Matching Radioactive Decay Processes with Emitted Particles

To determine the correct matching between List-I and List-II, we apply the laws of conservation of mass number ($A$) and atomic number ($Z$) for nuclear reactions.

A general radioactive decay process involving $n_{\alpha}$ alpha particles (${}^{4}_{2}\text{He}$), $n_{\beta^-}$ beta-minus particles (${}^{0}_{-1}e$), and $n_{\beta^+}$ beta-plus particles (${}^{0}_{1}e$) can be written as: ${}^{A}_{Z}\text{X} \rightarrow {}^{A'}_{Z'}\text{Y} + n_{\alpha} \left({}^{4}_{2}\text{He}\right) + n_{\beta^-} \left({}^{0}_{-1}e\right) + n_{\beta^+} \left({}^{0}_{1}e\right)$

From the conservation of mass number ($A$): $A = A' + 4 n_{\alpha} \implies n_{\alpha} = \frac{A - A'}{4}$

From the conservation of atomic number ($Z$): $Z = Z' + 2 n_{\alpha} - n_{\beta^-} + n_{\beta^+}$

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Analysis of List-I entries:

1. For Process (P): ${}^{238}_{92}\text{U} \rightarrow {}^{234}_{91}\text{Pa}$

   • Conservation of mass number: $n_{\alpha} = \frac{238 - 234}{4} = 1$
   • Conservation of atomic number: $92 = 91 + 2(1) - n_{\beta^-} + n_{\beta^+}$ $92 = 93 - n_{\beta^-} + n_{\beta^+} \implies n_{\beta^-} - n_{\beta^+} = 1$
   • This corresponds to one $\alpha$ particle and one $\beta^-$ particle ($n_{\alpha} = 1, n_{\beta^-} = 1, n_{\beta^+} = 0$).
   • Matching: $\text{P} \rightarrow 4$

2. For Process (Q): ${}^{214}_{82}\text{Pb} \rightarrow {}^{210}_{82}\text{Pb}$

   • Conservation of mass number: $n_{\alpha} = \frac{214 - 210}{4} = 1$
   • Conservation of atomic number: $82 = 82 + 2(1) - n_{\beta^-} + n_{\beta^+}$ $82 = 84 - n_{\beta^-} + n_{\beta^+} \implies n_{\beta^-} - n_{\beta^+} = 2$
   • This corresponds to two $\beta^-$ particles and one $\alpha$ particle ($n_{\alpha} = 1, n_{\beta^-} = 2, n_{\beta^+} = 0$).
   • Matching: $\text{Q} \rightarrow 3$

3. For Process (R): ${}^{210}_{81}\text{Tl} \rightarrow {}^{206}_{82}\text{Pb}$

   • Conservation of mass number: $n_{\alpha} = \frac{210 - 206}{4} = 1$
   • Conservation of atomic number: $81 = 82 + 2(1) - n_{\beta^-} + n_{\beta^+}$ $81 = 84 - n_{\beta^-} + n_{\beta^+} \implies n_{\beta^-} - n_{\beta^+} = 3$
   • This corresponds to three $\beta^-$ particles and one $\alpha$ particle ($n_{\alpha} = 1, n_{\beta^-} = 3, n_{\beta^+} = 0$).
   • Matching: $\text{R} \rightarrow 2$

4. For Process (S): ${}^{228}_{91}\text{Pa} \rightarrow {}^{224}_{88}\text{Ra}$

   • Conservation of mass number: $n_{\alpha} = \frac{228 - 224}{4} = 1$
   • Conservation of atomic number: $91 = 88 + 2(1) - n_{\beta^-} + n_{\beta^+}$ $91 = 90 - n_{\beta^-} + n_{\beta^+} \implies n_{\beta^+} - n_{\beta^-} = 1$
   • This corresponds to one $\alpha$ particle and one $\beta^+$ particle ($n_{\alpha} = 1, n_{\beta^-} = 0, n_{\beta^+} = 1$).
   • Matching: $\text{S} \rightarrow 1$

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Conclusion:

The correct combination is: $\text{P} \rightarrow 4, \quad \text{Q} \rightarrow 3, \quad \text{R} \rightarrow 2, \quad \text{S} \rightarrow 1$

Thus, the correct option is (A).