List- I shows different radioactive decay processes and List- II provides possible emitted particles. Match…

In $\alpha$ decay mass number decreases by $4$ unit and atomic number decreases by $2$ unit.

In ${\beta }^{-}$ decay mass number does not change but atomic number increases by $1$ unit.

In ${\beta }^{+}$ decay mass number does not change but atomic number decreases by $1$ unit.

So for

$\left(P\right){{ }^{238}}_{92}U{{\to }^{234}}_{91}Pa$.

Number of alpha particle $=\frac{238-234}{4}=1$. Due to alpha particle, atomic number should decrease by $2$ but it has decreased by $1$ only, therefore an additional ${\beta }^{-}$ decay is required, then net change in atomic number will be $-2+1=-1$. Therefore, $\mathrm{P}\to 4$.

$\left(Q\right){{ }^{214}}_{82}Pb{{\to }^{210}}_{82}Pb$.

Number of alpha particle $=\frac{214-210}{4}=1$. Due to alpha particle, atomic number should decrease by $2$, but there is no change in atomic number, which means two additional ${\beta }^{-}$ decay is required, then net change in atomic number will be $-2+2=0$. Therefore, $\mathrm{Q}\to 3$.

$\left(R\right){{ }^{210}}_{81}Tl{{\to }^{206}}_{82}Pb$.

Number of alpha particle $=\frac{210-206}{4}=1$. Due to alpha particle, atomic number should decrease by $2$, but atomic number has increased by $1$, which means three additional ${\beta }^{-}$ decay is required, then net change in atomic number will be $-2+3=1$. Therefore, $\mathrm{R}\to 2$.

$\left(S\right){{ }^{228}}_{91}Pa{{\to }^{224}}_{88}Ra$.

Number of alpha particle $=\frac{228-224}{4}=1$. Due to alpha particle, atomic number should decrease by $2$, but atomic number has decreased by $3$, which means one additional ${\beta }^{+}$ is required, then net change in atomic number will be $-2-1=-3$. Therefore, $\mathrm{S}\to 1$.