible emitted particles. Match each entry in List-$I$ with an appropriate entry from List-$II$, and choose the correct option.
List-$I$ List-$II$
($P$) ${ }_{92}^{238} U \rightarrow{ }_{91}^{234} \mathrm{~Pa}$ ($1$) one $\alpha$ particle and one $\beta^{+}$particle
($Q$) ${ }_{82}^{214} \mathrm{~Pb} \rightarrow{ }_{82}^{210} \mathrm{~Pb}$ ($2$) three $\beta^{-}$particles and one $\alpha$ particle
($R$) ${ }_{81}^{210} \mathrm{Tl} \rightarrow{ }_{82}^{206} \mathrm{~Pb}$ ($3$) two $\beta^{-}$particles and one $\alpha$ particle
($S$) ${ }_{91}^{228} \mathrm{~Pa} \rightarrow{ }_{88}^{224} \mathrm{Ra}$ ($4$) one $\alpha$ particle and one $\beta^{-}$particle
  ($5$) one $\alpha$ particle and two $\beta^{+}$particles

Answer

$\mathrm{Z}_1 Z^{A_4} \rightarrow Z_2 Y^{A_1}+N_{12} \mathrm{He}^4+N_{21} e^0+N_{3-1} e^0$

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 $

$\mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 \rightarrow 3 \beta^{-}$

$(\mathrm{S})$

${ }_{91} \mathrm{~Pa}^{22 k} \rightarrow_{k 8} \mathrm{Ra}^{224}$

$\mathrm{~N}_1=\frac{228-224}{4}=1 \alpha$

$\mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}$

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